VIII A diagram showing the structure of a molecule with atoms represented by spheres.
2
STEAM POWER
Experiment and mathematical reasoning seem to indi-
cate that the molecules of all materials are in constant
motion and that there are neutralizing attractive and repul-
sive forces acting between them. In solids the molecules are
apparently bound together in such a way that, although they
are in constant motion, the external pressure of the body tends to remain constant; in fact it requires the expenditure of force to cause a change of form. In liquids the molecular attraction is so altered that practically all
rigidity disappears and the shape assumed by the liquid is
determined by that of the surrounding surfaces, as, for
instance, the shape of the vessel containing the liquid. In
gases the molecules are still more free and actually tend to
move apart from each other and to move in all directions until it fills any closed containing vessel.
2. Energy. Nearly everyone has a conception of what
is meant by the term energy, but no one yet knows what
energy really is. It is defined as the capacity for doing work,
or the ability to overcome resistance. A man is said to be very
energetic or to be possessed of a great deal of energy when he
has the ability to perform a great amount of work or to
overcome great resistances. Matter is said to be posses-
sed of a certain amount of energy when it can do work or overcome resistance. Actually, matter is not known in any form in which it is not possessed of energy.
There are many different forms of energy. A body in
motion can do work and is said to be possessed of mechani-
cal energy. A body which we recognize as hot can do work at
the expense of the heat associated with it and is said to be possessed of heat energy. Light, sound and electricity
are all forms of energy.
Experiment and experience have never shown that energy
can be destroyed or created by man, but they have shown
that one form of energy can be converted into another form
under proper conditions. The first part of this experience is stated as a law known as the Law of the Conservation of
PHYSICAL CONCEPTIONS AND UNITS
**Energy.** This law states that "the total quantity of energy in the universe is constant."
**3. Units of Matter and of Energy.** When attempts are made to measure the amount of anything, some unit of measurement is adopted. Matter is measured in numerous ways, such as by weight, volume, length, etc., but all of measuring matter are by volume and by weight. Engineers in English-speaking countries use the cubic yard, the cubic foot or the cubic inch as units in measuring matter by volume and they use the pound, the ounce, the grain, etc. as units in measuring matter by weight.
Energy is measured in various units and, in general, there is a characteristic unit or set of units for each form in which it occurs. Thus the **foot-pound** is very commonly used for measuring mechanical energy; the **British thermal unit** for measuring heat energy; and the **joule** for measuring electrical energy. Some of these units will be defined and considered in greater detail in subsequent paragraphs.
**4. Work.** Work is done when a weight is raised against a resistance through a distance. Thus, work is done when a weight is raised against the resistance offered by gravity; work is done when a spring is compressed against the resistance which the metal offers to change of shape; work is done when a wheel is turned against another against the resistance offered by friction.
The **unit of work** is the quantity of work which must be done in raising a weight of one pound through a vertical distance of one foot. It is called the **foot-pound.** Thus, one foot-pound of work must be done in raising one pound one foot; two foot-pounds of work must be done in raising two pounds one foot or in raising one pound two feet.
If a weight of one pound were suspended from a spring balance as shown in Fig. 1, the balance would indicate a pull or force of one pound. No work would be
![]() A diagram showing a spring balance with a weight attached to it.
Fig. 1.
4
**STEAM POWER**
done by this force as long as the weight remained stationary, because no resistance would be overcome through a distance. If, however, the same weight were slowly or rapidly raised a vertical distance of a foot, one foot-pound of work would be done. A force or pull of one pound would then have overcome a resistance of one pound through a distance of one foot. In general:
$$\text{Work in ft.-lbs.} = \text{Resistance overcome in lbs.} \times \text{distance.}$$
$$= \text{Force in lbs.} \times \text{distance in ft.}$$
so that if a force of 10 lbs. pushes or pulls anything which offers a resistance of 10 lbs. while that something travels a distance of, say, 5 ft., the work done will be given by the expression,
$$\text{Work} = 10 \times 5,$$
$$= 50 \text{ ft.-lbs.}$$
A body in falling a certain distance can do work equal to its weight multiplied by the distance it falls because it could theoretically raise an equal weight an equal distance against the action of gravity, and the work done upon this second body would be equal to its weight multiplied by the distance through which it was raised.
It is very important to note that no work is done by a force if there is no motion; resistance must be overcome through a distance before any work may be done. Thus, a force of 1000 lbs. might be required to hold a man in position, that is to balance a resistance, but no work would be done if the body upon which the 1000-pound force acted did not move. Again, a weight of 50 lbs. held at a distance of 10 ft. above the surface of the earth would exert a downward push or pull equal to 50 lbs., on whatever held it in that position, but no work would be done until it was moved to that position. If allowed to fall through the distance of 10 ft., it could do $50 \times 10 = 500$ ft.-lbs. of work.
It is very convenient to represent graphically the action
PHYSICAL CONCEPTIONS AND UNITS 5
of forces overcoming resistances, that is, doing work. This is done by plotting points showing the magnitude of the force at the time that the body on which it is acting has traveled different distances. Thus, suppose a constant force of 10 lbs. pushes a body a distance of 15 ft. against a constant resistance of 10 lbs. The force acting on the body will have a value of 10 lbs. just as the body starts to move, a value of 10 lbs. when the body has moved 1 ft., a value of 10 lbs. when the body has moved 2 ft., and so on. This might be represented by points on squared paper as shown
![]()
A graph with axes labeled "Distance Traveled in Feet" on the x-axis (from 0 to 15) and "Force in Pounds" on the y-axis (from 0 to 20). The points representing the action of this force are plotted at intervals of 1 foot along the x-axis.
FIG. 2.
in Fig. 2 or by a horizontal line joining those points as shown in the same figure.
The work done by this force would be $10 \times 15 = 150$ ft.-lbs., according to our previous definition. But $10 \times 15$ is also the number of small squares under the line representing the action of this force in Fig. 2. The number of these small squares then must be a measure of the work done, but it is also a measure of the area under the line representing the action of the force, so that this area must be a measure of the work done. Each small square represents 1 lb. by its vertical dimension and 1 ft. by its horizontal dimension,
6
6 STEAM POWER
so that its area must represent 1 ft.\times1 ft.=1 ft.-lbs.
The total number of squares below the line equals $10 \times 15 = 150$, and since the area of each one represents 1 ft.-lbs.
the total area under the line represents $150 \times 1 = 150$ ft.-lbs.
It is not always convenient to choose such simple scales as those just used. Thus it might be more convenient to plot the action of this force as is done in Fig. 3. Here the height of a square represents 2 lbs., and the width represents 1 ft.; the area then represents $2 \times 1 = 2$ ft.-lbs.
There are $5 \times 15 = 75$ squares under the line and as each
![]()
A graph with a horizontal axis labeled "Distance traveled in Foot" ranging from 0 to 15 in increments of 2 (0, 2, 4, 6, 8, 10, 12, 14, 15). A vertical axis labeled "Force in Pound" ranges from 0 to 12 in increments of 2 (0, 2, 4, 6, 8, 10, 12).
Fig. 3.
represents 2 ft.-lbs, the total area under the line represents $2 \times 75 = 150$ ft.-lbs. as before.
This is a very useful property of these diagrams and the area under the line representing the action of the force always represents the work done, no matter what the shape of that line.
Thus, assume a force which compresses a spring a distance of 6 ins. Suppose that a force of 10 lbs. is required to compress the spring 1 ins., a force of 20 lbs. to compress it 2 ins., and so on up to a force of 60 lbs. to compress it 6 ins.
Starting with a force of zero, the force will have to gradually increase as the spring is compressed, as shown by the line in Fig. 4.
The area of each of the small squares will represent $\frac{1}{12} \times \frac{1}{12} = \frac{1}{144}$ ft.-lbs.
Under the line there is an area equal
| $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{12}$ |
| Force in Pound | Force in Pound | Force in Pound | Force in Pound | Force in Pound | Force in Pound | Force in Pound | Force in Pound | Force in Pound | Force in Pound | Force in Pound | Force in Pound | Force in Pound | Force in Pound | Force in Pound |
| Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot | Distance traveled in Foot |
| Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-lbs.) | Area under Line (ft.-bs)
PHYSICAL CONCEPTIONS AND UNITS
7
$\frac{6 \times 6}{2} = 18$ small squares, and the work done in compressing the spring must then be $18 \times \frac{10}{12} = 15$ ft.-lbs.
**5. Mechanical Energy**
A body which exists in such a position or location that it could do work by dropping or falling is said to be possessed of **potential mechanical energy**, or of mechanical energy due to position. As long as it remains in this position, it cannot do work at the expense of this energy, but if allowed to fall, it could do so. The
![]() Graph Showing Action of Spring.
Fig. 4. - Graph Showing Action of Spring.
work it could do would be equal to the product of its weight by the distance it could fall and the potential energy it possesses before starting to fall is measured by this work. Thus, a body weighing 40 lbs., located 10 ft. above the surface of the earth could do $40 \times 10 = 400$ ft.-lbs. of work in falling, and, therefore, it was possessed of 400 ft.-lbs. of potential energy before it starts to fall.
If in falling it raises a weight equal to its own (theoretically) through a distance equal to that through which it falls (theoretically), it will have used up 400 ft.-lbs. of energy in doing 400 ft.-lbs. of work upon the body mixed
8
STEAM POWER
and will no longer be possessed of that amount of potential energy. The body which has been raised will, however, have an equal amount of energy stored in it and will in turn be able to do 400 ft.-lbs. of work if allowed to fall a distance of 10 ft.
If the body assumed above falls through a distance of 10 ft. without raising another body or doing an equivalent amount of work in some other way, it acquires a high velocity. When it arrives at the bottom of the fall of 10 ft., it certainly does possess energy due to its potential energy which it had before dropping nor has it done work at the expense of that energy. Moreover, the energy could not have been destroyed because it is indestructible. The only conclusion is that it must still be possessed of this energy in some way. At the end of the fall it has lost its advantageous position but it acquired a high velocity, and expended energy so that if brought back up again no work upon which that brings it to rest equal to what it could have done in raising a weight as previously described.
At the end of its fall and before being brought to rest, the body is therefore said to be possessed of energy by virtue of its velocity, and this form of energy is called **kinetic mechanical energy**. The kinetic energy will be exactly equal to the potential energy which disappeared as the body fell.
Any body which is moving is possessed of kinetic energy because it can do work on anything which brings it to rest. This energy is expressed by the equation,
$$\text{Kinetic Energy in ft.-lbs.} = \frac{1}{2} \times W \times V^2,$$
in which
$W$-the weight of the moving body in pounds.
$V$-the velocity in ft. per second, and
32.2-a gravitational constant commonly represented by $g$.
PHYSICAL CONCEPTIONS AND UNITS 9
**6. Heat.** One of the most familiar forms of energy is heat, which manifests itself to man through the sense of touch. In reality every body with which man is familiar possesses an unknown amount of heat energy and it is assumed that this heat energy is in some way associated with the motions and relative positions of the molecules and their constituents.
For this reason heat is often described as molecular activity, and regarded as energy stored up in a substance by virtue of its molecular condition. Heat energy can be made to perform work in ways which will be discussed later and this is proof that it is a form of energy and not a material substance, as was once supposed.
Heat is observed and recorded by its effects on matter, producing changes in temperatures or volumes of objects; changes of internal stress; changes of state, as ice to water and water to steam; changes of temperature; and electrical and chemical effects.
Neglecting certain atomic phenomena not yet well understood, the probable source of all heat energy appearing on the earth is the sun. Heat, however, may be obtained from the sun by means of electrical energy; from chemical changes; from changes of physical state; from the internal heat of the earth.
**7. Temperature.** Man early realized that under certain conditions bodies felt "hotter" than under other conditions and gradually came to speak of the "degree of hotness" as the temperature of the body. It was only later that what was really measured as the "hotness" or intensity of heat or temperature of a body was the ability of that body to transmit heat to others and that it had no connection with quantity of heat.
Thus if the temperature of two adjacent bodies happened to be the same, one of them could not lose heat by transmitting it to the other, but if the temperature of one happened to be higher than that of another, the body at
10
**STEAM POWER**
higher temperature would always lose heat to the one at lower temperature.
As a means of measuring temperature certain arbitrary scales have been chosen. The centigrade scale of tempera-
ture, for instance, is based upon the temperatures of melting ice and boiling water. The difference between the tempe-
rature differences between boiling water at atmospheric pressure and melting ice at atmospheric pressure is arbitrarily called one hundred degrees of temperature, and the
temperature of the melting ice is called zero, making that of the boiling water 100 degrees.
Any body which has such a temperature that it will not give heat to, or take heat from, melting ice is said to be at a temperature of zero degrees centigrade, represented as $0^\circ$ C. Similarly, any body in such a condition that it will not give
heat to or take heat from water boiling under atmospheric
pressure is said to have a temperature of $100^\circ$ C. A body
with a temperature exactly half way between these two limits is said to have a temperature of $50^\circ$ C.
**8. Measurement of Temperature.** The temperatures of
bodies could be determined by bringing them in contact with
such things as melting ice and boiling water and determining
whether or not a transfer of heat occurred, but this would be a very cumbersome and unsatisfactory method. As a consequence, many other means have been devised for the
measurement of temperature.
One of the most common and convenient methods in-
volves the use of what are known as **mercury thermometers**.
These depend upon the fact that the expansion of mercury
with changing temperature is very uniform over a wide
temperature range. Thus, if mercury expands a certain
amount when its temperature is raised half way from tem-
perature to final boiling water, i.e., $100^\circ$ C., it will expand just
half as much when its temperature is raised half as high,
and one-quarter as much when its temperature is raised one-
quarter of the range from $0^\circ$ to $100^\circ$ C.
PHYSICAL CONCEPTIONS AND UNITS 11
The thermometer is made by enclosing a small quantity of mercury in a glass tube fitted with a bulb at one end, as shown in Fig. 5. The lower end of the thermometer is immersed in melting ice and the point on the stem which is reached by the top of the mercury column is marked and labeled $0^\circ$ C. The thermometer is then immersed in the steam from water boiling under atmospheric pressure and the point reached by the top of the mercury column is marked and labeled $100^\circ$ C. The distance between the two marks is then divided into one hundred parts and each represents the distance which the end of the column of mercury will move when its temperature changes one centigrade degree.
It is customary to extend this same scale below $0^\circ$ and above $100^\circ$, carrying it, on extensive thermometers, as far in each direction as the approximation to a constant expansion on the part of the mercury and to constant properties of the glass.
The temperature of a body can then be found by placing the thermometer in or in contact with that body and noting the point reached by the end of the mercury column. The division reached gives the temperature directly.
The centigrade scale described is the one commonly used by scientists throughout the world over, but engineers in this country more often use what is known as the Fahrenheit scale. This is so chosen that the temperature of melting ice is called $32^\circ$ F and the temperature of water boiling under atmospheric pressure is called $212^\circ$ F. There are
![]() Fig. 5.—Mercury Thermometer.
![]() Fig. 6.—Comparison of Centigrade and Fahrenheit Scales.
12
STEAM POWER
thus 180° on this scale for the same temperature difference
as is represented by 100° on the centigrade scale. The
relation between the two scales is shown diagrammatically
in Fig. 6. It is apparent that the temperature of a body
at 0° C. will be 32° F., and that of a body at 0° F. will be
-175° C.
Since 100 centigrade degrees are equal to 180 Fahren-
heit degrees, it follows that,
$$1^{\circ} \mathrm{C} = \frac{180}{100} \frac{9}{5}^{\circ} \mathrm{F}. \qquad (1)$$
and that
$$1^{\circ} \mathrm{F} = \frac{180}{100} \frac{5}{9}^{\circ} \mathrm{C}. \qquad (2)$$
Therefore, if $t_F$ and $t_C$ represent temperatures on the Fahren-
heit and centigrade scales respectively,
$$t_F = \frac{9}{5}(t_C + 32). \qquad (3)$$
and
$$t_C = \frac{5}{9}(t_F - 32). \qquad (4)$$
There is still another temperature scale of great impor-
tance. It is known as the absolute
scale and temperatures measured on this scale are spoken of as absolute tem-
peratures. At zero on this scale, which is located at -273°C or 273 centi-
grade degrees below centigrade zero, or, what is the same thing,
at -459.4°F, or 459.4 Fahrenheit degrees below centigrade zero. The
degrees used are absolute centigrades
of Fahrenheit, or conversely, so
that there are absolute tempera-
tures expressed above centigrade de-
grees above absolute
![]() Fig. 7.—Comparison of Absolute and Ordinary Temperature Scales.
degrees below absolute
PHYSICAL CONCEPTIONS AND UNITS
13
tures expressed in Fahrenheit degrees above absolute zero.
The relations between the various scales are shown dia-
grammatically in Fig. 7.
It is apparent from this diagram that,
$$T_F = (t_F + 460) \text{ (approximately)} \quad . . . \quad (5)$$
and that
$$T_C = (t_C + 273) \quad . . . \quad . . . \quad (6)$$
if $T_F$ and $T_C$ represent absolute temperatures and if the
number 430.4 is rounded out to 460, as is commonly done.
**9. The Unit of Heat Energy.** The unit used in the
measurement of heat energy in the United States is the
British Thermal Unit (abbreviated B.t.u.). It is defined as
the quantity of heat required to raise the temperature of one
pound of pure water one degree Fahrenheit. In order to
make this definition very definite, it is necessary to state
the temperature of the water before the temperature rise occurs,
because it requires different amounts of heat to raise the
temperature of a pound of water one degree from differ-
ent initial temperatures. For ordinary engineering pur-
poses, however, such refinements generally may be omitted.
Moreover, it must be pointed out that heat energy and
mechanical energy are mutually convertible, that is, the one
can be changed into the other. When such a change occurs
no energy can be lost since energy is indestructible, and it
follows that, if one form is changed into the other, there
must be just as much energy present after the change as there was before.
As will be used in measuring the two forms of energy
are very different and as it is often necessary to express
quantities of energy taking part in such conversions, it is
desirable to determine the relations between these units.
This was first accurately done by Joule, who showed that one
British thermal unit of heat energy resulted from the con-
14
**STEAM POWER**
version of 772 ft.-lbs. of mechanical energy. Later experimenters have shown that the number 778 more nearly expresses the truth than does the number 772 and the larger value is now known as Joule's Equivalent.
Expressed mathematically, the relation between the units is
$$1 \text{ B.t.u.} = 778 \text{ ft.-lbs.} \quad \quad (7)$$
$$1 \text{ ft.-lbs.} = \frac{1}{778} \text{ B.t.u.} \quad \quad (8)$$
**10. Specific Heat.** The specific heat of a substance is defined as that quantity of heat which is used up or recovered when the temperature of one pound of the material in question is raised or lowered one degree. Its numerical value depends upon the specific heat of water since the quantity of heat is measured in units dependent upon the amount required to raise the temperature of water. The specific heat of water is, however, very variable, as shown by the values given in Table 1, but it is therefore evident that exact numerical values of specific heats can only be given when the definition of the B.t.u. is exactly expressed.
The specific heats of all real substances vary with temperature and the values commonly used are either rough averages or those determined by experiments at one temperature. For most engineering purposes errors arising from this source may, however, be neglected.
From the definition of specific heat it follows that:
$$C = \frac{Q}{W(t_2 - t_1)} \quad \quad (9)$$
in which
$$C = \text{a mean or average specific heat over a range of temperature from } t_1 \text{ to } t_2,$$
and
$$Q = \text{the heat supplied to raise the temperature of } W \text{ pounds of material from } t_1 \text{ to } t_2.$$
PHYSICAL CONCEPTIONS AND UNITS
| TABLE I |
| Specific Heats of Water.* (Value at 5° F. taken as unity) |
| Temp. F. |
Spec. Ht. |
Temp. F. |
Spec. Ht. |
| 20 |
0.108 |
350 |
1.045 |
| 30 |
0.168 |
400 |
1.064 |
| 40 |
1.0045 |
450 |
1.086 |
| 50 |
1.0012 |
500 |
1.112 |
| 60 |
0.9970 |
550 |
1.117 |
| 70 |
0.9977 |
600 |
1.125 |
| 80 |
0.9973 |
650 |
1.128 |
| 90 |
0.9967 |
700 |
1.134 |
| 100 |
0.9967 |
750 |
1.140 |
| 120 |
0.9964 |
850 |
1.166 |
| 140 |
0.9986 |
575 |
1.152 |
| 160 |
1.0022 |
580 |
1.158 |
| 180 |
1.0038 |
590 |
1.165 |
| 200 |
| | | |
|---|
| | | |
|---|
| | | |
|---|
| | | |
|---|
| | | |
|---|
| | | |
|---|
| | | |
|---|
| | | |
|---|
| | | |
|---|
| | | |
|---|
| |
$\text{Q}_u = WC_{(t_t - t_i)} = \text{~}2.1 \times 12 \times 10^3 = \text{~}21 \text{~}B.t.u.$
$\text{Q}_u = W C_{(t_t - t_i)} = \text{~}2.1 \times 13 \times 10^3 = \text{~}23 \text{~}B.t.u.$
$\Delta Q = \text{~}9 \text{~}B.t.u.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\therefore ~\Delta h = \text{~}34.4^\circ F.$
$\\$
16
STEAM POWER
3. If the initial temperature of the silver in Prob. 2 were 150° F., what would be the final absolute temperature Fahr.
$$t_1 = t_0 + 34^{\circ}F = 150 + 34^{\circ}F = 184^{\circ}F \quad (approximately)$$
$$T_1 = 460 + 184 = 644^{\circ}F \quad Abs.$$
4. 100 lbs. of water in a 20-lb. tank of iron, both at 90° F., are placed in salt brine at 0° F. The water becomes ice at 32° F., and the temperature of the ice is lowered to 20° F., the brine being raised to 20° F.; Sp. ht. water: 10°; Sp. ht. ice: 0.5; Sp. ht. iron: 0.1; Wt. of brine: 100 lbs. The amount of water must be removed to convert liquid water at 32° F., to ice at the same temperature.
$$100(1(90-32)+143+5(32-26)+20\times101(32-20)-W)$$
$$= W\cdot0.8(20-0)$$
$$W = $84$ lbs. of brine.
11. Quantity of Heat. It is impossible to determine the total quantity of heat in or " associated with " a substance, because no means of removing and measuring all the heat contained in any real material have ever been devised. Since, however, the engineer is concerned with changes of heat content rather than with the total amount of heat entailed, this problem has little importance.
For convenience in figuring changes of heat content, it is customary to assume some arbitrary starting point or datum and to call the heat in the material in question zero at that point.
Thus, for example, if it were necessary to figure heat changes experienced by a piece of iron weighing 5 lbs., and having a specific heat of 0.1138, and the temperature of this iron never dropped below 40° F., under the conditions existing, this temperature might be taken as an arbitrary starting point above which to figure heat contents. If the iron were later found at a temperature of 75° F., " the heat content above 40° F." would be said to be
$$Q = CW(t_2 - t_1) = 0.1138 \times (5(75 - 40)) = 2.27 \text{ B.t.u.}$$
PHYSICAL CONCEPTIONS AND UNITS 17
This type of formula can only be used when the sub-
stance does not change its state between the limits of tem-
perature concerned. In the case of water which might change to steam during such a rise of temperature, it might be necessary to include other heat quantities in the cal-
culations, as shown in a later chapter.
**12. Work and Power.** Since steam engines are designed for the purpose of converting the heat energy contained in fuel into mechanical power which may be used to perform work, it will be necessary to consider the units used in measuring work and power.
**Work** was defined in a previous paragraph as the over-
coming of a resistance through a distance, by the application of a force. The product of force and distance, multiplied by the distance in feet through which the force acts, gives a product expressed in foot-pounds.
The amount of work performed in a unit of time is termed
**power,** which may be defined as the rate of doing work.
Therefore,
$$\text{Power} = \frac{\text{Force} \times \text{Distance}}{\text{Time (min. or sec.)}} \quad . . . \quad (10)$$
The unit of power used by steam engineers is the horse-
power, which is equivalent to the performance of 33,000
ft.-lbs. of work per minute, or 550 ft.-lbs. of work per second,
or 1,800,000 ft.-lbs. per hour. Therefore, the horse-power developed by any mechanism is
$$\text{h.p.} = \frac{\text{ft.-lbs. of work per min.}}{33,000} \quad . . . \quad (11)$$
Since 33,000 ft.-lbs. of work can be accomplished only by
the expenditure of 33,000 ft.-lbs. of energy and since one B.t.u. of energy is equal to 778 ft.-lbs., it follows that 33,000
ft.-lbs. of work must be the equivalent of $\frac{33,000}{778} = 42.41$ B.t.u.
It is customary to speak of power in terms of **horse-**
| Table header |
Table body |
| Column 1 |
Column 2 |
| This type of formula can only be used when the sub- |
|
| stance does not change its state between the limits of tem- |
|
| perature concerned. In the case of water which might change to steam during such a rise of temperature, it might be necessary to include other heat quantities in the cal- |
|
| culations, as shown in a later chapter. |
|
| 12. Work and Power. Since steam engines are designed for the purpose of converting the heat energy contained in fuel into mechanical power which may be used to perform work, |
|
| it will be necessary to consider the units used in measuring work and power. |
|
| Work was defined in a previous paragraph as the over- |
|
| coming of a resistance through a distance, by the application of a force. |
|
| The product of force and distance, multiplied by the distance in feet through which the force acts, gives a product expressed in foot-pounds. |
|
| The amount of work performed in a unit of time is termed |
|
| power, which may be defined as the rate of doing work. |
|
| Therefore, |
|
| \[ \text{Power} = \frac{\text{Force} \times \text{Distance}}{\text{Time (min. or sec.)}} \quad . . . \quad (10) \] |
|
| The unit of power used by steam engineers is the horse- |
|
| power, which is equivalent to the performance of 33,000 |
|
| ft.-lbs. of work per minute, or 550 ft.-lbs. of work per second, |
|
| or 1,800,000 ft.-lbs. per hour. Therefore, the horse-power developed by any mechanism is |
|
| \[ \text{h.p.} = \frac{\text{ft.-lbs. of work per min.}}{33,000} \quad . . . \quad (11) \] |
|
| Since 33,000 ft.-lbs. of work can be accomplished only by |
|
| the expenditure of 33,000 ft.-lbs. of energy and since one B.t.u. of energy is equal to 778 ft.-lbs., it follows that 33,000 |
|
18
**STEAM POWER**
**power-hours.** One horse-power-hour means the doing of work equivalent to one horse-power for the period of one hour, or the doing of work at the rate of 33,000 ft.-lbs. per minute for an hour. A horse-power-hour is therefore equivalent to 33,000,000 = 1,380,000 ft.-lbs. As 35,000 ft.-lbs. are equivalent to 42.41 B.t.u., it follows that 42.41×60 = 2546.0 B.t.u. are the equivalent of one horse-power-hour.
The number 2545 should be memorized as it is very often used in steam-power calculations. If an engine could deliver one horse-power-hour for every 2545 B.t.u. it received, it would be working without losses of any kind that is, all the heat which entered it was converted into the form of useful mechanical energy. It will be shown later that this is impossible even in the most perfect or ideal engine.
**REVIEW PROBLEMS**
1. Express 32° F. in degrees centigrade.
2. Express 150° F. in degrees centigrade.
3. Express 175° C. in degrees Fahrenheit.
4. Express the results of problems 1, 2 and 3 in absolute values.
5. What is the heat equivalent of 233,000 ft.-lbs. of work?
6. Find the heat supplied 10 lbs. of water when its temperature is raised from 100° F. to 150° F. The maximum heat over this range is 0.997.
7. Find the temperature change of 2 lbs. of lead (p. ht. 0.0314) when 20 B.t.u. are added to it.
8. How many B.t.u. must be abstracted to lower the temperature of 15 lbs. of water from 212° F. to 32° F., assuming the specific heat of water is 1.015.
9. Find the weight of water which will have its temperature tripled in value by the addition of 250 B.t.u., the final temperature being 150° F., assuming no heat loss.
10. The specific heat of a piece of wrought iron is 0.113 and of a given weight of water is 1.015. 1 cu. ft. of water weighs approximately 825 lbs. Find the increase in temperature of 4 cu. ft. of water when a certain temperature of 65° F., results from placing in the water a piece of iron weighing 3 lbs., at a temperature of 900° F.
![]() A diagram showing a steam engine with various parts labeled.
PHYSICAL CONCEPTIONS AND UNITS 19
11. Find the final temperature of the mixture, when 100 lbs. of iron (sp. ht. = 0.113), at a temperature of 120° F, are immersed in 300 lbs. of water (sp. ht. 1.001) at a temperature of 50° F.
12. If 1 lb. of silver (sp. ht. 10.5) at 800° F are immersed in water at 60° F, what will be its final temperature? Assume Sp. ht. water = 1.
What weight of water is necessary?
13. An engine is developing 10 horse-power. Express this in ft.-lb., in Btu., and in cal. What quantity of heat energy equivalent to this quantity of mechanical energy.
14. A pump raises 1000 lbs. of water 50 ft. every minute. How many horsepower does it develop? The equivalent horse-power.
15. An engine develops 1,980 ft.-lb., of work at the fly-wheel per minute.
(a) What horse-power developed?
(b) If this engine operated in this way for an hour, how many horse-power hours would it make available?
(c) What would be the equivalent of this number of horse-power hours in British thermal units?
![]() A diagram showing a cylinder with pistons and connecting rods, representing an internal combustion engine.
CHAPTER II
THE HEAT-POWER PLANT
13. The Simple Steam-Power Plant. The various pieces of apparatus necessary for the proper conversion of heat energy into mechanical power constitute what may be called a "Heat-Power Plant," just as the apparatus used in obtaining mechanical energy from moving water is called an hydraulic or water-power plant. These plants are distinguished as "Steam-Power Plants"; "Gas-Power Plants"; etc., according to the way in which the heat of the fuel happens to be utilized.
The apparatus around which the plant as a whole centers, that is, the apparatus in which heat energy is received and from which mechanical energy is delivered, is termed the engine or prime-mover. This heat engine may use steam generated in boilers and may require certain apparatus, such as condensers, pumps, etc., for proper operation; or it may use gas, generated in gas-producers requiring coolers, scrubbers, and other accessories, depending upon the class of fuel used and upon certain economic considerations. Again, the power-plant may simply contain an internal-combustion engine using natural gas, gasoline or oil, a type of plant which is now very common.
But whatever type of plant is used, a general method of operation is common to all. Heat energy in fuel is constantly fed in at one end of the engine, and mechanical energy is delivered at the other end. The steam-power plant will be briefly described in the following paragraphs, showing the cycle of events with the attendant losses through the system.
20
THE HEAT-POWER PLANT
21
![]() A detailed diagram of a steam power plant, labeled with various components and their functions.
Fig. 8.—Diagram of a Steam Power Plant.
22
STEAM POWER
In Fig. 8 is shown a simple steam-power plant which con-
verts into mechanical energy part of the heat energy, origi-
nally stored in coal, by means of a prime-mover called a
steam-engine. The main pieces of apparatus used in this
type of plant are the steam-boiler; the steam-engine; the
condenser; the vacuum pump; and the feed-pump. The
energy stream shows the various losses occurring through-
out the plant. These losses cause the “delivered energy”
stream to be only a small fraction of the total heat sent into
the system.
14. Cycle of Events. 1. Fuel is charged on the grate
under the boiler, where it is burned with the liberation of a
large amount of energy. Air is drawn or forced through
the grates in proper proportions to support this combustion.
The hot gases pass over the heating surfaces and take
path set by the baffle plates, so that the largest possible
amount of heating surface may be presented to the products
of combustion.
There are certain losses accompanying this operation,
such as radiation, loss of volatile fuel passing off unburned,
loss of fuel through the grate, and loss of heat through the
excess air which must always be supplied to insure com-
bustion.
2. That part of the heat in the gases which is not lost
by radiation from the boiler and in the hot gases flowing up
the stack passes through the heating surfaces of the boiler
to the water contained in it. Here it is absorbed, and the
total heat energy in the fuel passes through the heating
surfaces and serves to raise the temperature of the water
to the boiling point at the pressure maintained, and to con-
vert this water into steam according to the requirements.
3. Having obtained steam within the boiler, it is led
through a pipe line to a cylinder in which a piston is moved.
Of the heat stored in the steam is converted into mechanical
energy by the action of that steam against a piston. The
steam is then discharged, or exhausted, from the engine
THE HEAT POWER PLANT
at a much lower temperature and pressure than when it entered.
From 5 to 22 per cent of the available heat in the steam is converted into mechanical energy in the engine cylinder, and because of frictional and other losses occurring in the mechanism, only from 85 to 95 per cent of this energy is turned into useful work at the fly-wheel.
4. In some plants, known as non-condensing plants, the exhaust steam, which still contains the greater part of all the heat received in the boiler, is discharged to the atmospheres, where it cools and condenses. This is known as condensing plants, the exhaust steam is led to a condenser, where it is condensed by cold water, which absorbs and carries away the greater quantity of the heat not utilized in the engine. The condensed steam or "condensate" is then either discharged to the sewer or transferred by means of a pipe called a feed-pipe to another boiler, drawn by means of the feed-water-pump, raised to the original pressure of the steam, and returned to the boiler. Here it is again turned into steam and the cycle of operations outlined above is repeated. Naturally there is some loss due to evaporation and leaks throughout the system, so that the efficiency of such a plant cannot be very high.
The series of events just described constitutes a complete, closed cycle of operations, wherein the water is heated, vaporized, condensed and returned to the boiler, having served only as a medium for the transfer of heat energy from fuel to engine and the conversion of part of that energy within the cylinder. The water in such a case is known as working substance.
It is often more convenient to discard the working substance after it leaves the cylinder, as suggested above in the case of a non-condensing plant; or, as in the case of a gas engine, where a new supply of working substance must be supplied for each cycle, because the burned gases of the previous cycle cannot be used again.
24 STEAM POWER
**15. Action of Steam in the Cylinder.** In order to prepare for the more detailed discussion of the action of the steam in the engine cylinder, to be taken up in a later chapter, a brief outline of the events occurring within the prime-mover will be considered at this point.
Steam enters the cylinder through some kind of admission valve, and acts upon the piston, just as the latter has approximately reached one end of its stroke and is ready to return. The steam, being hot, causes it to expand behind the piston, thereby driving the latter out and performing work at the fly-wheel. At about 90 or 15 per cent of the stroke, the exhaust valve opens, and the steam begins to exhaust, the pressure within the cylinder dropping almost to atmospheric or to that mainta- ned by the atmosphere. As long as the steam has reached the end of its stroke. On the next or return stroke the remaining steam is forced out through the exhaust port, until, at some point before the end of the piston travel, the exhaust valve closes and the low-pressure steam trapped in the cylinder is compressed into the clearance space so that its pressure rises. Admission then occurs, and the cycle is repeated.
The diagram given in Fig. 9 illustrates the operation of steam within the cylinder. This diagram is plotted between pressures of steam within the cylinder as ordinates and corresponding piston positions as abscissas.
The method of obtaining such a diagram known as an indicator-diagram will be fully described in a later chapter. Since vertical ordinates represent pressure in pounds per square inch, and horizontal abscissas repre-
![]()
Diagram showing steam engine indicator diagram.
Fig. 9.—Steam Engine Indicator Diagram.
15
THE HEAT-POWER PLANT 25
sent feet moved through by the piston, the product of these two must be work. But the product of vertical by horizontal distances must also give area. Therefore, by
![]() (a) Diagram showing the relation between the energy supplied to the cylinder and the work done by the piston.
(a)
![]() (b) Diagram showing the relation between the energy supplied to the cylinder and the work done by the piston.
(b)
Fig. 10.—Hydraulic Analogy.
finding the area enclosed within the bounding lines of the cycle and multiplying this by a proper factor, the foot-pounds of work developed within the cylinder can be determined.
16. Hydraulic Analogy. The operation of heat-engines is analogous to that of water-wheels. A water-wheel de-
26
STEAM POWER
ve'ops mechanical energy by receiving water under a high head, absorbing some of its energy, and then rejecting the fluid under a low head. Similarly, the heat-engine receives heat energy at a high temperature (head), absorbs some of it by conversion into mechanical energy, and then rejects the rest at a low temperature (head).
This analogy can be carried still further. The water-wheel cannot remove all the energy from the water, nor can the heat-engine remove all the heat-energy from the working substance. There is a certain loss in the material discharged in both cases and this cannot be avoided.
This analogy is illustrated diagrammatically in Fig. 10 (a) and (b), in which the widths of the streams represent quantity of energy.
CHAPTER III
STEAM
17. Vapors and Gases. When a solid is heated, under the proper pressure conditions, it ultimately melts or fuses and becomes a liquid. The temperature at which this occurs depends upon the particular material in question and upon the pressure under which it exists. Ice, which is merely solid water, melts at 32° F., under atmospheric pressure, while iron melts at about 2000° F., under atmospheric pressure.
When a liquid is heated, it ultimately becomes a gas, similar to the air and other familiar gases. If this gas is heated to a very high temperature and if the pressure under which it is held is not too great, it very nearly obeys certain laws which are simple and which are called the laws of ideal gases.
When the material is in a state between that of a liquid and that in which it very nearly obeys the laws of ideal gases, it is generally spoken of as a vapor. This term is used in several different ways and with several different modifying adjectives which will be explained in greater detail in later sections.
18. Production of Steam. Of the many vapors used by the engineer, steam or water vapor is probably the most important, because of the ease with which it can be formed and also because of the tremendous field in which it can be used. It is generated in a vessel known as a steam boiler, which is constructed of metal in such a way that it can contain water, and that heat energy, liberated from burning fuel, can be passed into the water, converting part or all of it into water vapor, that is, into steam.
37
28
**STEAM POWER**
The properties of water vapor must be thoroughly understood before the steam engine and steam boiler can be studied profitably. Probably the easiest way of becoming familiar with these properties is to study the use made of heat in the generation of steam from cold water.
**19. Generation of Steam or Water Vapor.** For the purpose, a vessel is used which is cylindrical in form, fitted with a frictionless piston of known weight, as shown in Fig. 11, (a) and (b), the whole apparatus being placed under a bell-jar in which a perfect vacuum is maintained.
![]() A diagram showing a cylinder with a piston inside, connected to a vacuum pump. The piston is labeled (a) and (b). A pressure gauge is attached to the top of the cylinder.
Fig. 11.—Formation of Steam at Constant Pressure.
Assume one pound of water in the cylinder, with the piston resting on the surface of the liquid. There will be some definite pressure exerted by the piston upon the surface of the liquid, and its value will be determined entirely by the weight of the piston.
It is convenient in engineering practice to refer all vaporization phenomena to some datum temperature, and since the melting point of ice, $32^\circ$ F., is a convenient reference point, it is used as a standard datum temperature, in practically all steam-engineering work. Therefore, assuming the water in the jar to be at $32^\circ$ F., if heat is applied the temperature of the liquid will rise approximately $1^\circ$ F.
STEAM 29
for every B.t.u. of heat added, since the specific heat of water is approximately unity.
Experiment shows that for each pressure under which the water may exist some definite temperature will be attained at which further rise of temperature will cause the liquid will
![]()
A graph showing Pressure (in psia) on the y-axis and Temperature (in °F) on the x-axis.
The curve starts at the origin and rises steeply to the right, indicating a rapid increase in pressure with increasing temperature.
The y-axis ranges from 0 to 500 in increments of 100.
The x-axis ranges from 0 to 500 in increments of 100.
Fig. 12. —Pressure-Temperature Relations, Saturated Water Vapor.
begin to change to a vapor, that is, to vaporize. The temperatures at which vaporization occurs at different pressures are called the temperatures of vaporization at those pressures.
The temperatures of vaporization of water are plotted against pressure in Fig. 12. It should be noted that the values of vaporization temperature increase very rapidly for small pressure changes in the case of low pressures, but
30 STEAM POWER
that, for the higher pressures, the variation of temperature is very small for enormous variations of pressure. This fact is of great importance in steam engineering.
The temperatures of vaporization are tabulated with other properties of water in the "Steam tables" and are constantly referred to by engineers. An example of such a table is given on pp. 392 to 399.
Returning now to the apparatus under discussion, as heat is supplied, the temperature of the water will rise from 32° F. until it reaches the temperature of vaporization corresponding to the pressure at which the water is by the piston. When this temperature is reached vaporization will begin, and if sufficient heat is supplied, will continue without change of temperature until the water is entirely converted into vapor.
Up to the time at which vaporization starts the volume of the water will change very little, so that the piston will be raised only a negligibly small amount and practically no work will be done upon it by the water. On the other hand, when vaporization occurs the volume of the material will change by a very large amount and the piston will be driven out (raised) against the action of gravity. That is, work is done upon the piston by the water during the increase in volume which accompanies vaporization.
It is found that a very great quantity of heat is used up during the process of vaporization despite the fact that no temperature change occurs. This is described by saying that the heat supplied is supplied partly to raise the temperature of the steam in the steam engine and the quantity of heat is therefore spoken of as the latent heat of vaporization. It is assumed to consist of two parts, that used for separating the liquid molecules against their attractive forces and that used for doing the work which is done upon the piston as it is moved upward. The former is called the internal latent heat because it is used for doing internal or intermolecular
![]() A diagram showing a piston moving up and down in a cylinder containing water.
STEAM 31
work; the latter is called external latent heat because it is used for the doing of external work.
It is to be noted that the internal latent heat may be assumed to be tied up in some way within the molecular structure of the material and hence to be in the steam. The external latent heat, on the other hand, is used up fast as supplied for the purpose of driving the piston out against the pressure of the steam. When the piston has been raised to any point, the energy used in raising it is not in the steam, but is stored as potential energy in the piston. To get it back the piston must be allowed to drop. The term "external" is therefore well chosen; the external latent heat is not in sense in the steam; it is stored in certain parts of the piston.
After the constant temperature vaporization is complete, the further addition of heat will again cause a rise of temperature and a gradual increase of volume. Such raising of the temperature of steam already formed is called superheating and results in carrying the vapor nearer and nearer to the conditions which nearly obey the laws of ideal gases. Since an increase of heat causes incomplete superheating, the molecules of the vapor must move farther and farther apart as superheating progresses.
Vapor in the condition in which it is formed from the liquid and which has the same temperature as the liquid from which it forms is called saturated vapor. This term can be pictured mentally by thinking that a number of molecules of vapor are packed into a given space; the addition of heat to saturated vapor would cause superheating and the separation of the molecules so that fewer could be contained in a given space.
20. Heat of Liquid q or h. Returning once more to the situation described in the preceding section, heat was added to water initially at 32° F. until the temperature of vaporization corresponding to the existing pressure was attained. The heat added during this period is
32
STEAM POWER
called the heat of the liquid, and is usually designated by the letters $q$ or $h$. If the mean specific heat of water at constant pressure ($C_p$) for the temperature range under consideration were constant, and equal to 1, then, since
$$q = C_p (t_4 - 32)$$
in which $t_4$ is the temperature of vaporization, it would follow that, for this pressure
$$q = t_4 - 32 \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \\ q = t_4 - 32 = 240\ B.t.u.$$ (12)
Therefore, if water boils under a pressure of 50 lbs. at a temperature, read from the steam tables, of $281^\circ F$, it would follow that
$$q = 281 - 32 = 249\ B.t.u.$$ But the steam tables (see p. 394) for this pressure (50 lbs.) give $q = 250.1\ B.t.u.$ indicating, as was shown in Chap. I., that the specific heat of water does not remain constant, and for this case the mean value must have been approximately 1.004 as indicated by the following calculation.
$$q = C_p (t_4 - 32) = 250.1\times 249$$ so that
$$C_p = {250.1\over{249}} = 1.004 +$$
Hence it is always advisable to use the steam table values of $q$, except for very approximate calculations.
**21. Latent Heat of Vaporization, r or L.** The heat supplied during the period of vaporization has already been referred to as the latent heat of vaporization and has been divided into internal and external latent heats.
The internal latent heat is generally designated by $\rho$ or by $I$ and the external latent heat by the group of letters $APw$ or by $E$. The group $APw$ merely represents the prod-
STEAM 33
uct of pressure, $P$, by volume change during vaporization, $u$, and by the frictional $\frac{1}{2}v^2$, which is represented by $A$. The product of the first two terms gives external work in foot-pounds during vaporization, and this is 778 (Joule's Equivalent) converts it to heat units to correspond with the other values. It should be noted that $P$ in this expression stands for pressure in pounds per square foot.
The total latent heat of vaporization is generally designated by $r$ or by $L_0$, and it follows from what has preceded that
$$r = p + A P u \quad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (13)$$
The value of $r$ for atmospheric pressure, that is, for a temperature of vaporization of 212° F., is very often used in engineering and should be memorized. Its value is now generally taken as 970.4 B.L.U., though recent work would seem to indicate a value of about 972 as nearer the truth.
22. Total Heat of Dry Saturated Steam, $\lambda$ or H. The total heat required to convert a pound of water at 32° F. into a pound of saturated vapor at some temperature $t_s$ is called the total heat of dry saturated steam above 32° and is designated by $\lambda$ or by $H$. It is obviously the sum of the quantities which have just been considered, so that
$$\lambda = q + r + q + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p + A P u = q + r + p +
14
24
STEAM POWER
have a quality of 90 per cent. Quality expressed as a decimal fraction is designated by the letter $x$, so that if $x$ is said to be equal to 0.8 in referring to a certain sample of steam, it means that that steam sample consists of 80 per cent by weight of saturated steam and 20 per cent liquid at the same temperature.
Since the water in wet steam has the same temperature as the steam, it contains all the heat of the liquid which it would contain if it had been converted into vapor but it obviously contains no internal heat of vaporization. It follows that the total heat in a pound of wet steam (one pound of a mixture of saturated steam and water) with quality equal to $x$ is
$$\text{Heat per pound} = q + x \cdot r = q + xp + xAPu \quad . \quad (15)$$
The letter $\lambda$ should never be used in designating the total heat per pound of wet steam, as it has been chosen as the symbol of the total heat per pound of dry, saturated steam.
**24. Heat of Superheat.** When the temperature of saturated steam exceeds that of the saturation temperature, that is, when it is superheated, a very definite quantity of heat is required. The quantity required per pound per degree would, by definition, be the specific heat of the material in question.
If the specific heat of superheated steam were reasonably constant, the heat required to raise its temperature at constant pressure from saturation temperature to some higher value $t_2$ would be given by the expression
$$\text{Heat required per pound} = C_p(t_2 - t_s)$$
but superheated steam, as handled by the engineer, is generally comparatively near the saturated condition, and under these circumstances the values of the specific heat vary rapidly with temperature. A diagram showing the extent of these variations is shown in Fig. 13. It will be observed that for low pressures the specific heat is approxi-
STEAM 35
mately constant at a value below 0.5 for any given pressure,
but that for very high pressures it varies widely over a comparatively small temperature range. Thus at 600 lbs.
per square inch the specific heat changes from unity at
about 510° F, to 0.6 at about 550° F.
Practically, it is customary to use the type of equation
just given and to substitute a mean specific heat over the
required temperature range for the specific heat which can-
not be assumed constant without too great an error. The
![]() A graph showing the progressive values of \( C_p \) (specific heat) for water vapor with temperatures ranging from 100 to 900 degrees Fahrenheit.
Fig. 13.—Progressive Values of Specific Heat, \( C_p \), Water Vapor.
equation for heat required to raise the temperature from
\( t_1 \) to \( t_2 \) is then
Heat of superheat per pound = \( C_{pm}(t_2 - t_1) \). . . (16)
in which \( C_{pm} \) stands for the mean specific heat at constant
pressure over the temperature range from \( t_1 \) to \( t_2 \).
Values of mean specific heats of saturated steam are given in Table 17. The curves giving the mean specific heat between saturation temperatures and various higher temperatures at different pressures,
35
36 STEAM POWER
**25. Total Heat of Superheated Steam.** The total heat required to convert one pound of water at 32° F. into superheated steam at a temperature of $t_2$° F. under constant pressure conditions is obviously
$$\text{Total heat per pound} = q + r + C_{ps}(t_2 - t_0) \quad . \quad (17)$$
![]() A graph showing the variation of mean specific heat with temperature above saturation T°. The x-axis represents temperatures above saturation T°, ranging from 0° to 80°. The y-axis shows mean specific heat, ranging from 0.40 to 0.80. Curves are drawn for different values of $t_2 - t_0$, labeled as follows: 0°, 10°, 20°, 30°, 40°, 50°, 60°, 70°, 80°.
Fig. 14.—Variation of Mean Specific Heat, Water Vapor.
and representing the degrees of superheat ($t_2 - t_0$) by $D$, as is customary, this becomes
$$\text{Total heat per pound} = q + r + C_{ps}D \quad . \quad (18)$$
**26. Specific Volume of Dry Saturated Steam, Vor S.** The volume occupied by one pound of a substance is spoken of as the **specific volume** of that material. In the case of dry saturated steam there are as many specific volumes as there are pressures under which the steam can exist. These values are generally tabulated in steam tables and are represented by the letter $V$ or the letter $S$.
STEAM 37
The values of the specific volumes of steam at different pressures are given in Fig. 15. It is important to note the very gradual change of specific volume at high pressures and the very rapid change and enormous increase at low pressures. These facts have considerable influence on steam engineering practice.
![]() A graph showing pressure-volume relations for saturated water vapor. The x-axis represents specific volume (in cubic feet per pound) and the y-axis represents pressure (in pounds per square inch). The curve starts at a high value of specific volume and decreases rapidly with increasing pressure.
Fig. 15.—Pressure-Volume Relations, Saturated Water Vapor.
A curve giving properties of saturated steam is called a saturation curve, so that this name may be, and often is, applied to the curve given in Fig. 15.
The volume occupied at any pressure by half a pound of dry saturated steam will obviously be half that occupied by one pound of such material at the same pressure, and
38
STEAM POWER
The same statement can be made for any other fraction of a pound. It follows that if the small volume occupied by liquid water in wet steam be neglected, the volume occupied by one pound of steam (measured at 50 per cent quality) can be termed equal to that of one pound of dry saturated steam at the same pressure. A similar statement could of course be made for any other quality and a corresponding fraction.
Hence if one pound of " wet steam " at a given pressure is found to have such a volume that it would be indicated by point b in Fig. 16, the quality of this material must be given by the expression $x = \frac{a}{a + b}$ if the volume occupied by liquid water in the mixture be neglected.
27. Specific Density of Dry Saturated Steam, $\frac{V}{V}$ or $\delta$. The weight per cubic foot of saturated steam is spoken of as its specific density. The specific density is obviously the reciprocal of the specific volume and is therefore $\frac{1}{V}$.
Fro. 16.-Determining Quality from Volume.
28. Reversal of the Phenomena Just Described. If any process which has resulted in the absorption of a quantity of heat by a substance be carried out in the reverse direction, the same effect will be produced as before. It follows that a pound of dry saturated steam will give up the total latent heat of vaporization when condensed to liquid at the same temperature, and that the resultant pound of hot water will give up the total heat of the liquid if cooled to 32° F.
29. Generation of Steam in Real Steam Boiler. The steam boiler is equivalent to a vessel partly filled with water
STEAM 39
and fitted with means for supplying heat to the water and for carrying off the vapor formed. This is shown diagrammatically in Fig. 17. At first glance this would not seem to be at all similar to the cylinder and piston already constructed, but when we consider what is meant by the generation of steam is concerned. The flow of steam out of the steam-pipe is restricted to the extent necessary to maintain a high and constant pressure within the boiler; and, when in regular operation, steam is formed within the
![]() A diagram showing the formation of steam in a steam boiler.
Fig. 17.—Formation of Steam in a Steam Boiler.
boiler under this pressure just as fast as necessary to replace that flowing out.
By picturing the steam as flowing out in layers or lamina these lamina can be imagined as taking the place of the piston in the apparatus of Fig. 11, and each pound of steam formed will then push a piston before it exactly as was assumed in the latter case.
**30. Gauge Pressure.** The steam pressure in a boiler is commonly determined by means of an instrument called a **pressure gauge**. These instruments are almost always constructed about as shown in Fig. 18 (a) and (b). The Bourdon spring is a tube of elliptical section bent approximately into
40
**STEAM POWER:**
the arc of a circle. One end of this tube is connected directly to the pressure connection of the gauge and the other end is closed and connected to a toothed sector as shown.
When the pressure inside a tube of this character is increased, the tube has a tendency to unroll or straighten out, and in so doing it moves the toothed sector in such a way as to rotate the pointer or gauge hand and make its end move over the scale in the direction of increasing pressure. With diminishing pressure the tube again rolls up and rotates the hand in the opposite direction.
![]() (a) shows a Bourdon Pressure Gauge with a toothed sector attached to one end of a tube. The other end of the tube is connected to the pressure connection of the gauge.
![]() (b) shows a close-up view of the toothed sector and spring mechanism. The spring is compressed by the movement of the toothed sector.
FIG. 18.—Bourdon Pressure Gauge.
Instruments of this kind are so made and adjusted that the hand points to zero when the gauge is left open to the atmosphere. Under such conditions the pressure inside the tube is equal to that of the atmosphere and is not zero. The gauge therefore only indicates pressures above atmospheric on its scale, and the total pressure inside the boiler is really that shown by the gauge plus that of the atmosphere.
Pressures as indicated by the gauge are called **gauge pressures.** Pressures obtained by adding the pressure of
STEAM 41
the atmosphere to the reading of the gauge are known as absolute pressures. Then
$$Absolute\ Pressure = Gauge\ Pressure + Atmospheric\ Pressure$$
and
$$Gauge\ Pressure = Absolute\ Pressure - Atmospheric\ Pressure.$$
In accurate work the existing atmospheric pressure should be determined by means of the barometer, but for ordinary, approximate calculations and for cases in which no barometric data are available, it is customary to assume the pressure of the atmosphere to be equal to 14.7 lbs. per square inch. This is very nearly true, on the average, at sea level, but is generally far from true at higher elevations.
**PROBLEMS**
1. Determine by means of the steam tables the temperature, total heats, heats of liquid, internal and external latent heats, and the specific volumes of 1 lb. of dry, saturated steam under the following absolute pressures (lbs. per sq. in.): 15, 30, 75, 190 and 400.
2. Determine by means of the steam tables the heats of vaporization and total heats for 2 lbs. of dry saturated steam at the following temperatures in °F.: 108.3, 212 and 327.
3. Determine the volumes occupied by 1 lb. of dry saturated steam under each of the following pressures in lbs. per square inch absolute.
4. Determine the heats of the liquid, latent heats of vaporization and total heats for 1 lb. of saturated steam with a quality of 90% at the following absolute pressures: 25, 50, 75, 125.
5. Determine by means of the steam tables the volume of saturated steam with quality of 57% at a pressure of 125 lbs. per square inch absolute.
6. What space will be filled by 20 lbs. of dry saturated steam at a pressure of 150 lbs. per square inch absolute?
7. What space will be filled by 20 lbs. of saturated steam at a pressure of 150 lbs. per square inch absolute and with a quality of 90% if the volume occupied by the water present be neglected?
8. If it is many pounds of dry saturated steam at a pressure of 75 lbs. per square inch absolute will be required to fill a space of 10 cu. ft.?
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STEAM POWER
9. How many pounds of saturated steam with quality 96%, and at a pressure of 110 lbs. per square inch absolute will be required to fill a space of 8 cu. ft.?
10. Heat, or internal work, measured in B.t.u., is done when 1 lb. of water at the temperature of 212° F, is converted into dry saturated vapor at the same temperature?
11. How much heat is required in foot-pounds is done when 2 lbs. of water at a temperature of 312° F, are converted into 90% quality steam at the same temperature?
12. How much heat is required for doing internal work during the vaporization of 1 lb. of water under conditions that the total latent heat of vaporization is 82.7 B.t.u. and the external latent heat is 83.3 B.t.u.?.
13. What is the quality of steam containing 1000 B.t.u. above 32° F, per pound when under a pressure of 150 lbs. per square inch absolute?
14. Heat is added to 1 lb. of mixed steam and water while the pressure is maintained constant at 100 lbs. per square inch absolute. The percentage of steam in the mixture is increased thereby from
50% to 95%.
(a) How much heat was added?
(b) How much internal latent heat was added?
(c) How much external latent heat was added?
15. How much heat is necessary to completely vaporize 1000 lbs. of water at a temperature of 92° F, when pumped into a boiler in which steam is generated at a pressure of 150 lbs. per square inch gauge and the temperature of the feed water is 205° F, water is given as g steam table for a temperature of 92° F.
16. Find the amount of heat necessary to produce in a boiler, 300 lbs. of steam at a pressure of 150 lbs. per square inch gauge when the feed water has a temperature of 205° F.
17. What volume would be occupied by the material leaving the boiler in problem 16, neglecting volume occupied by water?
CHAPTER IV
THE IDEAL STEAM ENGINE
31. The Engine. If the cylinder and piston assumed in the discussion of the last chapter be imagined as turned into a horizontal position and fitted with a frame, piston
![]() A diagram of a simple steam engine, showing the cylinder, piston, connecting rod, crankshaft, crosshead, flywheel, and foundation or base.
Fig. 19. Simple Steam Engine.
rod, crosshead, connecting rod, crank shaft and flywheel as in Fig. 19, a device results which might be used as a steam engine for the production of power. By adding heat to, and taking heat from, the water and steam in the cylinder in the proper way and at the proper time, the water and steam can be made to do mechanical work upon the piston. The piston can transmit this work through the mechanism to the rim of the flywheel, and it can be taken from the rim by a belt connected to a pulley on a machine which is to be driven.
43
**STEAM POWER**
44
To make the analysis easier, a simplified type of engine will be assumed. It is shown in Fig. 20 and consists of the same cylinder, piston and piston rod as just described. A wire is fastened to the end of the piston rod and run back over a pulley in such a way that a weight fastened to the free end of the wire will be raised by the motion of the piston. The weight is made up of two parts, one large and one small. When both are on the wire the pull which they exert causes the piston to exert a high pressure upon whatever is con-
![]()
A diagram showing a simplified steam engine.
- Cylinder labeled "Cylinder"
- Piston labeled "Piston"
- Piston Rod labeled "Piston Rod"
- Wire labeled "Wire"
- Pulley labeled "Pulley"
- Large weight labeled "Large weight"
- Small weight labeled "Small weight"
- Pressure labeled "Pressure"
- Volume labeled "Volume"
Fig. 20—Simplified Steam Engine.
tained in the cylinder. When only the small weight hangs on the wire, the piston exerts a much lower pressure upon the material in the cylinder.
Imagine that the piston and the walls of the cylinder are made of some ideal material which will not receive or conduct heat. Assume also that the cylinder is fitted with a permanent head which is a perfect conductor of heat. These conditions are of course ideal but are assumed for the sake of simplicity.
Assume further that, when one pound of water is con-
THE IDEAL STEAM ENGINE
45
ained in the cylinder and the piston is driven into the cylinder by the two weights until the space between the piston and the cylinder head is just large enough to contain the pound of water, the piston exerts a high pressure equal to $P_1$ pounds per square inch against the water. The volume of this water and the pressure upon it can be represented by the points of the $PV$ diagram, Fig. 20.
32. Operation of the Engine. With conditions as described in the preceding paragraphs, imagine a flame or other source of heat at high temperature to be brought into contact with the contents of the cylinder so as to pass heat into them. This will raise the temperature of the water within to the temperature of vaporization and ultimately vaporize it. As the water vaporizes it will push the piston out of the cylinder just as described in the last chapter and a horizontal line such as in Fig. 20 will represent the increase of volume (vaporization) at constant pressure $P_1$. The point where this line intersects with the vertical line representing one pound of dry saturated vapor at pressure $P_1$. Obviously, the steam, as it is formed, does work in driving out the piston against the resistance offered by the weights which must be raised.
If a stop is provided which will prevent the movement of the piston beyond some point corresponding to the point $b$, it will be possible to remove the larger weight when that point is reached and the high pressure steam will hold the piston and rod hard against the stop. If now some cooling medium is applied, such as a large piece of ice held against the conducting head of the cylinder or water running over that head, heat will be abstracted and a partial condensation of steam will occur. The steam will then will occur. As condensation progresses the pressure will drop because there will be less and less steam, by weight, in a given volume. Such a process would be indicated by the line bc which represents a drop of pressure, while the volume
46
STEAM POWER
contained within the cylinder walls between head and piston remains constant.
When some point $c$ is reached, the steam pressure will have been reduced to a value equal to that exerted by the small weight, and the piston will be driven in toward the cylinder head while the heat absorbing medium continues to remove heat from the piston and cylinder by conduction and radiation. The combination of piston motion and heat absorption will be so regulated that the pressure remains constant at $P_2$ during this process, because the weight will move the piston inward just as fast as necessary to maintain a constant pressure. If sufficient heat is absorbed, the pound of material with which the piston is loaded will all be condensed or liquefied and will just fill the volume $V_a$.
The heat absorbing body may now be removed and an infinitesimal motion of the piston toward the head would serve to raise the pressure on the liquid water from $P_2$ to $P_1$, so that the volume $V_a$ may be taken equal to the volume $V_b$ and the work done must be equal to $W_{ab}$. It would then represent an increase of pressure at constant volume. This might be caused by hanging a weight of the larger size on the wire when condition $d$ was reached.
Having brought the material, or working substance, back to conditions originally shown in Fig. 3, at high temperature, it can again be brought in contact with the end of the cylinder and the entire cycle carried through once more. There is obviously no reason why it could not be repeated as often as desired.
**30. Work Done by the Engine.** If the device just described is used as a steam engine, it must actually make mechanical energy available, that is, it must convert into mechanical energy part of the energy supplied it. It is now necessary to see whether it does.
Water vaporizing and increasing in volume as from $V_a$ to $V_b$ was shown in the last chapter to do work upon the piston confining it. Work has been shown to be equal to
THE IDEAL STEAM ENGINE
(total force × total distance) and in this case if $L$ represents the distance in feet traveled by the piston, the work done by the steam upon the piston while the latter moves from $a'$ to $b'$ must be
Work done on piston = total force × distance
$= P_1 \times area of piston \times L \ldots ft.-lbs.$
But the product of area of piston in square feet by distance traveled in feet is equal to the piston displacement or volume swept through by the piston, that is $(V_s - V_a)$ cubic feet. Therefore
Work done on piston = $P_1 (V_s - V_a)$ ft.-lbs.
$(19a)$
$= \frac{P_1 (V_s - V_a)}{778}$ B.t.u. $(19b)$
The first form of this expression $P_1 (V_s - V_a)$ is very obviously represented by the area under the line $ab$ in Fig. 20 and this area therefore represents the work done by the steam upon the piston during the change of volume at constant pressure represented by that line. While the steam is supplying this amount of energy to the piston or doing this amount of work upon the piston, the latter does an equal amount of work upon the steam, and a frictionless mechanism is assumed. In such a case the total weight hung on the wire multiplied by the distance raised would therefore give the same result in foot-pounds as that just obtained.
It should be noted that Eq. (19) is merely an expression of the total work done during vaporization, that is, an expression of amount of heat which is used for the doing of external work. It is the exact equivalent of the external latent heat previously discussed. In fact, the group of symbols $APu$ is really a condensation of Eq. (19) formed by putting $A$ for $\frac{1}{778}$ and $u$ for $(V_s - V_a)$.
48
STEAM POWER
The line $cd$ also represents a change of volume at constant pressure and the same type of formula as applied to $ab$ will express the work done during this process. In this case, however, the piston is being pushed into the cylinder by the steam, and energy is being supplied to push the piston in. This energy is equal to the weight of the small weight (pounds) multiplied by the distance it falls (feet). The piston is therefore doing work upon the steam, and the amount is
Work done on steam = $\frac{P_2(V_2 - V_1)}{78}$ ft.-lbs. . . . (20)
= $\frac{P_2(V_2 - V_1)}{78}$ B.t.u. . . . (21)
The first form of expression also represents the area under the line of and this area therefore represents the work done by the piston upon the steam mixture in the cylinder during the process represented by $cd$.
No work can be done by steam on piston or by piston on steam during the processes represented by $bc$ or $da$ because both the weights and the piston are stationary during these changes and it has already been shown that work in any direction is negative.
The total work done upon the piston by the steam is therefore represented by the area $abcd$ and this amount of energy is used in raising the two weights through a vertical distance equal to the piston travel. Some of this energy, or its equivalent, will have to be returned at an instant later than that under which it was expended, as the work shown by the area $cdef$ upon the steam. It is returned by the small weight dropping through a distance equal to the travel of the piston. The net mechanical energy made available by carrying through the series of processes is therefore represented by the area $(abdf)-(cdef)=(abcd)$ or the area enclosed by the four lines representing the
THE IDEAL STEAM ENGINE 49
pressure and volume changes experienced by the working substance during one cycle of events. It is equal to the work done in raising the larger weight a vertical distance equal to the travel of the piston.
This net energy made available is obviously
$$\text{Energy made available} = P_1(V_s - V_a) - P_2(V_s - V_d)$$
$$= (P_1 - P_2)(V_s - V_d) \text{ ft-lbs.} \quad (22)$$
$$= (P_1 - P_2)(V_s - V_d) \quad \text{B.t.u.} \quad (23)$$
Since this amount of energy is made available while one cycle of events is being carried out and since the cycle can be repeated time after time if sufficient heating and cooling mediums are available, any quantity of mechanical energy can be produced from heat energy by repeating the cycle a sufficient number of times. This would correspond to picking up a number of the larger weights which were slid on to the wire at the lower elevation and slid off at the higher.
The repetition of cycles would correspond, in a real engine, to running at such a speed that the required number of cycles would be produced in a given time to make available the amount of mechanical energy required.
Or, the power made available per cycle could be increased. This is easily seen from inspection of Eq. (23). Increasing the value of either of the pressures $P_1$ or $P_2$ will obviously increase the amount of energy made available. The value of $(P_1 - P_2)$ can be increased by raising the initial pressure $P_1$ or by lowering the final pressure $P_2$. The value of $(V_s - V_a)$ may be increased by using more than one pound of material, thus increasing both the volume $V_s$ of the saturated steam and the volume $V_a$ of the liquid water, but getting a greater numerical value for $(V_s - V_a)$. This would correspond in a real case to using a larger cylinder and therefore a larger engine.
50
STEAM POWER
ILLUSTRATIVE PROBLEM
An engine of the type described is to work with a maximum pressure of 100 lbs. per square inch absolute and a minimum pressure of 15 lbs. per square inch absolute. The cylinder is to be one inch in diameter and the stroke of the steam is to be dry and saturated at the point b of the cycle.
Find: (a) the amount of mechanical energy made available per cycle; (b) the amount of energy made available per minute if 150 cycles are produced per minute; (c) the horse power of the engine.
It will first be necessary to find the piston displacement required and the distance between the piston and cylinder table to accommodate the pound of water in liquid form. The steam tables give the volume of one pound of dry saturated steam at 100 lbs. per square inch absolute, which is 7.36 cu.in. The pound of water may be taken as 0.017 cu.in. The values of the various volumes and pressures will therefore be
$$V_a = V_b = 0.017 \text{ cu.in.}$$
$$V_a - V_b = 1.429 \text{ cu.in.}$$
$$P_a - P_b = 100 \times 144 = 14,400 \text{ lbs. per sq.ft.}$$
$$P_a - P_b = 15 \times 144 = 2160 \text{ lbs. per sq.ft.}$$
(a) Using Eq. (22) the amount of mechanical energy made available per cycle will be
$$(F_1 - F_2)(V_a - V_b) = (14,400 - 2160)(1.429 - 0.017) = 12,340 \times 1.412 = 54,002 \text{ SS ft.lbs.}$$
(b) If 150 cycles are produced per minute, the total amount of mechanical energy made available per minute must be
$$150 \times 54,002 \text{ SS} = 8,100,300 \text{ ft.lbs.}$$
(c) The horse power must then be
$$\frac{8,100,300}{33,000} = 245 +.$$
34. Heat Quantities Involved. It is a very simple matter to determine the quantity of heat which must be supplied to produce the process ab, and the quantities of
THE IDEAL STEAM ENGINE 51
heat which must be removed to produce the processes $bc$ and $cd$. This can be done by making use of the known properties of water and steam as given in the steam tables.
The water at $P$ must be at the temperature of vaporization corresponding to pressure $P_2$ since it has just been formed by condensation from steam under that pressure. It therefore contains the heat of the liquid corresponding to that pressure. If it is to be vaporized at pressure $P_1$, it must first be raised to the higher temperature corresponding to that pressure. The amount of heat required to do this will obviously be the difference between the heat of the liquid at the temperature corresponding to $P_2$ and the heat of the liquid at the temperature corresponding to $P_1$. These can be found in the steam tables.
The latent heat of vaporization at $P_1$ must then be added to cause the increase of volume shown by $ab$. This can also be found in the steam tables for any given case.
The quantity of heat which must be removed to produce the processes represented by $bc$ and $cd$ can be found similarly. The method used in these cases is not quite as obvious as in the preceding cases.
Assuming that it is possible to find the heat supplied, $Q_1$, and the heat removed, $Q_2$, it is obvious that the energy made available in mechanical form, per cycle, must be equal to $(Q_1-Q_2)$ B.T.U., since this is the amount of heat energy which has disappeared and since it cannot have been destroyed. This may be put in the form of an equation, thus
Energy made available = $Q_1 - Q_2$. . . . (24)
If the proper substitutions are made in this formula and it is then simplified, it becomes
Energy made available = $(APu)_{P_1} - x_i(APe)_{P_1}$ B.T.U., (25)
52 STEAM POWER
in which
\((APu)_{P_1}=\) the external latent heat at pressure \(P_1\);
\((APu)_{P_2}=\) the external latent heat at pressure \(P_2\), and
\(x_s=\) quality at point \(c\), which can be found from
the ratio of \(dc\) to \(de\).
Numerical substitution in this equation for any given case will show that it gives exactly the same values as would be obtained by the use of Eq. (23).
It is to be noted particularly that the energy made available is actually less than the external latent heat at the higher pressure, while the heat supplied must be equal to the total latent heat at the lower pressure. An inspection of the steam tables will show that the external latent heat for ordinary steam pressures forms a very small fraction of even the total latent heat, and therefore the mechanical energy made available for a given expenditure of heat energy is very small in the case under discussion.
35. Efficiency. The term efficiency is used in engineering as a measure of the return obtained for a given expenditure. It may be defined in any one of the following ways:
| Efficiency |
Expenditure made to obtain that result |
| Result |
Effort |
| Output |
Input |
(26)
In the case of a heat engine, the useful result is the mechanical energy obtained by the operation of the engine, while the expenditure made is the heat which is supplied. For this case efficiency may therefore be defined by the expression
THE IDEAL STEAM ENGINE
Engine efficiency = $\frac{\text{Mechanical energy obtained per cycle}}{\text{Heat supplied per cycle}}$
$= \frac{E}{Q_1} \quad \ldots \quad \ldots \quad \ldots \quad (27)$
$= Q_1 - Q_2 \quad \ldots \quad \ldots \quad \ldots \quad (28)$
in which $E$ stands for mechanical energy obtained,
$Q_1$ stands for heat supplied, and
$Q_2$ stands for heat rejected.
In the case of the type of steam engine just considered, this efficiency would have a value between 6 and 8 per cent for ordinary pressures. That is, the engine would produce in mechanical form only to 6 to 8 per cent of the energy supplied it in the form of high temperature heat. Moreover, it would be impossible to build an electrically perfect engine; a real engine built to operate upon this cycle would probably give efficiencies of the order of 2 to 3 per cent. The reasons for this great discrepancy will be discussed in a later chapter.
36. Effect of Wet Steam. In what has preceded, it was assumed that the volume of steam was completely vaporized along the line $a b$, so that dry, saturated steam existed in the cylinder at $b$. It might, however, be assumed that vaporization was incomplete at the upper right-hand corner of the cycle, so that this point occurred at a point on the left of $b$ and with a quality $x$ at that point less than unity.
Under such conditions, the cylinder would not have to be so big, since the maximum volume attained by the steam would be smaller than in the preceding case. The work done per cycle would obviously be smaller in quantity, because the area enclosed within the lines of the cycle would be smaller. It can also be shown that the efficiency would also be lowered by lowering the quality at $b$.
![]() A diagram showing a steam engine cycle with wet steam.
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STEAM POWER
37. Application to a Real Engine. The engine which has been described in the preceding paragraphs could easily be converted into the counterpart of a real engine by substituting connecting rod, crank shaft and flywheel for wire, pulley and belt, and steam for air in the cylinder as in this chapter. It could then be made to do work in just the same way as has been described; some of the energy made available during the outstroke would be used for overcoming resistance at the shaft, that is, doing useful work, and some would be used to drive the flywheel which would speed up slightly. The energy which remains available on the steam during the return stroke would be obtained by allowing the flywheel to slow down and thus deliver sufficient kinetic energy to drive the piston back against the low-pressure steam. The cycle and the efficiency would thus, theoretically, be exactly the same as those just investigated.
Great difficulty would, however, be met in a real engine if the steam had to be formed and condensed within the cylinder, and another method which gives the same results is therefore used. Steam is generated in a boiler and allowed to flow into the cylinder and push out the piston just as the steam did in the model engine in this chap- ter as previously described. When the piston reaches the end of its outstroke the inlet valve is closed and the exhaust valve is opened, allowing some of the steam to blow out into a space in which a lower pressure exists. As the piston stands still at the end of its stroke while the pres- sure drops down to atmospheric pressure by precess per- tuation, but by a different method. The piston then returns and drives the remaining steam out of the cylinder at a constant pressure theoretically equal to that of the space into which the steam is being forced or exhausted. The line of $d$ is thus produced and the closure of the exhaust valve and opening of the admission valve when $d$ is reached will start the cycle over again.
THE IDEAL STEAM ENGINE
55
In order to get more work out of a given size of cylinder and to obviate the necessity of giving back energy which has already been given out, engines are generally made to take steam on both sides of the piston. They are then known as double acting engines. In this case the steam admitted on one side of the piston would supply the energy necessary both for overcoming the resistance due to the load and for driving out the low-pressure steam on the other side of the piston. On the return stroke conditions would be just reversed.
38. Desirability of Other Cycles. The cycle of operations described in preceding paragraphs is the most inefficient of all those actually used, that is, it gives the smallest return for a given amount of heat supplied. This is because some of the heat supplied is converted into mechanical energy and part of that energy must be returned to complete the cycle. All of the internal latent heat and all of the heat of the liquid supplied along $ab$ pass through the engine without conversion and are exhausted.
Therefore, cycles which differ from that described in any way so as make it possible to convert into mechanical energy some of the internal latent heat and possibly some of the heat of the liquid should be highly desirable as they ought to yield a larger return of mechanical energy for the same total amount of heat supplied. Two such cycles are commonly used; they may be described as the Carnot cycle and as the Rankine cycle respectively.
The former is used in steam turbines, the latter in most reciprocating steam engines. The rectangular cycle which has just been described is used in duplex pumps and similar apparatus.
39. The Complete-expansion Cycle. This cycle, which is also known as the Clausius and as the Rankine cycle, starts just the same as that already described. This is shown in Fig. 21. The pressure on, say, a pound of water
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STEAM POWER
is raised from $P_2$ to $P_1$ and its temperature is raised from that of vaporization at $P_2$ to that of vaporization at $P_1$. After this it is vaporized, giving the increase of volume.
![]()
A graph showing the process of steam expansion. The x-axis represents pressure (P) and volume (V), while the y-axis shows temperature (T). The curve starts at a high point on the left side of the graph, representing the initial state of the steam. It then descends sharply, indicating the decrease in pressure and increase in volume as the steam expands. The curve continues to descend, reaching a lower point on the right side of the graph, representing the final state of the steam after expansion.
Fig. 21.—Complete Expansion Cycle or Clausius Cycle.
shown by ab. The supply of heat is then stopped. The cylinder of the engine is made larger than in the preceding type so that when the point $b'$ is reached the piston can travel still further, and it is allowed to do so, that is, the
THE IDEAL STEAM ENGINE 57
high-pressure steam is allowed to push it further out. This can be pictured by imagining the steam to act like the compressed spring shown in the figure and to push the piston in much the same way as does the spring. The line $b_1c_1$ shows the decreasing pressure exerted on the piston by the spring as the latter expands so as to get longer and longer. Because of the properties of a spring this is a straight line. The line $bc$ shows the decreasing pressure exerted on the piston by the steam as the latter expands so as to occupy greater and greater volumes. Because of the properties of steam this line is curved instead of straight.
Now, if we imagine that all the heat supplied to the expanding steam during the process $bc$ and the amount of this work will be indicated by the area under the line $bc$ as shown in the figure. This work must have been done by the expenditure of energy on the part of the steam and since no energy was added after the point $b$ was reached the work must have been done at the expense of heat energy contained in the steam itself. It has already been shown that the heat along $bc$ in the figure is equal to the sum of the heat of the liquid and the internal latent heat, but some of this heat must obviously be used for the doing of work along $bc$ instead of being entirely rejected to the cooling medium as in the preceding cycle without "expansion."
The expansion of the steam continues until the "back pressure" $P_2$ is reached. The cooling medium may then be imagined to be brought into use and to abstract such heat of vaporization as may remain in the steam besides absorbing the equivalent of the work done on the steam by the returning piston, thus giving the process shown by the line $cd$.
If the expansion line $bc$ of the cycle just described could be carried out within walls constructed of such material that it would not give heat to nor take heat from the steam, it is obvious that any heat energy lost by the steam during the expansion could be lost only by conver-
58
STEAM POWER
sion into mechanical energy. An expansion of this kind is called an **adiabatic expansion**.
In the figure, the curve of adiabatic expansion is shown in its correct position with respect to the saturation curve and it is obvious that for an adiabatic expansion, during with dry, saturated steam, the quality decreases as the expansion progresses.
**Comparison with Cycle without Expansion.** The heat supplied is the same in both of the cycles just considered when they operate between the same two pressures, but the mechanical energy obtained in the case of the complete expansion cycle is greater than that obtained by the cycle described. The mechanical energy obtainable with the cycle first described is represented by the area $abcd$ while that obtainable with the complete expansion cycle with the same heat supply $Q_1$ is represented by the same area $abcd$ plus the additional area $bde$. The efficiency of the complete expansion cycle is therefore very much higher than that of the cycle without expansion.
For conditions similar to those giving a theoretical efficiency of about 6 per cent without expansion, the complete expansion cycle will give a theoretical efficiency of about 12 per cent and this figure can be doubled by expanding at a lower pressure.
The cylinder required for the production of the complete expansion cycle would be much larger than that required for the other cycle if both used the same weight of steam per cycle. The proportion would be in the ratio of the volume shown at $c$ in Fig. 21 to the volume shown at $a$. But the complete expansion cycle would make available more mechanical energy per unit weight of steam than would the other, so that the difference in the size of cylinders would not be so great if both were required to make available the same amount of mechanical energy per cycle.
**40. The Incomplete-expansion Cycle.** The shape of this cycle is shown in Fig. 22. It is just like the complete
THE IDEAL STEAM ENGINE
59
expansion cycle down to the point $c$. The cylinder in which it is produced has a smaller volume than that used for the complete expansion cycle so that the piston arrives at the end of its stroke before it has opened up volume enough to enable the steam to expand all the way down to the lowest pressure (terminal or back-pressure). When the point $c$ is reached in the real engine, the exhaust valve is opened and enough steam is admitted to reduce the pressure to the back pressure $P_b$. The piston then returns and drives out the remainder of the steam as shown by the line $de$.
In the ideal method assumed in the preceding treat- ment, the heat absorbing medium could be brought into use at $e$, allowing sufficient time to reduce the pressure from $P_a$ to $P_b$ so that the piston remained stationary at the end of its stroke. The latent heat of vaporization remain- ing in the steam at $d$ would then be absorbed as the piston was driven back from $d$ to $e$.
Comparison with Other Cycles. The incomplete expan- sion cycle is intermediate between the two previously dis- cussed cycles and can be appreciated readily by an inspection of Fig. 22. In this figure the area $abc'd'e$ represents the mechanical energy obtainable with the cycle without expansion; the area $abc'e$ represents the energy obtainable from the same quantity of steam with complete expansion; and the area $abcd'e$ represents the energy obtainable from the same quantity of steam with incomplete expansion.
The later the point at which the exhaust valve is opened, point $c$, the more nearly do efficiency and energy obtain- able approach the values for the complete expansion cycle.
![]()
A diagram showing a partial expansion cycle.
a - Initial state
b - Point where steam starts to expand
c - Point where piston reaches end of stroke
d' - Point where exhaust valve opens
d - Point where steam is exhausted
e - Point where piston returns
f - Final state
Fig. 22—Incomplete Expansion Cycle.
60 STEAM POWER
The earlier the point at which the exhaust valve is opened, the more nearly do efficiency and energy obtainable approach the values for no expansion.
Despite the lower efficiency of the incomplete expansion cycle, owing to its connection with Fig. 22 it is universally used on all reciprocating engines excepting those which make no pretense to economy and use no expansion. The less efficient cycle is used for the simple reason that complete expansion in a reciprocating engine does not pay commercially. For complete expansion the cylinder must be larger in the ratio of $V_1$ to $V_2$, as shown in Fig. 22 and the cost obtained by such expansion is a very small part of the total. In most cases it would not be great enough to overcome the friction of the engine, not to mention paying interest on the necessarily higher cost of the larger cylinder and accompanying parts.
It will be shown in a later chapter that the steam turbine can economically expand the steam completely and the complete expansion cycle is therefore used with such prime movers.
CHAPTER V
ENTROPY DIAGRAM
41. Definitions. In Chapter III temperature, pressure and volume were discussed as criteria determining the condition of water and steam. Other things may be used in determining the condition of such materials. One which is particularly useful from an engineering standpoint is known as entropy and is designated by the Greek letter $\phi$.
For convenience in dealing with steam, we have a characteristic value of entropy just as there is a characteristic value of temperature, pressure, volume, heat above 32° F., etc. These values of entropy are given in the steam tables in just the same way as the value of temperature, pressure, volume, heat above 32° F., etc. are given.
The entropy of the liquid given for any particular pressure is the amount of heat experienced by one pound of the liquid when its temperature is raised from 32° F. to the temperature of vaporization corresponding to that particular pressure. It might be spoken of as the entropy of the liquid above 32° F., just as $q$ is spoken of as the heat of the liquid above 32° F. It is represented by $\phi_0$.
The change of vaporization for any particular pressure is the change of entropy experienced by one pound of the material while changing from water at the temperature of vaporization to dry saturated steam at constant pressure. It corresponds to the latent heat of vaporization and is designated by $\phi_1$.
The total change of saturated steam at any pressure is the sum of $\phi_0$ and $\phi_1$ and therefore is the total change of entropy experienced by a pound of material in changing
61
62 STEAM POWER
from water at 32° F. to dry saturated steam at the particu-
lar pressure in question.
The entropy of superheat at any pressure and tempera-
ture is the change of entropy experienced by a pound of
dry saturated steam at that pressure when superheated
to that particular temperature. It is designated by $\phi_s$.
The entropy of superheated steam at any pressure and
temperature is the total change of entropy experienced by
one pound of material when changed from water at 32° F.
![]()
A graph showing temperature-entropy diagrams.
(a) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(b) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(c) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(d) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(e) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(f) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(g) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(h) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(i) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(j) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(k) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(l) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(m) A graph with a horizontal line labeled "Absolute Temperature Scale" from 0°F to 250°F. Below this line, a vertical line labeled "Water Steam Regimes" extends from 0°F to 250°F. Below this line, a vertical line labeled "Superheated Steam Regimes" extends from 0°F to 250°F. The area between these two lines represents the range of temperatures for which the substance can exist as either water or superheated steam.
(n) A graph with a horizontal line labeled "Absolute Temperature Scale" from
ENTROPY DIAGRAM 63
this chart are plotted above 32° F, as datum temperature.
The water line or water curve is obtained by picking out of the steam tables the values of $\phi$, entropy of the liquid, for different pressures and plotting them against the absolute temperature corresponding to those pressures. Obviously, at each point on this curve, the absolute temperature corresponding to 32° F., i.e., about 492° F., abs.
The saturation curve or dry steam curve is obtained by picking out of the steam tables the values of $\phi_+$ + $\phi_-$ for different pressures and plotting against corresponding absolute temperatures.
The area of vaporization is obviously shown for each different temperature (or pressure) by the distance between the water curve and the saturation curve, since the former is distant from the vertical axis by an amount equal to $\phi_+$, while the latter is distant an amount equal to $+\phi_-$.
Superheating lines are drawn by picking from the steam tables the values of $\phi$ for steam superheated to different temperatures at one particular pressure and plotting against the proper temperatures. There will be as many superheating lines on the diagram as one chooses pressures for which to plot them. Only one is shown in this figure.
One very useful property of this diagram follows from the fact that points on its surface indicate the condition of the material. For instance, if the temperature-entropy, or $T-\phi$, values of the material at a given condition should plot to the left of the liquid line, the material must be in the liquid condition; if they plot between the liquid line and the saturation curve, the material must be a mixture of liquid and saturated vapor. If they plot to the right of the saturation curve, the material must be dry, saturated steam; or if they plot to the right of the saturation curve, the material must be superheated steam. This all follows directly from the definition of entropy above 32° F., as plotted in these dia.
64 STEAM POWER
grams. The various regions, or fields, into which the dia-
gram divides in this way are shown in Fig. 23 (a).
Another very useful property of this diagram follows
from the fact that area represents heat just as area on a pressure-temperature diagram represents work.
Thus the area under the line $ab$, for instance, represents
the heat required to raise the temperature of one pound
of water from 32° F. to the temperature at $b$. Similarly
the area under the line $bc$ represents the heat required to
change a pound of water at the temperature at $b$ to a pound
of dry steam at the same temperature. The area under
line $bd$ represents the heat required to superheat this pound of saturated steam
at constant pressure up to the temperature shown at $d$
is similarly represented by the area under the line $cd$.
In this connection, it should be noted that this diagram
is plotted above absolute zero of temperature just as the
pressure-volume diagram is plotted above absolute zero of
pressure. The areas in question therefore extend down
to the absolute zero of temperature. In order to indicate
this in Fig. 23 (b), a large part of the chart is supposed
to have been broken out, so that the lower end of the diagram
could be moved up into view. In Fig. 23 (a), the bottom
of the diagram is drawn at temperatures below 32° F., and
this is indicated by pointing $T > O$ opposite the horizontal axis.
The various areas hatched in Fig. 23 (b) indicate the
various quantities of heat previously discussed. It should
be understood that the areas represent the heat quantities
only for the particular pressures which corresponds to the
temperatures indicated by $T_0$. For a higher pressure,
line $bc$ would be higher and the area proportionately larger;
for a lower pressure the line $bc$ would be lower and
the areas smaller.
ENTROPY DIAGRAM 65
ILLUSTRATIVE PROBLEM
Starting with liquid at a temperature $T_L$ corresponding to the temperature of vaporization at a pressure of 50 lbs. per square inch absolute, assume the liquid raised to the temperature of vaporization under the same pressure, then condensed into liquid and then completely vaporized. Determine the various changes of entropy and indicate them on a $T_s$-chart.
The specific entropy of the liquid, $s_0$, is equal to 0.4113 for water about to vaporize under 50 lbs. per sq. in. absolute, and 0.4743 for water about to vaporize under a pressure of 100 lbs. per sq. in. absolute. The difference, that is, $0.4743 - 0.4113 = 0.063$ is the change in entropy experienced by the liquid when its temperature is raised from $T_L$ to $T_v$. This is the higher value. These values are shown in Fig. 24.
The following values give entropy of vaporization, $s_v$, at 100 lbs. per square inch absolute as 1.1277.
Adding this value to entropy above 25 F. of the liquid at saturation temperature under 100 lbs. pressure gives $0.4743 + 1.1277 = 1.602$.
The total change in entropy of dry, saturated steam at 100 lbs. per square inch absolute. These values are all indicated in the preceding table.
The total change of entropy experienced by the material in changing from water at the temperature of vaporization under 50 lbs. pressure to dry, saturated steam at 100 lbs. pressure is obviously $s_v - s_0 = 1.602 - 0.4113 = 1.19$. This is shown in Fig. 24.
**43. Quality from $T_s$-chart** The entropy change experienced by steam in the process of vaporization is directly proportional to the addition of heat. Thus, when half the latent heat has been added to one pound of material, the entropy change is $\frac{1}{2} s_v$. In general, if a fraction $x$ of the latent heat has been added, the entropy change has been $x s_v$ during the process. Therefore, if the temperature entropy condition of a pound of material should plot at a
![]() A graph showing changes in entropy (S) with respect to temperature (T). The x-axis represents temperatures ranging from approximately -25°F to 25°F, while the y-axis represents entropy values ranging from approximately 0 to 2 units.
Fig. 24
66
STEAM POWER
point such as $c$ in Fig. 25, it follows that the material is a mixture of water and steam and that a fraction of the pound equal to $\frac{bc}{bd}$ is steam, the rest being water. But, by definition, the fraction $\frac{bc}{bd}$ is $x$, the quality of the material.
The temperature-entropy chart is very useful when used in connection with this property of showing quality. Thus, in Fig. 25, the area under $bc$, down to absolute zero temperature, represents the fraction of the latent heat of vaporization per pound which must be added to give a pound the quality $x$.
For convenience in use, constant quality lines are generally drawn on temperature-entropy charts. Such lines are shown in Fig. 26. Each line is obtained by plotting the temperature entropy conditions for a given quality at different pressures. For the steam tables for a given pressure. The numerical value of $\phi_0$ is then multiplied by the fraction representing the particular quality desired. This product is added to $\phi_0$ giving the total entropy above $32^\circ$ F. for quality 0.9 at the particular pressure chosen. The same
![]() Fig. 25.—Quality from Temperature-Entropy Chart.
![]() Fig. 26.—Constant Quality Curves.
ENTROPY DIAGRAM
67
![]()
A graph showing an entropy diagram with various lines labeled with temperatures (T) and pressures (P). The x-axis represents temperature (T), ranging from 0 to 100, and the y-axis represents pressure (P), also ranging from 0 to 100. The lines include saturation curves, constant volume, constant quality, and constant heat lines.
Fig. 37. -Constant Volume, Constant Quality, and Constant Heat Lines.
68
STEAM POWER
process is repeated with the same value of the quality, but with different pressures, until enough points have been secured to make it possible to draw a smooth line through them.
**44. Volume from T-φ-chart.** Since quality changes at any given temperature, or pressure, are accompanied by volume changes, it is possible to find a series of values for the quantity of water which will occupy one pound when that pound occupy the same volume at different temperatures. Having found the quality which will be necessary at a number of different temperatures, the total entropy above 32° F, can be found for each case and these values can then be plotted on the T-φ-chart. Connecting the points so obtained would give a curve known as the Constant Heat Line. Several of these constant volume lines are shown in their correct positions in Fig. 27. It will be observed that, for each volume, the quality must increase as temperature (and pressure) increases in order to maintain a constant value for the volume occupied by one pound of mixture.
**45. Heat from T-φ-chart.** Equations for obtaining the total heat content of steam and superheat of steam were given in an earlier chapter. By means of these equations, it is possible to find a succession of values for quality and superheat which will give a pound of material any chosen heat content at different pressures. If the corresponding values of temperature and entropy are found and plotted on the T-φ-Chart, Constant Heat Line results. Several of these lines are shown in Fig. 27.
**46. The Complete T-φ-Chart for Steam.** A very complete, graphical representation of the properties of water and steam can be procured by combining in one diagram all of the lines discussed in preceding paragraphs. Such a diagram is usually called the T-φ-Chart or simply the T-φ-chart for steam. An example of such a diagram is given in Fig. 28.
This chart is very useful, as it enables one to solve by
TEMPERATURE-ENTROPY DIAGRAM
TO ACCOMMODATE STEAM POWER (Pumping 110 and T.C.L. service)
(For pumping 110 and T.C.L. service)
Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
![]() A graph showing temperature-entropy diagram with various lines and labels.
Fig. 28.
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| Temperature-Entropy Diagram for Steam Power (Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping 110 and T.C.L. Service) | Return (with modifications) From Pumping to Pumping Out Energy (With 6 & 8 mm.)
(For Pumping
70
STEAM POWER
inspection many of the most difficult problems which arise in the theory and practice of using steam. As an example, assume that it is desirable to know what will happen to water at the temperature of vaporization when subjected to about 24 lbs. per square inch absolute has its volume increased indefinitely at constant temperature. The initial condition of the water would be shown on the water curve of Fig. 28 at the point at which the 700° absolute temperature line crosses. Increase of volume at constant temperature would be indicated by a straight line parallel to this line from this point. Obviously, vaporization will occur at constant pressure (because the temperature is constant) and the quality will change from zero to unity at which the saturation curve will have been reached. Further increase of volume will result only in the production of superheated steam, since the line representing the material will run out into the superheated steam field. It is also interesting to note that the pressure on the material will have to be decreased as the volume increases in the superheated steam region, as is evidenced by the fact that the horizontal line representing the assumed process cuts lower and lower pressure lines as it is extended to the right in the superheated field.
Note also that the intersections of this horizontal line with constant volume and constant heat lines afford the means of determining volume and heat above 32° F. at different stages of the assumed process.
PROBLEMS
1. Determine from the steam tables the change of entropy experienced by one pound of water when its temperature is raised from 32° F. to the temperature of vaporization under a pressure of 100 lbs. per square inch absolute.
2. Determine from the steam tables the entropy change experi- enced by one pound of water when its temperature is raised from 42° F. to the temperature of vaporization under a pressure of 150 lbs. per square inch absolute.
ENTROPY DIAGRAM
5. Determine the entropy change experienced by one pound of water when its temperature is raised from the temperature of vaporization corresponding to 150 lbs. per square inch to that corresponding to 150 lbs. per square inch by subtracting the value found in Prob. 1 from that found in Prob. 2.
6. Determine the amount of heat required to raise one pound of material completely vaporizing at a temperature of 32° S.F.
7. Plot a T-e-chart for one pound of water. Start by plotting entropy of the liquid for various temperatures; then plot entropy of saturated steam at various pressures on the saturation line, and several lines showing change of entropy during vaporization.
8. Determine from a T-e-chart the quality which would be attained by one pound of steam if it experienced a change which carried it from the condition of dry saturated steam at 150 lbs. per square inch absolute to that of wet saturated steam at 100 lbs. per square inch absolute, and determine what would plot as a vertical line on the T-e-Chart.
9. Assume a pound of mixed water and steam to have a quality of 80%, at a pressure of 200 lbs. per square inch absolute. Determine from the T-e-Chart the heat above 32° per pound of mixture and the volume occupied by the mixture. Determine also how much heat would be required to carry this material drops to 200 lbs. per square inch absolute at constant entropy. How does the heat above 32° F. change during such a process?
10. Assume a pound of mixture as in Prob. 7, but with a quality of 30% at a pressure of 200 lbs. Find all quantities called for in that problem.
11. Assume a pound of material as in Prob. 9, and 8 above, but superlucated 200° at a pressure of 200 lbs. per square inch absolute. Determine all quantities called for in Prob. 7.
12. Choose a point on the T-e-Chart at which a constant volume line intersects the saturation curve. Determine the change of quality, entropy or heat above 32° F., if the material drops to half pressure at constant volume.
![]() A graph titled "ENTROPY DIAGRAM" with axes labeled T (temperature) and e (entropy). The x-axis ranges from approximately -150 to +150, while the y-axis ranges from approximately -150 to +150.
CHAPTER VI
TEMPERATURE ENTROPY DIAGRAMS OF STEAM CYCLES
47. Complete Expansion Cycle. This cycle was considered in Chapter IV and the $P-V$-diagram was given there as Fig. 21. The diagram of this cycle drawn to $T_{0}$-coordinates is shown in Fig. 29. The same letters are used to represent corresponding points in the two diagrams.
The entropy change during the heating of the liquid is shown by the part of the liquid line between $d$ and $a$, and the heat supplied during that process is represented by the area below this line, measuring clear down to the absolute zero of temperature.
The entropy change during vaporization is represented by the line $ab$ and the heat supplied during the process is shown by the total area under that line.
The adiabatic expansion of the steam is represented by the line $bc$, such an adiabatic change fortunately being a constant entropy process and therefore easily drawn in this diagram. Obviously no heat is received or removed during this process, as there is no area under the line $bc$.
The entropy change during condensation is represented by the area under the horizontal line representing the working substance during this process is represented by the area under that line.
72
TEMPERATURE ENTROPY DIAGRAMS
**48. Area of Cycle Representative of Work.** It will be remembered that area under a line in the $PV$-diagram represents work in foot-pounds. That diagram, however, gives no information of heat supplied and rejected and it is not possible to obtain any direct idea of efficiency from it. In this respect, the $T_e$-diagram is much better. Area under the lines $da$ and $ab$ in Fig. 29 represents heat supplied the working substance. Area under the line $cd$ represents heat rejected by the working substance. The difference between these two, or the area enclosed within the lines of the cycle, must represent the heat converted into mechanical energy per cycle.
This diagram therefore shows directly by areas the heat supplied, the heat rejected, and the heat converted into mechanical energy. Further, the ratio of the area representing heat converted into work, and the area representing heat supplied must be the ratio of the cycle.
Remembering also that if the lines of the cycle are drawn upon a $T_e$-chart such as that given in Fig. 28, all volume changes, heat contents and qualities at different points are shown without further work, it becomes evident that this form of representation is decidedly convenient and far superior to the $PV$-diagram.
**49. Modifications for Wet and Superheated Steam.**
The complete expansion cycle is supposed to represent an idealization of what happens in a real prime mover. In real cases, however, the steam may arrive at the prime mover wet or superheated and it is desirable to investigate the method of representing such conditions as they occur in practice.
Wet steam corresponds to incomplete vaporization, i.e., a quality less than unity at the upper right-hand corner of the cycle. This might be shown for a given case by the location of the point $b'$ in Fig. 20. The cycle would then be $ab'cd'$ and a smaller amount of work would be obtained per pound of working substance as evidenced by the smaller area enclosed within the lines of the cycle.
74 STEAM POWER
In the case of superheated steam, superheating occurs at constant pressure after vaporization is complete. This would be shown by the location of the upper right-hand corner of the cycle at some point $b'$ on the constant pressure line which extends out from $b$. The cycle is now represented by $ab'c'd$ and evidently has a different shape than it had in the preceding cases. Obviously the area enclosed within the lines of the cycle is greater than it was before and therefore more mechanical energy is obtained per pound of steam.
50. Incomplete Expansion Cycle. The only difference between the incomplete and complete expansion cycles is
![]() Fig. 30.—T-φ-diagram, Incomplete Expansion Cycle.
![]() Fig. 31.—T-φ-diagram, Cycle Without Adiabatic Expansion.
the termination of the expansion in the former by means of a constant volume line. This is shown to T-φ-coordinates in Fig. 30 in which the incomplete expansion cycle is drawn in heavy lines over the one in which expansion continues to the back pressure.
The constant volume line is seen to cut off a corner, thus reducing the area representing heat converted into work. The heat supplied in each case is measured by the area under the lines $ea$ and $ob$. The efficiency of the cycle with incomplete expansion can therefore be seen to be less than that of the other cycle by simple inspection of the diagram.
If the adiabatic expansion is terminated at a higher
TEMPERATURE ENTROPY DIAGRAMS 75
pressure, as by the constant volume line $c'd''$ in Fig. 30, still more of the work area is lost, but the same quantity of heat is supplied, and therefore the efficiency is still lower than when the expansion terminated at $c$. Obviously as the temperature decreases, the cycle becomes less idealized moves nearer and nearer to $a$ as shown in Fig. 31, the cycle becomes less and less efficient. If the constant volume line starts at $b$, there is no adiabatic expansion and the cycle becomes that previously considered as having a rectangular shape in the $PV$-diagram. This cycle has the shape indicated by $abc$ in the $T_4$-diagrams of Fig. 31. Obviously this is least efficient of all as was previously shown by other means.
**51. Effect of Temperature Range on Efficiency.** It has already been stated (see p. 26) that heat engines receive heat at a high temperature, convert some of it into mechanical force, and then give off heat at a low temperature. Inspection of the $T_4$-diagram shows this very clearly, and, remembering that the area of the cycle measures the heat converted, these diagrams also show how raising the upper temperature (or pressure) or lowering the lower temperature (or pressure) will increase the efficiency. It can be seen readily that lowering the lower temperature will increase the efficiency more than increasing the efficiency than raising the upper temperature.
**PROBLEMS**
1. Draw a complete expansion cycle to $T_4$-coordinates for the following conditions (using $T_4$-diagram for steam to get values):
- weight of working substance: 1 lb.
- initial pressure: 125 lbs. absolute;
- quality at beginning of adiabatic expansion: 100%;
- back pressure: 10 lbs. absolute.
2. Determine the following values for cycle drawn in Prob. 1:
(a) Entropy of liquid at beginning of vaporization;
(b) Entropy at end of adiabatic expansion;
(c) Quality at end of adiabatic expansion;
(d) Volume at end of adiabatic expansion;
(e) Entropy at end of condensation.
76
**STEAM POWER**
3. Show by measuring the area on $T_e$-diagrams, the increase of efficiency resulting from the use of an initial pressure of 175 lbs. absolute and from the use of a terminal pressure of 2 lbs. absolute in place of the values given in Prob. 1.
4. Compare the work and efficiency of an incomplete expansion cycle with conditions as in Prob. 1 with a complete expansion cycle with same pressures but with a temperature of 500° F. at the beginning of the adiabatic process.
5. Draw an incomplete expansion cycle to $T_e$-coordinates for the same pressures as in Prob. 1, but with adiabatic expansion ending at a pressure of 15 lbs. absolute.
6. Compare work and efficiency of the two cycles of Probs. 1 and 5 above.
7. Draw a cycle without expansion for the conditions of Prob. 1 to $T_e$-coordinates and compare the work area with that obtained in Probs. 1 and 8.
CHAPTER VII
THE REAL STEAM ENGINE
52. Operation of Real Engine. In previous chapters the ideal steam engine was considered and several cycles upon which it might be operated were discussed. Real engines are built to operate on the same cycles, but because of certain practical considerations they only imperfectly approximate the ideal performance.
Real engines must be built of iron and steel for practical reasons and these metals absorb, conduct and radiate heat so that certain characteristics between the working substance of the engine and certain heat losses occur in practical operation. These were eliminated in the ideal case by simply assuming ideal materials not possessed of the characteristics of real metals.
It is also practically impossible to generate steam in the cylinder of real engine as was assumed to be done in the ideal case. Heat is not always obtained by the combustion of fuels, and the higher the temperature attained the better can the liberated heat be utilized in the generation of steam. To subject the cylinder to such high temperatures and to control the heating and cooling as necessary to produce a number of cycles in rapid succession would lead to serious overheating and breakdowns. It has been found best to generate the steam in a boiler which is properly equipped for that purpose and then to transmit it with its contained heat to the engine, which is constructed in such a way as to utilize that heat to the best advantage. If the steam is to be condensed, as assumed in the ideal cases, it has also been found best to remove it from the
77
78
STEAM POWER
cylinder and to condense it in a separate piece of apparatus properly constructed for that purpose.
![]() A diagram showing the arrangement of a non-condensing engine. The diagram includes labels such as "Steam Pipe," "Condenser," "Non-Condensing Plant," and "Steam Plant." The diagram also shows a steam cylinder with a piston and connecting rods.
The entire arrangement which results from these practical modifications in the case of a non-condensing engine
Fig. 32 - Diagramatic, Non-Condensing Plant.
THE REAL STEAM ENGINE 79
is shown in Fig. 32. Steam is generated within the boiler at some constant pressure $P_1$ and at the proper instant the admission valve at one end of the cylinder is opened, allowing steam to flow in and drive the piston outward. If there is no friction, the motion would be represented by some such line as $ab$ at a height $P_1$ on the $PT$-diagram of Fig. 33. Closing of the valve after the piston had moved part way out would cut off all further flow of steam, and, with continued motion of the piston, the steam within the cylinder would expand.
If no internal overpressure occurred, this expansion $bc$ would be adiabatic as in the real case.
It will be observed that the two lines on the $PT$-diagram thus far produced represent equally well the corresponding two lines of the complete or incomplete-expansion cycles. The heat supplied in the boiler is that assumed as supplied in the cylinder under ideal conditions of perfect expansion, and the work under the line $ab$ is equal to the external work done during vaporization just as in the ideal case. If difficulty is experienced in connection with the statement regarding external work, it is only necessary to picture the process of expansion in terms of a piston which has formed in the boiler does the external work equivalent to $APu$ by pushing the pound previously generated ahead of it as a piston, and that this motion communicated along the pipe from layer to layer results in pushing an equivalent weight (and volume) into the cylinder against the resistance offered to the piston's motion.
When at point $c$, at the end of its stroke at the point $c$, the opening of the "exhaust valve," connecting the interior of the cylinder with the space in which the pressure $P_2$ lower than $P_1$ is maintained, will permit some of the steam to blow out of the cylinder with the piston standing stationary at the end of its stroke. This would
![]() A diagram showing a PT-diagram with lines ab and bc.
Fig. 33
80
STEAM POWER
give a constant volume change equivalent to the corresponding line in the incomplete-expansion cycle.
The return of the piston from $d$ to $e$, with the exhaust valve still open, would force the remainder of the steam out of the cylinder, and maintain the pressure $P_2$ at which the pressure $P_3$ is maintained. The result, so far as the diagram is concerned, is obviously the same as in the ideal case, and if the steam were condensed within a proper vessel into which it exhausted (instead of being exhausted to atmosphere), the result would also be the same so far as the shape of the diagram is concerned. The pressure $P_2$ might, however, be maintained at a lower value, thus giving a greater temperature range.
The pressure rise $ca$ within the cylinder would result directly from the opening of the admission valve and the admission of steam for the next cycle. But, if the working substance is to be returned to starting conditions as steam, it must be cooled by heat transfer. In this case, therefore, $ca$ must also be raised to $P_1$ and its temperature to a corresponding value. The pressure is raised in the case of condensing operation by means of the boiler feed pump, which picks up the condensed steam (condensate) and forces it into the boiler. The temperature of the working substance is raised by passing it through a heater where heat is absorbed directly from the heated water in the boiler.
When operating non-condensing the working substance exhausted during the last part of each cycle is really thrown away by allowing it to mix with the atmosphere, but an equivalent quantity of water is fed to the boiler by the boiler feed pump, and passes through the place where it is exhausted to atmosphere. This method of operating does not approximate the ideal as closely as does the condensing method, but the discrepancy is not very great.
53. Losses in Real Installations. The diagram given in Fig. 33 was obtained by assuming the absence of certain practical losses and is considerably modified when real
THE REAL STEAM ENGINE 81
apparatus is used. Thus the real engine, as shown in connection with Fig. 9, has clearance and operates with compression so that the clearance is filled with steam at a pressure indicated by the point $a'$ in Fig. 34 when the admission valve opens.
There is also always some drop of pressure along the steam pipe so that the pressure at the cylinder is lower than at the boiler. Further, the admission valve can never be made to give such a large opening into the cylinder that there is not a measurable drop of pressure in flowing through it. Of these actions the highest pressure attained within the cylinder as indicated at point $a$ in Fig. 34 is always lower than the boiler pressure $P_1$.
As the piston of a real engine moves out it acquires a higher and higher velocity until it reaches a point near mid-stroke where its velocity is zero. The flow through the valve with increasing velocity if it is to follow up the piston and fill the cylinder, but this usually necessitates greater pressure drops as the piston moves out, so that the admission line generally slopes downward instead of being horizontal. There is also another phenomenon which causes a fine line to appear on the cylinder, cylinder head and piston is in contact with comparatively low-temperature steam during the latter part of each cycle, and therefore acquires a lower temperature than that of the steam about to enter. Therefore, when the high-pressure steam enters the cylinder it gives up heat to the water at its companion side of the valve and is finally dry saturated, thus resulting in great deal of condensation. Such condensation is called initial condensation.
As the steam condenses after flowing into the cylinder and forms water occupying a negligibly small volume,
![]() Fig. 34.—Theoretical and Real Indicator Diagrams.
82
STEAM POWER
it follows that steam must flow into the cylinder at a pro-
portionately greater rate in order to fill the space vacated by the piston. But this results in an increased pressure drop and therefore would give a sloping admission line.
When the piston has finally been driven out as far as desirable by the action of high-pressure steam, the admis-
sion valve is closed. The admission valve cannot be closed suddenly; the closure is more or less gradual in all cases. As the opening becomes smaller it becomes increasingly more difficult for the steam to flow through and into the cylinder so that the pressure continues to drop at an increasing rate until the valve is finally closed. This gives the rounded cut-off shown at the point b.
The term "valve cutting" or "cutting off" admission is generally said to be due to throttling or wire drawing, these terms being intended to convey the idea that the steam has to squeeze its way through the inlet openings with correspond-
ing loss of pressure.
When admission has finally been completed, it leaves
the end of the cylinder filled with a mixture of steam and water at steam temperature, and this mixture then expands as shown by the line bc. At the beginning of the expansion the steam generally has higher temperature than that of the surrounding walls and it therefore continues to give heat to those walls. Were the expansion adiabatic it would follow that the line bc would figure that the
steam must not only convert heat into work, but must also supply heat to the walls, it condenses more rapidly than in the ideal case and its pressure and volume changes follow some such law as that indicated by the upper part of the curve bc.
As expansion continues, the pressure and temperature of the steam drop until some point is reached at which
the temperature has become equal to that of the walls.
Further expansion with drop of pressure and temperature
results in reducing the temperature of the steam below
THE REAL STEAM ENGINE S3
that of the walls, and then the direction of heat transfer is reversed, the hot walls giving heat to the cooler steam at an increasingly rapid rate. This heat causes re-evaporation of some of the water formed before and thus tends to increase the volume occupied by the material in the cylinder, with the result that the lower part of the expansion curve approaches and generally crosses the curve which would have been obtained by adiabatic expansion in non-conducting apparatus.
In many real engines the re-evaporation is so great that the steam is entirely dried and sometimes superheated before the exhaust valve opens.
The exhaust valves in steam engines are always opened before they reach the end of the stroke, as it is found necessary to do this if excessive losses are not to occur due to the difficulty of forcing the large volume of low-pressure steam through the exhaust passages. When opened early enough, the steam flows out in such quantity before the end of the stroke that the "back pressure" during the entire period of opening is less than that above the space into which the engine is exhausting.
During all of the exhaust period, the steam is probably at a lower temperature than the walls to which it is exposed and re-evaporation probably continues in most cases until the closure of the exhaust valve. It seems probable that this loss of heat is greater when air is admitted to the exhaust valve is approximately dry, but little really known regarding the quality of the clearance steam.
The rise of pressure during compression has two beneficial effects: It helps to bring the moving parts to rest gradually, and it raises the temperature of the clearance steam so that a portion of this clearance space values nearer that of the entering steam.
Remembering that area on a $PV$-diagram represents work, it is easily seen that throttling losses and rounding of corners due to slow valve action (which cause a loss of
84
STEAM POWER
diagram area) result in a loss of work. The fact that condensation also causes a great loss is easily shown. A given quantity of steam entering the engine with its supply of heat can, in the ideal case, do a certain amount of work at the expense of that heat. In the real case part of the heat is stored in the walls during the early part of the cycle, so that it is not available for the working steam. It re-evaporates from the walls and carried out into the exhaust as unutilized heat during the later part of the cycle. The phenomenon can be pictured by imagining the steam as dropping some of its heat into a pocket in the walls of the cylinder when entering the engine and then packing it up again and carrying it out without leaving out what the next charge of steam will have to fill this pocket again.
The net result of condensation and re-evaporation is the obtaining of less work from a given quantity of steam than should be obtained, or the use of more steam than theoretically necessary for a given quantity of work. This effect is shown graphically by the two adiabatic expansion lines of Fig. 34.
The initial condensation in real engines which are supplied with saturated steam generally amounts to from 20 to 50 per cent of all the steam supplied, so that it is evident that anything which will prevent part or all of this loss should do much to improve the steam consumption of engines. This subject will be discussed in more detail in later paragraphs and various methods of decreasing losses from this source will be considered.
54. Clearance. The term clearance is used in a two-fold sense: (a) to refer to mechanical clearance or the linear distance between the two nearest points of cylinder head and piston face when they are at the end of their stroke, and (b) to refer to volumetric clearance or the volume enclosed between the face of the valve, the cylinder head and the face of the piston when the latter is at the end of its stroke.
THE REAL STEAM ENGINE S5
The former is generally given in inches and varies from a very small fraction of an inch in the best engines to an inch or more in cheap and poorly designed engines. It is indicated by $a$ in Fig. 35.
The clearance between piston and cylinder is expressed as a percentage of the piston displacement or volume swept through by the piston. It varies from 2 per cent or less in the best engines to as high as 15 per cent in the cheaper and less economized machines. The clearance between the parts designated by $c$ in Fig. 35.
55. Cushion Steam and Cylinder Feed. It is customary to imagine the steam operating within an engine cylinder to consist of two parts, the **steam feed** and the **cylinder feed**. The former is that part of the total which is contained in the clearance space before the admission valve opens and serves to cushion the reciprocating parts of the engine. The cylinder feed is the steam which enters through the valve for each cycle.
If the same cycle is produced time after time so that all temperature effects are repeated at regular intervals and so that all events occur at the same points in successive cycles, the quantity of steam retained in the clearance volume will be the same for successive cycles. It is impossible to measure the quantity of this steam directly and indirect methods are therefore adopted for that purpose.
It is often assumed that the steam is dry and saturated when compression begins, as at the point $e$ in Fig. 34. With this assumption, the weight of cushion steam can be determined by dividing the volume occupied, that is,
![]() Fig. 35 — Mechanical and Volumetric Clearances.
86
STEAM POWER
$V_{r}$, by the volume occupied by one pound of dry saturated steam at the same pressure. Thus,
$$\text{Cushion steam} = \frac{17}{\text{Sp.v.d. at pressure } P} \text{ lbs.} \quad (20)$$
The weight of cylinder feed can be very accurately determined by condensing and weighing the steam leaving the engine in a given time and dividing by the number of cycles performed during the same period. It can also be determined by metering the steam entering the engine or by measuring the water fed to a boiler supplying only the engine in question. An approximate determination of the quantity of water indicated may be made by means of an indicator diagram by determining what is known as the diagram water rate. This will be considered in detail at a later point.
When cushion-steam and cylinder-feed have both been determined, the weight of steam contained in the cylinder between cut-off and release can be found by adding the two quantities. Thus,
$$W = W_{c} + W_{k} \quad (30)$$
in which
$W$ = total weight of steam expanding in cylinder per cycle;
$W_{c}$ = weight of cylinder feed per cycle; and
$W_{k}$ = weight of cushion steam per cycle.
The volume which the mixture would occupy if dry and saturated at any given pressure can be determined by multiplying $W$ by the specific volume for that particular pressure.
56. Determination of Initial Condensation. The loss due to initial condensation is so important that it is customary to determine the amount of this loss when studying engines. This can be done with fair accuracy by means of the indicator diagram.
THE REAL STEAM ENGINE
To make such a study it is necessary to know the total weight of material in the engine cylinder at the point of cut-off. This weight may be determined by any of the methods just given. With the known weights, the volumes which would be shut off at different pressures if dry and saturated can be determined by multiplying by the specific volumes at the various pressures. Plotting these points on a $PT$-diagram and connecting them will give a saturation curve for the material in the cylinder such as the curve shown in Fig. 36.
By plotting the curves on the indicator diagram obtained from the engine and then comparing distances such as $ab$ and $ac$ as explained in section 26 of Chapter III
![]() Fig. 36.
![]() Fig. 37.
the quality of the steam within the cylinder at all pressures between cut-off and release can be determined. The weight of initial condensation (up to the point of cut-off) must be the total weight of water shown as existing within the cylinder at that point minus any water brought in by the steam if it was not dry when entering the engine.
Should the saturation curve cross the expansion curve, as shown in Fig. 37, it is evident that the steam occurs in volume greater than the specific volumes toward the end of the expansion; the steam within the cylinder must therefore be superheated during this part of the cycle.
Many formulas have been devised for giving the quantity of initial condensation. They are all based upon the results of experiment and generally only give reliable
88 STEAM POWER
values for cases similar to those used in developing them.
One formula of this sort which has been very widely tested
and been found to give reliable results within its field of
applicability is that devised by Robert C. H. Heck and
explained in his books on the steam engine. The formula is
$$m = \frac{c}{\sqrt[3]{N}} \sqrt{\frac{a\theta}{pe}}, \ldots \ldots \ldots (31)$$
in which
$m$=the fraction representing initial condensation; for
ordinary cases it is the fraction of the cylinder
feed which is condensed during admission, but when
compression is very high and when great weights of
steam are admitted, it may be assumed that all the
portion of all the material within the cylinder which
exists in liquid form at the time of cut-off;
$c$=a coefficient, which varies between 0.25 and 0.30 with
type of engine. May be assumed at 0.27 for average
work;
$N$=engine speed in revolutions per minute (R.P.M.);
$s$=a constant for any engine, equal to nominal surface in
square feet divided by nominal volume in cubic
feet. The nominal surface is the area of the inner
walls and the ends of a cylinder with diameter equal
to the internal diameter of the cylinder and with
a length equal to the stroke of the engine. The
nominal volume is the cubic contents of such a
cylinder;
$$s = \frac{12}{D} \left( \frac{2D + h}{S} \right),$$
in which $D$ and $S$ represent diameter and
stroke of engine in inches;
$\theta$=temperature function obtained from Table II as
there indicated.
$p$=the absolute pressure in cylinder in pounds per square
inch just after completion of cut-off;
THE REAL STEAM ENGINE
$$e = \text{cut-off ratio}, \quad \text{that is, ratio of cylinder volume opened up by cut-off has just been completed to the total piston displacement.}$$
TABLE II
FOR FINDING VALUES OF $e$ FOR USE IN HECK FORMULA
$$F = 1 - e$$
and $a_1$ and $a_2$ are chosen from table for highest and lowest pressures existing in cylinder.
| p |
a |
p |
a |
p |
a |
p |
a |
| 1 |
175 |
15 |
210 |
50 |
269.5 |
30 |
321.5 |
389 |
230 |
441 |
| 2 |
175 |
15 |
210 |
50 |
277.7 |
300 |
332.5 |
370 |
244.7 |
447.5 |
| 3 |
183 |
25 |
220 |
60 |
299.5 |
310 |
343.5 |
390 |
259.5 |
454.5 |
| 4 |
183 |
25 |
220 |
60 |
301.5 |
320 |
355.5 |
400.5 |
276.5 |
467.5 |
| 5 |
191 |
| pn+1 | an+1 | pn+2 | an+2 | pn+3 | an+3 | pn+4 | an+4 | pn+5 | an+5 | pn+6 | an+6 |
|---|
<| Table of Values of e for Use in Heck Formula. |
| pn+1 | an+1 | pn+2 | an+2 | pn+3 | an+3 | pn+4 | an+4 | pn+5 | an+5 | pn+6 | an+6 |
|---|
<| Table of Values of e for Use in Heck Formula. |
| pn+1 | an+1 | pn+2 | an+2 | pn+3 | an+3 | pn+4 | an+4 | pn+5 | an+5 | pn+6 | an+6 |
|---|
|
| Methods of Decreasing Cylinder Condensation. | Before discussing methods of decreasing the loss due to cylinder condensation it will be well to consider what things may be expected to determine the extent of such loss.
The condensation is due directly to the transfer of heat from one body to another at lower temperature, and anything which tends to increase the total amount of heat thus transferred is likely to decrease the loss.
It is therefore evident that the ratio of steam condensed to steam supplied will be greatest when:
(a) The time of contact is greatest;
(b) The ratio of surface exposed to volume enclosed is greatest,
and
(c) The temperature difference is greatest.
The time of contact can be controlled to a certain extent by controlling the speed of the engine and, with other things equal, the higher the speed the lower should be the condensation.
The ratio of surface exposed to steam to the volume occupied by steam has a great influence on the amount of
90 STEAM POWER
condensation which occurs. The surface of the clearance space, including the interior surfaces of all ports or passages leading to the valves, seems to have the greatest influence, and the clearance space which is most nearly a short cylinder without any angles may be expected to give the least initial condensation.
The size of the engine is also important in this connection. The area exposed does not increase as rapidly as does the volume enclosed when the diameter of a cylinder is increased, and therefore large cylinders give smaller ratio of surface to volume and therefore a smaller percentage of steam condensed. Larger engines thus have a decided advantage over small engines.
The shape of the cylinder also has an effect. The longer the cylinder with respect to its diameter the more favorable its performance.
The point at which cut-off occurs is also intimately connected with the condensation loss. In a given cylinder with a given clearance the total condensation within the clearance space may be assumed practically constant if speed and temperature remain about the same. But if the cut-off is made later larger quantities of steam are admitted per stroke, and hence a fraction of the total cylinder feed will be lost by decreased condensation.
The temperature differences depend on upper and lower pressures, that is, on the pressure range. The inner surfaces of the walls follow as rapidly as possible the temperature changes of the steam within them. Thus their average temperature is somewhere between the upper and lower temperatures in the cylinder. For a given upper steam pressure and therefore temperature, the lower pressure be reduced, the average wall temperature also will be reduced, and therefore the differences between the temperature of the entering steam and the average temperature of the walls will be increased with a resulting increase in condensation loss.
THE REAL STEAM ENGINE 91
The methods of decreasing this loss can now be con-
sidered. They are given below under separate heads
with brief explanation when necessary.
(a) Clearance spaces should be properly designed so that they are not too large.
(b) The proportions of cylinder (diameter and stroke)
and the speed of the engine should be so chosen that the
condensation loss is reduced to a minimum.
(c) The engine should be so proportioned that when
delivering its rated power the cut-off occurs at such a point
as to make the percentage of cylinder condensation the
minimum possible.
(d) The cylinder should be surrounded by spaces filled
with air or by materials which are poor conductors of heat
so as to decrease loss by radiation, because all heat lost in this
way must be supplied by the condensation of steam within
the cylinder. Such metallic parts as cannot be “lagged”
in this way should be covered with materials which will
radiate less heat than dull surfaces under like conditions.
(e) The cylinder may be surrounded by a steam jacket,
that is, a space filled with steam similar to that supplied
to the cylinder. The use of such a jacket sometimes results
in a considerable saving and at other times in a great loss.
The amount of steam required for this purpose appears to be the determining factors, and most long-stroke cylin-
ders operating at low rotational speed and with small pressure
ranges are jacketed.
(f) The engine may be compounded, that is, the expan-
sion of the steam may be made up in two or more
cylinders working in series. This results in decreas-
ing the pressure range in each of the cylinders and effects
a decided saving under proper conditions. Compounding
will be considered in detail in a later chapter.
(g) The engine may be supplied with superheated steam.
If the steam is sufficiently superheated it can give up part
or all of its superheat to heat the cylinder walls, and thus no
92 STEAM POWER
condensation need occur. Heat interchanges between metal and superheated steam also appear to be less rapid than is the case when the boiler contains water, so that a saving results from the source also.
Tests made with saturated and superheated steam indicate that from 7° to 10° of superheat are generally required to prevent 1 per cent. of initial condensation. Results differ greatly with the character of the engine, with its economy on saturated steam, with its valve gear, etc. Superheaters are usually employed only in large engines, any well-designed engine, but higher temperatures usually require specially constructed engines. Superheats as high as 150° F. above saturation temperature are now quite common, and there seems to be a tendency to consider a value of 160° F. as high as the highest that is commercially advisable under ordinary conditions.
58. Classification of Steam Engines. Steam engines, are classified on many different systems, the one used in any particular case being determined largely by circumstances. The principal methods of classification are indicated in the following schedule:
| CLASSIFICATION OF STEAM ENGINES |
| On the basis of rotative speed |
Low speed
Medium speed
High speed |
| On the basis of ratio of stroke to diameter |
Long stroke
Short stroke |
| On the basis of slide valve |
Double valve
Balanced slide valve
Multiported slide valve
Piston valve |
| On the basis of corliss valve |
Drop cut-off
Positively operated |
| On the basis of poppet valve |
|
THE REAL STEAM ENGINE 93
| On the basis of position of longitudinal axis |
Vertical
Inclined
Horizontal |
| On the basis of number of cylinders in which steam expands |
Single expansion or Simple engine
Multi-expansion engine |
|
Compound expansion
Triple expansion
Quadruple expansion |
| On the basis of cylinder arrangement |
Single cylinder
Tandem compound
Cross compound
Duplex |
| On the basis of use |
Stationary engines
Portable engines
Locomotive engines
Marine engines
Horsepower engines |
**59. Rotative Speeds and Piston Speeds:** High-speed engines operate at a comparatively high rotative speed and are characterized by a short stroke in comparison with the diameter of the cylinder, the stroke generally being equal to, or less than, the diameter. The piston speed, by which is meant the feet travelled by the piston per minute, generally falls between 500 and 700.
It is not considered advisable to allow piston speeds of stationary engines to exceed about 750 feet per minute for ordinary constructions and the great majority of engines give much lower values. The piston speed will obviously be given by the formula
$$S = 2LN \quad \quad \quad \quad \quad \quad \quad \quad (32)$$
in which
$$S = \text{piston speed in feet per minute};$$
$$L = \text{stroke in feet};$$
$$N = \text{revolutions per minute},$$
94
STEAM POWER
and it is evident from this formula that as the rotative speed is increased the piston speed will increase unless the length of stroke is proportionately decreased. As a result, high-speed engines have short strokes in com-
![]()
A graph titled "HIGH SPEED ENGINES" shows the relationship between cylinder diameter (inches) and stroke (inches). The x-axis ranges from 0 to 28 inches, with increments of 2 inches. The y-axis ranges from 6 to 35 inches, with increments of 5 inches.
**Fig. 38.** —Proportions of High Speed Engines.
comparison with their cylinder diameters and slow-speed engines have long strokes.
The characteristic relations between cylinder diameter and stroke, rotative speed and piston speeds of high-speed engines are given in Fig. 38.
High-speed engines are generally fitted with some
| Diameter of Cylinder (Inches) |
Stroke (Inches) |
| 0 |
6 |
| 2 |
10 |
| 4 |
14 |
| 6 |
18 |
| 8 |
22 |
| 10 |
26 |
| 12 |
30 |
| 14 |
34 |
| 16 |
38 |
| 18 |
42 |
| 20 |
46 |
| 22 |
50 |
| 24 |
54 |
| 26 |
58 |
| 28 |
62 |
| Cyl. Dia. | Piston Speed (R.P.M.) |
| 0.75 inch diameter cylinder. | 175 R.P.M. |
| 1.0 inch diameter cylinder. | 150 R.P.M. |
| 1.25 inch diameter cylinder. | 130 R.P.M. |
| 1.5 inch diameter cylinder. | 110 R.P.M. |
| 1.75 inch diameter cylinder. | 90 R.P.M. |
| 2 inch diameter cylinder. | 70 R.P.M. |
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Piston Speed (R.P.M.)
Cyl. Dia. = Pistion
```json
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"y": [10]
},
{
"x": [4],
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"y": [18]
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{
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{
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```
THE REAL STEAM ENGINE
95
R.P.M.
| | |
|---|---|
| 135 | 100 |
| 100 | 80 |
| 75 | 60 |
| 50 | 40 |
| 25 | 20 |
Piston Speed (ft. per Min.)
| | |
|---|---|
| 200 | 150 |
| 150 | 120 |
| 100 | 90 |
| 75 | 80 |
| 50 | 70 |
| 25 | 60 |
Stroke (in.) in Inches
| | |
|---|---|
| 12 | 14 |
| 16 | 18 |
| 20 | 22 |
| 24 | 24 |
| 28 | 26 |
| 32 | 28 |
| 36 | 30 |
| 40 | 32 |
| 44 | 34 |
Diameter of Cylinder (Inches)
| | |
|---|---|
| 12 | 14 |
| 16 | 18 |
| 20 | 22 |
| 24 | 26 |
| 28 | 30 |
| 32 | 34 |
Fig. 20.
Slow Speed Engines
Fast Speed Engines
96
STEAM POWER
form of balanced slide valve, and are controlled by what is called a shaft governor. They are very compact, having small weight and occupying small space in comparison with the power developed.
Slipper valves, which are, in general, the most economical and are characterized by low rotational speed, long stroke, and elaborate valve gear. The weight per horse-power is high, and they generally occupy a great deal of space. The Corliss engine is the best known and most widely built engine of this type.
The characteristic relations of cylinder diameter to stroke, rotational speed and piston speed for slow-speed engines are given in Fig. 39.
Medium-speed engines generally operate at rotational speeds between 150 and 250 R.P.M. They are generally fitted with the better forms of multipported and balanced slide valves, with poppet valves, or with a positively operated Corliss type of valve gear.
**60. The Simplest D-slide Valve Engine.** The simplest and cheapest type of reciprocating steam engine manufactured is shown in part section in Fig. 40 with the principal parts labelled. The cylinder, piston, steam chest and valve are sectioned in order to show the internal construction.
This engine, like most steam engines, is double acting, that is, a cycle is produced on each side of the piston during every revolution. Steam is admitted and expanded on one side of the piston while steam is being exhausted on the other side. The control of admission and exhaust is effected by the slide valve and will be considered in detail in later sections.
The mechanical energy made available by the steam operating in an engine cylinder is not developed at a uniform or constant rate, but fluctuates, during each revolution, above and below the amount required to overcome the constant resistance at the shaft due to the work the engine
THE REAL STEAM ENGINE 97
is doing. If no provision were made to prevent it, this would result in a very variable rate of rotation during each revolution. When the energy made available was in excess of the demand it would be used in accelerating the moving parts of the engine and the speed of the latter would increase. The reverse would occur when the supply did not equal the demand.
The fly-wheel is used to prevent violent fluctuations
![]() A diagram of a simple D-slide Valve Engine.
Quintin Fleetwood and Brothers
Max Revolving Gear
Bulldog
Excess
Fig. 40.—Simple D-slide Valve Engine.
of this kind. It is made with a comparatively heavy rim and a great deal of energy must be supplied to accelerate it to any appreciable extent in a short time. Similarly it can give out a considerable store of energy when slowing down. The fly-wheel therefore serves as a sort of reservoir which excess energy can be stored temporarily and from which it can later be withdrawn when a deficiency exists.
The fly-wheel thus acts as a damper to variation of rotational
98
STEAM POWER
speed during each revolution, minimizing but not entirely eliminating such variation. It may also serve as a belt wheel, as shown in the illustration.
The governor controls the steam supply to the cylinder in such a way that enough heat will be supplied to make available the power demanded at the shaft. Were more supplied, the excess would be absorbed by the moving parts and the engine speed would increase; were less supplied the engine speed would decrease.
**61. Engine Nomenclature.** The meanings of several terms used in describing engines are not self-evident, their definitions depending merely on accepted usage. Some of these terms and their illustrations are illustrated in Fig. 41.
The crank end of a horizontal engine is called the front end of the engine, so that the cylinder head nearest the crank is called the front head and the stroke of the piston toward the crank is known as the forward stroke. The forward stroke of the piston is also spoken of as the outstroke, particularly in connection with single acting engines. The stroke away from the crank is correspondingly designated as the return or the instroke.
**62. Principal Parts of Engines.** The parts of engines may be roughly divided into stationary and moving, such as frame, cylinder, cylinder and valve chest covers, etc., which
![]() Fig. 41.—Engine Nomenclature.
Left Hand Engine
Right Hand Engine
Bolt Backward
Bolt Forward
Right Hand Engine
Left Hand Engine
THE REAL STEAM ENGINE 99
are stationary, and piston, piston rod, crosshead, fly-wheel,
etc., which are all moving parts when the engine is in opera-
tion. The moving parts are often divided into reciprocal-
![]()
A diagram showing a steam engine with a piston, connecting rod, and flywheel.
Fig. 42.—Frame for Small Vertical Engine.
ing and rotating parts. Thus the piston and all connected
parts through and including the crosshead, and the valve
and many connected parts in the case of slide valve engines,
Cylinder linked against
cylindrical
surfaces
Jaws for
main bearings
Flat cross head guides
![]()
A diagram showing a steam engine with a cylinder, piston, connecting rod, and flywheel.
Fig. 43.—Frame for Medium Speed Center Crank Engine.
all reciprocate when the engine is in operation. The
shaft, fly-wheel, eccentric sheaves and governor constitute
the principal rotating parts.
100
STEAM POWER
Some engines also have oscillating parts, such as the valves in Corliss types, which rock back and forth in the arc of a circle, and the rocker arms in various forms of valve gear. These arms rocking through a short arc are about a fixed pin near one end.
The principal parts and their functions are briefly considered in the following paragraphs:
(a) The Frame. The frame of the engine, sometimes known as the bed, serves to support the other parts, to tie
![]() (a) Cylinder bolted to frame with large boss.
![]() (b) Jaws for main bearing.
![]() (c) Boxed cross head guide.
Fig. 44.—Frame for Slow Speed Engine of Corliss Type; Side or Overhung Crank.
them together in their proper relations and to fasten the whole structure to whatever foundation is used. The cross-head guides and the seats for the main bearings are incorporated in the frame.
The frame is commonly made of cast iron in the form of a hollow box which is properly ribbed to give the necessary stiffness.
Examples of frames are shown in Figs. 42, 43 and 44.
THE REAL STEAM ENGINE 101
(b) The Cylinder and Steam Chest. The cylinder and steam chest are generally incorporated in the same casting, and surfaces of covered cavities in this casting are finished
![]()
A diagram showing the components of a steam engine. The top part shows a steam pipe connected to a cylinder. Below that, there is a steam chest cover with a valve on it. The bottom part shows a cylinder with a piston rod extending out of it.
Fig. 45.
![]()
A diagram showing the internal components of a steam engine. The top part shows a steam chest cover with a valve on it. Below that, there is a cylinder with a piston rod extending out of it. The bottom part shows the cylinder head and the cylinder.
Fig. 46.
to form the cylinder in which the piston operates and the seat or seats upon which valves rest and move.
The cylinder may be single walled with flanges on the end to receive the cylinder head, as illustrated in Fig. 40
102
STEAM POWER
(plain D-slide valve), in which case a thin sheet-metal jacket is fastened around it and the space between filled with heat-insulating material. Or, the cylinder may be cast with double walls, the space between the two being used as a heat-insulator.
Some cylinders are fitted with a liner, which is a plain cylinder pressed into place within the cylinder casting and forming the bore of the working cylinder. This prac-
![]() A diagram showing the internal components of a steam engine, including a cylinder, piston, connecting rod, and other parts.
Fig. 47.—Section of Atlas Medium Speed Engine, Showing Balanced Slide Valve.
Tice is common on the larger types, the liner being used so that when wear has occurred it can be replaced cheaply, instead of it being necessary to reboil or even replace the main casting.
Examples of cylinder construction are shown in Figs. 45, 46, 47, 48, 49 and 50.
(c) The Piston. The function of the piston is two-fold. It must prevent the leakage of steam by it from one end of the cylinder to the other, and it must receive the
THE REAL STEAM ENGINE
103
![]()
A diagram showing the internal components of a steam engine. The diagram includes labels such as "Main Cylinder," "Steam Cylinder," "Guide," "Air Space," "Valve," "Steam Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port," "Valve Port,", and so on.
Fig. 48.—Section of Cylinder, Cylinder and Guides, and Half-section of Crosshead.
103
104
STEAM POWER
pressures exerted by the steam and transmit them to the other parts of the mechanism as it moves.
![]() A diagram showing the internal structure of a Corliss cylinder. The top part shows the steam port, surface of the piston, exhaust port, and chambers for admission valves. The bottom part shows the steam connection and lagging in place.
Fig. 49.—Corliss Cylinder Casting.
![]() A diagram showing the internal structure of a Corliss cylinder with lagging in place. The top part shows the steam port, surface of the piston, exhaust port, and chambers for admission valves. The bottom part shows the steam connection and lagging in place.
Fig. 50.—Corliss Cylinder with Lagging in Place.
Leakage of steam is prevented by the use of piston rings, which are metal rings fitted into grooves in the circular surface of the piston and pressed out against the cylinder
THE REAL STEAM ENGINE 105
![]() Piston
Fig. 51.
![]() Piston
Fig. 52.
![]() Follower Plate
Piston Rings
![]() Piston Body at Stroke
End of Piston Rod
![]() Concentric 3/4"
Pin to Lock Nut on Rod
![]() Shell Rings
Fig. 53.—Buil-up Piston Used in Large Engines.
106
STEAM POWER
walls by spring action. They may be made of one piece of metal turned into a ring of slightly larger diameter than the cylinder, cut through and sprung into place, or they may be made in pieces as shown in Fig. 51, and pressed out against the wall by small helical or leaf springs.
The piston itself may consist of a solid disk of metal fitted with a hub and a short cylindrical part with grooves for the rings, as shown in Fig. 52 (a) and (b), or it may be an elaborate built-up structure, as shown in Fig. 53.
![]() Fig. 54.—Crosshead and Pin.
(d) The Piston Rod. The piston rod is a plain circular steel rod fitted with such shoulders and threads at the ends as are necessary for the fastening of the piston and the cross-head. Examples of such fastenings are given in Figs. 51, 52, 53 and 55.
In large horizontal engines the piston rod sometimes extends through the piston and rear cylinder head, and the rear end of this rod is known as the tail rod crosshead. The extension of the rod is known as the tail rod and the auxiliary crosshead as the tail rod crosshead.
THE REAL STEAM ENGINE 107
Such constructions are used when the weight of the piston is so great that it would cause serious cylinder wear if not supported more perfectly than is possible with the ordinary overhung arrangement.
![]() Fig. 55.—Single Slipper Crosshead.
![]() Fig. 56.—Crosshead with Adjustable Slippers.
(e) The Crosshead and Guides. The crosshead and guides are used for the purpose of supporting the piston and its rod and guiding them in a straight line. The crosshead also serves to connect the piston rod and the connecting rod through which the forces are transmitted to the crank pin.
108
**STEAM POWER**
Crossheads are generally cast in the form of imperfectly shaped boxes and carry slippers which are faced with anti-friction metals where they come in contact with the guides. The slippers may be fixed to the crosshead by means of pins, as shown in Fig. 40, or they may be turned and operate in bored guides, as shown in Figs. 48, 54, 55 and 56. Provision is generally, though not always, made for taking up wear of guides and slippers by setting the slippers further out from the body of the casting.
With the type shown in Figs. 40 and 54 this can be
![]() A diagram showing a crosshead with a slipper attached to it.
Fig. 57.—Solid End Connecting Rod; for Overhung Cranks only.
done by the insertion of thin sheets of metal or paper (known as mains) between the body of the crosshead and the slipper. In the type shown in Fig. 56 the slippers are finished with inclined surfaces where they come in contact with the main casting, and the adjustment is made by wedging the slippers apart by the use of the adjusting bolts shown.
The wrist-pin end of the connecting rod enters the crosshead casting and is held in place by means of the wrist pin, about which it oscillates when the engine is in operation.
THE REAL STEAM ENGINE 103
(f) The Connecting Rod. This rod connects the re-
ciprocating crosshead with the rotating crank shaft and
transmits the forces from one to the other. It consists
of a body or shank and two ends or heads. The ends may
![]() Illustration of a connecting rod with a bolted strap end.
Fig. 58.—Connecting Rod with Bolted Strap Ends; May Be Used
with Center or Side Crank Construction.
be "closed" or "solid" as shown in Fig. 57; they may be made with a strap bolted in place as shown in Fig. 58; or the crank-pin end may be made of two half boxes bolted together to form a "marine end" as shown in Fig. 59.
The ends are always made adjustable so that wear can be taken up, thus preventing noisy operation due to hammering between the ends and the pins at times when the direction in which forces act is reversed. With solid and strap types this adjustment is generally made by means of wedges similar to those shown in Figs. 57 and 58.
![]() Illustration of a marine end connecting rod.
Fig. 59.—Marine End Connecting Rod.
Crank End Crosshead End
110
STEAM POWER
With the marine type shims are used between the two halves, and the diameter of the hole formed by the latter is decreased by the removal of shims of the required thickness.
(g) The Shaft. The crank shaft itself is generally made of steel, but the counterbalances are often of cast iron. It may be one forging throughout or may be built up by shrinking the various parts together. Multicrank shafts of large size are generally of built-up construction, the crank pins being forged into the crank arms and the latter shrunk on to the pieces of shaft.
The counterbalance weights are used to balance the
![]() Crank 2nd
Count.Jam
Jam
Fig. 60.—Crank Shaft, Center Crank.
![]() Counterweights or Crank Arms
Fig. 61.—Center Crank.
![]() Counterweight
Fig. 62.—Center Crank.
THE REAL STEAM ENGINE 111
centrifugal effect of the crank pin, part of the crank arms and part of the connecting rod, all of which rotate off center. In some engines part of the unbalanced effect of the reciprocating parts is also imperfectly balanced by these counter-balances.
Various types of shafts are shown in Figs. 60, 61, 62, 63, and 64.
(a) Bearings. Bearings are distinguished as main and as outboard bearings. Main bearings are those carried by
![]() Crank Pin
![]() Journal
![]() Crank Pin
![]() Outboard Bearing
Fig. 63.—Crank Shaft and Disc, Overhung Crank.
![]() Overhung Crank
Fig. 64.—Overhung Crank.
the frame of the engine and outboard bearings are carried by separate pedestals or by pedestals fastened to a plate which is in turn fastened to the frame. Center-crank engines have two main bearings, and side-crank engines only have one at the other end of the shaft being supported by an overhung crank.
The bearings of steam engines are generally formed of habbit-lined boxes carried within jaws machined in a frame, or in a separate pedestal, and held in place by a bearing cap. The boxes are made in two, three or four parts to allow for adjustment to compensate for wear and
112
STEAM POWER
to give a certain degree of flexibility. Adjustment for wear is either made by means of wedges or by means of screws which force the various parts of the boxes toward the shaft. An example of a three-part bearing with screw adjustment is shown in Fig. 65. The parts of a three-part bearing with wedge adjustment are shown in Fig. 66.
Bearings are often lubricated by rings or chains, and they are then known as ring- or as chain-oiling bearings.
![]() A diagram showing a three-part main bearing with screw adjustment.
Fig. 65.—Three Part Main Bearing with Screw Adjustment.
In the ring-oiling bearing one or more metal rings of large diameter hang loosely on the shaft within the bearing and dip into an oil reservoir below the shaft. Rotation of the shaft causes the ring bearings to rotate and they carry oil up from the reservoir and spill it out over the shaft within the bearing. Chain bearings are similar except that chains are substituted for rings.
(i) Fly-wheels. The function of the fly-wheel has already been considered and need not be discussed further. The wheel is constructed with a heavy rim joined to a hub
THE REAL STEAM ENGINE 113
by six or eight arms. In the smaller sizes the wheel may be cast in one piece, but best practice calls for a split hub in that case to partly equalize certain casting strains which result from unequal thicknesses of metal in different parts of the wheel. The two-part wheels are made in two parts both for the purpose of partly avoiding casting strains and for the purpose of facilitating handling and shipping.
![]() Fig. 66.—Three Part Bearing Showing Wedge Adjustment.
![]() Fig. 67.
A two-part wheel with the rim sections joined by prisoner links shrunk in place and the hub fastened with bolts is shown in Fig. 67.
**PROBLEMS**
1. A given engine has a piston displacement of 3 cu.f.t. and a clearance volume of 3%. Compression begins when 85% of the exhaust steam has left the cylinder. At this point, the cylinder at that time is 16 lbs. per square inch absolute. Determine the weight of the cushion steam on the assumption that this steam is dry and saturated at the beginning of compression.
114
STEAM POWER
2. Assume the engine described in Prob. 1 to cut-off at 1 stroke and with a pressure inside the cylinder equal to 115 lbs. per square inch absolute. Find the weight of cylinder feed if the quality of the material in the cylinder at the time of cut-off is 75%.
3. Find the piston speed of an engine with a stroke of 2 ft. and a rotation of 600 RPM.
4. Show by means of Heck's formula that initial condensation increases with pressure range.
CHAPTER VIII
THE INDICATOR DIAGRAM AND DERIVED VALUES
63. The Indicator. The ideal steam engine cycle was described in Chapter IV, and the sort of diagram which would be obtained from a real engine was shown in Chapter VII; but the means by which such diagrams are obtained from operating engines was not given.
Indicator diagrams showing the pressure and volume changes experienced by steam in the cylinders of real
![]()
A diagram of an indicator diagram. The main components are labeled:
- Cylinder
- Point Holder
- Lever
- Clutch to hold Card
- Drum
Fig. 68.
115
116
STEAM POWER
engines are obtained by means of an instrument known as an indicator. The operation of obtaining such diagrams is known as *Collecting the engine*.
An external form of indicator is shown in Fig. 68 and a section through the instrument is given in Fig. 69. The method of connecting an indicator to the
![]() A diagram showing the internal components of an indicator, including pistons, rods, cylinders, and a drum with a cord attached. The diagram also shows how the indicator is connected to the cylinder of a steam engine.
Fig. 69.
cylinder of a steam engine and one method used for driving it are illustrated in Fig. 70.
The indicator is intended to draw a diagram showing corresponding pressures and volumes within the engine cylinder and must, therefore, contain one part which will move in proportion to pressure variations and another which will move in proportion to volume changes. The one may
INDICATOR DIAGRAM AND DERIVED VALUES 117
be called the pressure-measuring and the other the volume-measuring device.
The pressure-measuring device generally consists of a piston, such as shown in the figure, working with minimum friction against a cylinder and fitted with a spring which will resist what may be called outward motion (upward in the figure). The cylinder containing this piston is coupled to a short pipe connected with the clearance space of the engine and, whenever the indicator cock in this connection is open, the steam acting on the engine piston
![]() A diagram showing a steam engine with an indicator attached. The indicator has a piston that moves up and down, and a pencil mechanism that traces a line on a drum.
Fig. 70.—Method of Attaching and Operating an Indicator,
will also act on the indicator piston. Steam of any given pressure will drive the indicator piston out against the action of the spring until the pressure exerted by the spring is equal to that exerted on the face of the piston. The indicator piston will thus move out different distances for different pressures, and, through the piston rod and pencil mechanism, will make marks on the drum at different heights corresponding to different steam pressures. The pencil mechanism is so arranged that the point traces a straight vertical line on the drum as the indicator piston moves in and out.
Springs are made to certain definite scales, thus there
118 STEAM POWER
are, for instance, 10-lb., 25-lb., 50-lb. and 100-lb. springs.
The number which is known as the scale of the spring designates the steam pressure in pounds per square inch which is required to move the pencil point 1 inch against the action of the steam. In order to measure the steam on the indicator, a steam pressure of 50 pounds per square inch acting on the indicator piston would drive the pencil up a distance of half an inch, a pressure of 100 pounds per square inch would give 1 inch of motion and so forth.
The volume-measuring device is of an ingenious kind. It simply consists of a cylinder containing by the engine piston at the time when a given steam pressure existed in the cylinder and the volume occupied by the steam can be calculated from piston position and cylinder dimensions. The position of the piston is indicated by connecting the cord wound around the drum to some part of the engine which moves with the piston, such as a pedal. This cord is commonly used for this purpose, and, since the motion of this member is generally much greater than the circumference of the drum, it is necessary to use a reducing mechanism of some sort. This mechanism must be very accurate, so that it may move the drum as nearly as possible in proportion to the motion of the piston.
The pencil point moves up and down as the pressure within the cylinder varies, and the drum rotates under the point in proportion to the motion of the engine piston, so that the combination of the two motions brings the pencil point to successive positions on the drum which indicate accurately corresponding values of steam pressure and piston position. By cutting out a piece of paper, known as a card, on the drum and pressing the pencil point upon this paper, the successive positions occupied by the pencil point will be recorded in the form of a series of curves and straight lines.
If the drum is rotated with the lower side of the indica-
tor piston connected to atmosphere, the pencil will trace
INDICATOR DIAGRAM AND DERIVED VALUES 119
a horizontal line. This is known as the atmospheric line and is used as a reference for locating the pressure scale. If the indicator cylinder is then connected with the engine cylinder and the drum is rotated by the reducing mechanism, a diagram similar to that of Fig. 71 will be drawn upon the card. The atmospheric line indicates the height assumed by the pencil when atmospheric pressure acts on the piston at zero displacement. The drawing atmospheric pressure (barometer reading) and the scale of the spring, a line at a height representing zero pressure can be drawn on the card. This line is indicated in Fig. 72.
The length of the card between the lines $a$ and $b$ is proportional to the length of the engine stroke and there-fore to the piston displacement, that is, to the volume swept through by the piston. Knowing the clearance volume of the engine as a percentage or fraction of the piston displacement, this fraction of the length of the diagram can be laid off from the oral zero diagram to give a line of zero volume. This line is also indicated in Fig. 72.
With the line of zero pressure and the line of zero volume drawn in, all values of steam pressure and volume occupied by steam can be read directly from the diagram, and thus obtain a picture of what occurs within the real engine cylinder.
The indicator diagram is used for a number of purposes, the more important being:
![]() Fig. 71.
![]() Fig. 72.
Atmospheric Line
Pressure Scale
Line of Zero Volume
Atmospheric Line
Volume Scale
120
STEAM POWER
(1) The determination of the energy made available within the cylinder, that is, the indicated horse-power, I.h.p.
(2) The determination of the amount of initial condensation and of heat interchanges between walls and cylinder.
(3) The determination of what is known as the digram water rate.
(4) The study of the operation and timing of valves.
The second one of these uses has already been considered in Chapter VII, the others are treated in succeeding sections.
**64. Determination of I.h.p.** The lines of the indicator diagram show by their height the pressures or forces acting on the engine piston as it moves. But the product of force by distance is equal to work and these lines can be used therefore to determine the total work done by the steam upon the piston.
In Fig. 73 is shown the upper part of the diagram, the curved lines representing the pressure acting in pounds per square inch which acted on the left face of the piston while it moved outward. If the average pressure could be determined over and multiplied by the area of the piston face, this product would be the average total force acting on the piston. Multiplying this by the distance traveled would give the work done by the steam upon the piston. Expressed in the form of an equation,
$$E_0 = p_0 \times A \times L \text{ ft.-lbs.,} \quad \ldots \quad (33)$$
in which
$$E_0 = \text{work done upon piston by steam during outstroke;}$$
$$p_0 = \text{mean pressure (in pounds square inch) acting on piston during outstroke;}$$
INDICATOR DIAGRAM AND DERIVED VALUES 121
$a=$ area of piston face in square inches; and
$L=$ stroke of piston in feet.
For the instroke shown in Fig. 74, the work done by the piston on the steam is given by the similar expression:
$$E_1 = p_0 \times a \times L \text{ ft.-lbs.,} \quad \ldots \quad (34)$$
![]() Fig. 74.—Negative Work Area.
in which $E_1$ and $p_0$ represent work done and mean pressure respectively.
The net work done by the steam upon the piston per cycle is then,
$$E_{\text{cycle}} = E_0 - E_1 = (p_0 - p_1) aL \text{ ft.-lbs.,} \quad \ldots \quad (35)$$
The values of $p_0$ and $p_1$ can be found directly from the diagram by dividing the areas $A_0$ and $A_1$, respectively by the length $l$ and then multiplying by the scale of the spring, giving
$$p_0 = \frac{A_0}{l} \times \text{scale of spring},$$
and
$$p_1 = \frac{A_1}{l} \times \text{scale of spring},$$
122 STEAM POWER
so that,
$$p_0 - p_k = \frac{A_0 - A_1}{l} \times \text{scale of spring} \quad . . . \quad (36)$$
$$= \frac{\text{area of diagram}}{l} \times \text{scale of spring} \quad . . . \quad (37)$$
The value of $p$ evidently can be determined very simply from the indicator diagram, and the work per cycle can be found when $p$ is known by substituting in the following equation, obtained by putting $p = p_0 - p_k$ in Eq. (35),
$$E_{\text{cycle}} = p \times a \times L \times n \text{ ft.-lbs.} \quad . . . \quad (38)$$
The pressure $p$ is known as the **mean effective pressure** and is often represented by M.E.P.
If $n$ cycles are produced per minute, the net work done by the steam upon the piston per minute will be
$$E_{\text{min}} = p \times a \times L \times n \quad . . . \quad (39)$$
which is generally rearranged to read,
$$E_{\text{min}} = pLan \quad . . . \quad (40)$$
in which form the group of letters forming the right-hand member is easily remembered.
Since 33,000 foot-pounds per minute are equivalent to one horse-power, it follows that the power made available as shown by the indicator diagram, that is, the **indicated horse-power**, must be,
$$\text{l.h.p.} = \frac{pLan}{33,000} \quad . . . \quad (41)$$
in which
$$p = \text{mean effective pressure in pounds per square inch;}$$
$$L = \text{stroke of piston in feet;}$$
$$a = \text{area of piston in square inches;}$$
$$d = (\text{diam. cyl. in inches})^2 \times 7854p;$$
$$n = \text{number of cycles per minute}.$$
INDICATOR DIAGRAM AND DERIVED VALUES 123
If an engine cylinder takes steam on one side of the piston only, that is, if the cylinder is *single acting*, the number of cycles produced per minute is equal to the number of revolutions per minute, but it should be noted that for other arrangements this is not necessarily true. In the case of double-acting engines which receive steam at both ends of the cylinder, the number of cycles produced is equal to twice the number of revolutions.
It should also be noted that the symbol $a$ represents the area of the piston face upon which the steam acts. If a piston rod extend from the face of the piston to and through the cylinder head (as is always the case at the crank end), then the area of the piston head must be subtracted from that of the piston to obtain the area on which the steam really acts. When a tail rod is used, a correction must be made for each side of the piston.
In the case of double-acting engines the indicated horsepower may be determined in two ways: It may be figured separately for the two ends of the cylinder, or the values for the area and pressure may be averaged for the two ends and the value of $a$ chosen equal to twice the revolutions per minute. The former is generally the more accurate method.
It will have been observed that the area of the indicator diagram must be determined before the mean effective pressure can be found. This area is generally measured by means of an instrument known as a *planiometer*, and this is the most accurate method. It occasionally happens, however, that a pliniometer is not available when the value of the indicated horse-power is desired. Under such circumstances it is possible to determine approximately the area of the indicator diagram can be made by the method of ordinates.
For this purpose the length of the diagram is divided into an equal number of parts, usually ten, as shown in
124
STEAM POWER
Fig. 75 and vertical lines $y_1$, $y_2$, $y_3$, etc., are drawn at the center of each of the parts into which the diagram has been divided. The mean ordinate or height is then found from the equation,
$$y_m = \frac{y_0 + y_1 + y_2 + y_3 + \text{etc.}}{\text{number of vertical lines}} \quad \quad (42)$$
and the mean effective pressure is then determined by multiplying $y_m$ by the scale of the spring.
An indicator diagram similar to that shown in Fig. 76 is occasionally obtained. The small loop on the end represents negative work, since the pressure of the steam which does work upon the piston is lower than that which resists the return of the piston. When using a planimeter, this area is automatically subtracted from that of the rest of the diagram, but care should be taken to see that this is also done when the method of ordinates is used.
ILLUSTRATIVE PROBLEM
1. Determine the L.H.p. of a double-acting steam engine, having a cylinder 8 ins. diameter, length of stroke, 12 ins., running at 100 R.P.M., the mean effective pressure (M.E.P.) on the piston being 45 lbs. Neglect the area of the piston rod.
$$\text{L.H.p.} = \frac{p \times A}{(x \times o)} \times \text{L.s.} \times \text{L.e.} \text{ ft. per min.}$$
$$= \frac{33,000 \times 8 \times 7.854}{(45 \times x \times o)} \times \frac{1}{(x \times 100 \times 2)} \text{ ft. per min.}$$
$$= 33,000$$
INDICATOR DIAGRAM AND DERIVED VALUES 125
$$= \frac{2260 \text{ lbs.} \times 200 \text{ ft. per min.}}{33,000} = 14 \text{ nearly}.$$
2. The L.h.p. of a double-acting engine is 14, the R.P.M. = 100; M.F.R. = 1. Find the diameter of the cylinder neglecting area of piston rod.
First determine the area of the piston from the formula
$$\text{L.h.p.} = \frac{p \text{Lan}}{33,000} \text{ or } a = \frac{33,000}{p \times L \times n},$$
$$a = \frac{23,000 \times 14}{45 \times 1 \times 100 \times 2} = 51.4 \text{ sq.in.} = \frac{\pi d^2}{4},$$
$$d = \sqrt{\frac{51.4}{784}} = \sqrt{65.4} = 8 \text{ ins. (approx.)}.$$
65. Conventional Diagram and Card Factors. It is often necessary to approximate the mean effective pressure obtained in the cylinder of an engine when no indicator diagram are available. This can be done by assuming that when an engine is being designed to carry a certain load and it is desired to determine the necessary cylinder dimensions and speed. If the probable mean effective pressure can be determined, the dimensions and speed can be found from the equation,
$$\text{l.h.p. per cylinder end} = \frac{p \text{Lan}}{33,000},$$
by rewriting it
$$\text{l.h.p.} = \frac{p \text{Lan} \times 0.7854 d^2}{33,000},$$
from which
$$d^2 = \frac{33,000 \text{ l.h.p.}}{0.7854 p \text{Lan}},$$
(43)
Since $n$ is equal to revolutions per minute for one cylinder end, the product of $L$ by $n$ must be equal to half the piston
126
STEAM POWER
speed of the engine, and a proper value of this product can be chosen for substitution in the equation. If a proper value for $p$ can then be predicted the only unknown remaining will be the diameter $d$, and this can be found by solving the equation.
The prediction of the mean effective pressure is made either by drawing upon recorded experience in the form of values obtained in similar engines previously constructed or by means of what is known as a conventional indicator diagram.
The conventional diagram is drawn with upper and lower pressures equal to those expected in the case of the real engine, and all expansions and compressions are drawn as rectangles or hyperbolas. The equation of the ordinary hyperbola is
$$P_1V_1 = P_4V_4 \quad \ldots \quad \ldots \quad (44)$$
in which subscript 1 indicates initial conditions and sub- script $a$ represents any later conditions with the same material in the cylinder. This law is assumed because it is the simplest and, as a rough average, gives values as close to those actually attained as do any of the more complicated laws.
The diagram may be drawn as nearly as possible like the one which the engine may be expected to give or it may be drawn with increasing accuracy until it gets closer to it more and more from the approximation to an actual indicator diagram. In any case, the mean effective pressure is determined from this diagram and this value is then multiplied by a corrective factor, the value of which has been determined by experiment. The corrective factor is called the "corrector factor" and its value is such that the ratio of the area of the diagram the engine would really give to the area of the conventional diagram used.
The simplest form of conventional diagram is drawn
INDICATOR DIAGRAM AND DERIVED VALUES 127
by neglecting the clearance volume and has the shape shown in Fig. 77. The upper line is drawn horizontal at a height representing the highest pressure expected and of such a length (compared with the length of the diagram) as will approximately represent that portion of the stroke at which work is to occur in the real engine. The expansion curve is then drawn in as a rectangular hyperbola and extended until the end of the diagram is reached. The next line is drawn vertical and the lower limit of the area $A_1$ is made horizontal at a height representing the pressure expected in the space into which the engine is to exhaust.
This simple diagram can be divided into the three areas shown and the value of the work represented by these areas can be determined from the equations given below, the first and last of which should be self-evident from what has preceded. The equation for the work represented by area $A_2$ can be determined very easily by means of integral calculus. The equations are,
$$A_1 \text{ represents } P_1 V_1 \text{ ft.-lbs.,}$$
$$A_2 \text{ represents } P_1 V_1 \log_{\frac{V_2}{V_1}} = P_1 V_1 \log r \text{ ft.-lbs.,}$$
and
$$A_3 \text{ represents } P_3 V_2 \text{ ft.-lbs.,}$$
in which $P$ represents pressure in pounds per square foot and $V$ represents volume in cubic feet.
The total area is then equal to the sum $A_1 + A_2 - A_3$ and the net work is equal to a similar sum of the right-Fig. 77—Conventional Indicator Diagram.
![]() A diagram showing a conventional indicator diagram with labeled areas A1, A2, A3, and V1, V2, V3.
Fig. 77—Conventional Indicator Diagram.
128 STEAM POWER
hand members given above. The net work must also equal the mean-effective pressure $P_w$ multiplied by the total volume change, so that
$$P_w V_2 = P_1 V_1 + P_2 V_1 \log r - P_2 V_2 \quad . . . \quad (45)$$
and
$$P_w = P_1 \frac{V_1}{V_2} + P_2 \frac{V_1}{V_2} \log r - P_2 \quad . . . \quad (46)$$
$$= P_1 \left( \frac{V_1}{V_2} \log r \right) - P_2 \quad . . . \quad (47)$$
and substituting $\frac{1}{r}$ for $\frac{V_1}{V_2}$ this gives
$$P_w = P_1 \left( \frac{1 + \log r}{r} \right) - P_2 \quad . . . \quad (48)$$
The ratio $\frac{V_2}{V_1} = r$ is called the **ratio of expansion** and its reciprocal $\frac{1}{r}$ is known as the **cut-off ratio**. By the use of this ratio the volume terms can be disposed of and the equation above is obtained. This equation then gives the mean effective pressure in terms of upper and lower pressures and the fraction of the stroke at which cut-off is desired in the real engine and no cylinder dimensions need be known.
Since pressures in steam-engine practice are usually given in pounds per square inch, the equation for mean effective pressure is more useful in the form
$$p_w = p_1 \left( 1 + \log r \right) - p_2 \quad . . . \quad (49)$$
in which $p_1$ and $p_2$ and $p_w$ are expressed in pounds per square inch absolute. For convenience in the use of this
INDICATOR DIAGRAM AND DERIVED VALUES 129
equation the values assumed by the bracketed quantity are given for various conditions in Table III.
| r |
1+logr |
r |
1+logr |
r |
1+logr |
| 1.0 |
1.00 |
6.0 |
0.465 |
16.0 |
0.236 |
| 1.8 |
0.97 |
7.0 |
0.421 |
28.0 |
0.225 |
| 2.0 |
0.847 |
8.0 |
0.385 |
38.0 |
0.216 |
| 2.5 |
0.766 |
9.0 |
0.355 |
49.0 |
0.208 |
| 3.0 |
0.716 |
10.0 |
0.331 |
63.0 |
0.200 |
| 3.3 |
0.644 |
11.0 |
0.309 |
71.0 |
0.192 |
| 4.0 |
0.597 |
12.0 |
0.290 |
82.0 |
0.186 |
| 4.5 |
0.554 |
13.0 |
0.274 |
94.0 |
0.183 |
| 5.0 |
0.522 |
14.0 |
0.256 |
110.0 |
0.174 |
5.5
|
The values of the mean effective pressures obtained from this form of diagram are very much higher than are to be expected from real engines with the same initial and terminal pressures and the same nominal ratio of expansion. They are therefore corrected by multiplying by the proper diagram factor as selected from Table IV. It is obvious from the range of values given that the selection of a proper value for the factor depends largely on experience, but such experience is quickly gained by contact with real engines and a study of the practical diagrams.
TABLE IV
DIAGRAM FACTORS
Simple slide-valve engine.................. 55 to 99%
Simple Corliss engine....................... 83 to 99%
Compound slide-valve engine................. 55 to 88%
Compound Corliss engine..................... 75 to 88%
Triple-expansion engine...................... 58 to 78%
66. Ratio of Expansion.—The ratio of expansion used above is sometimes called the apparent ratio. It is not the
130
**STEAM POWER**
real ratio of expansion for an engine with clearance. For such an engine the real ratio of expansion is
$$r = \frac{V_2 + V_{\text{eff}}}{V_1 + V_{\text{eff}}} \dots \dots \dots \dots (50)$$
in which the symbols represent the volumes indicated in Fig. 78.
The numerical values of $r$ and $T$ are often very different and care should be used in distinguishing between them. The diagram: factors referred to in Table IV are for idealized conventional cards without clearance as shown in Fig. 77.
**ILLUSTRATIVE PROBLEMS**
1. Given an engine with a stroke of 24 ins. and cut-off occurring at $\frac{1}{4}$ stroke. Steam pressure of 160 lbs. per square inch and back pressure of 16 lbs. Assume diagram factor = 80%. Neglecting clearance, find the probable M.E.P.
M.E.P. = $p(\frac{1+log r}{r}) - p_2 = 160(\frac{1+log 3}{3}) - 16$
= $160 \times 7 - 16 = 112.0 - 16 = 96$ lbs.
Hence probable M.E.P. = $80 \times 96 = 76.8$ lbs.
2. A given double-acting engine indicates 75 L.h.p. under the following conditions:
Cut-off at 20%; steam pressure, 140 lbs. per square inch absolute; piston speed, 600 ft. per minute; back pressure, 2 lbs. per square inch absolute.
Assume a diagram factor for this type of engine equal to 85%; and neglecting clearance, find a convenient size of the cylinder (diameter and stroke).
![]() A graph showing the relationship between volume and pressure in a steam engine.
Fig. 78.
77
**Indicator Diagram and Derived Values**
**Solution.**
$$r = \frac{1}{20} = 5;$$
$$p_m = p\left(\frac{1 + \log r}{r}\right) - p_0 = 140\left(\frac{1 + \log 5}{5}\right) - 2 = 140(5.22) - 2$$
$$= 73.1 - 2 = 71.1 \text{ lbs. per sq.in.}$$
Diagram factor = 80%. Hence probable M.E.P. = 71.1 × .85 = 60.4 lbs.
Therefore, since
$$I.h.p. = \frac{pLm}{g} = \frac{75 \times 33,000}{32,000} = 60.4 \times 600 = 68.3 \text{ sq.in.}$$
$$d = 91 \text{ ins. (approx.)};$$
and since $$2Ln = 600,$$ assume $$L = 1 \text{ ft},$$
hence $$n = 300 \text{ R.P.M.}$$
The engine is rated 9.5 × 12 ins., running at 300 R.P.M.
**67. Determination of Clearance Volume from Diagram.**
It was shown in a preceding paragraph that the clearance volume of a cylinder must be known in order to draw the line of zero volumes on the indicator diagram. This volume can be determined accurately for any real engine by weighing the quantity of water required to fill the clearance space, but this procedure is often impossible and unattractive, though approximate, method is often resorted to.
This method is graphical and depends upon the assumption of the law of expansion and compression. As in the case of the conventional diagrams, expansion and compression are assumed to follow a rectangular hyperbola.
It is a property of this curve that diagonals such as $ac$ and $bd$ drawn for rectangles with their corners on the
132
STEAM POWER
curve all pass through the origin of coordinates as shown in Fig. 79.
If two points $a$ and $c$ are selected on the expansion curve of a real diagram and a rectangle is drawn upon them as shown in Fig. 80, the diagonal $bd$ extended will pass through the origin of coordinates, if the expansion follows the assumed law. The point at which this diagonal cuts the vertical line of zero expansion is therefore the point through which the vertical line of zero expansion is to be drawn.
If the original assumption were correct, this construction would give the same point when different locations of the points $a$ and $c$ were chosen and when used on the
![]() Fig. 79.—Rectangular Hyperbola.
![]() Fig. 80.
compression as well as on the expansion line. In reality it will generally give as many different locations for the origin as are chosen for the rectangle itself. It is customary to construct this rectangle of fair size and to locate it near the center of the expansion curve.
**68. Diagram Water Rate.** As was shown in an earlier chapter, part of the steam supplied an engine is generally condensed upon the cold metal walls surrounding it. The indicator diagram therefore shows the volumes occupied by the mixture which exists in liquid form in the cylinder, but, since the volume occupied by the liquid is negligible, it may be assumed to show the volumes occupied by the part of the mixture which exists in vaporous form.
INDICATOR DIAGRAM AND DERIVED VALUES 133
Assuming that the vapor is saturated, the volume occupied by one pound at various pressures can be found from the steam tables and, therefore, the weight existing in the cylinder can be calculated. The weight of steam determined on the diagram is called indicated steam, the diagram steam or the diagram water rate.
The diagram water rate is generally determined for a point such as $z$ in Fig. 81 just after cut-off, though some engineers prefer to use a point nearer the lower curve of the expansion curve. The volume occupied by the steam contained in the cylinder at point $z$ is equal to $V_z$ and its weight can be determined by dividing this volume by $V_s$ for the existing pressure $P_s$. Thus, calling the weight of steam in the cylinder $w_s$
$$w_s = \frac{V_z}{V_s} \cdot \cdot \cdot \cdot \cdot \cdot (51)$$
This quantity of steam is a mixture of cylinder feed and clearance or cushion steam and the weight of the latter must therefore be subtracted from $w_s$ to obtain the weight of cylinder feed $w_c$. Assuming the cushion steam dry and saturated at the point $k$, the weight of cushion steam is
$$w_c = \frac{V_r}{V_s} \cdot \cdot \cdot \cdot \cdot \cdot (52)$$
so that the weight of cylinder feed per cycle as shown by the diagram at the point $z$ is
$$w_c = w_s - w_k = \frac{V_r}{V_s} - \frac{V_z}{V_s} \cdot \cdot \cdot \cdot (53)$$
The formula is generally modified to give the steam consumption per indicated horse-power hour, instead of
134 STEAM POWER
per cycle, and it is also expressed in different terms as a matter of convenience.
For this purpose let
$$y_{\text{ac}} = \text{clearance volume divided by piston displacement per stroke}$$
$$= \frac{l_1}{l_p};$$
$$y_2 = \text{piston displacement to point } z \text{ divided by piston displacement per stroke}$$
$$= \frac{l_2}{l_p};$$
$$y_k = \text{piston displacement to point } k \text{ divided by piston displacement per stroke}$$
$$= \frac{l_k}{l_p};$$
$$a = \text{area of piston in square inches};$$
$$p = \text{mean effective pressure in pounds per square inch};$$
$$L = \text{stroke in feet};$$
$$n = \text{number of cycles per minute}.$$
The piston displacement is then $$\frac{a}{144}L$$ cubic feet and the volumes at $z$ and $k$ are given by
$$V_z = \left( y_2 \times \frac{aL}{144} \right) + \left( y_a \times \frac{aL}{144} \right),$$
and
$$V_k = \left( y_k \times \frac{aL}{144} \right) + \left( y_a \times \frac{aL}{144} \right).$$
Substituting these values in Eq. (53) gives the cylinder feed per cycle as
$$w' = \frac{aL}{144} \left( y_2 + y_a - y_k + y_a \right) V_z - V_k.$$ (54)
Multiplying by the number of cycles per hour ($60\times n$) and dividing by the indicated horse-power, $$\frac{p_{\text{indicated}}}{35,000}$$ gives
INDICATOR DIAGRAM AND DERIVED VALUES 135
the diagram water rate, or steam shown by the diagram per L.h.p. hour as
$$w_{g} = \frac{13.750}{p} \left( \frac{y_1 + y_2}{V_1} - \frac{y_2 + y_3}{V_2} \right) \dots (55)$$
in which form the equation involves only values which can be determined directly from the diagram without any knowledge of the engine dimensions.
The value obtained for $w_g$ will vary as the location of points $z$ and $k$ are varied because of the quality changes occurring during expansion and compression, and it is obvious that the diagram water rate is in no sense an accurate measure of the real water rate. It is, however, very useful for comparison with the real water rate, the ratio giving an indication of the loss by condensation.
Average values for real water rates are given in Chapter XI.
ILLUSTRATIVE PROBLEM
Given the diagram shown in Fig. S2 and the following data from an actual test, find the diagram water rate for point e, and for point $a$.
Double-acting steam engine:
Average piston area = 28.9 sq. in.;
Length of stroke = 8 in.;
R.P.M. = 257;
L.h.p. = 8.75;
M.E.P. = 31.6 lbs.;
Clearance = 15%; Beginning of compression = 10%
Weight of clearance per hour = 371 lbs.;
Quality at throttle = 05%;
Sp. vol. at $e$ = 7.8;
Sp. vol. at $a$ = 12.57;
Sp. vol. at $K$ = 38.4 cu.ft. per lb.;
Assume $z_k$ = 100%.
![]() Fig. S2.
135
136 STEAM POWER
**Solution.** Substitution in Eq. (55) gives
$$\begin{align*}
(W_d)_c &= \frac{13.750}{p} \left( \frac{q_c + q_a}{V_c} + \frac{q_b + q_a}{V_b} \right) \\
&= \frac{13.750}{p} \left( 0.29 + 0.13 \right) \left( 0.29 + 0.13 \right) \\
&= 24.38 \text{ lbs. per L.H.p. per hour at point } c.
\end{align*}$$
$$\begin{align*}
(W_d)_a &= \frac{13.750}{p} \left( \frac{q_a + q_b}{V_a} + \frac{q_a + q_b}{V_b} \right) \\
&= \frac{13.750}{p} \left( 0.29 + 0.13 \right) \left( 0.29 + 0.13 \right) \\
&= 21.83 \text{ lbs. per L.H.p. per hour at point } a.
\end{align*}$$
Real water rate = $\frac{371}{8.75} = 40.2$ lbs.
**69. T-φ-diagram for a Real Engine.** In Chapter VI the T-φ-diagrams of the various ideal cycles were given and attention was called to the fact that these diagrams were particularly useful, because they showed certain things which were not apparent from the more common PV-diagrams.
It has been customary for many years to draw T-φ-diagrams for real engines by “transferring” the PV-diagram to T-φ-coordinates, and various analytical and graphical methods have been developed for this purpose.
There are certain unavoidable errors in all the methods used for drawing these diagrams, but the φ-curve is the only one of all the lines finally obtained which has any claim to accuracy. Even this curve is generally incorrectly interpreted, because a knowledge of the exact weight of clearance steam is necessary for an accurate interpretation and such knowledge is never available.
Under these circumstances it is unnecessary to consider in this book the rather complicated details involved in the construction of T-φ-diagrams purporting to show the behavior of steam in real engines.
INDICATOR DIAGRAM AND DERIVED VALUES 137
**70. Mechanical and Thermal Efficiencies.** The method of obtaining the indicated horse-power from the indicator diagram has been given in preceding paragraphs. In the real engine this power is not all made available at the shaft, because some of it is lost in overcoming the engine against its own frictional resistance. Calling the power lost in this way the **friction horse-power**, it follows that
$$\mathrm{L.h.p.} = \mathrm{F.h.p.} + \mathrm{D.h.p.}, \quad \ldots \quad (56)$$
in which
$\mathrm{L.h.p.}$ = indicated horse-power determined from the real indicator diagram;
$\mathrm{F.h.p.}$ = friction horse-power, i.e., power required to overcome the frictional resistance;
$\mathrm{D.h.p.}$ = developed horse-power, i.e., power made available at shaft.
The **developed horse-power** is therefore always less than the indicated horse-power. The better the construction of the engine the smaller the friction loss, and the measure of this loss is usually given in the form of an efficiency. It is called the **mechanical efficiency**, and is defined by the equation
$$\mathrm{Mech.\ eff.} = \frac{\mathrm{D.h.p.}}{\mathrm{L.h.p.}}, \quad \ldots \quad (57)$$
Values of mechanical efficiency range from about 80 per cent in the case of poorly designed and poorly adjusted horizontal engines to about 95 per cent in the case of the best vertical designs.
The efficiency determined by dividing energy made available by heat supplied is known as the **thermal efficiency**. There are two possible thermal efficiencies, one based on the indicated power and the other on the developed power. The former is called the **thermal efficiency on the indicated horse-power** or the **indicated thermal efficiency**; the other
138 STEAM POWER
is known as the thermal efficiency on the developed horse-power or the developed thermal efficiency. Obviously
Dev. ther. eff.=Mech. eff.×Indie. ther. eff. . . . (58)
The heat supplied may be assumed in two different ways; it may be taken as the total heat above 32° F., in the steam supplied the engine, or it may be taken as this value less the heat of the liquid corresponding to exhaust temperature. In practice, however, when the general idea is reasonable to assume that the exhaust steam can be condensed to water at the same temperature and that this water can be pumped to the boiler with the heat of the liquid corresponding to this temperature. This is practically parallel to the assumption made in treating the theoretical cycles.
The thermal efficiencies are then
Indie. ther. eff.
$$= \frac{\text{L.h.p.} \times 2545}{\text{Heat above exhaust temp. supplied per hour}}$$ (59)
and
Dev. ther. eff.
$$= \frac{\text{D.h.p.} \times 2545}{\text{Heat above exhaust temp. supplied per hour}}$$ (60)
Values of the indicated thermal efficiency range from about 5 per cent in ordinary practice with small engines to about 25 per cent in the best large engines. Values as low as 1 per cent are not uncommon with small, poorly designed and poorly operated engines.
The actual performance of the cylinder of an engine is sometimes compared with the ideal possibilities as indicated by the Clausius and the Rankine cycles. The ratio of the work obtained in the real engine to that which could be obtained from the same quantity of heat with a Rankine
INDICATOR DIAGRAM AND DERIVED VALUES 139
or Clausius cycle is a measure of the performance of the real cylinder. This ratio is variously designated as cylinder efficiency, indicated efficiency, relative efficiency, etc. Its values range from less than 40 per cent to over 80 per cent, the highest recorded value being just over 88 per cent.
PROBLEMS
1. Using Table I, Chapter I, plot the specific heat of water between the range of temperatures of 20° F and 300° F, for the intermediate values given. By the ordinate method for finding the mean height of an indicator diagram, determine the mean of average steam heat over this range.
2. A double-acting engine is required to give 50 L.h.p. under the following conditions:
- Cylinder diameter = 16 in.
- Steam pressure = 150 lbs. per square inch absolute;
- Back pressure = 16 lbs. per square inch absolute;
- Piston speed = 540 ft. per minute.
If the diagram factor for this type of engine is 75%, find the diameter of the cylinder and select the stroke and R.P.M.
3. Assume a single-acting engine with 10-in. diameter and 12-in. stroke, having a diagram factor of 85%. The clearance volume is between 10% and 50% of stroke. Assume also the pressures, speed, and card factor as given in Prob. 2. Find the probable L.h.p. at different loads.
4. Given an 18×24-in. engine running at 120 R.P.M.
- Back pressure = 2 lbs. per square inch absolute;
- Clearance = 10%
- Cut-off angle = 30°
Diagram factor = 85%
Supposing cut-off to remain constant, find the L.h.p.'s corresponding steam pressure of 50, 90, and 130 lbs. per square inch absolute.
5. Find the weight of dry steam which must be supplied per L.h.p. hour for each of the previous problem, assuming the quality of steam to be 80%. Assume that the steam is to be 30 lbs. absolute and that steam is dry and saturated at end of compression.
6. Find the quality of steam at cut-off in a cylinder, in which the piston displacement is 0.1278 cu.ft.; clearance = 10%; cut-off at 25% stroke; steam pressure at cut-off, 115 lbs. per square
140
**STEAM POWER**
inch absolute, and weight of steam in the cylinder at cut-off = 0.012 lb.
Note. Quality = $\frac{\text{Actual vol.}}{\text{Weight} \times \text{Sp. vol.}}$ for the given pressure.
7. The piston displacement of a certain engine is 0.2 cu.f.t.
What would be the quality of the steam if the relative quality is 90%, and pressure is 25 lbs. per square inch absolute, and clearance is 10%, and release occurs at 95% of the stroke?
8. Find the quality of cushion steam in a 6X6 in. engine in which clearance is 10%, and back pressure is 14.7 lbs. per square inch absolute, and return stroke; back pressure is 14.7 lbs. per square inch absolute, and the quality of the cushion steam at the beginning of compression is 95%.
9. Find the pressure and quality at the end of the compression line of the previous problem, assuming it to be adiabatic.
10. At the beginning of the stroke of a certain double-acting, and cuts off at 15% of the stroke at a pressure of 120 lbs. per square inch absolute. It has a steam consumption of 35 lbs. per L.h.p.-hour. The compression begins at 60% of the return stroke with a quality of 95%, and back pressure is 3 lbs. per square inch absolute. Clearance = 10%.
If this engine delivers 27 H.P., and has a mechanical efficiency of 90%, what is its thermal efficiency?
11. In the previous problem, assume release to occur at 90% of the stroke with an absolute pressure of 30 lbs. per square inch.
What is the quality at this point?
12. A certain engine uses one horse-power hour at the shaft for every 20 lbs. of steam supplied. The steam has an initial pressure of 150 lbs. absolute and is dry and saturated when it arrives at the cylinder, but the back pressure against which steam is exhausted is 4 lbs. absolute.
(a) Find the thermal efficiency of this engine on the developed or shaft power basis.
(b) If the mechanical efficiency of the engine is 90%, what is the value of the thermal efficiency on the indicated horse-power?
CHAPTER IX
COMPOUNDING
71. Gain by Expansion. The cycle which gives a rectangular $P^{\prime}V$-diagram is the least economical of all the ideal cycles described in Chapter IV. This comes from the fact that none of the heat stored in the steam is converted into work when this cycle is used. Thus, if the cylinder shown by full lines in Fig. 83 operate on this cycle and be of such size that it will receive just one pound of steam per cycle, it makes available an amount of work represented by the area $abefd$. The positive work done by the steam upon the piston is the equivalent of the external latent heat of vaporization while no use is made of the heat stored in the steam. This stored heat is removed as heat during the condensation and exhaust, which give the lines $bc$ and $cd$.
If a piece be added to the cylinder as indicated by the dotted lines, the same quantity of steam will make more heat available by expanding after cut-off, as shown by the curve $be$, than by using the heating represented by the area $abefd$ instead of by the smaller area $abcd$. But the heat supplied is the same in both cases, namely that required to form one pound of steam at the pressure $P_1$, so that the use of a large cylinder and the incomplete ex-
![]() Fig. 83.
141
142
STEAM POWER
pansion cycle results in the development of more work than can be obtained with the rectangular cycle from the same amount of heat.
Obviously, it would be theoretically advantageous to add still more to the length of the cylinder and allow the expansion to continue to back pressure, giving the complete expansion cycle as shown in Fig. 84, thus obtaining the maximum quantity of work at the expense of the heat stored in the steam supplied to the cylinder. Practically, it is found inadvisable to continue the expansion to such a degree in reciprocating steam engines, because at low pressures the volume increases very rapidly for small pressure drops. Thus a great increase is necessary in the size of the cylinder if the last part of the expansion is to be completed, but the amount of work obtained is comparatively small, as shown by the small height of the long line on the diagram. This may result in an actual loss, because the increased friction losses of the very large cylinder may more than balance the small increase of net work gained by its use. It thus results that, in every real reciprocating engine, there is some point beyond which it is not economical to carry the expansion, and the incomplete expansion
![]() Fig. 84.
d
COMPOUNDING 143
cycle is therefore approximated in such engines rather than the cycle with complete expansion.
Viewing the matter from another angle, a cylinder of a certain size may be assumed as shown in Fig. 85. The use of the rectangular cycle $abed$ in this cylinder will make available the maximum capaci- tity of work obtainable with the upper and lower pressures chosen. If cut-off be made to occur earlier as at $b'$, the expansion $b'e'$ will result in a loss of the quantity of work obtained per horse-power by the area $b'be'$, but the steam used per horse-power will be less, so that there will be a gain in steam economy. Putting the cut-off still earlier will cause a still greater loss of work obtained from a cylinder of the chosen size, but theoretically will give no further economy of steam.
Summing up, it may be said that the greater the ratio of expansion the greater should be the economy in the use of steam on a theoretical basis.
The lower pressure is set in real engines by the pressure in the space into which the engine is to exhaust. If the engine is to be operated condensing, the atmospheric pressure determines the lowest possible exhaust pressure; if the engine is to be operated condensing, the exhaust pressure is set by the lowest pressure which can be economically maintained at the condenser.
There is thus a real limit to the extent to which expansion can be carried in any engine with a given initial pressure. A certain drop must exist at the end of the diagram, for reasons already explained, and an expansion line drawn backward from the top of the line representing this drop will give the earliest possible cut-off
![]() Fig. 85.
144
STEAM POWER
which can be used in the engine with a given initial pressure.
The ratio of expansion can be further increased, how-
ever, by raising the initial pressure as indicated by the
diagram shown in Fig. 80, and the limit in this direc-
tion would come with the inability of materials of
construction to withstand the resulting strains.
There are conclusions drawn from the facts
developed above which must all be modified in the case of
real engines, because of
the effect of cylinder condensation. This has been shown
to increase as the cut-off is made earlier and as the
pressure (and therefore the temperature) range in a cylin-
der is increased. There is, therefore, a limit beyond
which it is not advisable to carry the ratio of expansion in
a single cylinder.
Experience has shown that the best commercial results
are obtained from simple engines, that is, those expanding
the steam at constant pressure. The reasons for this are:
(a) the non-condensing, (b) the initial pressure is between 80 and 100
pounds per square inch for the simpler forms of valves and up
to 125 lbs. with the better forms of valves, and (c) the point
of cut-off is at about $\frac{1}{4}$ stroke with the simpler valves and at
from $\frac{3}{4}$ to $\frac{1}{2}$ stroke with the better forms of valves. These
values of cut-off correspond to nominal expansion ratios of
5 and 7 respectively and to lower values when clearance
is taken into account.
**72. Compounding.** If the ratio of expansion is to be
increased above the values just given, some means must
be used for the reduction of loss by condensation. This
loss can be reduced by decreasing the surface exposed to
![]() Fig. 86.
P V
COMPOUNDING 145
high-temperature steam and by decreasing the temperature range in a cylinder. Both of these results can be achieved by what is known as compounding.
Assume that it is deemed advisable to produce a cycle similar to that shown in Fig. 87 (clearance neglected) and that in order to obtain high steam economy (low water rate) the ratio of expansion chosen is very much greater than four. No great economy would result from such excessive expansion in a single cylinder, in fact there would be a well-defined, unavoidable loss. But suppose that the high-pressure steam is admitted to a small cylinder such as that shown and is expanded to the point $f$, then exhausted as shown by $fg$ into the larger cylinder. The cycle produced is the same as that which would have been obtained by expanding entirely in one cylinder, but the surface of the clearance space of the high-pressure (H.P.) cylinder, which is exposed to high-pressure steam at $f$, is thus cooled down to a temperature lower than the size required to hold the steam when fully expanded and, moreover, the lowest temperature to which it is subjected is that corresponding to the pressure at $f$ instead of the much lower temperature corresponding to the pressure at $a$.
The condensation which would occur in the H.P. cylinder would obviously be less than that which would result from
![]()
A graph showing a curve labeled "g" with points labeled "a", "b", "c", "d", "e", and "f". The x-axis is labeled "v" and ranges from approximately 0.1 to 0.6. The y-axis is labeled "g" and ranges from approximately 0.1 to 0.6.
High Pressure Cylinder
Low Pressure Cylinder
Fig. 87.
145
146
STEAM POWER
the use of one large cylinder and, remembering that the greater part of the heat given up during condensation is received again by the steam during exhaust, it is obvious that upon using this method all of the heat can again be given to the low-pressure cylinder wall. Thus although there are two cylinders in which condensation and re-evaporation occur, and although the sum of the heat given to the walls of the high-pressure cylinder and that given to the walls of the low-pressure cylinder might be greater than that given to the walls of a single cylinder under similar conditions, the use of two cylinders results in considerable saving because loss in the high-pressure cylinder is practically wiped out by the exhaust of the heat concerned into the low-pressure cylinder.
If this is by radiation and conduction from the high-pressure cylinder to the exhaust, the result of the use of two cylinders is particularly to limit the loss by radiation and re-evaporation to that occurring in the low-pressure cylinder. As the ratio of expansion in this cylinder is in the neighborhood of that common in simple engines, or even less, and as the temperature range is small, the net loss is also small.
It is obvious that the smaller the surface of the high-pressure cylinder can be made, and the smaller the temperature range in such a cylinder, the smaller will be the net loss by cylinder condensation and re-evaporation. A saving should therefore be effected by using more than two cylinders, and it is not inconceivable that five or more might be used. The result of using five cylinders is shown in Fig. 88, and it is evident that the clearance surfaces exposed to
![]() Fig. 88.
COMPOUNDING
147
high temperatures, the temperature ranges per cylinder and the ratios of expansion per cylinder are all small. The gain in economy should therefore be correspondingly great.
There are two limits to the possible multiplication of cylinders in this way:
(1) As the number increases the radiating surface and therefore the heat lost by radiation increases. The extent of this effect can be appreciated by noting that every cylinder with the exception of the low-pressure cylinder is really an unnecessary addition, because the cycle could be produced entirely by one large high-pressure cylinder. On the other hand, the surfaces of cylinders which reach high temperature are small as compared with that which would be exposed to this temperature if the entire cycle were produced in the low-pressure cylinder.
(2) As the number of cylinders is increased, the first cost, the complexity and the cost of fabrication and attendance increase more rapidly than does each installation, some number will be found beyond which the interest on the investment and the added cost of operation and maintenance would more than balance the saving of fuel.
The second limit mentioned is the more important consideration. In many cases one cylinder under ordinary operating conditions in stationary power plants expansion in two cylinders generally gives the most economical results. The total ratio of expansion is generally between 7 and 16, that is, the volume of steam at release in the L.P. cylinder is from 7 to 16 times the volume at cut-off in the H.P. cylinder. For large pumping stations and large marine installations expansion in three cylinders is generally considered too economical and only those ratios of expansion of 20 or more are used. Four and five cylinders have been used, but the resultant gains do not seem to warrant any extensive installation of such units.
Engines using more than one cylinder for the expansion
148
STEAM POWER
of steam in the way just described are called **multi-expansion engines**, or **compound engines**, and the use of multi-expansion is spoken of as compounding. Custom has almost confined the use of the term compound engine to those in which only two cylinders are used in series as indicated in Fig. 80, and such engines are often spoken as 2x engines.
Engines in which three cylinders are used in series are called **triple-expansion** or 3x engines. With four and five cylinders in series the engines are known as **quadruple** or 4x and **quintuple** or 5x, respectively.
In the case of triple-expansion engines of large size,
![]() Diagram showing a triple-expansion engine with three cylinders: one low-pressure cylinder (L.P.) and two high-pressure cylinders (H.P.). The steam flows from the boiler through the L.P. cylinder, then into the H.P. cylinders, and finally out of the L.T. cylinder.
Fig. 89.
![]() Diagram showing a quadruple-expansion engine with four cylinders: two low-pressure cylinders (L.L.) and two high-pressure cylinders (H.H.). The steam flows from the boiler through the L.L. cylinders, then into the H.H. cylinders, and finally out of the L.T. cylinder.
Fig. 90.
the volume of the low-pressure cylinder required generally becomes so great that it is found economical to use two low-pressure cylinders instead of one. The flow of steam in such an engine is represented diagrammatically in Fig. 90. This type is known as a **four-cylinder, triple-expansion engine**.
All multi-expansion engines are generally operated condensing, and the choice of type is determined partly by the character of work to be done and partly by economical considerations. In all cases the boiler pressure must be chosen to suit the type of engine used. The pressures ordinarily used with the different types are given in Table V.
COMPOUNDING
149
TABLE V
| Type of Engine. |
Boiler Pressure. |
| Simple. |
80 to 125 |
| High-speed compound. |
100 to 170 |
| Low-speed compound. |
125 to 200 |
| Triple expansion and higher. |
125 to 235 |
**73. The Compound Engine.** The term compound engine will be used hereafter in the commercial way as referring to a 2 stroke. Such engines may roughly be divided into two types, receiver and no-receiver engines. The latter are often called Woolf engines, after the man who first used this construction.
A **receiver engine** has a vessel known as a receiver located between the two cylinders and so connected with them that the high-pressure cylinder exhausts into the receiver and the low-pressure cylinder draws steam from the receiver. By using a receiver the cylinders are made independent of each other so far as steam events are concerned; the high-pressure cylinder can exhaust at any time with reference to the events occurring in the low-pressure cylinder.
A **Woolf type** has practically no receiver, the high-pressure cylinder exhausting directly into the low-pressure cylinder through the shortest convenient connecting passage. As the high-pressure cylinder must exhaust directly into the low-pressure cylinder it follows that cut-off must not occur in the latter until compression starts in the former; i.e., very near the end of the stroke.
An engine of this type having a large size would give a horizontal exhaust line for the high-pressure cylinder and a horizontal admission line for the low-pressure cylinder, since the small amount of steam given to or taken from the
150
STEAM POWER
receiver would have no appreciable effect upon the pressure within that vessel. Neglecting throttling losses, the high-pressure and low-pressure cards would therefore fit together as originally indicated in Fig. 86.
With receivers of finite size there are pressure changes during exhaust by the high- and admission to the low-pressure cylinders which cause a deviation from the ideal curves, so that certain throttling losses, so that the lines representing these events are not horizontal nor do they exactly coincide.
A diagrammatic arrangement of the Woolf engine is given in Fig. 91 with idealized diagrams obtained by
![]() A diagram showing the arrangement of a Woolf engine and its idealized diagrams.
Fig. 91,
assuming hyperbolic expansions, no clearances, and no throttling losses. The two idealized curves make angles with each other in such engines, but they move in the same direction, as shown in the figure, or in opposite directions.
The ideal diagram would be that shown at (a) by the lines $abcdCA$. The idealized high-pressure diagram is $abed$ and the idealized low-pressure diagram is $ABCD$. The exhaust line $BC$ of the high-pressure diagram and the admission line $AC$ of the low-pressure diagram are produced at the same time. Corresponding points on these two lines represent the common pressures assumed by the steam not yet exhausted from the high-pressure cylinder, the steam in the small connecting passage and the steam
COMPOUNDING 151
already admitted to the low-pressure cylinder. As the movement of the low-pressure piston opens up volume faster than the high-pressure piston closes up volume, the volume occupied by the steam continues to increase as the high-pressure piston moves out, and its pressure therefore decreases.
The two diagrams are shown back to back at (b) in the figure and the horizontal line $xX$ connects corresponding points on the exhaust of the high pressure and the admission of the low pressure.
Compound engines are also divided into two types on the basis of cylinder arrangement. When the axes of both cylinders coincide as shown in Fig. 92 they are called **tandem compounds**. When the axes are parallel as shown in Fig. 89, the engines are spoken of as **cross-compound engines**.
74. Cylinder Ratios. The idealized diagrams of a compound engine with infinite receiver volume are shown in Fig. 93 by $abcd$ and $ABCD$. The height of the high-pressure exhaust line is the same as that of the low-pressure admission line, but the value of the receiver pressure $p_a$ is zero. The value of the receiver pressure is determined by the point chosen for cut-off in the low-pressure cylinder. Thus if cut-off in the low-pressure cylinder is made to occur earlier, as at some point $c'$, the admission line for this
![]() A diagram showing a compound engine with two pistons moving in opposite directions. The left piston is labeled "High Pressure Piston" and the right piston is labeled "Low Pressure Piston". A horizontal line xX connects corresponding points on the exhaust of the high pressure and the admission of the low pressure.
Fig. 92.
![]() A diagram showing a cross-compound engine with two pistons moving in parallel directions. The left piston is labeled "High Pressure Piston" and the right piston is labeled "Low Pressure Piston". A horizontal line xX connects corresponding points on the exhaust of the high pressure and the admission of the low pressure.
Fig. 93.
152 STEAM POWER
cylinder must move up to $B'C'$ and the receiver pressure must rise correspondingly. The exhaust pressure in the high-pressure cylinder would also rise an equal amount.
Changing the point of cut-off in the low-pressure cylinder also produces a change in the area of the diagram. In Fig. 92, the work area of the high-pressure diagram is obviously decreased, while that of the low-pressure diagram is increased. In a simple engine the area of the diagram becomes smaller the earlier the cut-off, and it should be noted that just the reverse of this occurs in the low-pressure cylinder of a compound engine.
It is evident that the choice of the receiver pressure or of the point of cut-off in the low-pressure cylinder determines the relative areas of the high-pressure and low-pressure diagrams and it also determines the relative size of the two cylinders. The diagram of Fig. 93 shows that late cut-off in the low-pressure cylinder calls for a larger high-pressure cylinder than early cut-off.
The ratio of the piston displacement of the low-pressure cylinder to that of the high-pressure cylinder is called the cylinder ratio. Designating this ratio by $R$, and using other symbols as in Fig. 93,
$$R = \frac{V_2}{V_1} \quad \ldots \quad \ldots \quad \ldots \quad (61)$$
The cylinder ratios chosen for real compound engines vary greatly in different designs and no given ratio has been proved the best for a given set of conditions. Normal practice gives the average values listed in Table VI, but cylinder ratios as high as 7 have been used with excellent results.
**TABLE VI**
| Cylinder Ratios for Compound Engines |
|
|
|
|
| Cylinder ratio. |
2 |
3 |
4 |
4 |
| Initial pressure (gage) non-condensing. |
100 |
120 |
|
|
| Initial pressure (gage) condensing. |
100 |
120 |
150 |
|
COMPOUNDING 153
The cylinder ratio to be used in a given case may be determined by any one of several considerations or by a combination of them, the latter being more often the case.
Thus it may be deemed desirable to obtain the same amount of power at different speeds, or to have the same mean pressures; or to have cut-offs occur at the same fraction of the strokes; or to have the same total load on the two piston rods during admission; or to obtain the maximum possible uniformity of running effort at the crank. The consideration of equal work is generally regarded as the most important.
To Determine the Mean Pressures. The idealized diagram for a compound engine with clearance, with incomplete expansion in both cylinders, and without compression are given in Fig. 94. The nominal total ratio of expansion would be $L_{a} + L_{b}$, but the total ratio of expansion taking account of clearance is
$$\frac{L_{a} + C_{l}}{L_{a} + C_{l'}} \quad . . . . . . (62)$$
and the cylinder ratio is
$$R = \frac{L_{a}}{L_{a'}} \quad . . . . . . . (63)$$
The mean effective pressures can be found from each of the diagrams in the ordinary way and the indicated horse-power of each cylinder determined therefrom. The indicated horse-power of the engine is then equal to the sum of the two values for the two cylinders.
It is often convenient to refer the mean effective pressure of all cylinders to the low-pressure cylinder as though this were the only cylinder acting. In the simple form of diagram, such as that shown in Fig. 93, it is obvious that this could be obtained by measuring the area $AbcDEA$,
![]() Fig. 94.
dividing by the length $AE$ and multiplying by the scale of the spring, just as though the diagram were all produced in one cylinder with the piston displacement equal to $V_L$. In the case of the diagrams given in Fig. 94 a similar method could be employed, but in this case the low-pressure cylinder could be determined separately and then the equivalent pressure which would give the same result on the low-pressure piston could be determined analytically.
Assume for this purpose that the mean effective pressure of the high-pressure is equal to $p_{H}$ pounds per square inch, that the mean effective pressure of the low-pressure cylinder is equal to $p_{L}$ pounds per square inch, and that the area of all cylinders of a multi-expansion engine are generally equal, so that the piston areas are in the same ratio as the cylinder volumes (piston displacements). In the case of a $2x$ engine, therefore, the area of the low-pressure piston is $R$ times as great as that of the high-pressure piston, and the pressure required on the low-pressure piston to do the same work as that done by pressure $p_H$ on the high-pressure piston will be $\frac{p_H}{R}$.
In the case of a $2z$ engine therefore the total M.E.P. referred to the low-pressure cylinder is
$$p_E = \frac{p_H}{R} + p_L. \quad (64)$$
This mean effective pressure acting on the low-pressure piston only would give the same indicated horse-power as is obtained with the two cylinders of the engine.
In designing compound engines it is customary to determine first size of the low-pressure cylinder as though it were to do all the work expected of the engine by receiving steam at the highest pressure available and exhausting it at the lowest. The mean effective pressure which would thus be assumed to exist is the referred value $p_R$ just explained. Having found the size of the low-pressure cylinder
COMPOUNDING
and the value of the referred M.E.P. the size of the high-pressure cylinder can be determined so that the work done by each cylinder will be just half of the total for which the engine is being designed. This size will have to be such that the mean effective pressure referred to the low-pressure cylinder (i.e., $p_{H} + p_{L}$) is equal to half the total mean effective pressure referred to that cylinder. That is, the size will have to be so chosen that:
$$\frac{p_H}{R} = \frac{p_L}{2} \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots \quad \ldots $$
(65)
ILLUSTRATIVE PROBLEM
A double-acting compound engine is capable of developing 500 L.h.p. The stroke is 18 ins.; revolutions per minute, 175; mean effective pressure referred to L.P. piston, 45 lbs. per square inch; cylinder ratio, 3:1. Find cylinder diameters.
From:
$$\text{l.h.p.} = p_{L}e_{L} = 33,000$$
$$a_{L,p} = 500\times33,000 = 165,000$$
so that
$$d_{L,p} = 78.5 = 30\text{ ins. (approx.)},$$
with the cylinder ratio equal to 3:1,
$$a_{H,p} = 200\text{ sq.ins.},$$
$$d_{H,p} = 28.5 = 16\text{ ins. (approx.)},$$
76. Combined Indicator Diagrams.
When a compound engine is indicated, the diagrams of the two cylinders as drawn by the indicator are not directly comparable. The scales of pressure and volume are different on the two diagrams, and correction must be made for this fact before the
156 STEAM POWER
diagrams can be compared. It is customary to do this and to draw the average high-pressure and low-pressure diagrams on the same set of coordinates in order to determine how well they approximate the ideal diagram that would be obtained with a cylinder operating between the extreme limits of pressure.
Diagrams approximating those that would be obtained from high- and low-pressure cylinders are shown at (h) and (i) respectively, in Fig. 95, and the result of drawing
![]()
A diagram showing two expansion curves labeled x₁ and x₂, along with a combined diagram labeled SS' and S'S'. The combined diagram includes saturation curves.
Fig. 95.
both to the same scales is shown at the left of this figure. The curves x₁ and x₂ show the variations of quality along the two expansion curves.
Drawing of two diagrams to the same scales in this way is known as combining the diagrams and the result is known as a **combined diagram**.
The curves SS' and S'S' added to the combined diagram are **saturation curves**. They do not, in general, form a continuous curve, because of the different quantities of steam contained in the two clearances and because any
COMPOUNDING 157
moisture in the high-pressure exhaust is generally removed in the receiver. The volumes occupied by clearance steam at initial pressures are indicated by the points $b'$ and $B'$ respectively. The lengths $b'S$ and $B'S$ approximately represent the volumes that would be occupied by cylinder feed when in each cylinder if dry and saturated.
A representative diagram of a triple-expansion engine is shown in Fig. 96. The heavy lines give diagrams constructed so as to represent as nearly as possible what may be expected to occur in the cylinders of such an engine assuming perfect adiabatic and hypobaric expansions and compressions. The dotted diagrams indicate the shapes that would be drawn by indicators applied to the real cylinders. The discrepancy between these due to overlapping of events, one cylinder suddenly starting to draw from a receiver while another is exhausting. It will be observed that the dotted diagrams do not contain any of these sharp angles, but that they generally outline forms on the average of them.
The curve $a$ is a receding hyperbola, taken as a continuation of the assumed hyperbolic expansion line of the high-pressure cylinder. The failure of the expansion lines of the other cylinders to fall upon this curve is explained by quality changes, different quantities of clearance steam in the different cylinders and withdrawal of moisture from the receiver to maintain admission to the following cylinder.
![]()
A diagram showing a triple-expansion engine with dotted lines representing hypothetical expansion and compression paths.
Fig. 96.
158
# STEAM POWER
## PROBLEMS
**1.** Find the size of the cylinders of a double-setting compound engine, which is to give 600 H.P., when using steam at a pressure of 150 lbs. per square inch absolute, and having a back pressure of 2 lbs. per square inch absolute. The cylinder ratio is to be 4, and the total ratio of expansion 12; piston speed 750 ft. per minute, and R.P.M. equal 80%. The fuel factor is 80%.
**2.** Given a 200 H.P. compound Corliss engine with cut-off in the H.P. cylinder at 60% stroke. Ratio of expansion is 7; clearance equals 10%; fuel factor is 75%; pressure at the H.P. cylinder is 165 lbs. absolute. Find:
(a) Cylinder ratio;
(b) Theoretical and actual M.E.P.;
(c) Determine size of four engines, and select the best one.
Note.
$$R = \frac{1+4\% CI}{(CO)+4\% CI}$$
**3.** Given a compound engine 18×40 ins., having a stroke of 28 ins. Steam pressure is 165 lbs. per square inch absolute; cut-off in H.P. cylinder occurs at 62% stroke; clearance equals 10%; back pressure equals 5 lbs.; R.P.M. equal 150. Find
(a) Cylinder ratio;
(b) Ratio of expansion;
(c) Actual M.E.P.;
(d) L.H.p.
![]() A diagram showing a compound steam engine with cylinders and pistons.
CHAPTER X
THE D-SLIDE VALVE
77. Description and Method of Operation. The simple D-slide valve, shown in place in Fig. 97, is so named because of the similarity of its section to the letter D. It is located in the steam chest, rides back and forth upon its seat and
![]() A diagram showing the internal components of a D-slide valve.
Fig. 97.
serves to connect the two ports alternately with steam and exhaust spaces respectively in order to give the necessary distribution of steam.
The valve has to perform the following functions for each end of the cylinder during each revolution of the engine:
(1) It connects the proper port to the steam space or
159
160
STEAM POWER
steam chest at such a time that steam can enter the cylinder as the piston moves away from the head.
(2) It shuts off this port and thus cuts off the supply of steam when the piston has completed a certain definite fraction of its stroke.
(3) It connects the port with the exhaust cavity shortly before the piston reaches the end of the stroke, thus effecting "exhaust" or "release"; and
(4) It shuts off the port again when the piston has completed the proper fraction of the next stroke, thus trapping in the cylinder the steam which is compressed during the remainder of the stroke.
![]() Engine Crank
![]() Main Connecting Rod
![]() Connecting Rod
![]() Operating Valve
FIG. 98.
It is obvious that the valve must be reciprocated upon its seat and that its motion must be connected with that of the piston in such a way so that the proper phase relation may be retained. This could be effected by the system shown diagrammatically in Fig. 98, a small crank operating on the end of a connecting rod giving the valve its short stroke just as the main crank fixes the longer stroke of the piston. Such an arrangement would however, be very inconvenient because the center line of the cylinder would be located too far from the center line of the cylinder.
It is customary to use what is known as an eccentric for the purpose of operating the slide valve. The parts and arrangement of an eccentric, together with an illustration of the way in which it is mounted on the shaft of
![]()
A diagram showing the parts of an eccentric mechanism.
Top left: A strap with a circular hole in the center.
Top right: A strap.
Bottom left: A sheave.
Bottom right: A sheave.
Fig. 99.—Parts of Eccentric.
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162
STEAM POWER
![]() A diagram showing the internal components of a steam engine, including the crankshaft, connecting rod, and flywheel. The diagram is labeled with "Economie" at the top left, "Plan View" at the bottom right, and "Crank Shaft" at the top center.
Fig. 100.—Economie on Horizontal Engine.
THE D-SLIDE VALVE
163
an engine are shown in Figs. 99, 100 and 101. The motion it gives the valve is exactly the same as that imparted by the crank first assumed, and it can easily be shown that it is the exact equivalent of the former.
Assume, for example, a crank such as that shown in Fig. 98 with a length of arm or throw equal to $a$. If the crank pin is made larger while other parts of the crank remain the same, as shown in Fig. 102, the crank mechanism will be found to give the same motion which it would impart to a connecting rod is not changed. If this process of enlarging the pin be continued until the pin has become large enough to surround the shaft and if the crank arm be then renewed so that what was the crank pin be fastened directly on the shaft, an
![]() A crank mechanism.
Fig. 101.—Eccentric on Vertical Engine.
(a)
(b)
Fig. 102.—Equivalence of Crank and Eccentric.
until
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STEAM POWER
![]() Fig. 103.
![]() Fig. 104. - Slide Valve without Lap.
Valves and piston both start to move toward right.
Piston at mid-stroke, valve at end of stroke and about to return.
Valve just closing left hand steam port to steam. Piston about to start on return stroke.
Piston at mid-stroke, valve wide open and about to return.
![]() Fig. 105.
THE D-SLIDE VALVE 165
eccentric results. It is the exact equivalent of the original crank; its center, which is the center of the crank pin, revolves about the center line of the shaft in a circle with a radius $a$ just as in the original mechanism.
The D-slide Valve is possible to place a short crank (short arm) upon a large diameter shaft without having to cut the shaft away as shown in Fig. 103, and it is therefore very useful for driving valves.
78. Steam Lap. The simplest form of D-slide valve would be to place shorter edges of the ports when in its central position as shown in Fig. 104. The crank drives it (that is the crank equivalent to the eccentric which would probably be used in a real case) would have to be located 90° ahead of the engine crank in the direction of rotation, as can easily be seen by consulting Fig. 105, which illustrates the mechanism in various critical positions. The steam admitted during this time is only full stroke admission, thus producing a rectangular cycle which has already been shown to be very inefficient as a means of obtaining work from the heat used in forming steam.
If cut-off is to occur before the end of the stroke, the edge of the valve must return and close the port before the piston reaches the end of its stroke. But since the crank mechanism does not permit the valve to remain stationary in any one position, cut-off will only occur if the valve over-travelled, as shown in Fig. 106, and this would unfortunately result in connecting the working end of the cylinder to exhaust and in admitting steam to the other side of the piston at such a time as to oppose the piston's motion. The solution of the difficulty lies in making the valve longer, so that when in its central position it overlaps the outer edges of the ports as shown in Fig. 107.
![]() A diagram showing a steam lap valve mechanism.
Steam Lap Valve Diagram
166 STEAM POWER
The amount of overlap of the outer edge is called the out-
side lap, and when steam is admitted by the outer edges of
the valve, as in the case under discussion, it is also called
the steam lap.
With such an arrangement the valve must be drawn
out of its central position by the amount of the lap when
the piston is at the end of its stroke as shown by $a$ in Fig.
![]() A diagram showing a steam engine valve with a side view on the left and a top view on the right. The side view shows the valve in its central position, while the top view shows it drawn out to the outside lap.
Fig. 107.--Steam and Exhaust Lap.
![]() A diagram showing a steam engine valve with a side view on the left and a top view on the right. The side view shows the valve in its central position, while the top view shows it drawn out to the angle $\alpha$. The angle $\alpha$ is indicated by a line extending from the center of the valve to the edge of the valve.
Fig. 108.--Lap $a$ and Lap Angle $\alpha$.
108 in order that steam may be admitted just as the
piston starts to move. It follows that the crank driving
the valve must be more than 90° ahead of the engine
crank and that it must be ahead by the angle required
to move the valve a distance equal to the outside lap.
This angle, represented in the figure by $\alpha$, is called the lap
angle.
79. Lead. In real engines it is further desirable to start
the admission of steam just before the piston arrives at
the end of its stroke. This assists in bringing the moving
parts to rest, raises the pressure in the clearance to full
value before the piston starts, and gives a valve opening
through which steam can escape during part of the
stroke, thus reducing wiredrawing and loss of area at
the top of the diagram. If the valve is open before the
piston reaches the end of its stroke, the crank driving it
must be shifted still further ahead of the engine crank.
It must be shifted ahead by an angle which will draw
the valve through its entire opening before it reaches
the opening of valve with the piston at the end of its stroke as shown by $b$ in Fig. 100. The angle required, indicated
THE D-SLIDE VALVE 167
by $\beta$, is known as the angle of lead, and the width of the steam opening with engine crank on dead center, i.e., the distance $b$, is known as the lead. The lead varies from less than $\frac{1}{4}$ in. on small engines and with low speeds up to over 1 in. on large engines and with very high speeds.
80. Angle of Advance. The eccentric or valve-operating crank must be ahead of the engine crank by an angle equal to $90^\circ +$ angle of lap $\alpha +$ angle of lead $\beta$, as can be seen
![]() A diagram showing the relationship between the eccentric (E), the valve operating crank (VOC), and the engine crank (EC). The eccentric is shown ahead of the engine crank by an angle equal to $90^\circ +$ angle of lap $\alpha +$ angle of lead $\beta$. The valve operating crank is shown ahead of the eccentric by an angle equal to $90^\circ - \delta$. The engine crank is shown at the bottom.
Fig. 109. —Lead $b$ and Lead Angle $\beta$.
by an inspection of Fig. 109. The sum of $\alpha$ and $\beta$ is called the angle of advance and will be represented by $\delta$. This is the number of degrees in excess of $90^\circ$ by which the eccentric leads the engine crank.
Fig. 109 shows that cut-off in an engine fitted with a valve slide valve must occur when the engine crank has turned through an angle equal to $180^\circ - 2\alpha$, because the valve will then have returned to the closed position. Apparently, cut-off can be made to occur at any point in the stroke by properly choosing the value of $\alpha$, but it will be discovered later that the exhaust events set a limit to increase in the value of this angle and hence do not permit of cut-off occurring earlier than a certain fraction of the stroke.
81. Exhaust Lap. Inspection of Fig. 105 will show that the simple valve without lap originally assumed will give no compression, because the cylinder end is connected to the exhaust cavity for the entire stroke. Inspection of all
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STEAM POWER
the changes which have been suggested in the subsequent paragraphs will show further that if the inner edges of the valve are kept in the original positions the exhaust events will be considerably distorted in the case of a valve having steam lap and lead.
This trouble may be remedied by moving the inner edges of the valve closer together, making the exhaust cavity in the valve shorter and giving inside lap as shown in Fig. 107 by b. When the inner edge of the valve control exhaust, as shown in the valve under discussion, this inside lap is also called exhaust lap.
The length of the valve, the lap and the lead are generally chosen so as to give the desired arrangement of admission and cut-off and then the exhaust edges are so arranged as to give sufficient space for expansion and compression. In some forms this necessitates the use of an exhaust cavity in the valve such as that shown in Fig. 110.
The amount by which the edges of the valve fail to meet the inner edges of the port is spoken of as negative inside lap. This dimension is indicated by c in the figure.
It should be noted particularly that all measurements of lap are made with the valve central on its seat and that the measurement of lead is made with the piston at the end of its stroke, i.e., with the engine crank on dead center.
**82 The Biligran Diagram.** The action of all slide valves could be studied by means of drawings of the actual mechanism, as has been done in preceding paragraphs, but such a method is time and space consuming. Numerous diagrams such as the Elliptical, the Sweet, the Zeuner and the Biligran have been developed for the purpose of simplifying and expediting such studies and when properly understood, they are very convenient. The scope of this book does not permit a discussion of all of these diagrams
![]() A diagram showing a slide valve with various parts labeled.
THE D-SLIDE VALVE
and, since the Bilgram diagram is probably the most generally applicable, attention will be confined to it.
The construction of this diagram is illustrated in Fig. 111. The point $O$ represents the center of the engine crank shaft and the two circles drawn about this point as a center represent respectively the paths traveled by the
![]() A diagram showing the construction of the Bilgram diagram for a D-slide valve. It includes two circles, one labeled "Crank Circle" and another labeled "Eccentric Circle". A line AB connects points on these circles, representing the path of the valve pin relative to the engine crank pin.
![]() A larger diagram showing the Bilgram diagram in more detail. It includes three circles: one labeled "Crank Circle", another labeled "Eccentric Circle", and a third labeled "Stylographic Circle". Points M, O, and X are marked on each circle. Lines connecting these points show the relative positions of the valve pin and the engine crank pin at different stages of the stroke.
Fig. 111.
pin of the valve crank and the pin of the engine crank. These circles are drawn to any convenient scales.
The diagram is constructed essentially diagrammatically such a way that the line OM represents the head end dead center position of the crank and in all subsequent paragraphs the relative positions shown by the small sketch in Fig. 111 will be assumed. The cylinder will be assumed to the
170 STEAM POWER
left of the shaft and the engine will be assumed to run “over.”
With the crank in position $OM$, the eccentric (equivalent crank) must be in the position $OB$, ahead of the crank by an angle $90^\circ + \alpha + \beta - 90^\circ = k$. The valve must then be displaced to the right of its central position by an amount represented by the distance $DB$, if a small correction for “angularity” of the valve connecting rod be neglected. As rotation continues, horizontal distances corresponding to this line will always give the instantaneous valve displacement. For a given $OB$, for instance, the valve displacement will be $DQ$.
If the angle $\alpha$ is now laid off above $OX$, locating the point $Q$ as shown, a perpendicular $QE$ dropped upon $OX$ from this point will equal in length the line $DH$, and will therefore show the valve displacement when the crank is in head end dead center. Thus $OM$ and $OE$ are parallel because the triangles $QOE$ and $BOD$ are similar and have the sides $OQ$ and $OB$ equal to the radius of the same circle.
The perpendicular $QE$ is really a perpendicular dropped upon the extension of the line representing the crank position, and it is a general property of this diagram that a line starting at any point on one side of a straight line and passing through any chosen crank position (or an extension of that line) will show by its length the displacement of the valve when the crank is in the chosen position. Thus assume the engine crank to rotate through the angle $\gamma$ to the position $OM'$. The eccentric will have rotated to $B'$ and the valve displacement with respect to $B'$ will be $D'B'$. A perpendicular drawn from $Q$ upon $OX'$, the extension of the crank position, gives $QE$ equal to $D'B'$ and hence representing the valve displacement to the same scale.
This construction drawn for different crank positions $OA$, $OA_1$, $OA_2$, etc., is shown in Fig. 112, the dash-dot radial lines about $Q$ representing the various values of the valve displacement. The number of each of these
THE D-SLIDE VALVE 171
lines indicates the crank position to which it corresponds.
It will be seen that the displacement increases in value
until the crank position $OM_3$ is reached, after which it
decreases again.
![]() A diagram showing the steam lap circle, eccentric circle, crank circle, and steam lead. The crank position $OM_3$ is indicated by a point on the eccentric circle.
Fig. 112.
Since the opening to steam is equal to the displace-
ment minus the lap, as shown in Fig. 109, the actual amount
by which the valve is open for any crank position can be
found by subtracting from the corresponding valve dis-
placement the amount of lap possessed by the valve. For
head end dead-center position, the displacement is equal
to lap plus lead, and is shown by $QE$ in Fig. 112. Subtract-
172 STEAM POWER
ing the lead $EF$, the remainder $FQ$ gives the lap of the valve.
A circle drawn about $Q$ with radius equal to $QP$ (or a circle drawn about $Q$ and tangent to the line $L$) will cut off all the lines representing valve displacement the amount represented by the lap of the valve. The remainders, that is, the parts of the lines radiating from $Q$ in Fig. 112 which are outside of the lap circle, must then represent the amounts by which the valve port is actually open.
It will be observed that the valve is open by the amount of the line $OM_1$ through the center, position $O M$. The crank position for which the valve displacement is just equal to the lap, and hence at which the valve is just beginning to open, can be found by drawing a tangent through $O$ to the lower side of the lap circle and then extending it to give the crank position $O A$ in Fig. 112.
As the crank rotates from this position, the valve opens wider until, when position $O M_2$ is reached, the greatest valve opening exists. Further rotation results in partial closure of the valve and, when the crank has finally rotated into position $OC$, the valve has just closed, that is, cut-off has occurred, the displacement being just equal to $O Q$, its steam lap.
Thus, as shown in Fig. 112 far developed, indicates crank positions for admission and cut-off and the values of valve displacement and valve openings for all intermediate crank positions.
ILLUSTRATIVE PROBLEM
A certain valve has an external steam lap equal to 14 ins. The lead is 4 in. and the throw of the eccentric is 20 ins. (a) Construct such parts of the Bilgram diagram as are necessary to indicate "head end" crank positions for admission, maximum valve opening at admission, and minimum amount of valve opening at various crank positions between admission and cut-off. (c) Determine the value of the angle of advance.
THE D-SLIDE VALVE
173
Draw a circle with radius equal to the eccentric throw, 2J
in., using any convenient scale. This circle is designated by $aeb$
in Fig. 113. Draw a line parallel to the horizontal axis of any
convenient size. Draw in the horizontal diameter $ac$ and extend
as shown. On the right-hand side of the circle draw the line $ef$,
![]() A diagram showing a circle with various lines drawn on it, including a line parallel to the horizontal axis labeled "ac", and another line labeled "ef". The diagram also includes labels such as "Steam Lap Circle", "Eccentric Circle", and "Drain Circle".
Fig. 113.
parallel to the horizontal axis and a distance above it equal to the
lead, $\frac{1}{4}$ in., to the same scale as that chosen for eccentric circle.
The steam lap circle must have its center $Q$ on the upper right-
hand quadrant of the eccentric circle, and it must be tangent
to the line $ef$. With compass points set at $\frac{1}{4}$ in., 14 in. to scale,
Therefore, with compass points set the proper distance apart, find
the center $Q$, about which a 11-in. radius circle will just be tangent
to the line $ef$, and draw the steam lap circle.
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STEAM POWER
The crank position at admission is found by drawing the line $AO$ so that, if extended, it is tangent to the lower side of the steam lap circle.
The crank position at cut-off is found by drawing the line
![]() A diagram showing the crank position at admission (A) and cut-off (C). The crank position at maximum valve opening (M) is also shown. The diagram includes labels such as "Steam Lap," "Cut-off," "Maximum Valve Opening," and "Crank Position." A line labeled "L-E Inside Lap" connects points on the diagram.
Fig. 114.
$M^{III}O$ in such position that it is tangent to the upper part of the steam lap circle.
The crank position for maximum valve opening is found by drawing the line $MO$ in such position that a line through $OO$ will be perpendicular to it. The amount of valve opening at this
THE D-SLIDE VALVE
175
crank position is shown by the length of the part of this per-
pendicular line outside of the steam lap circle, i.e., the distance
\(Oy\) interpreted according to the scale chosen for eccentric and
steam lap circles.
When the crank is in position \(M'O\), the length of \(h_1\), interpreted to scale, gives the amount by which the valve is open to steam.
When the crank is in position \(M''O\), the length of \(j_k\), inter-
preted to scale, gives the amount by which the valve is closed to steam.
The angle indicated by \(\alpha\) is equal to the angle of advance
because it is a property upon which the construction of this
diagram is based.
83. Exhaust and Compression. The exhaust edge events can be shown on the Bilgram diagram by a method similar to that used for the steam edge events. The direction in
which valve displacements occur are indicated in the upper
part of figure 10 in which the crank and eccentric circles
have been drawn to scale so that they coincide. In-
spection of the small sketch in the lower part of the figure will show that head end release must occur when the valve has traveled a distance equal to the inside lap to the left of its central position. A crank position \(OR\) drawn tangent
to the lower part of a circle about \(Q\) with radius equal to
the inside lap, will therefore be a crank position at re-
lease. Clockwise rotation from this point will result in a wider opening to exhaust until position \(OM_1\) is reached,
after which the valve will begin to close. Final closure
will occur when the crank reaches position \(OK\), the exten-
sion of which is tangent to the top of the exhaust lap circle.
At that time the valve will have returned (moving from left to right) and we still have to move a distance equal to
the exhaust lap before attaining a central position.
ILLUSTRATIVE PROBLEM
Given the exhaust lap of a D-slide valve equal to 1 in.; the
steam lap \(h_1\) inside than outside equal 2 fins; and the
lead \(\frac{2}{3}\) in. Find the angle of advance, the maximum port opening to steam and to exhaust, and the crank positions of cut-off, release,
compression and admission for the head-end of the cylinder.
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STEAM POWER
Draw the eccentric (and crank) circle with a radius equal to 2 ins., and draw the horizontal diameter as in Fig. 115.
Draw a horizontal line in the upper right-hand quadrant at a distance of $\frac{1}{4} + 1\frac{1}{4}$ ins. above the horizontal diameter. Locate the point $Q$ at intersection.
![]() A diagram showing the construction of a steam engine's eccentric and crank positions. The eccentric circle has a radius of 2 inches, and the horizontal diameter is drawn through its center. A line is drawn horizontally above this diameter at a distance of $\frac{1}{4} + 1\frac{1}{4}$ inches, intersecting the upper right-hand quadrant at point Q.
Fig. 115.
Draw the steam lap circle with a radius $\frac{1}{4}$ in. and the exhaust lap circle with a radius $\frac{3}{4}$ in.
The angle of advance is the angle between $OQ$ and the horizontal.
The maximum opening to steam is given by the distance $OA = \frac{3}{4}$ in. The maximum opening to exhaust is given by the distance $OB = 1\frac{1}{4}$ in.
The crank positions shown are obtained by drawing lines
THE D-SLIDE VALVE 177
tangent to the lap circles. $A$ represents admission; $C$, cut-off; $B$, release; and $K$, beginning of compression.
The piston positions corresponding to these events are given to reduced scale by vertical projection.
**84. Diagram for Both Cylinder Ends.** The complete diagram for the head end cylinder is shown in Fig. 114 with all critical crank positions marked. The positions for the crank end of the cylinder can be found in a similar way by constructing a diagram in which the point $Q$ and the lap circles are located in the opposite quadrant. The resulting
![]() A diagram showing the positions of the piston during the stroke of a double-acting valve. The diagram is divided into four quadrants, each representing a different position of the crankshaft. The top left quadrant shows the admission position, where the piston is at its lowest point and the valve is open. The top right quadrant shows the cut-off position, where the piston is at its highest point and the valve is closed. The bottom left quadrant shows the release position, where the piston is at its lowest point and the valve is open again. The bottom right quadrant shows the compression position, where the piston is at its highest point and the valve is closed again.
Fig. 116.
diagram for both cylinder ends, with laps the same for both ends of the valve, is given in Fig. 116.
**85. Piston Positions.** The valve events might be studied entirely in conjunction with crank-pin positions, but it is more convenient and customary to consider them in connection with piston positions. A series of diagrams corresponding to different crank-pin positions could be found by drawing the mechanism to scale for each different position as shown in Fig. 117 for piston positions 1 and 2.
It is obvious that this would involve a great deal of work and that, if drawn to large scale, it would consume a great
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STEAM POWER
deal of space. Further, it is convenient to be able to locate relative piston positions on the line which serves as the horizontal diameter of the crank circle of the Bilgram diagram.
The method used depends upon the fact that the motion of the crosshead is exactly the same as that of the piston, so that if the movement of the cross-head end of the connecting rod can be followed, it will be equivalent to following the motion of the piston itself.
It should also be noted that the distance $a$ of the crank circle must be equal to the stroke of the engine.
Assume now, that the point $b$ in Fig. 117 be taken to represent the position of the piston when it is really in position 1. When the cross-head has moved to position 2, the cross-head will have moved from $a$ to $a'$ and the crank pin from $b$ to $b'$. If with $a'$ as a center the connecting rod be swung down to its horizontal position, its right-hand end will arrive at the point $c$. The distance $bc$ must then represent the distance that crosshead (and piston) have moved from dead-center position because $ab$ and
![]() A diagram showing a crankshaft and connecting rod. The crankshaft is shown in a circular arc, with points labeled 'a', 'b', 'c', and 'd'. The connecting rod extends from 'b' to 'c'. The diagram illustrates how the motion of the crosshead is equivalent to the motion of the piston.
Fig. 117
THE D-SLIDE VALVE
$a'e$ both represent the length of the connecting rod and $e$ must therefore be as far to the right of $b$ as $a'$ is to the right of $a$. The point $c$ may therefore be taken to represent piston position when the connecting rod is in the position of $a'e$.
In general, if the horizontal diameter of the crank shaft be taken to represent the stroke of the engine, the piston position corresponding to any crank position can be found by taking a radius equal to the connecting-rod length (to the same scale as the circle) and striking an arc from the
![]() A diagram showing a crankshaft with a connecting rod attached. The crankshaft has a horizontal diameter representing the stroke of the engine. A line is drawn from the center of the crankshaft to a point on the circumference, representing the piston position when the connecting rod is at its maximum length. Another line is drawn from this point to a point on the circumference, representing the piston position when the connecting rod is at its minimum length. The two lines intersect at a point, which represents the piston position when the connecting rod is at its average length.
Fig. 118.
crank-pin position, using a center on the horizontal line on the cylinder side of the crank circle.
An approximate method is also used for finding the piston position. Instead of projecting down from the crank-pin position with an arc, such as $b'e$ in Fig. 117, a vertical line through the crank-pin position is used. Such a line would give $e'$ as the piston position when $e$ is really correct. This method gives results very close with a connecting rod of infinite length. For ordinary lengths of rod, however, the results are far from correct. The error is said to be due to the angularity of the connecting rod.
The effect of the angularity of the connecting rod is shown in Fig. 118 for different positions. On the outstroke the piston is always farther ahead than the rectilinear pro-
179
180
STEAM POWER
jection would indicate. On the return stroke the piston is always behind the position indicated by rectilinear projection.
![]()
A circle with a line drawn from point A to point B, which is tangent to the circle at point C. The line AB is labeled "b". The line AC is labeled "a". The line BC is labeled "c". The line AD is labeled "d". The line AE is labeled "e". The line AF is labeled "f". The line AG is labeled "g". The line AH is labeled "h". The line AI is labeled "i". The line AJ is labeled "j". The line AK is labeled "k". The line AL is labeled "l". The line AM is labeled "m".
![]()
A graph with a horizontal axis labeled "Time" and a vertical axis labeled "Indicator Diagram". The graph shows a curve that starts at the left side of the graph, rises to a peak, then falls back down to the left side of the graph. The peak of the curve is labeled "Beginning of Compression". The left side of the graph is labeled "Release". The right side of the graph is labeled "Adiabatic Expansion".
Fig. 119.
86. Indicator Diagram from Bilgram Diagram. Since the piston positions corresponding to different crank posi-
Since the piston positions corresponding to different crank posi-
THE D-SLIDE VALVE
181
tions can be determined, it is a comparatively simple matter to construct the indicator diagram which theoretically would be given by an engine fitted with a valve of certain dimensions. It is necessary to assume the upper and lower pressures to assume the form of the expansion and compression curves. These are generally taken as rectangular hyperbolas.
The method of constructing an indicator diagram from the Bilgram diagram is shown in Fig. 119. The crank-pin positions for admission (A), cut-off (C), release (R) and beginning of compression (B) are first found. These pin positions are then plotted on to a horizontal diameter by means of arc with radius equal to the connecting-rod length and with centers on the line $MN$ produced to the left. The intersections $a$, $c$, $r$ and $k$ indicate the piston positions at which the corresponding events occur. These are then projected vertically downward to intersect the proper pressure lines and the card is drawn through the intersections.
Diagrams constructed in the same way, but for both head and crank ends, are given in Fig. 120. A symmetrical valve was assumed, that is, one built exactly alike on head and crank ends. The diagrams show that such a valve cannot be used satisfactorily in a reciprocating engine because of the effect of the angularity of the connecting rod. It is most evident in the case of cut-off. The cut-off in this case occurs just before three-quarter stroke for the head end and just after half stroke for the crank end of the cylinder. All other events are distorted in the same way, but the actual lengths of the variations are not as great as in the case of the cut-off, but they are still very considerable.
The effect of the angularity of the connecting rod upon the diagrams can be remembered easily if it is noted that all valve events occur later with respect to piston position on the outstroke and earlier on the instroke than they would with a connecting rod of infinite length.
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STEAM POWER
It is possible to "equalize" the cut-offs, that is, make them occur at the same fraction of the stroke by using unequal steam laps at opposite ends of the valve, but this will result in still further distortion of admissions, as can be seen by constructing a Belyeu diagram. Similarly, the compressions can be equalized by the use of
![]() A diagram showing the relationship between the admission and compression curves of a steam valve. The diagram shows how unequal exhaust laps can be used to equalize the cut-offs and reduce distortion of the release events.
Fig. 120.
unequal exhaust laps, but this results in distortion of the release events.
Various linkages have been developed which are so arranged that they distort the motion of the valve to just the extent necessary to counterbalance the effects of the angularity of the connecting rod. The scope of this book does not, however, permit a discussion of such valve gears.
THE D-SLIDE VALVE
87. Limitations of the D-slide Valve. The simple valve discussed in the preceding paragraphs has numerous limitations and is therefore only used on small and cheap engines, or in cases where economy in the use of steam is not of importance. The steam, after passing over the outside edges as previously considered, is pressed to its seat by the live steam acting over its entire upper surface. This pressure is practically unbalanced, as the greater part of the lower surface of the valve is subjected to the low pressure of the steam being exhausted. As a result the friction to be overcome in moving the valve is very great and there is an appreciable loss from this source.
Further, the steam must pass through ports which form part of the cylinder clearance and which are alternately exposed to live and to exhaust steam with results previously discussed. These ports can be decreased in length by increasing the length of the valve, but this in turn increases the areas exposed to high pressure and hence increases the friction loss.
It can be shown by means of the Bilgram diagram that, if a cut-off engine than about $\frac{3}{4}$ stroke is desired, the area of admission, the amount of steam lap and the size of the eccentric must all be made very great. This results not only in large friction losses, but also in very early release and compression, because of the great angle of advance. As a result, slide valves of the simple D type are seldom used when a cut-off earlier than $\frac{5}{6}$ to $\frac{3}{4}$ stroke is desired. It should be remembered in this connection that the simple engine generally gives its best economy with a cut-off of about $\frac{3}{4}$ stroke.
The drawing of lines representing the opening of the valve to steam as in Fig. 112 will show that this simple valve is further handicapped by the very slow opening and closing of the steam ports, causing a great amount of wire drawing with a corresponding loss of diagram area. In order to get an adequate opening to steam the valve
![]() A diagram showing a simple D-slide Valve and its operation.
184
STEAM POWER
must also be given a great displacement and, since this occurs under great pressure, it results in great friction loss.
The unbalanced feature can practically be overcome by rolling up the valve and ports about an axis parallel to the length of the cylinder. This gives rise to a valve known as a piston valve, shown diagrammatically in Fig. 121.
It can also be partially overcome by using a balance plate or ring of some kind between the top of the valve and the inside of the steam-chest cover, arranged that live steam is excluded from the greater part of the upper surface of the valve. Valves of this type are generally called balanced slide valves and are used on many high- and medium-speed engines.
The valve travel required for obtaining a given opening
![]() Fig. 121.—Piston Valve.
can be decreased and the rate of opening and closing can be increased by the use of multiported constructions. These are so arranged that two or more ports open or close at the same time, so that the total movement required for a given opening is divided by the number of ports and the rate of opening and closing is maintained in the same proportion. One simple type of double-porled valve is shown in Fig. 122.
When several ports are used the valve often becomes
![]() Fig. 122.—Allen Double Ported Valve.
THE D-SLIDE VALVE 185
a rectangular frame crossed by a number of bars and is known as a **gridiron valve**, because of its appearance. Such valves are often combined with balance plates and give very satisfactory results.
A number of designs of slide valves have been developed for the purpose of making cut-off independent of the other events. Many of these use a separate cut-off valve which either controls the steam supply to the main valve or else rides on the main valve itself, thus controlling ports in that valve. Devices of the latter type are called **riding cut-off valves**. They are either driven by separate eccentrics, or by linkage from the eccentric controlling the main valve, the linkage being so arranged as to give the proper relative motion between main and auxiliary valves so as to allow the auxiliary valve to be actuated so as to give the desired admission, release and compression and the cut-off is then taken care of by proper adjustment of the cut-off valve.
**88. Reversing Engines.** It was shown in one of the early paragraphs of this chapter that the eccentric may be set 90°-45° ahead of the crank ahead meaning in the direction of rotation. To cause the engine to revolve in the opposite direction, that is, to "reverse" the engine, it is therefore only necessary to shift the relative positions of eccentric and crank so that the eccentric leads the crank by 90°+45° in the new direction of rotation. This corresponds to shifting ahead (in first direction of rotation) through an angle equal to 180°-2×45° or shifting backward through an angle equal to 180°+2×45°, as can be seen by inspection of Fig. 109.
In practice it is generally more convenient to use two eccentrics, one set properly for rotation in one direction and the other set properly for rotation in the opposite direction. A diagram showing this arrangement is given in Fig. 123. This figure is drawn for a vertical engine and in such position that the engine is on crank-end dead center.
186
STEAM POWER
The point $P$ represents the position of the center of the crank pin; the point $f$ represents the position of the equivalent crank (center of eccentric) which drives the valve for "forward," "ahead" or clockwise rotation; and the point $b$ represents the position of the equivalent crank which drives the valve for "backing," "reverse," or counter-clockwise rotation.
Fig. 123.
The real mechanism, in one of its numerous forms known as the Stephenson Link Gear, is shown in perspective in Fig. 124. The forward eccentric corresponds to $f$ of Fig. 123 and the backing eccentric corresponds to $b$ of that figure. The eccentric rods are fastened to opposite ends of a curved "link" and move the valve through a "link block" fastened to the end of the valve stem. In the position shown in the figure the link is in such position that the forward eccentric operates practically directly on the valve stem so that the valve motion is practically entirely governed by that eccentric. If the reverse shaft were to be rotated clockwise into the backing position, the "suspension rod" would pull back until the eccentric rod of the backing eccentric was directly under the valve stem. Under such conditions the valve motion would be controlled almost entirely by the backing.
![]() Fig. 124.—Stephenson Link Gear.
8
THE D-SLIDE VALVE
eccentric and the engine shaft would rotate counter-clockwise.
If the mechanism were so set that the link block occupied a position on the link between the ends of the two eccentric rods, the valve motion would be controlled by both eccentrics and would be a compromise between the motions given by either eccentric separately. It is characteristic of this gear that the cut-off is latest when either one or the other eccentric is fully "in gear" and that it becomes earlier as the link block approaches the center of the link. With the link block at its center, the cut-off of the link does not occur at all, i.e., the cut-off occurs at zero stroke.
There are numerous other forms of link gears, the best known being the Gooch, the Allan and the Porter-Allen. There are also numerous reversing mechanisms known as radial gears in which the motion of the valve is controlled by peripherally arranged teeth on a wheel meshing with the desired valve motion. The valve motion is obtained indirectly through the radius rod from an eccentric, from the crank, or from the connecting rod. The limits of this book do not permit a detailed discussion of these forms.
89. Valve Setting. From what has preceded it will be evident that in order to obtain correct valve action and its seat and driving mechanism be correctly designed, but also that the various parts must be correctly connected up in order that the valve may move in its proper place relation with respect to the piston.
Adjusting the mechanism in such a way that the proper phase relations are established is known as setting the valve. This can be done with fair accuracy by a simple study of the mechanism in various positions, as will be shown below, but it is always advisable to check the setting by means of indicator diagrams taken after the setting is completed. Such diagrams will often show errors of such character or size that they cannot be determined by measurement on an engine which is not operating.
188
STEAM POWER
Before beginning operations it is always advisable to go over the entire engine carefully and to eliminate excessive lost motion at all pins and bearings in order that the relative positions of parts obtained while setting the valve may approximate those which will be obtained when the engine is in operation. The effect of lost motion will be appreciated after a study of Fig. 125. Assume that all parts of the mechanism are tight except the crank-pin end of the connecting rod as shown. If the engine is rotated by hand,
![]() A diagram showing a crankshaft with a connecting rod attached, and two views of the same crankshaft with the connecting rod removed. The first view shows the crankshaft in its normal position, and the second view shows the crankshaft in its opposite position.
Fig. 125.
for instance, by turning the fly-wheel, the crank will pull the piston mechanism and the piston will be drawn into the position shown in the upper half of the figure when the crank has turned through an angle $\alpha$. On the other hand, when the engine is operating under steam, the piston will push the crank pin around and will occupy a position such as that shown in the lower half of the figure when the crank has been turned through an angle $2\alpha$. It will thus be seen that the piston can occupy two very different positions for the same crank position, and a valve setting based upon the conditions shown in the upper part of the figure might be
THE D-SLIDE VALVE 189
very incorrect when used under the conditions shown in the lower part of the figure.
Lost motion in any part of the mechanism can produce analogous results and it is therefore necessary to remove as much of this lost motion as possible by adjusting the valve. It is practically impossible to eliminate all lost motion, as there must be sufficient clearance at all bearing surfaces to accommodate a film of oil, and this alone would make necessary the taking of indicator diagrams for the checking of valve settings, even if it were possible to set perfectly by measurement for stationary conditions.
In general, two adjustments can be made in setting a plain slide valve. The length of the valve stem or eccentric rod can be changed and the eccentric can be shifted around the shaft. It is necessary to understand the effects of each of these adjustments.
Changing the length of the valve stem is equivalent to shifting the upper end of the piston away from or towards moving the engine as shown in Fig. 126. In this figure the valve is shown in its central position by full lines. The lap is the same at both ends. If, now, the valve is worked to the right upon its stem until it reaches the dotted position, the head-end lap will have been decreased and the crank-end lap will have been increased by the same amount. This would make admission earlier and cut-off later for the head end and admission later and cut-off earlier for the crank end. Obviously, the effects of changing the length of the valve stem are opposite to the two effects of the eccentric.
Shifting the eccentric about the shaft simply changes the time relation between valve motion and piston motion; it does not alter the valve motion itself. If difficulty is experienced in realizing the truth of this statement, it is only necessary to draw several Bilgram diagrams for
![]() Fig. 126.
190 STEAM POWER
same valve, but with different angles of advance, and then to construct indicator diagrams for both cylinder ends in every case, it will be discovered that shifting the eccentric ahead of the direction of rotation of the piston will make all events occur earlier with respect to piston power for both ends of the cylinder.
In setting a plain slide valve which is built symmetrical about a central axis, i.e., same inside and outside lap at each end, it is first necessary to adjust the length of the valve stem. This can be done by placing a cover so as to expose the valve and then rotating the engine slowly by hand and observing the distance traveled by the valve on each side of its central position. This is conveniently done by observing the distance between the outer edge of the steam port and the outer edge of the valve when the valve is fully open at each end. As the valve travels further towards one end than it does towards the other end, with reference to the port edges, the valve stem must be shortened; if it travels further toward the crank end the stem must be lengthened.
In making these adjustments it is advisable to turn the engine only in the direction in which it is going to rotate, so that any change in the valve mechanism will have approximately the same effect as when the engine is operating.
When the length of the valve stem is correctly adjusted, the eccentric must be set on the shaft as to give the proper angle of advance. This is commonly done by shifting it about three degrees until this angle has been obtained. In order to determine the value of the lead it is necessary to be able to set the engine on each dead center. This can be done approximately by turning the engine until the crosshead has come to either end of its stroke, but it will be found by trial that the fly-wheel and shaft can be turned through a very large angle at each end of the stroke without causing an appreciable motion of the crosshead,
THE D-SLIDE VALVE
191
so that this method is not very satisfactory for the purpose of adjusting the eccentric. It is customary, therefore, to work in such a way as to give a more accurate determination of the eccentric.
The engine is rotated until the crosshead has been brought near one end of its stroke, as shown in Fig. 127, and a mark is then scribed across the crosshead and guide as at $ab$. An arc $xy$ is then marked on the fly-wheel by means of a tram such as that shown, the end $c$ being placed at point $P$ on some solid part of foundation or floor. The engine is then rotated, clockwise in the figure, until the crosshead has reached the end of its stroke and returned to such a point that the marks on crosshead and guides again coincide, as shown by the dotted pencil line in the figure. The arc $x'y'$ is then scribed on the fly-wheel with the tram, the end $c$ again bearing on the point $P$. A point $z$ is then found by bisecting the arc $cf$ and when this point is brought under point $d$ of the tram the crank will obviously be at crank-end dead center and the piston at the crank end.
![]() Fig. 127.
At point $P$ on some solid part of foundation or floor. The engine is then rotated, clockwise in the figure, until the crosshead has reached the end of its stroke and returned to such a point that the marks on crosshead and guides again coincide, as shown by the dotted pencil line in the figure. The arc $x'y'$ is then scribed on the fly-wheel with the tram, the end $c$ again bearing on the point $P$. A point $z$ is then found by bisecting the arc $cf$ and when this point is brought under point $d$ of the tram the crank will obviously be at crank-end dead center and the piston at the crank end.
192
STEAM POWER
![]() A diagram showing two curved lines labeled "C. E. Card" and "H. E. Card". The top line is labeled "(a) Perfect Cards for Slide Valve Type."
![]() A diagram showing two curved lines labeled "C. E." and "H. E.". The top line is labeled "(b) Actual Card; Small Engine. Center Line of Valve on Center Line of Seat; Eccentric Advanced to Give Normal Lead of 0.05 inch. Engine Running Over."
![]() A diagram showing two curved lines labeled "C. E." and "H. E.". The top line is labeled "(c) Same Setting as (b) except Engine Running Under."
Fig. 128.
THE D-SLIDE VALVE
193
![]()
A diagram showing the angular advance of eccentric increase. Valve stem length same as in (b) and (c). Lead 0.375 inch.
(d) Angular Advance of Eccentric Increased. Valve Stem Length Same as in (b) and (c). Lead 0.375 Inch,
![]()
A diagram showing the angular advance of eccentric decreased so as to give negative lead of 0.5 inch. Length of Valve Stem Unchanged.
(e) Angular Advance of Eccentric Decreased so as to Give Negative Lead of 0.5 Inch. Length of Valve Stem Unchanged.
![]()
A diagram showing the length of valve stem changed; Angle of Advance as in (b).
(f) Length of Valve Stem Changed; Angle of Advance as in (b).
Fig. 128.
104
**STEAM POWER**
of its stroke. A point on the fly-wheel diametrically opposite to $z$ is next found, so that when it is brought under point $d$ of the tram the engine will be on head-end dead center.
It is probable that more accurate results are obtained by rotating the engine in a direction opposite to that in which it rotates under steam, because lost motion is then taken up in the eccentric, and the valve is closed. When the whole process of valve-setting is considered it is questionable whether this is the correct direction of rotation. Opinion and practice differ in this respect. In the end, the setting should be checked by the taking of indicator diagrams, so that effects of incorrectable lost motion may be finally eliminated.
With the dead-center points found the engine is placed on, say, head-end dead center, and the eccentric shifted until the valve is open to steam by the desired lead. The eccentric is then fastened in this position and the engine turned to the opposite dead center. Because of angularity of connections and tolerances in manufacture seat dimensions, it generally will be discovered that the valve is not now open to steam by the same amount as at the other end. If it is desired that it should be, the valve can be shifted on its stem about half of the distance by which it is out and the eccentric can then be swung so about the shaft to take up the remaining difference. This effect should then be checked by putting the engine on the opposite dead center.
Valves may be set for equal leads as above, or for equal cut-offs or for any sort of a compromise desired. In any case the procedure is about the same. The length of the valve stem is adjusted, then the eccentric position is adjusted, and then refinements are effected by small changes of both adjustments. Remember always that increasing the length of the valve stem changes events at opposite cylinder ends in opposite directions, while shifting the eccentric changes all events in the same direction.
The effects of various adjustments are shown by the
THE D-SLIDE VALVE 195
indicator diagrams given in Fig. 128. These diagrams were taken from a small, slide-valve engine and serve very well to show the way in which the indicator discloses poor adjustments.
**PROBLEMS**
1. Given: angle of advance, $30^\circ$; throw of eccentric, $\frac{1}{2}$ ins.; lead, $\frac{1}{2}$ in.; maximum exhaust-port opening, 1 in.; find the steam lap, maximum opening to live steam, and the exhaust lap.
2. Given: steam lap of $\frac{4}{7}$ in.; lead of $\frac{1}{2}$ in.; exhaust lap of $\frac{1}{2}$ in.; and the angle of advance equal to $30^\circ$. Find the valve travel (throw of eccentric) and maximum port opening to steam and to exhaust.
3. An engine has an eccentric throw of $\frac{1}{2}$ ins.; a steam lap of 1 in.; a lead of $\frac{1}{2}$ in.; and a maximum exhaust lap of 1 in. Find the return stroke. Assume a connecting rod of infinite length and find the angle of advance, the exhaust lap, and the maximum port openings to steam and to exhaust.
4. Given: eccentric travel, 3 ins.; steam lap, $\frac{4}{7}$ in.; exhaust lap, $\frac{1}{2}$ in.; and lead, $\frac{1}{2}$ in.; find maximum port opening, angle of advance, and piston positions at cut-off, release, compression, and admission for both ports. Find also the length of the connecting rod equal to 44 times the length of the crank.
5. It is required to build an engine having a steam-port opening of 1 in., a steam-lap opening of 1 in., a lead of $\frac{1}{2}$ in., and a maximum port opening of 1 in. on each side of the crank. Cut-off must occur at $\frac{3}{4}$ strokes and release at 65% of the stroke. Find the inside and outside lap, the throw of the eccentric and the fraction of stroke completed by the beginning of compression.
![]() A diagram showing an indicator diagram for a slide-valve engine.
CHAPTER XI
CORLISS AND OTHER HIGH-EFFICIENCY ENGINES
90. The Trip-cut-off Corliss Engine. The slide valve has certain limitations which can be partly, but never wholly, overcome. In most slide-valve gears, for instance, the various events occur more slowly than is desirable, and this is particularly true when ideal valves would open suddenly to full opening when necessary and would close as suddenly and as wide as possible at the proper time, and such action would give minimum throttling loss and rounding of corners of the diagram. Engines fitted with such ideal valves would therefore give indicator diagrams with maximum work area as shown by the dotted line in Fig. 129, the full line showing the type of diagram obtained with the ordinary slide valve.
Again, the simpler forms of slide valve involve the use of long ports connecting with the clearance space within the cylinder, thus adding greatly to the clearance surface exposed and to the cylinder condensation. These ports serve for both admission and exhaust, and their walls are therefore cooled partially by the admission air, with the result that excessive condensation occurs during admission.
Many attempts have been made to devise valve gears which should not be subject to the limitations of the
![]() Fig. 129.
196
HIGH-EFFICIENCY ENGINES
157
simple slide valve. Some of these have resulted in the development of the more complicated slide valves de-
scribed in the last chapter, but such designs generally leave much to be desired. One of the earliest and most success-
ful solutions was that of Corliss, who developed what is
known as the trip-cut-off Corliss gear.
The long combined steam and exhaust ports are elimi-
nated by the use of four valves, two for steam and two for
exhaust. These are rocking valves and are located top and
bottom, at the extreme ends of the cylinder, with their
longitudinal axes perpendicular to those of the cylinder,
as shown in Fig. 130. The other two valves are located below so as to drain out water of condensation.
Details of valves of this type are shown in Fig. 130.
These valves may each be regarded as an elementary
slide valve which has a cylindrical instead of a flat face,
and which is oscillated about a center near the face instead
of being reciprocated, i.e., oscillated about a center at an
infinite distance.
The valves are operated as shown in Fig. 131 by short
links from a wrist-pivot potted on the side of the cylinder
and rocked back and forth about its center by means of an
eccentric operating through the linkage indicated. The
location of the wrist-pivot is such that the various links are so chosen that the valves travel at high velocity
when opening and closing, that they open very wide, and
that they close only far enough to prevent leakage and then
remain practically stationary until about to open again.
Throttling losses are thus decreased and wear caused by
useless motion after closure is minimized.
The operation of the steam linkages in this gear is
effected positively by the linkage already explained, but
they are closed differently. For opening, the steam link
rotates the bell crank B in Fig. 132 and thus raises the
latch C. The hook on the end of one of the arms of this
latch engages the steam arm which is fastened on the end
198
STEAM POWER
(8)
(9)
Fig. 130.-Details of Corrosion Valves.
![]() A diagram showing three different types of steam valves with their respective symbols below them.
HIGH-EFFICIENCY ENGINES 109
![]() Fig. 131.--Cylinder Engine Gear.
200
STEAM POWER
of a rod which is slotted into the end of the valve. The valve is thus drawn further open as the wrist plate revolves, until the tripping end $D$ of the latch strikes the cam indicated by $E$. This throws the hook out of engagement and thus disconnects the valve from the driving mechanism. The
![]() A detailed diagram of a Corliss Trip-Cut-off Gear.
Fig. 132.—Details of Corliss Trip-Cut-off Gear.
valve is closed by the action of a dash pot, one form of which is shown in Fig. 131. As the steam arm rises during the opening of the valve it draws up the plunger or piston of the dash pot, leaving a partial vacuum beneath it; and, when the
![]() A diagram showing a dash pot with a plunger.
Fig. 133.
valve is released by unhooking of the latch, atmospheric pressure drives the plunger down and thus causes cut-off to occur. The action of a dash pot is found to be unsatisfactory when the speed of the engine exceeds about 125 R.P.M. and most Corliss engines with trip-cut-off operate
HIGH-EFFICIENCY ENGINES 201
at still lower speeds. Under such circumstances the cut-off is very rapid as compared with the piston speed, and the diagram shows a comparatively sharp corner at this point. A set of diagrams obtained from a large Corliss engine operating at 80 R.P.M. is given in Fig. 133, and it is obvious that little throttling occurs.
Because of the low speed at which these engines operate the stroke can be made long with respect to the diameter without attaining a prohibitive piston speed. The economy mentioned in Chapter VII as resulting from the use of long strokes can thus be obtained in these engines. An idea of the saving in steam effected by the partial elimination of throttling and condensation losses by means of the Corliss valves is given by the curves in Fig. 134 (a) and (b), which give average performance.
The position of the cam which determines the time at which cut-off occurs is controlled by the governor of the engine. When moved in the direction taken by the steam arm it causes cut-off to occur later. Variation of the point of cut-off is used in both and in many other engines to control the admission of steam during each cycle so that the engine may make available the quantity demanded at the shaft, as will be explained in a later chapter. It is therefore desirable that the range of cut-off should be as great as possible, but it has been found very difficult to design trip-cut-off gears which will give a cut-off later than about 0.4 stroke if steam and exhaust valves are operated from the same eccentric. Later cut-off causes poor timing of the exhaust events.
This has led to the introduction of Corliss engines with two eccentrics and two wrist plates per cylinder. One set operates the steam valves and the other the exhaust valves. With this arrangement the range of cut-off is unlimited.
91. Non-detaching Corliss Gears. Because of the low speed at which trip-cut-off Corliss engines are operated,
202
STEAM POWER
![]() A graph showing approximate steam consumption of various types of engines. The x-axis represents different types of engines (e.g., 1st-class, 2nd-class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Approximate Steam Consumption of Various Types of Engines
| Type of Engine | Steam Consumption |
|----------------|-------------------|
| 1st-class | 150-200 |
| 2nd-class | 100-150 |
| 3rd-class | 80-100 |
| 4th-class | 60-80 |
| 5th-class | 40-60 |
Steam Consumption per Hour
| Type of Engine | Steam Consumption |
|----------------|-------------------|
| 1st-class | 150-200 |
| 2nd-class | 100-150 |
| 3rd-class | 80-100 |
| 4th-class | 60-80 |
| 5th-class | 40-60 |
Steam Consumption per Hour
| Type of Engine | Steam Consumption |
|----------------|-------------------|
| 1st-class | 150-200 |
| 2nd-class | 100-150 |
| 3rd-class | 80-100 |
| 4th-class | 60-80 |
| 5th-class | 40-60 |
Steam Consumption per Hour
| Type of Engine | Steam Consumption |
|----------------|-------------------|
| 1st-class | 150-200 |
| 2nd-class | 100-150 |
| 3rd-class | 80-100 |
| 4th-class | 60-80 |
| 5th-class | 40-60 |
Steam Consumption per Hour
| Type of Engine | Steam Consumption |
|----------------|-------------------|
| 1st-class | 150-200 |
| 2nd-class | 100-150 |
| 3rd-class | 80-100 |
| 4th-class | 60-80 |
| 5th-class | 40-60 |
Steam Consumption per Hour
| Type of Engine | Steam Consumption |
|----------------|-------------------|
| 1st-class | 150-200 |
| 2nd-class | 100-150 |
| 3rd-class | 80-100 |
| 4th-class | 60-80 |
| 5th-class | 40-60 |
Steam Consumption per Hour
| Type of Engine | Steam Consumption |
|----------------|-------------------|
| 1st-class | 150-200 |
| 2nd-class | 100-150 |
| 3rd-class | 80-100 |
| 4th-class | 60-80 |
| 5th-class | 40-60 |
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for different types of engines. The x-axis represents different types of engines (e.g., first class, second class, etc.), and the y-axis shows the corresponding steam consumption in pounds per hour. The lines represent different types of engines with their respective steam consumption values.
Steam Consumption per Hour
![]() A graph showing steam consumption per hour for不同类型的发动机。x轴表示不同的类型（例如，第一类、第二类等），y轴显示每小时的蒸汽消耗量。线条代表不同类型发动机的相应蒸汽消耗值。
HIGH-EFFICIENCY ENGINES
| APPROXIMATE STEAM CONSUMPTION |
| VARIOUS TYPES OF ENGINES |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| (U)Bipole-Rush (Very High) (Rush (Very High) |
(1)Bipole-Engine with Variable Bypass (Variable Bypass) |
(2)Bipole-Engine with Variable Bypass (Variable Bypass) |
| Fig. 134a—Steam Consumption per Lb. Per Hour |
| Steam Consumption per Lb. Per Hour | Rated Horse Power | 600 | 800 | 1000 | 1200 | 1400 | 1600 | 1800 | 2000 | 2200 | 2400 | 2600 | 2800 | 3000 | 3200 |
|---|
(U)(A)(C)(D)(E)(F)(G)(H)(I)(J)(K)(L)(M)(N)(O)(P)(Q)(R)(S)(T)(U)(V)(W)(X)(Y)(Z)
BIPOLAR RUSH HIGH EFFICIENCY ENGINE
WITH VARIABLE BYPASS
VARIABLE BYPASS
VARIABLE BYPASS
VARIABLE BYPASS
VARIABLE BYPASS
VARIABLE BYPASS
VARIABLE BYPASS
VARIABLE BYPASS
VARIABLE BYPASS
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VARIABLE BYPASS
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VARIABLE BYPASS
VARIABLE BYPASS
VARIABLE BYPASS
VARIABLE BY PASS BRASS PLENTY OF VARIATION IN THE VARIOUS TYPES OF ENGINES. |
|---|
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they are necessarily large, heavy and costly and efforts have been made to design gears which shall possess the advantages of the original Corliss mechanism without the limitation as to speed.
In many models the Corliss valves are retained and are located in the ends of the cylinder as just described or in
![]() A diagram showing two views of a steam engine valve mechanism. The top view shows a single-ported, double-ported, and triple-ported Corliss valve. The bottom view shows a single-ported Corliss valve in a cylinder head.
Fig. 135.—Non-detaching Corliss Valves Located in Cylinder Head.
the cylinder heads as shown in Fig. 135. In some the wrist plate and the connecting links are also retained, but in others they are eliminated. In all engines of this type the admission valves are closed positively, the closure being effected by the same linkage that opens the valves to admit steam. Quick action is obtained by the arrangement of the operating mechanisms, the centers of rotation and the
HIGH-EFFICIENCY ENGINES
205
lengths of links being so chosen that the valve travel is small when the valves are closed, that it is rapid when the valves are opening and closing, and that the valves remain practically open during most of the time that steam is being admitted.
The advantages of such clearance and short and separate ports are attained in these arrangements and the operation of the valves is almost as perfect as that of the trip-cut-off gear. Engines fitted with these modified Corliss gears are operated at speeds considerably higher than those permissible with the older arrangement, and they may be classed with medium-speed engines.
Engines of this type have been commercially as four-valve engines, but as this name applies equally well to the ordinary trip-cut-off gear and to others which will be described later, it is best to use some other designation. The term non-detaching Corliss engines seems to best describe them and is apparently gaining in favor.
Non-detaching Corliss engines generally give diagrams intermediate between those obtained with the long-seated, trip-cut-off gear, and with the ordinary side-valve engines with the simpler forms of valves, though the latter designs very closely approximate the performances of the trip-cut-off Corliss engine.
92. Poppet Valves. Attention has already been called to the fact that the use of highly superheated steam is very effective in lessening or even eliminating initial condensation. Experience has shown that large valves and valves with sliding surfaces such as poppet valves and Corliss valves do not work well with highly superheated steam. The large castings warp so that contact surfaces do not remain true and the lack of moisture which acts as a seal with saturated steam leads to excessive leakage. Difficulties has also been experienced with the lubrication of these sliding types of valves when using highly superheated steam.
![]() A diagram showing a Corliss engine with a poppet valve.
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An old form of valve known as the poppet valve has recently been adopted by some builders as a solution of the difficulties met in the case of highly superheated steam. This form of valve in four-valve arrangement, combined with design in which short ports and symmetrical castings are used, yields very economical engines which can be safely used with a degree of superheat not obtainable in the case of the sliding and oscillating forms of valves.
![]() Pic. 130a.-Lentz Poppet Valve Engine.
![]() Fig. 130b.-Cross-section, Lentz Engine.
HIGH-EFFICIENCY ENGINES 207
Sections of a modern type of poppet valve engine are shown in Figs. 136 (a) and 136 (b), and details of the admission valve and its operating mechanism are given in Fig. 137 (a) and (b). The valves are all double-seated (double-ported or double-beat), that is, they seat at both ends and are made hollow so that the steam passes both around the outside of the valve and through the valve as shown by the arrows in Fig. 137 (b). This results in large area for passage of steam and in quick opening and
![]() A diagram showing the operation of a poppet valve engine. The admission valve and operating mechanism are depicted in detail.
Fig. 137a.—Admission Valve and Operating Mechanism, Lenzt Engine.
closing, as in the case of gridiron valves, with small actual movement of the valve.
The valves are opened positively by eccentrics operating through cams and rollers as shown in Fig. 136 (b) and they are closed by springs as rapidly as the return motion of the cam permits. The eccentrics are mounted on a horizontal lay shaft which is located to one side of the engine, with its axis parallel to that of the latter, and which is driven by a belt from the main driving shaft of the engine.
Since this valve arrangement gives short steam and exhaust ports, permits the use of small clearance, and
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gives fairly rapid opening and closing of valves with little throttling when open, it gives good economy when used with saturated steam. By adding superheat the economy is still further improved. The water rate of one of these engines is shown for one load in Fig. 134 (a). A simple, Lentz non-condensing engine is reported to have given a consumption of 16.13 lbs. of steam per horse-power hour with 92.7° superheat and a pressure of 153 lbs., and this figure was considered low. Lowering the compounding, higher superheat, lower back pressure, etc.
**93. The Una-flow Engine.** A very interesting modification of the steam engine, known as the Una-flow Engine, has recently appeared. In this design an attempt is made to decrease the duty to condensation in a very original way, and the results are so excellent that the design makes possible very great economy.
In the ordinary forms of engine the entire wall of the cylinder is subjected to the cooling action of the lowest temperature steam during the entire exhaust stroke, and in double-acting types these cooled walls are immediately brought into contact with the high-pressure steam issuing on the other side of the piston, as well as coming into contact later with the next charge of high-pressure steam. The una-flow design minimizes this action by admitting steam at the ends of the cylinder, exhausting it at the center of length of the cylinder, and compressing the steam caught in the clearance space before admitting maximum pressure thus heating the clearance walls. The heating of the clearance walls is further effected by partly jacketing the head with live steam on its way to the admission valve and the jacket is sometimes extended along the cylinder to the point at which cut-off normally occurs.
One form of this engine is shown in Figs. 138 and 139. The steam entering the cylinder head by means of passages up to a double-seated poppet valve, flows into the cylinder until cut-off occurs and then expands until the piston
HIGH-EFFICIENCY ENGINES 209
uncovers the exhaust ports. The steam is exhausted until the returning piston again covers these ports, after which the material trapped within the cylinder is compressed as indicated in the diagrams. The ideal sought is to maintain a temperature which the expanding steam will have when reaching it and thus to minimize thermal interchanges and loss.
![]() Section of Una-flow Engine Cylinder.
Tests of these engines show that very great ratios of expansion can be used in a single cylinder without the excessive losses customary when such ratios are attempted in the ordinary counter-flow type. It is thus possible to obtain good economy with one una-flow cylinder expanding from a high pressure to a vacuum; conditions which would involve the use of compounding with ordinary construction.
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The results of tests on one of the first Una-flow engines built in this country are shown in Fig. 134 (a) and (b). In comparing with the curves it should be noted that two of the tests were run with high superheat.
**94. The Locomobile Type.** In the effort to improve the economy of small steam plants the Germans developed a form of plant now known as the Locomobile Type. The
![]() A diagram showing a cross-section of a Locomobile Type steam engine.
Fig. 139.
name came from the fact that these plants, as originally made, were mounted on wheels and intended for portable use by agriculturists and contractors. Their economy in the use of fuel proved so great that they have since been built for stationary use in sizes running well toward 1000 horsepower.
A Locomobile of American construction known as the Buckeye-mobile is illustrated in Fig. 140, which shows a longitudinal section of the plant. The tandem compound
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![]() A detailed diagram of a high-efficiency engine, showing various components such as pistons, cylinders, connecting rods, and valves. The diagram includes labels for different parts like "Pistons," "Cylinders," "Connecting Rods," "Valves," and "Exhaust Valve." There is also a section labeled "Jaw for Blowing Tubes" and "Jaw for Blowing Tubes" on the right side of the diagram.
FIG. 140
140
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engine is mounted on top of an internally fired boiler with the engine cylinders located in the flues which lead the products of combustion away from the boiler.
The steam generated in the boiler is passed through a superheater suspended in the smoke box. The flow of steam is from the rear toward the front of this superheater (counter flow) so that the hottest steam comes in contact with the hot walls of the superheater first, and then a pipe contained within the flue to the high-pressure cylinder, which is jacketed by the hot flue gases and in which the loss of heat to metal is thus minimized. From the high-pressure cylinder the steam passes to a receiver contained in the smoke box, the receiver serving as a reheater to evaporate any water remaining in the steam and to superheat the steam admitted to the low-pressure cylinder. From the low-pressure cylinder, the steam passes through a feed-water heater in which it raises the temperature of the boiler feed and then it passes to atmosphere or to a condenser. Boiler-feed pump and condenser pump, if used, are integral parts of the plant, being driven directly from the main shaft.
It will be observed that every precaution is taken to guard against initial condensation, and to minimize loss of heat in flue gases and in exhaust steam leaving the plant. The high economies achieved are due to such facts alone.
Small plants of this type have given an indicated horse-power hour on a little over one pound of coal when operated condensing, whereas the best large compound reciprocating engine plants seldom do better than about 1.75 lbs. of coal per I.h.p. and often use 2 or more pounds when operated condensing.
CHAPTER XII
REGULATION
95. Kinds of Regulation. There are two distinctively different kinds of regulation referred to in connection with reciprocating steam engines, one of which may be called fly-wheel-regulation and the other governor-regula-
tion or governing.
The remarkable effect of the fly-wheel has already been referred to. The turning effort exerted at the crank pin by the action of steam on the piston or pistons of an engine is not constant, and the angular velocity of the engine shaft is therefore constantly varying during each revolution. It is the function of the fly-wheel to damp these variations so that they do not exceed the allowable maximum for any given set of operating conditions. The efficiency of the fly-wheel in this respect is measured by the coefficient of fly-wheel regulation $k_w$ which is defined by the equation
$$\delta w = \frac{V_{\text{max}} - V_{\text{min}}}{V} \quad \quad (60)$$
in which
$$V_{\text{max}} = \text{maximum velocity attained by a point on fly-wheel rim or other revolving part};$$
$$V_{\text{min}} = \text{minimum velocity of the same point}, \text{and}$$
$$V = \text{mean velocity of the same point}$$
$$= \frac{V_{\text{max}} + V_{\text{min}}}{2} \approx \text{approximately}.$$
Governor-regulation is absolutely different. Its function is to proportion the power made available to the instantaneous demand. The fly-wheel takes care of variations
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occurring during the progress of one cycle, while governor regulation varies the work value of successive cycles.
**96. Governor Regulation.** If the effect of engine friction is neglected, the power delivered at the shaft of the engine will vary directly with the indicated horse-power. Such an assumption is accurate enough for the discussion which follows.
The indicated horse-power of a given engine is determined entirely by the value of the mean effective pressure and the number of cycles produced in a given time, since these are the only variables in the formula for indicated horse-power. The power made available by an engine might therefore be varied by varying the mean effective
![]() Fig. 111.—Throttling Governing.
![]() Fig. 112.—Cut-off Governing.
pressure, or by varying the number of cycles produced in a given time, or by a combination of both processes.
All of these possibilities are used. In ordinary stationary power plants, the mean effective pressure is generally varied. In the case of pumping engines working against a constant head, but required to deliver different quantities of water at different times, the number of cycles per minute is generally altered by changing the speed at which the engine operates. In locomotive and hoisting practice the number of cycles per minute (speed) and the mean effective pressure are varied as required to meet the instantaneous demands.
These variations may be effected manually as by the driver of a locomotive, in which case the engine may be said to be manually governed. Or, they may be brought
REGULATION 215
about mechanically, as in the case of most stationary power-plant engines, in which case the engine may be said to be **mechanically governed**. In some instances a combination of manual and mechanical governing is used.
**97. Methods of Varying Mean Effective Pressure.** The mean effective pressure increases and decreases with the area of an indicator diagram of constant length, so that the mean-effective pressure can be varied by changing the method which will change the area of the diagram. Two methods are in use and they are illustrated in Figs. 141 and 142. The first causes a variation in area by changing the value of the initial pressure. This is generally done by changing the opening of a valve in the steam line just outside of the steam chest. It is called **throttling governing**, and the valve is called a throttling or throttle valve. The latter name is given to a valve near the cylinder located near the engine, which is used to shut off the supply of steam entirely when the engine is not in operation.
The second method, illustrated in Fig. 142, is known as **cut-off governing**. The variation of cut-off determines the amount of steam admitted to the cylinder per cycle and is used to measure out the quantity required for the load which appears at any instant. Cut-off governing is used on most marine locomotives and is exclusively used in large reciprocating engine power plants.
**98. Constant Speed Governing.** Most engines used for such purposes as the operation of mills and the driving of electrical and centrifugal machinery are required to run at practically constant speed irrespective of the load. They are furnished with mechanical governors which so regulate the power made available that there shall never be any appreciable increase or decrease in speed, but would respectively cause an increase or a decrease in speed.
These mechanical devices always contain some sort of tachometer which moves whenever the speed of the engine exceeds or falls below the proper value. The tachometer
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is so connected to the valve gear that it decreases the power-making ability of the engine whenever the speed starts to increase and it increases the power-making ability if the speed decreases.
Since the valve gear must have a different position for each lead in order that it may throttle or cut off as necessary to suit that load, it follows that the tachometer which controls the position of the valve gear must also have different positions for different loads. But tachometers assume constant speeds depending on their speed, and therefore different loads can only be obtained if they are used with the engine to which it is connected operate at different speeds for different loads.
Constant-speed governing is therefore an anomaly. The device is supposed to maintain constant speed irrespective of load, but may be operated at any desired speed, as the load varies, in order that it may maintain the valve gear in the different positions required to handle the different loads. All so-called constant-speed engines have their highest speed when carrying no load, and the speed gradually decreases to a minimum as the load increases to a maximum. The total variation is generally between 2 and 4%.
The efficiency of such an engine is thus determined by means of the coefficient of regulation, $\delta_{\mathrm{g}}$, which is defined by the equation
$$\delta_{\mathrm{g}} = \frac{n_2 - n_1}{n}$$ (67)
in which
$n_2$ = highest rotational speed attained by the engine;
$n_1$ = lowest rotational speed attained by the engine, and
$n$ = mean speed
$$= \frac{n_2 + n_1}{2}$$ approximately.
99. Governors. The mechanical devices which are used for controlling the power-making ability of an engine
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as described above are known as governors. There are many varieties and only a few of the more prominent can be described.
(a) The Pendulum Governor. One of the earliest forms of governor used on steam engines is illustrated in Fig.
143. It is often called a fly-ball governor. This governor is driven by gearing, chain or belt from the engine, and the weights assume some definite position for each different speed, thus drawing the collar to different positions. The valve gear is connected to this collar and is moved correspondingly. A similar governor is shown in Fig. 131, which also indicates the way in which the collar is connected to the valve gear in the Corixa type of engine. The governor rods are moved by the weights so that they in turn alter the position of the knock-off cam, and thus vary the time at which cut-off occurs. As the speed increases due to a decrease of load, the governor weights and collar move up, and this shifts the cams so as to produce earlier cut-off and decrease power-making ability.
(b) Sliding Valves. On medium- and high-speed engines fitted with some form of slide valve it is found best to use what are known as shaft governors. They are generally carried within the fly-wheel of the engine, operate in a plane passing through the rim of the wheel at right angles to the shaft, and operate upon the eccentric in such a way as to vary the cut-off with speed (and load) changes.
![]() Fig. 143.
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One simple form of such a governor is shown in Fig. 144. The eccentric is not mounted directly upon the engine shaft, but is carried by a pin $P$ in the fly-wheel and revolves with it to swing back and forth across the shaft, about $P$ as a center. Its position at any time is determined by the position of the governor weights $W$, which draw the eccentric down (in the figure) as they move out.
The center of the eccentric is indicated by a heavy dot in the figure, and it will be seen that this center would travel in the arc of a circle about $P$, as shown dotted. If the path of the eccentric is drawn on a Bilgram diagram, it will be found that this motion is equivalent to decreasing the length of the eccentric crank and increasing the angle of advance, resulting in earlier cut-off as the weights move out with increasing speed and decreasing load. Other events will also be changed as the eccentric swings, and some of these changes are technically undesirable.
Numerous designs have been developed in which the eccentric is so guided as to produce various sorts of relations between the different steam and exhaust events. All can be divided into two classes, those in which the eccentric swings about a fixed point $Q$, and those in which the path of the eccentric is guided to move in a straight line. All can be studied by plotting the path of the eccentric center (path of $Q$) on the Bilgram diagram.
The Rites Inertia Governor is a form of shaft governor so designed as to act very quickly with change of speed,
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and to be very powerful, so that it can shift heavy parts.
It is shown in place in the wheel in Fig. 145. With changes in speed it acts like a governor of the type just described, swinging (with increasing speed) about a fixed point $P$ in the wheel, and at the same time being acted upon by the action of the centrifugal effect $C$ and against the action of the spring. This motion shifts the center of the
![]() A diagram showing a mechanical device with various labeled components such as "W", "E", "P", "S", "C", "G", "O", "D", "A", "B", "C", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z". The diagram includes a central gear mechanism with a governor bar, a spring, and other components.
Fig. 145.
eccentric from $E$ toward $e$, giving the desired variation in cut-off.
Superposed upon this action is that of inertia. Assume the wheel and governor to be rotating clockwise at a given constant speed. If the engine speed is suddenly increased, the wheel will move faster, but the governor bar will tend to continue rotating at the same speed because of its inertia. It will thus lag behind the wheel, rotating about $P$ and bringing about an earlier cut-off. The position thus assumed will later be maintained by centrifugal effect if the new speed
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is maintained. The particular advantage resulting from using inertia in this way is speed of action. In many forms of governor the inertia of the moving parts actually resists the efforts of the governor to assume the new position required by changed load and speed.
CHAPTER XIH
THE STEAM TURBINE
100. The Impulse Turbine. One of the oldest of modern water wheels is the tangential or impulse wheel shown diagrammatically in Fig. 146. Water flowing from a reservoir above the wheel passes through a nozzle and the
![]() Tangential or Impulse Wheel.
Fig. 146.—Tangential or Impulse Wheel.
jet, moving at high velocity, strikes buckets on the rim of the wheel and causes the latter to revolve. Theoretically the velocity of the water in the jet would be
$$v = \sqrt{2gh} \text{ feet per second}, \quad \quad (68)$$
in which
$$g = \text{gravitational constant, } 32.2, \text{ and}$$
$$h = \text{head in feet as shown in the figure}.$$
The kinetic energy possessed by the moving water would be
$$K = \frac{\text{w} v^2}{g 2}, \quad \quad (69)$$
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in which $w$ represents pounds of water discharged per second and $g$ and $v$ have the same meanings as above.
If the buckets of the wheel could reduce the velocity of the water to zero they would absorb all of this kinetic energy (which is equal to $\frac{1}{2}mv^2$) but the buckets and the bearings of the wheel would make all of it available at the shaft for the doing of useful work.
Any fluid moving at velocity $v$ and striking buckets in the form of a jet would possess kinetic energy in quantity given by Eq. (60) and would drive the wheel in the same way. Steam is used instead of water with exactly the same results, and steam is so used in what are known as impulse steam turbines.
Experience shows that steam will flow at high velocity from any opening made in the steam once on the boiler or from any other enclosed pipe connected to such a boiler. This is commonly said to be due to the high pressure within the boiler, the spectator piecing the process as the driving force for the steam by the high-pressure steam within the boiler, just as though the part leaving were a solid piston and were driven out as is the piston of an engine during admission, as shown in Fig. 147.
An hydraulic analogy is given in Fig. 148.
The tube shown is supposed to be fitted with a piston upon it assumed to be possible to exert any desired pressure upon the piston. Any such pressure exerted is the exact equivalent of some given head of water and the resultant jet velocity would be given by Eq. (68) by substituting for $h$ the head in feet equivalent to the pressure exerted upon the piston.
When an "elastic" fluid such as steam is being con-
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idered it is, however, necessary to take account of other factors. The steam within the boiler exists at a high pressure; after issuing it exists in the atmosphere at a lower pressure. But low-pressure steam contains less heat than does steam at high pressure, and this difference must exist in some form of energy and could not possibly have been destroyed during the flow.
Experiment shows that steam after flowing into the atmosphere from a boiler in this way has exactly the same characteristics as though it had expanded adiabatically behind a piston through the same temperature range, excepting for the fact that it has very high velocity, which would not have been explained before by experiment. Experiment further shows that, if small losses are neglected, the kinetic energy possessed by a jet of steam is exactly equal to the energy which would be turned into work if that steam acted on a piston as in an ordinary engine.
A complete picture of the process of flow can then be made by supposing that the steam expands out in the form of a piston driven by high-pressure steam, as before, and adding to this idea that this piston expands adiabatically as it travels from the region of high to that of low pressure. This expansion liberates heat contained within the piston or plug of steam and this heat is used in imparting additional velocity to the moving steam which is giving up this heat.
The result of using such a jet upon a theoretically perfect tangential or impulse wheel would be to rob the jet of all this energy. But the energy possessed per pound of steam in the jet is just the same as shown under the upper lines of Fig. 20, i.e., approximately equal using one pound of steam. The area under these upper horizontal line of the $PV$-diagram of the case as shown in Fig. 21 may be assumed to represent the work done upon one pound of steam (flowing out) by another pound which is being evaporated and pushing out the first in order to make room for itself. The
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STEAM POWER
area under the expansion curve in the $PV$-diagram represents the energy converted into velocity energy by the adiabatic expansion of the flowing steam. The lower horizontal line represents the negative work during condensation of water at the lowest pressure and temperature, and the left-hand line represents the pumping of this water back into the boiler and the raising of its temperature to the value
![]() A diagram showing the components of an impulse turbine. It includes a nozzle, a wheel, buckets, a turbine shaft, and a steam drum. The diagram is labeled with "Nozzle," "Wheel," "Buckets," "Turbine Shaft," and "Steam Drum." A text box below the diagram reads: "Fig. 149.—Early Form of Impulse Turbine.""
maintained within the boiler. The complete expansion cycle is therefore the cycle upon which the impulse steam turbine operates and, as a matter of fact, it is the theoretical cycle of all steam turbines.
The early impulse turbine would therefore be acted upon by a jet which possessed available kinetic energy represented by the area of the complete expansion cycle. If the buckets could entirely remove this energy, that is, could reduce the velocity of the jet to zero, the same amount
THE STEAM TURBINE 225
of energy could theoretically be made available at the shaft of the turbine.
An example of a simple form of impulse steam turbine is given in Fig. 149, in which the essential parts of an early form of the turbine are shown. The wheel, the diaphragm and nozzles are all enclosed within a casing. The space on one side of the diaphragm is connected to the steam pipe and that on the other is in communication with the space into which the exhaust steam is to be exhausted.
Another form of impulse turbine is shown in Fig. 157. It will be described later.
101. Theoretical Cycle of Steam Turbine. It was shown in the preceding section that the steam turbine operates on the complete expansion cycle. If a turbine could remove from the steam passing through it and convert into mechanical form all of the energy which is theoretically possible, it would have a theoretical efficiency equal to the mechanical energy represented by the area of the $PY$-diagram of the complete expansion cycle. The area of the corresponding $Te$-diagram would show the same quantity measured in thermal units. The theory of these two curves can therefore be studied by means of these two diagrams.
In Fig. 150 is shown the $Te$-diagram of the complete expansion cycle for several different conditions. The figure $abcd$ represents conditions when the steam is dry and saturated at the beginning of the adiabatic expansion $cd$. Constant quality lines are designated by $x$ and $x'$. It is obvious that by the time the steam has expanded down to the pressure at $d$ it will have a quality
![]() Fig. 150.
226
STEAM POWER
less than unity. If, therefore, it be in the form of a jet issuing from a nozzle and having a high velocity by virtue of its adiabatic expansion, the jet will really be a mixture of steam and water.
If the steam is superheated at constant pressure as shown by $c'$ before passing through the nozzle, it is evident from the figure that the jet issuing from the nozzle will contain less water than in the preceding case, because the condition of the material in the jet after adiabatic expansion will be as shown at $f$ instead of as shown at $d$. The cycle in such a case would be represented by a quadrant indicated by the arc $efd$, representing just that much energy converted into mechanical energy per pound of steam or other unit for which the diagram happened to be drawn.
If superheating had been carried to the point indicated by $g$ before expansion, the jet would obviously issue from nozzles in the form of a vapor stream as shown by the point $h$ in the figure. In that case the cycle would be $e'gha$, and superheat would have to be removed from the low-pressure steam to bring it to the conditions indicated at $t$ before condensation could begin.
If desired, the $P_{v}$-diagrams for such cycles can be drawn very easily. The line $bc$, or $b'$ or $b_{0}$ is a horizontal line in the $P_{v}$-diagram. The line $cd$, or $c'$ is similarly horizontal and the line $ab$ is vertical. The adiabatic expansion is represented by a curved line in the $P_{v}$-diagrams, but can be drawn easily because the necessary data are obtainable from the $T_{e}$-diagram, in which this expansion is represented by a straight line.
ILLUSTRATIVE PROBLEM
Draw the $P_{v}$-diagram for a steam turbine receiving one pound of steam at a pressure of 200 lbs. absolute, with a temperature of 500 F., and discharging it at a pressure of 0.5 lbs. absolute.
First, locate on a $T_{e}$-chart for steam the point representing the condition of steam at 200 lbs. pressure with a temperature
THE STEAM TURBINE 227
of 500° F., and draw a vertical line extending downward until it cuts the horizontal temperature line corresponding to
![]()
A graph showing temperature (°F) on the y-axis and pressure (lb.) on the x-axis.
The curve starts at the origin (0°F, 0 lb.) and rises steeply, reaching approximately 300°F at around 10 lb. pressure.
At 200°F, the pressure is about 15 lb.
At 300°F, the pressure is about 25 lb.
At 400°F, the pressure is about 35 lb.
At 500°F, the pressure is about 45 lb.
At 600°F, the pressure is about 55 lb.
At 700°F, the pressure is about 65 lb.
At 800°F, the pressure is about 75 lb.
At 900°F, the pressure is about 85 lb.
At 1000°F, the pressure is about 95 lb.
At 1100°F, the pressure is about 105 lb.
At 1200°F, the pressure is about 115 lb.
At 1300°F, the pressure is about 125 lb.
At 1400°F, the pressure is about 135 lb.
At 1500°F, the pressure is about 145 lb.
At 1600°F, the pressure is about 155 lb.
At 1700°F, the pressure is about 165 lb.
At 1800°F, the pressure is about 175 lb.
At 1900°F, the pressure is about 185 lb.
At 2000°F, the pressure is about 195 lb.
The curve has a slope that increases with temperature, indicating a higher rate of increase in pressure with increasing temperature.
Legend:
- Temperature (°F)
- Pressure (lb.)
Fig. I3L
Fig. I3L
0.5 lb. pressure. This is practically at 540° F. absolute, or about
80° F.
228
**STEAM POWER**
Second, take from the steam table the volumes of one pound of steam at any, 200 lbs., 140 lbs., and 100 lbs. absolute pressure when superheated to the values shown by the vertical line. These will be about 2.75 cu.ft., 3.58 cu.ft., and 4.67 cu.ft., respectively.
Plot these values with corresponding pressures on a PV-chart as shown in Fig. 150.
Third, take from the $T_e$-chart the pressures at which the vertical line intersects different volume lines on the wet steam region and plot these pressures on the $T_e$-chart.
Fourth, draw a smooth curve, as shown, through all points so determined.
Fifth, draw horizontal top and bottom lines and a vertical line at the left of the diagram. This vertical line should be to the right of the pressure axis by an amount representing the volume of one pound of water, but the volume is so small that it cannot be drawn to any appreciable scale.
**102. Nozzle Discharge.** It was stated in preceding sections that the energy which would be converted into work by the introduction and adiabatic expansion of steam behind a piston is converted into kinetic energy when steam flows out of an orifice or nozzle and that an ideal impulse turbine could absorb all this kinetic energy from the jet, bringing it to rest and making the energy available in the form of useful power at its shaft. It is, therefore, of interest to determine the velocity which a jet will acquire under different conditions.
This could be done by evaluating the area of a diagram, such as that of Fig. 151, and then putting this value in place of $K$ in Eq. (69) and solving for $r$, but it can be done much more simply by considering the following two ways. The heat energy which can be converted into kinetic energy of the moving jet and which can later be converted into useful work by the turbine wheel is represented by the area enclosed within the lines of the complete expansion cycle when drawn on the $T_e$-diagrams. That is to say, it is equal to the heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plus heat supplied minus heat rejected plusheat supplye
THE STEAM TURBINE 220
that representing the heat rejected, that is, $Q_1 - Q_2$, so that
$$K \text{(in B.t.u.)} = Q_1 - Q_2 \dots \dots (70)$$
The values of $Q_1$ and $Q_2$ can be found very readily by plotting the points $c$ and $d$ upon a $T\theta$-chart for steam and observing the constant heat lines upon which they fall,
![]() Fig. 152.
or they can be obtained even more conveniently from what is known as a Moller Chart for steam. In this chart, entropy above $32^\circ$ F. is plotted against heat above $32^\circ$ F., as shown in Fig. 153. An adiabatic expansion on this chart is shown by a horizontal line, since this shows a constant entropy change just as a vertical line on the $T\theta$-chart shows a constant entropy change.
If a point is found in this chart giving conditions corresponding to those at point $c$ in Fig. 152, the value of $Q_1$
$$Q_1 = \text{value of heat at point } c$$
![]() STEAM POWER
| MOLLER CHART |
| HEAT-ENERGY CHART FOR SATURATED AND SUPERHEATED STEAM |
| at various pressures. |
| at various temperatures. |
| at various pressures. |
| at various temperatures. |
| at various pressures. |
| at various temperatures. |
| at various pressures. |
| at various temperatures. |
| at various pressures. |
| at various temperatures. |
230
THE STEAM TURBINE
can be read directly under that point on the horizontal axis. A horizontal line drawn from that point to the terminal-pressure line will give the point corresponding to $d$ of Fig. 152 and the value of $Q_2$ can be read on the horizontal scale. The difference between the two readings gives the value of the kinetic energy $K$ or of the mechanical energy which an ideal turbine could make available, but the expression will be in British thermal units and not in foot-pounds.
This value of the kinetic energy, i.e., $K = Q_1 - Q_2$, may then be placed in Eq. (60), giving,
$$K = 778(Q_1 - Q_2) = \frac{v^2}{2g} \text{ ft.-lbs.}, \quad \ldots \quad (71)$$
since $Q_1$ and $Q_2$ refer to one pound of steam, or
$$K = 778(Q - Q_2) = \frac{wv^2}{2g} \text{ ft.-lbs.}, \quad \ldots \quad (72)$$
when $w$ represents the number of pounds of steam flowing per second.
Solving either Eq. (71) or Eq. (72) for $v$ gives,
$$v = \sqrt{778 \times 2g(Q_1 - Q_2)}$$
$$= \sqrt{778 \times 64.4(Q_1 - Q_2)}$$
$$= 224V(Q_1 - Q_2) \text{ feet per second}. \quad (73)$$
The design of a nozzle consists simply in choosing such sections that the desired amount of steam may flow through it with the desired pressure drop, as the velocity obviously is determined by that pressure drop. This is very conveniently done by working in terms of one pound of steam, since all formulas and charts are generally given on that basis, and then multiplying the cross-sectional areas found by the number of pounds of steam required.
Assume for instance that it is desired to design a nozzle
232 STEAM POWER
to pass one pound of steam per second with an initial pressure of 100 lbs. per square inch abs., and a terminal pressure of 60 lbs., the steam being initially dry and saturated.
The Mollier chart shows that $Q_1$ is equal to about 1187 B.t.u. per pound of steam, while $Q_2$ is equal to about 1147 B.t.u. The velocity with which a jet would issue from a theoretically perfect nozzle under these conditions may then be found by using Eq. (73). This gives
$$v = \frac{224 \cdot 1187 - 1147}{1416} \text{ feet per second}$$
$$v = 1416 \text{ feet per second}$$
The shape of the entrance end of the nozzle is generally made such that the steam will enter it without great disturbance and the shape beyond that point is determined by methods which will be explained below. The cross section at the entrance is so shaped that the steam will give the required quantity at the velocity found above to be equal to 1416 feet per second. This is easily done by determining the volume of steam discharged.
Drawing the adiabatic expansion on the Ta-chart will give the quality at the end of the expansion; or, the quality can be found by multiplying the specific volume of a pound of steam at 60 lbs. pressure must have to give it a heat content of 1147 as found above. With the quality known the terminal volume per pound can be found by multiplying the quality by the specific volume at terminal conditions. Thus for the case under discussion the quality will be about 0.967, and the terminal volume will be about 0.967 cubic feet, the volume to be passed per second, per pound of steam is 0.967 x 0.967 = 0.934 cu.ft approximately. If the velocity is 1416 feet per second the area per pound of steam must be 6.94 + 1416 = 0.0049 sq.ft.
The exact shape of the nozzle is determined by deciding upon the way in which pressure, or velocity, or volume
THE STEAM TURBINE
shall change as the steam passes through it. Suppose, for instance, that a nozzle is to be constructed of the length shown by $ab$ in Fig. 154, and that the pressure is to vary along its length as shown. Assume also that the nozzle is to pass 10 lbs. of steam per second. Taking initial pressure as 100 lbs. and terminal as 60 lbs., the conditions
![]() A graph showing Pressure (lbs per sq. ft.) on the y-axis and Length of Nozzle (inches) on the x-axis.
![]() A graph showing Velocity (ft. per sec.) on the y-axis and Length of Nozzle (inches) on the x-axis.
Fig. 154—Nozzle Design.
will be the same as in the problem above. The discharge area will have to be $10 \times 0.0409$ sq.ft.; or 0.049 sq.ft.
The area at the plane $x_2$ must be that required to pass the steam when it has the velocity resulting from expansion from 100 down to 64 lbs., just as though the nozzle ended at that point. This can be found just as the terminal area was found above. Similarly the sections at $x_1$ and $x$ can be found by figuring velocity and area for expansions to
$$\text{Area} = \frac{\text{Velocity}^2}{2g}$$
234 STEAM POWER
74 and 90 lbs., respectively. If the various areas required are determined in this way, the nozzle will have a longitudinal section about as shown by the dotted lines in the figure and the variation of velocity will be about as shown by the broken line.
If the shape of a nozzle is determined in the same way for a case in which the terminal pressure is less than about 0.58 of the initial pressure, the nozzle will be found to have a very different shape. This is shown in Fig. 155. The nozzle is known as an expanding nozzle, and the smallest section is known as the neck. The pressure $P_n$ in the neck is always equal to about 0.58 $P_1$ and the velocity in the neck is always equal to just over 1400 ft./sec. It is therefore the section at the neck which determines the quantity of steam which a nozzle will discharge if expanding to a pressure equal to or lower than 0.58 $P_1$.
103. Action of Steam on Impulse Blades. It has been stated that the steam acting in an impulse type of turbine delivers energy to the wheel of the turbine by giving up its kinetic energy. In an ideal turbine the steam jet would be brought to rest and would thus give up all of its kinetic energy.
In real turbines it is impossible to bring the jet to rest, as practical design problems prevent it. There is therefore always a loss in real machines because of the residual or terminal velocity of the steam as it leaves the wheel. Thus let the black section in Fig. 156 represent the section of a bucket or blade sticking out radially from the rim of a
![]() Fig. 155—Expanding Nozzle
THE STEAM TURBINE
235
wheel, the wheel revolving about the axis indicated by the dot dash line but located behind the plane of the paper. If minimum loss by eddying is to be experienced at the point at which the steam jet enters the blade, it must enter the blade along a tangent to the curve of the inside of the blade at the entrance edge. This direction is shown by the line marked $r_0$ in the figure.
Were the bucket stationary, the steam jet would move as shown by $v_0$, but as the bucket moves ahead, and, so to speak, runs away from the jet, the steam must really travel in a direction such as that indicated by $v_1$. The velocity of the bucket in the direction indicated by $v_1$. The conditions governing the flow of steam into a bucket are the same as those governing the speed with which and direction in which an individual runs toward and jumps upon a moving vehicle. It is evident that he checks when he is moving ahead at the same rate as the vehicle at the instant when he gets on board. His motion must therefore be made up of two, one toward the vehicle and the other in the direction of the vehicle's travel.
In the case of steam flowing onto a blade as shown in Fig. 156, the various velocities are so related that when drawn down to a single line, the velocity of the steam, $v_a$ and the real or absolute velocity of the blade, $v_b$, form two sides of a triangle of which the closing side represents $r_v$, the velocity of the steam relative to the bucket. The value and direction of $r_v$ is obviously obtained from $v_a$ by geometrically subtracting the velocity of the bucket.
![]() Fig. 156
236
**STEAM POWER**
After entrance, the steam flows around the inner curve of the blade and is finally discharged with the same relative velocity as that with which it entered, and at an angle set by the tangent to the inner curvature of the discharge edge of the wheel. The steam leaving the bucket has been moving ahead with the same velocity as the bucket during the entire time that it was in contact with the bucket, it is also moving ahead with a velocity $v_5$ when it leaves the wheel. Its real or absolute velocity is then $e_a$, which is found by combining $r_2$ and $v_5$ as shown in the figure.
The kinetic energy possessed by the jet when entering the blade is equal to $\frac{w r^2}{2g}$ ft.-lbs., and that which it possesses when leaving is $\frac{w e^2}{2g}$. Obviously, the energy removed while passing over the blade is $\frac{w e^2}{2g} - \frac{w r^2}{2g}$. If the blade were theoretically perfect, it would be so constructed that $r_e$ would be zero and all of the kinetic energy would then be removed. This is practically impossible in a real mechanism, and there is always a loss due to the residual velocity $e_a$. The best that can be done is to so choose the angle of jet admission to the wheel, i.e., to choose $e_a$, with respect to the steam that the actual numerical value of $e_a$ is made as small as possible.
Designs usually work out in such a way that this occurs when the blade velocity is equal to about 0.47 of the absolute velocity of the steam.
104. Double-Action Impulse Turbine. The expanding nozzle already described was first used by De Laval in an impulse type of turbine. The essential elements of this device are shown in Fig. 157. The nozzles are arranged at such an angle to the plane of the wheel that the steam jets strike radially arranged blades at the proper angle to enter without much loss. The blades deflect the jets as shown and
THE STEAM TURBINE 237
absorb the greater part of their kinetic energy, so that the steam flows away from the wheel with low absolute velocity.
As many nozzles are used as are required to make avail-
able the amount of energy desired at full load, and pro-
vision is made for shutting off one or more nozzles by hand when conditions do not warrant the use of all. Governing for ordinary variations of load is effected by throttling the steam flowing to the nozzles in use, thus altering the initial pressure as necessary.
![]()
A diagram showing the components of a De Laval Impulse Turbine. The diagram includes:
- Nozzle
- Steam In
- Steam Out
- Turbine Shaft
- Babbitts
- Wheel
- Surround Ring
FIG. 157.—Single Stage, De Laval Impulse Turbine.
238
STEAM POWER
A section through the wheel and casing of such a tur-
bine directly connected to a centrifugal pump is given
in Fig. 15B. The steam flows into the live steam space
through a throttle valve controlled by the governor; the
valve control sections are shown shaded in Fig. 15B.
From the live steam section the steam flows through nozzles
not shown, into the exhaust steam space, thus acquir-
ing a high velocity. The buckets of the wheel are located
just in front of the discharge ends of the nozzles and the
steam moving at high velocity must pass through them
before moving on toward the exhaust outlet.
106. Gearing and Speeds. It has been stated that the
most efficient operation with ordinary designs is obtained
when the blade speed is equal to about 0.47 of the absolute
steam velocity or, roughly, half the velocity of the impul-
sing jet. To get high economy in the use of steam, large
pressure drops are used and very high jet velocities result.
When the buckets of a turbine operate at peripheri-
cal speeds equal to about 0.47 of velocity, certain diffi-
culties is often met. The stresses induced in the wheel
structure by centrifugal effects become so high as to offer
serious difficulties in design, or the rotative speed of the
unit becomes too high for direct connection to the machine
which is desired.
One method of partly overcoming the latter difficulty
is to operate the turbine at or near the theoretically desir-
able speed and transmit the power to the driven machine
through gears which decrease the rotative speed to the
necessary extent. This method was used with all of the
early De Laval turbines which were of comparative small
capacity. In recent years, however, this method has been largely
abandoned for marine propulsion and other purposes for which large units are used.
It is only a partial remedy in the ease of large units,
however, as the gears necessary for the desired reduction
and the size of the turbine wheels would both become
excessive.
THE STEAM TURBINE
239
![]()
A detailed diagram of a steam turbine, showing various components such as the steam inlet, exhaust, and turbine blades. The diagram includes labels for different parts of the turbine, such as "Steam Inlet," "Exhaust," "Turbine Blades," and "Distributor." The diagram also shows the flow of steam through the turbine.
Fig. 158.—Section of Da Laval Single Stage Impulse Turbine.
240
STEAM POWER
Another and very common method is known as com-
ounding or staging. This may be of two varieties. The
pressure drop in one stage may be limited to that
which will give a reasonable specific speed, and a number of
such stages may be put together in series on one shaft.
This would give one set of nozzles and a wheel for each
stage, the steam discharged from one wheel with very
low velocity expanding to a lower pressure through the
nozzles of the next stage and impinging upon the
wheel of the next stage with the resultant high velocity.
Such an arrangement is known as pressure staging or
pressure compounding, and is extensively used in large
turbines.
The pressure staging method is illustrated in Fig. 150
as applied to the De Laval type of impulse turbine. The
combined increase in diameter of the wheels and increase
in length of blades gives the necessary larger area to pass the larger volumes of steam as the drop of pressure
continues from stage to stage.
Instead of staging on a pressure basis, staging on a veloc-
ity basis may be used. In such a case the drop in pressure
through one set of nozzles is great and the utilizable veloc-
ity high. The steam then passes through a set of stationary
blades which reduce its velocity to zero, and then a second
wheel, in passing through which it gives up still more of
its kinetic energy with a corresponding further decrease
of velocity. If the velocity still possessed by the steam
warrants it, a second set of stationary guide vanes and a
third set of moving buckets can be supplied for further-
reducing it and by carrying this velocity, staging through
a sufficiently great number of stages any initial velocity
THE STEAM TURBINE 241
could be absorbed theoretically without the use of wheels with high peripheral speeds. Practically, losses due to
![]() A diagram of a steam turbine, showing various components such as the governor, steam pipes, and turbine blades.
Fig. 109.—Ablatelage De L'usine Steam Turbine.
friction, eddying and other sources limit the number of velocity stages to two or three.
242
STEAM POWER
![]() A detailed diagram of a steam turbine, labeled with various components such as "First Wheel," "Second Wheel," "Third Wheel," "First Diaphragm," "Second Diaphragm," "Third Diaphragm," and "Fourth Wheel." The diagram also includes labels for different parts of the turbine, such as "C" and "G." The diagram is annotated with arrows indicating the flow of steam through the turbine.
Fig. 100.—Early Form of Curtis Turbine.
THE STEAM TURBINE 243
Velocity staging is combined with pressure staging in the Curtis type of turbine. A section through part of an early design of vertical turbine of this type is shown in Fig. 160. The turbine illustrated had four pressure stages and one velocity stage.
Many varieties of impulse turbines have been developed and all of the larger ones employ several wheels and sets of nozzles and diaphragms to obtain the necessary staging. The same result has been obtained in some of the smaller models by discharging the steam from nozzles into a set of buckets, allowing a fraction of the kinetic energy, catching it at discharge and returning it for another passage through the buckets, and so on until the greatest practical fraction of the kinetic energy has been absorbed.
108. The Reaction Type. If high-pressure steam or other fluid be forced into a device arranged as shown in Fig. 161 and free to revolve about a vertical axis, the jets blowing out of the nozzles will cause the mechanism to revolve in the direction indicated by the arrow. This reaction is said to be due to the reaction of the jets, and the mechanism therefore constitutes a simple form of reaction turbine. By increasing the number of nozzles any amount of steam could be discharged and therefore any amount of steam could be used.
This multiplication of nozzles can, however, be more conveniently accomplished by fastening radial vanes to the periphery of a wheel as shown in Fig. 162, the space between any two vanes constituting a nozzle through which
![]() Fig. 161.
Elementary Reaction Turbine.
![]() Fig. 162.
244
the steam can discharge. By mounting such a wheel within a casing as shown in Fig. 163 it forms a simple reaction turbine. One of the characteristic differences between the impulse and the reaction types of turbines is that in the former the pressure of the jet is increased by the passage through the nozzle, while in the latter it is decreased.
In the impulse type the nozzles are fastened into a stationary part of the turbine and the drop of pressure occurs entirely within the nozzles. The wheels are therefore immersed in a space in which the pressure is constant. In the reaction type, on the other hand, the nozzles are carried on the wheel and there must be a higher pressure on one side of the wheel than there is on the other. Since there must also be mechanical clearance between the blade tips and the interior of the wheel, it follows that this type may be handicapped by considerable leakage which does not exist in the impulse type, excepting as some of the jet may "spill" over the ends of the blades.
The difference of pressure on the two sides of the wheel also causes a tendency toward motion of the wheel along the shaft, i.e., of the wheel and shaft, in a direction away from the high-pressure side.
Many unsuccessful efforts have been made to design efficient reaction turbines, but no pure reaction type has yet been commercialized. The turbines commonly called reaction turbines are really combinations of reaction and impulse types.
One example of what is commercially called a reaction turbine is shown in Fig. 164. Alternate rings (or rows) of stationary and movable blades guide the steam as it expands from the high pressure at one end to the low pressure at the other. The stationary blades project inward from the interior surface of the stationary casing and the movable blades project outward from the external surface
![]()
A diagram showing a reaction turbine with a single row of stationary blades (b) and a single row of moving blades (f). The stationary blades are located near the center of rotation (r1), while the moving blades are located further out (r2).
Fig. 163
THE STEAM TURBINE
245
![]()
A detailed diagram of a steam turbine.
Key components labeled:
- Governor
- Governor Valve
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Governor Spring
- Gover
246 STEAM POWER
of the cylindrical rotor. The rotor blades act like those of an impulse turbine in partly reversing the direction of jets of steam while they react them with comparatively high velocity, but they also act as variable nozzles at a reaction turbine since the steam in passing through them expands and acquires kinetic energy, giving a reaction on discharge.
The stationary blades serve to redirect the steam so that it strikes the next set of moving blades at the proper angle and they also serve as nozzles by which velocity energy is acquired. This is shown diagrammatically in Fig. 165, in which $S$ denotes stationary, and $M$ moving blades.
The Parsons type, illustrated in Fig. 164, may be described as a multistage type in which impulse and reaction are utilized in conjunction.
The balance pistons shown in the figure are used to balance the end thrusts caused by the difference in pressure existing on opposite sides of the wheels in the case of reaction turbines. Each piston has a diameter that it presents a surface equal to the blade surface acted upon by one of the unbalanced pressures, and by connecting across as shown in the figure a high degree of balance is secured.
The overload valve is used to admit high-pressure steam into the high-pressure blades for carrying excessive overloads. The larger area of the passages through these blades permits an abnormal amount of high-pressure steam
![]() Fig. 165.
THE STEAM TURBINE
247
to pass, thus giving a high load-carrying capacity with decreased economy.
**107. Combined Types.** The clearance at the ends of the stationary and moving blades in the Parsons type of turbine permits considerable steam to leak by, as previously explained. This clearance must have almost the same length (measured from blade tip to opposing metal) in all stages in order to insure freedom from rubbing, but it is more strained in the high-pressure stages than in the low. The high-pressure blades are much shorter than the low-pressure blades and a leakage length of a certain amount is therefore equal to a greater fraction of the total blade length. The density of the high-pressure steam is also much greater than that of the low-pressure steam that many more molecules pass through a given area in a given time. In discussions of this character, it should not be forgotten, however, that leakage area is determined by the dimension already referred to multiplied into a circumference and that the circumference is much greater at the lower end.
Because of these and other reasons many manufacturers have come to the conclusion that the impulse type is best for the high-pressure end of the turbine and the reaction type for the low-pressure end. Many such combinations have been produced and they are giving very good results.
**108. Economy of Steam Turbines.** In general, the economies of steam turbines and reciprocating engines are about the same under normal conditions of load and under the best conditions. It is probable that very large turbines have a slight advantage over reciprocating engines (as generally built) in the matter of economy and the reverse of this statement appears to be true for most small units, although very economical turbines have been produced.
The turbines, however, generally give a flatter water-rate curve than does a reciprocating engine; that is, for
248
STEAM POWER
loads each side of the most economical the steam per horse-power hour does not increase above the value attained at most economical load as rapidly in the case of turbines as it does in the case of most reciprocating engines. With a very variable load, therefore, or with a load which is far removed from the rated value, the turbine probably gives a better average performance than does the reciprocating engine. This is particularly true when large turbines are used.
It has been shown that the turbine operates on the complete-expansion cycle and it will be remembered that the reciprocating engine operates on a cycle with incomplete expansion. The turbine is therefore able to make better use of very low-pressure steam than can the piston type.
Trial on a T9 or Mollier chart will show that a turbine receiving steam at 300 lb. pressure and expanding it down to a vacuum of from 28 to 29 ins. should make available as much work as one receiving steam at a high boiler pressure and expanding down to atmospheric. In other words a drop of 100 lbs. or more above atmospheric pressure makes no more energy available than does a drop of about 150 lbs. above atmospheric pressure. For this reason the pressure the more heat is converted into work by a given pressure drop. A small decrease in back pressure (terminal or condenser pressure) is therefore very effective in the case of turbines. Tests show that an increase of one inch of vacuum will cause an increase of economy of from 3 to 10 per cent, depending upon the type of turbine and upon other factors.
Experience has shown that reciprocating engines are fully the equal of turbines in the high-pressure ranges, in many cases they are even superior, but the turbine is far superior in the low-pressure region and in cases where very great ratios of expansion are to be used. Advantages have been taken of this fact by building large turbines for high-pressure steam by constructing mixed plants, recipro- cating engines being used for expanding down to the neigh-
THE STEAM TURBINE
borhood of atmospheric pressure and turbines expanding the steam exhausted by these engines to the lowest vacuum which can be maintained economically. This system has been found particularly useful for increasing the capacity of a reciprocating-engine plant. The capacity of such a plant can often be almost doubled without any increase in boiler capacity by simply inserting turbines into the exhaust lines between the engines and the condensers, and then arranging the pressures so that the turbines carry the expansion from about 16 lbs. absolute down to a vacuum of from 28 to 29 ins. Turbines used in this way are called low-pressure or exhaust-steam turbines.
Superheat is also very effective in bettering turbine economy, every ten degrees of superheat generally causing a saving of about 1 per cent in the weight of steam required per horse-power.
The steam turbine is generally cheaper than the reciprocating engine of like capacity if the conditions of operation permit the use of the high rotating speed characteristic of the turbine. It is therefore extensively used for direct connection to blowers, centrifugal pumps and electrical machines. Most of the large electric generating stations which have been installed within the past few years have used turbines to drive the generators, and single units direct connected to 20,000 K.W. generators are now numerous. Units rated at 45,000 K.W. are also being installed and units of still larger capacity have been designed.
PROBLEMS
1. A steam turbine produces one horse-power hour at its shaft for every 30 lbs. of steam supplied. The initial pressure is 200 lbs. absolute and the steam is superheated 200° F. The turbine exhausts at 15 lbs. absolute.
Find the thermal efficiency on the assumption that heat of liquid at exhaust temperature is chargeable to the turbine.
2. Develop a complete expansion cycle for one pound of material used under the conditions of Prob. 1 and find the energy
250 STEAM POWER
made available per cycle. From this value determine the number of pounds of material theoretically required per horse-power hour and compare with the value given in Prob. 3.
3. Find the additional quantity of energy which would theoretically be required to raise the steam in above problems if the back pressure could be lowered to 1 lb. absolute.
4. Develop a complete expansion cycle from an initial pressure of 225 lbs., initial temperature of 600 deg. F., and final pressure of $\frac{1}{4}$ lb. absolute. Assume that this is to be divided up into six parts, each making available the same quantity of energy. Find the pressure at each stage and state that this is most easily done with the help of the Mollier chart.
5. A steam turbine receives steam at a pressure of 225 lbs. per square inch absolute and a temperature of 600 deg. F., and exhausts into a condenser in which a pressure of $\frac{1}{4}$ lb. per square inch absolute is maintained. The turbine is direct connected to an electric generator and produces a K.W.-hour on 12 lbs. of steam. If a K.W.-hour is worth $10$ dollars, what is the efficiency of the combination?
6. Develop a complete expansion cycle for the conditions of Prob. 5, assuming that 1000 pounds of steam which would be required theoretically to develop energy equivalent to 1 K.W.-hour. Compare with the value given in Prob. 5.
7. Determine the velocity attainable by expanding steam in one step from the initial to the final condition of Prob. 5 above. What would be the value of the kinetic energy of such a jet per pound of steam flowing?
8. Determine the area of nozzle required to discharge 1000 lbs. per horse-power hour, initial condition being 100 lbs. per square inch absolute, and dry saturated steam; final pressure being 2 lbs. absolute.
9. Determine velocity and kinetic energy of jet in Prob. 8.
CHAPTER XIV
CONDENSERS AND RELATED APPARATUS
109. The Advantage of Condensing. The lowest pressure which can be used in a steam-engine cylinder, that is the exhaust pressure, is determined by the pressure prevailing in the space into which the steam is exhausted. With a given initial pressure the amount of work which can be ob-
![]() Fig. 166.
tained theoretically from a given weight of steam increases as the exhaust or back pressure decreases, as shown by the areas of the two diagrams in Fig. 166, and experience has shown that at least a part of this theoretical increase can be obtained in actual engines. It is therefore desirable to ex-
haust into a space in which the lowest possible pressure exists when the work obtained per pound of steam is the only consideration.
The most available space into which an engine can
254
252
**STEAM POWER**
exhaust is that surrounding the earth and already occupied by the earth's atmosphere. The pressure in this space is approximately equal to 14.7 lbs. per square inch at sea level and is due to the weight of the atmosphere. Since the surface of the earth slopes downward, its pressure decreases as one moves upward, its weight also decreases and atmospheric pressure therefore averages less than 14.7 lbs. per square inch at high altitudes and has a greater average value at points below sea level.
If it is desired to exhaust into a pressure lower than atmospheric, a vessel must be designed having such an abnormal pressure within some sort of vessel must be devised. It is the purpose of a condenser and its associated apparatus to make available a space in which such a low pressure can be maintained. Its method of operation will be considered in later sections.
There is another advantage which may be obtained by the use of a condenser. It often happens that the water available is not well adapted to use in boilers. It may be salt water as in marine practice, or it may contain a number of undesirable gases and solids in solution as often occurs in stationary practice. Some types of condensation apparatus are so arranged that they remove these impurities from the engine water before it enters into liquid water without admixture and can therefore be returned to the boiler as practically pure water, thus largely eliminating the troubles that would ensue from the use of poor feed water.
**110. Measurement of Vacuums.** It appears that some non-volatile liquids, that is liquids that did not vaporize, could be found in such a way that it contained no gases in solution. If a long tube were inserted in a vessel filled with such a liquid and had its upper end connected with some form of vacuum pump which could remove air from its interior, as shown in Fig. 167, liquid would rise in the tube as the air was removed. Removal of air would result in lowering the pressure within the tube, but the constant
![]() A diagram showing a long tube inserted into a vessel containing a non-volatile liquid. The upper end of the tube is connected to a vacuum pump.
CONDENSERS AND RELATED APPARATUS 253
atmospheric pressure on the liquid surface outside the tube would then force liquid up the tube to such a height that the pressure $p_a$ of air in the tube plus the pressure due to the column of liquid of height $h$ within the tube just equals the pressure due to the atmospheric pressure on the surface of the liquid in the vessel. If the pump could remove all of the air from the tube, liquid would rise to such a height that the pressure exerted by it on a plane passing through its surface at this height was equal to that of the external atmosphere.
The same result could be attained by using a tube closed at one end, filling it with the liquid, and allowing it to rest in the liquid as shown in Fig. 167.
If the tube were long enough, the liquid would drop to some such point as shown, under which conditions the half-column of liquid would be at atmospheric pressure. This would only be true if the liquid did not volatilize and did not contain gases in solution; with these assumptions the space above the liquid in the tube would contain absolutely nothing. This space would be said to be perfectly vacuous, or a perfect vacuum.
A device of this character used to measure the pressure of the atmosphere and is known as a barometer. Mercury is used as the liquid because its high density makes it possible to use a short tube and because it may be considered
![]() A diagram showing a barometer. The top part is a U-tube with a small tube attached to one leg. The bottom part is a container filled with mercury.
Fig. 167.
254 STEAM POWER
as non-volatile at ordinary temperatures. The average atmospheric pressure at sea level, equal to 14.7 lbs. per square inch approximately, can support about 30 ins. of mercury, so that this figure is generally taken as the standard atmospheric pressure. A barometer reading of less than 14.7 lbs. would be shown by a barometer reading of less than 30 ins.; a greater atmospheric pressure by more than 30 ins. Corresponding values of atmospheric pressure and barometer reading are given in Table VII.
To this have also been added the altitudes to which different pressures existed. For example, a pressure of 14.7 lbs. existed at sea level and there were no variations of atmospheric pressure excepting those due to change of elevation. Values of this type can only be roughly approximate, because local barometric variations are constantly occurring and the sea-level atmospheric pressure varies both sides of 14.7 lbs.
**TABLE VII**
| Barometer, Inches of Mercury | Atmospheric Pressure Found at Sea Level | Altitude Feet (Approximate) |
|---|---|---|
| 25.00 | 12.27 | 4750 |
| 26.00 | 12.76 | 3250 |
| 28.50 | 13.01 | 3250 |
| 27.00 | 13.25 | 3250 |
| 27.50 | 13.49 | 2250 |
| 28.00 | 13.74 | 1300 |
| 28.50 | 13.98 | 450 |
| 29.00 | 14.23 | Ses level |
| 29.25 | 14.35 | --450 |
| 29.50 | 14.47 | --900 |
| 29.75 | 14.60 | --900 |
| 30.00 | 14.72 | Ses level |
| 30.25 | 14.84 | --900 |
| 30.50 | 14.96 | --900 |
| 30.75 | 15.09 | --900 |
| 31.00 | 15.21 | --900 |
(Negative signs mean distance below sea level.)
CONDENSERS AND RELATED APPARATUS 255
The exact value of standard atmospheric pressure at sea level is taken at 29.621 ins. of mercury, which is equal to 14.606 lbs., per square inch and corresponds to the 760 mm. of mercury, used by scientists as standard.
A tube with both ends open and arranged as shown in Fig. 167 can be used to measure the degree of vacuum existing in the space to which its upper end is connected, and may be employed for this purpose in the following simple, using mercury as the liquid. The extent to which the pressure is lowered in the top of the tube is indicated by the height to which the mercury column rises and this height in inches is used as a measure of the vacuum. Thus if a perfect vacuum were created and if the atmospheric pressure were reduced to zero, the mercury would show about 30 ins. of mercury above the level in the reservoir. If the vacuum were less perfect the gauge would show a shorter column.
It should be noted that the reading of the vacuum gauge does not give the pressure existing in the various space but gives the amount by which the pressure has been reduced below that of the atmosphere, the difference being expressed in inches of mercury. By subtracting this reading from the existing atmospheric pressure expressed in the same units, the absolute pressure in the partially vacuous space (expressed in inches of mercury) is obtained.
It is obvious therefore, that a vacuum-gauge reading of say 20 ins. of mercury represents a partial vacuum of the same absolute pressure. With a barometer reading of 28 ins. it would represent a perfect vacuum; with a barometer reading of 30 ins. it would represent a partial vacuum, the absolute pressure in the partially vacuous space being equal to 30 ins. of mercury.
114. Conversion of Readings from Inches of Mercury to Pounds per Square Inch. It is often necessary to convert readings of pressure in inches of mercury into pounds per square inch. This can be done with sufficient accuracy
256
**STEAM POWER**
under ordinary circumstances by multiplying the inches of mercury by the constant 0.4908. Thus,
Barometer in inches × 0.4908 = atmospheric pressure in pounds per square inch . . . (74)
and
(Barometer in inches - vacuum gauge in inches) × 0.4908 = absolute pressure in partially vacuous space in pounds per square inch. . . (75)
**ILLUSTRATIVE PROBLEM**
A vacuum gauge constructed like that shown in Fig. 167 reads 27 ins. when the barometer reads 20.5 ins. What is the absolute pressure in the partly vacuous space of mercury?
The absolute pressure is equal to $29.5 - 27 = 2.5$ ins. of mercury, which is equal to
$$2.5 \times 0.4908 = 1.227 \text{ lbs. per square inch}.$$
**112. Principle of the Condenser.** A perfect vacuum could be created in any closed vessel with impenetrable walls if a pump could be devised which could remove all material contained within that vessel. Or, any degree of vacuum can be maintained in any partially closed vessel by fitting it to a pump which can remove all material flowing into the vessel as fast or faster than it enters, raise the pressure of this material to atmospheric or higher and discharge it.
The latter principle is made use of in real condensers, a pump of some form, or an equivalent, removing from the condenser the material exhausted by the engine and leakage from the atmosphere, and discharging it at atmospheric pressure, thus maintaining a partial vacuum within the desired vacuum. If the condenser and connections can be made leak proof, the pump or equivalent has to handle only the material exhausted from the engine.
A steam engine exhausts a mixture of steam, water and gases, the gases being a mixture of those originally
CONDENSERS AND RELATED APPARATUS 257
dissolved in the boiler-feed water and air which leaks into those parts of the system in which a partial vacuum is main-
tained. If the pump had to handle the same volume of material as is exhausted by the engine, no gain of work would result from condensing, because the pump would have to do at least as much work in raising the pressure of this exhaust steam as in raising the air, as could be obtained by allowing it to expand in the engine.
Steam, however, occupies a much larger volume than water at the same temperature and pressure. Thus steam at 212° F. occupies a volume of about 26.70 cu.ft. per pound, but water at the same temperature and pressure occupies only 0.0163 cu.ft. per pound; at a temperature of 120° F., which is often used in condensers, the specific volume of steam is about 203 and that of water only 0.0102. Therefore, if the steam is condensed after exhaust from the engine and before entering the pump to be discharged to atmosphere, the pump work is greatly reduced. The work required to compress the steam is negligible in comparison with the volume of steam exhausted, and the work of pumping it is almost negligible in compar-
ison with the work it made available in the engine.
Gases contained in the exhaust steam cannot be lique-
fied and must be pumped as gases. The work required to pump them can, however, be reduced by lowering their temperature as far as possible.
The condenser equipment may be regarded as consist-
ing of a combination of a partially closed vessel and some form of pump. The vessel is so constructed that a low temperature can be maintained within it and that large quantities of heat can be removed from it for the purpose of cooling down the exhaust steam and condensing the contained gases. This is generally done by using large quantities of cool water.
The absolute pressure within the condenser is made up of two parts. The two parts are, (a) that due to the
258
STEAM POWER
water vapor, since the space over the condensed water will always be filled with saturated steam at the same temperature (approximately) as that of the water, and $(b)$ that due to any gases present.
The pressure of the saturated steam (water vapor) can be found from the steam tables opposite the temperature existing in the condenser and it is the pressure which would exist in the condenser of an ideal system in which no gases were mixed with the working substance. The pressure of the gases can be found by subtracting from the total measured pressure in the condenser the pressure exerted by the water vapor when seen on the steam table. The pressures exerted by the water vapor and gases are spoken of as partial pressures, since their sum makes up the total pressure within the condenser.
The presence of gases causes a two-fold loss. First, it increases the pressure against which the engine has to exhaust, thus raising the back-pressure line on the diagram and decreasing its work. Second, it increases the work which must be done by the pump which otherwise would only pump the condensate and such saturated water vapor as accompanied it.
113. Types of Condensers. The condensers actually used in steam plants can be roughly divided into two types, as
$(a)$ Contact condensers and
$(b)$ Non-contact condensers.
In the first type the water which is used for condensing and cooling is intimately mixed with the exhaust from the engine within the condensing vessel, and the resultant mixture of condensing water, condensate and gases is drawn out of this vessel and discharged to atmosphere by the pump.
In the second type condensing water flows on one side of metal surfaces of some sort and the exhaust is led over the other side, the heat being transmitted through the
CONDENSERS AND RELATED APPARATUS 250
metal. In condensers of this type the condensate and gases are not mixed with the condensing water and the condensate can therefore be returned to the boiler as feed water with the advantages already mentioned.
**114. The Jet Condenser.** One of the earliest forms of contact condensers which is still very widely used for moderate pressures is that commonly known as the jet condenser. The principle of operation of the jet condenser is shown in Fig. 160. Water, under pressure, entering as indicated, is broken up into fine streams and carried into the exhaust from the engine. The resultant mixture flows downward into the neck of the condensing vessel or and is removed by some form of pump. This pump handles gases, vapors and water and is known as a vacuum pump, a wet-cumsum pump, or a wet-air pump, the term wet signifying that it handles the water as well as the gases.
The pressure within such a condenser head would be theoretically equal to that corresponding to the temperature of the resultant mixture if no gases were present. In practice the pressure of the water vapor would roughly correspond to the average temperature of the gas in the vessel and there would be a partial pressure due to gas as well. This gas would consist of that brought over by the engine exhaust plus that released from the condensing water under the low pressure within the condenser.
![]() Fig. 160.—Jet Condenser.
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STEAM POWER
Details of a complete jet condenser and of the method of connecting it to an engine are given in Fig. 170. The atmospheric relief valve is installed in all condensing systems and is arranged to open automatically to atmosphere if the pressure within the system rises to atmospheric, that is, if the " vacuum is lost."
![]() A diagram showing the operation of a steam condenser with a jet condenser. The diagram includes a water pump, a steam pipe, a condenser, and a steam pipe. The diagram also shows the connection between the steam pipe and the condenser.
Fig. 170.—Jet Condenser and Method of Connecting to Engine.
With the jet condenser the pressure might start to rise because of slow action or even stoppage of the pump. As the condenser head filled up the rising water would ultimately entirely cover the jet and condensation would then practically cease. In the arrangement shown in Fig. 170 there is an additional safety device which breaks the vacuum in the exhaust system if the water in the head
CONDENSERS AND RELATED APPARATUS 261
rises above a certain height, thus preventing the external atmospheric pressure from forcing this water back along the exhaust pipe and into the cylinder, an event which would probably result in a wrecked engine.
The condenser shown in Fig. 3 is known as a parallel-flow type, because everything within the condensing vessel flows in the same direction. The gases and vapors handled by the pump theoretically have the same temperature as that of the mixture with which they flow out at the bottom of the condenser head. The temperature of this mixture therefore determines the temperature of the gases and vapors pumped.
There are numerous forms of contact condensers which more or less closely resemble the types of jet condensers just described. They are occasionally all classed as jet condensers, but more often are given distinguishing names.
One very common form of contact condenser is generally known as a barometric condenser. This consists essentially of a condenser head, similar to that used with the jet condenser already described, and a tail pipe or barometric pipe which partly or wholly takes the place of the wet-vacuum pump by removing part or all of the mixture formed within the condenser. One model of such a condenser is shown in Fig. 4.
The exhaust from the engine enters the head through the large pipe shown and divides into two parts, one part passing down through the center of the head and the remainder flowing downward in the annular space $A$. The condensing or injection water enters as shown and is divided by the spraying nozzles injected into the upper exhaust, which is in the central tube of the condenser. The mixture thus formed flows downward and finally meets the discharge from the lower end of the annular space $A$, which is then condensed. The mixture of injection and condensing water together with such gases as have been entrapped, then flows downward into the tail pipe, which is
262
STEAM POWER
over 34 ft. in length and which dips into the " hot well" at its lower end. As atmospheric pressure can only sup-
![]() A detailed diagram of a barometric condenser.
To Vacuum or Steam
Air Cooler
Exhaust
Spraying Cone
Inertial Water
Drain
Tail Pipe
Fig. 171. -Barometric Condenser.
port a column of water about 34 ft. high, the tail pipe forms an automatic wet-vacuum pump, water flowing from it as rapidly as it accumulates within it.
CONDENSERS AND RELATED APPARATUS 263
![]()
A diagram showing the method of connecting the steam engine to the condenser.
Fig. 172—Method of Connecting Hydrometric Condenser to Steam Engine
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STEAM POWER
Experience has shown that the maintenance of a high vacuum with this type of condenser depends upon the ex-
tent to which gases are removed from the condenser head.
These gases are generally called air; as the greater part of
them is water vapor, they are removed by the action of air,
as is
trapped by the descending mixture rises through the
hollow spraying cone, then
through the air cooler and
flows out through the pipe
indicated to the vacuum or
dry-seal pump. The air in
rising through the center of
the spraying cone is cooled by the water flowing around it, and it is further
cooled by coming into con-
tact with water as it works its way through the air cooler. This results not only in lowering its tem-
perature, but also in caus-
ing the condensation of a
great deal of water vapor accompanying it.
This condensed vapor col-
lects in the space surround-
ing
the air cooler and
flows down into the head
through the drain shown. The vacuum pump, therefore,
handles cool gases containing little water vapor and prac-
tically no liquid water. It is sometimes called a dry-air
pump or dry-vacuum pump for this reason.
The entrainer shown in the exhaust system in Fig. 172 is so shaped that water collecting in the exhaust piping and flowing into the entrainer is picked up by the exhaust steam and carried into the condenser.
![]()
A diagram showing the operation of a Barometric Condenser. The atmospheric relief valve allows air to escape when necessary. The exhaust pipe carries hot gases away from the condenser. The entrainer collects water vapor and liquid water from the exhaust piping and pumps it into the condenser.
FIG. 173.—Baropanath Barometric Condenser.
CONDENSERS AND RELATED APPARATUS 265
The flow of steam and injection water in this condenser is parallel, but the material on its way to the dry-vacuum pump flows upward and the cooling water flows downward so that counter-current flow is used in this part of the appa-
![]()
A diagram showing the operation of a baradwanath condenser. The left side shows the ordinary setting, with the exhaust pipe leading to the condenser. The right side shows the syphon method, with the water from the tank or flume being drawn into the condenser by the vacuum created.
SYPHONING ITS WATER FROM TANK OR FLUME
BARADWANATH CONDENSER ORDINARY SETTING
![]()
(a) A diagram showing the ordinary setting of a baradwanath condenser. The exhaust pipe leads to the condenser.
(b) A diagram showing the syphon method of a baradwanath condenser. The water from the tank or flume is drawn into the condenser by the vacuum created.
Fig. 174.
ratus. This has the advantage of bringing the air leaving the condenser into contact with the cooling water just as it enters and therefore when it has its lowest temperature.
A somewhat similar condenser, arranged so that it requires no pump, is shown in Figs. 173 and 174 (a) and (b).
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STEAM POWER
Exhaust and injection water mix as shown, the quantity of injection water being regulated by the hand wheel on top of the condenser. The mixture flows downward through the narrow neck and the velocity attained in this part of the tail pipe is so high that all air and similar gases are swept along with the current.
For starting, the valve T in Fig. 174 (b) is opened, allowing water to flow into the lower part of the tail pipe. This creates a partial vacuum and atmospheric pressure when forces water up the injection pipe and into the condenser head. The valve T is then closed and the condenser continues to supply its own water. Because of this action this type is often called a siphon condenser. By supplying a circulating pump as in diagram 174 (a) it can be converted into a barometric condenser similar to the type already discussed except for the fact that it requires no air pump.
The exhaust or tail pipe of any barometric condenser can be replaced by a hand of a pump, and centrifugal pumps are often used for this purpose. When large quantities of gas are to be handled, as when a dry-air pump is not used, the centrifugal pump must be specially designed.
![]() Fig. 175.—Westinghouse-Leblanc Air Pump.
174
CONDENSERS AND RELATED APPARATUS 267
A recently developed type of condenser in which the barometric tube is replaced by a centrifugal pump and in which a separate air pump of a rotary type is used is illustrated in Figs. 175, 176 and 177. It consists essentially of the condensing head and well, combined with a centrifugal tail pump and a rotary air or vacuum pump as indicated.
![]() A diagram showing the components of a condenser system.
**Fig. 176.—Westinghouse-Leblane Condenser.**
In Fig. 177, Injection water entering through nozzles in the head meets the exhaust, and the resultant mixture flows down into the well through the large nozzle shown. The liquid is continuously removed from the bottom of this well by the centrifugal tail pump and discharged to the hot well. The air and associated vapors are drawn down the air pipe and discharged by means of the device shown in
175
208
STEAM POWER
Fig. 175. Water enters the central part of this pump as indicated in Fig. 176 and is discharged through the stationary nozzles $N$ and the moving vanes $V$ shown in Fig. 175. The water is thus caused to form a series of "pistons" which move rapidly downward and which trap small plugs or lamina of "air" between them and the "air" to the atmosphere. The connection marked $P$ is used for priming at starting when necessary.
In small units the centrifugal tail pump may be omitted and the design so remodeled that all the injection water passes through the rotary air pump which discharges the entire mixture from the condenser just as it discharges the air and associated vapors in the larger sizes.
**L16. Non-contact Condenser.** The typical surface condenser is the best-known example of non-contact condenser. It consists essentially of a large cylindrical or rectangular vessel whose exhaust is discharged and through which pass numerous bronze or alloy tubes which carry the condensate to its surfaces of which act as the condensing and cooling surfaces.
One form of surface condenser mounted above the pumps which serve it is shown in Fig. 178. The exhaust enters at the top of the rectangular shell and works its
![]() Diagram showing a steam engine with a water pump attached to it.
**Fig. 177.—Westinghouse-Leblanc Condenser.**
CONDENSERS AND RELATED APPARATUS 269
![]()
A detailed drawing of a factory's water-tube boiler.
The diagram shows the following components:
- Water Tube
- Steam Drum
- Air Preheater
- Air Blower
- Fan
- Condenser
- Water Out
- Condensate Pump
- Circulating Pump
- Steam Turbine
- Steam Generator
Fig. 170—Whirlpool Admiralty Type Surface Condenser.
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STEAM POWER
way down over the water-cooled tubes. The condensate, mixed with gases and vapors, is drawn from the bottom of the shell by the wet-vacuum pump and discharged to the hot well.
The condensing water is forced through the tubes of the condenser by means of the reciprocating circulating pump, entering the lower tubes at the right-hand end in the figure, making two " passes" through the condenser and leaving at the top. Because of the path of the water a condenser of this type is sometimes called a two-pass or double-flow condenser.
With this arrangement illustrated, the steam which condenses upon the upper tubes falls as a rain from tube to tube until it finally settles at the bottom and is drawn off. The outer surfaces of the lower tubes are therefore practically covered with water and this has two disadvantages. First, these tubes carry the coolest circulating water and they therefore cool the condensate coming in contact with them. Second, the water that flows through them is unnecessarily heated. Cooling of the condensate means a lower hot-well temperature than would otherwise be obtained, but if the condensate is to be used for boiler feed, the temperature of water in the hot well should be maintained as high as possible. In such cases, boilers will have to be heated to boiler temperature with a corresponding expenditure of heat. Second, tubes which are being used to cool water covering them are of little use as condensing surface, and hence a large part of the surface in such a condenser is comparatively inactive.
The ideal arrangement would carry away the liquid condensate as fast as possible leaving the tubes first entered by the condensate at its coldest as the final condensing and cooling surfaces, thus bringing gages and non-condensible vapors into contact with the coolest surfaces just before entering the vacuum pump. Numerous designs which approximate this ideal have been developed recently and
CONDENSERS AND RELATED APPARATUS 271
they give better results than do the earlier and simpler forms. The improvement is shown by the values of con-
densing surface per developed horse-power of engine. In early designs it was customary to supply 2J sq.ft. of tube
surface per horse-power. Some of the most recent
installations are giving better vacuum with only 1 sq.ft.
per horse-power.
One of the newer models passes the condensate through
a set of tubes so located that the engine exhaust strikes them before impinging on any tubes carrying condensing water. This results in a partial condensation of the exhaust
and raising the temperature of the water entering the
tubes to very near that of the exhaust, thus heating the
boiler feed to a temperature practically corresponding
to the exhaust temperature of the engine.
Surface condensers are commonly operated with a
vacuum of from 24 to 26 ins. of mercury when used with
reciprocating engines, but with steam turbines of 28 to 29 ins.
when receiving the exhaust of steam turbines. When
operated at the lower vacuum wet-pump vacuums are gen-
erally used, but the best types of dry-air pumps must be
installed in combination with well-designed condensers
when the higher vacuums are sought.
**11. Maximum Condenser Load.** Condensers. The weight of circulating water required varies with the type of condenser and with the conditions of operation, such as initial temperature of water, vacuum desired, etc. It can be determined approximately by calculation and the values thus found must then be increased by such factors as experience has shown to be necessary.
In each case the injection water and the condensate are discharged as a mixture and therefore have the same average discharge temperature.
Let $t_1$ = initial temperature of injection water in F.*,
$t_2$ = temperature at which mixture is discharged in F.*,
![]() A diagram showing a schematic of a condenser system.
272
STEAM POWER
\lambda = \text{total heat above 32° F. of steam as exhausted};\\
W = \text{pounds of injection water per pound of exhaust steam}.
Assuming the exhaust steam to be dry saturated, each pound of steam in condensing to water at a temperature of $t_2$ degrees must give up an amount of heat equal to $\lambda$ minus the heat of the liquid at $t_2^\circ$ or roughly $(\lambda - (L_f + L_g))$ B.t.u. This heat is supplied by the injection water, while its temperature rises from $t_1$ to $t_2$ degrees. Each pound of water can then absorb approximately $(t_2 - t_1)$ B.t.u., and the pounds of injection water per pound of steam will be
$$W = \frac{\lambda - (L_f + L_g)}{t_2 - t_1} \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \\ W =\frac{\lambda - (L_f + L_g)}{t_2 - t_1}.\quad\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\quad(78)$$
The value of $t_2$ would be that corresponding to the absolute pressure in the condenser if it were not for the air and similar gases which exert some pressure. It is generally 10 or more degrees F. below the temperature corresponding to the vacuum. Values of $t_2$ in the neighborhood of 110° to 125° F. are customary with reciprocating engines and values as low as 80° are used with high vacuum engines.
The weight of water used per pound of steam as given by Eq. (78) will vary between about 15 for very low initial and moderate discharge temperature to about 50 with average initial and moderate discharge temperature. Experience shows that it is necessary to add 10 per cent or more to the values of $W$ obtained from equation (78) to obtain the weight of water which will probably be used.
**ILLUSTRATIVE PROBLEM**
Find the quantity of water theoretically required per pound of steam condensed in a contact condenser in which a vacuum of 25.5 ins. of mercury is maintained when the barometer reads
CONDENSERS AND RELATED APPARATUS 273
29.5 ins. of mercury. The initial temperature of the water is 60° F.
The absolute pressure in the condenser is $29.5 - 25.5 = 4.0$ ins. of mercury and the steam tables give for this pressure, $\lambda = 1115.0$ and $t_0 = 120$. Substituting in Eq. (78) gives
$$W = \frac{1115.0 - 120 + 32}{120 - 60} = 15.5 \text{ approximately}.$$
**117. Weight of Water Required by Non-contact Condensers.** In the case of non-contact condensers there is no definite relation between the discharge temperature of the cooling water and that of the condensate. Experience shows that the discharge temperature of the circulating water is generally from 15 to 20 degrees lower than the temperature corresponding to the vacuum.
The temperature of the condensate (hot-well temperature) is generally 15 or more degrees below that corresponding to the vacuum, but good design makes the hot-well temperature very closely approximate that corresponding to the vacuum.
Assuming
$t_1 =$ initial temperature of injection water in F.;,
$t_2 =$ final temperature of injection water in F.;,
$t_c =$ temperature at which condensate is discharged, i.e., hot-well temperature, in F.;,
$\lambda =$ total heat above 32° F. of steam as exhausted,
and
$$W = \text{pounds of injection water per pound of exhaust steam}.$$
The weight of water which must be circulated per pound of steam can be found as in the case of the contact condenser. It is given by
$$W = \frac{\lambda - t_c + 32}{t_2 - t_1}. \quad \quad \quad (79)$$
274
STEAM POWER
Values in the neighborhood of 25 lbs. of water per pound of steam are common with low vacuums and 50 or more pounds are often used with vacuums over 28 ins. of mercury.
ILLUSTRATIVE PROBLEM
A surface condenser receives circulating water at a temperature of 65° F. and discharges it at a temperature of 80° F. To maintain a vacuum of 28.0 ins. of mercury at a barometer at 29.5, and the temperature of the condensate discharged to the hot well is equal to 55° F. Find the quantity of circulating water theoretically required.
This vacuum corresponds to an absolute pressure of 20.5-28.0=1.5 ins. of mercury. Assuming this all due to steam (neglecting the effect of air), the value of W may be found from the steam table as 1100.1 B.L.U. Substitution in Eq. (79) then gives
$$W = \frac{1100.1 - 83 + 32}{80 - 65} = \frac{169.9}{15} \approx 11.3 $$
approximately.
118. Relative Advantages of Contact and Surface Condensers. The contact types are as a rule much cheaper than the surface condensers, and they are less subject to serious depreciation, the tubes of surface condensers often corroding seriously in short intervals of time. On the other hand, the injection of the cooling water into the condensing space in contact types results in the introduction of large quantities of dissolved gases, and much of this material is liberated under the reduced pressure, thus tending to increase the condenser pressure, that is, decrease the vacuum. When pumps are used to remove this gas, they must work against the resistance offered by the pump itself, and also against the resistance offered by the corresponding pump in a surface condenser, and the work of pumping this water out of the vacuum into the atmosphere combined with the additional work required of the pump which handles the "air" may partly balance the advantage of lower first cost of the contact type.
CONDENSERS AND RELATED APPARATUS 275
A surface condenser must always be installed where it is desirable to use the condensate as boiler feed, and it is generally used when very high vacuums (low absolute pressures) are required in the turbine. The surface condenser is at a serious disadvantage, however, since it is required to handle the exhaust of reciprocating engines. The exhaust from such engines always contains large quantities of lubricating oil carried out of the cylinder, and unless this material is separated before the exhaust enters the condenser, it deposits on the surfaces of the tubes and decreases the conductivity of these surfaces. Such oil can be eliminated to a great extent before the exhaust enters the condenser by means of oil separators, which are generally made up of a series of baffles upon which the steam impinges and upon which the oil is caught.
**118. Cooling Towers.** The large quantity of circulating water that may be condensed here is often an item of great, economic importance. When such plants are located near a river or near tide water, the circulating water can generally be procured for the cost of pumping. When they are located in the middle of cities or in regions where water is scarce, the cost of water may be excessive or it may be unavailable. In either case a continuous supply equal to the demand of the condensers.
In such cases the condensing water is often circulated continuously, being cooled after each passage through the condensers. This cooling is generally done by exposure of a large surface to the air. The resulting evaporation cools some of the water, and part of it is condensed. The heat of vaporization cools the remainder so that it can be used again. This sort of cooling may be effected by running the water into a shallow pond of large area, or by spraying it into the air over a small pond or reservoir or by passing it through a cooling tower.
Cooling towers are large wood or metal towers generally filled with some form of baffling devices. The hot water
276
STEAM POWER
is introduced at the top and spread into thin sheets or divided up into drops as it descends. Air enters at the bottom and flows upward, cooling the water by contact and by the partial evaporation which results. The circulation may be forced by fans located in the base of the tower, or the difference of temperature between the air inside and out, in which case a stack is fitted to the top of the tower; or the circulation may be forced by fans located in the base of the tower. In the latter case the apparatus is called a *forced-draught cooling tower*.
![]() A diagram showing a steam power plant with a cooling tower.
CHAPTER XV
COMBUSTION
120. Definitions. Certain substances are known to chemists as compounds, because they can be separated by chemical processes into simpler substances. Thus water and many of the most familiar materials known to man are compounds which can be separated into two or more simpler materials.
Those substances which cannot be further broken up by the processes used in separating compounds are called elements; they are regarded as elemental, as the stones of which the compounds of nature are built up. About seventy-five of these elements are known, but many of them are comparatively rare. Pure mercury contains all elements; the oxygen and nitrogen which are mixed to form the greater part of the atmosphere are elements; carbon, which forms the greater part of most fuels, is an element.
In many cases the combination of elements to form compounds is very useful, and some of these combinations are used by the engineer for the purpose of obtaining heat in large quantities. When the elements which occur in fuels, such as coal, wood and petroleum, combine with oxygen, the process is spoken of as combustion. The quantity of heat liberated when a pound of any material combines with oxygen (burns) is called its heat value. This is a measure of the material.
Fuels contain a great number of elements, but only three of these ordinarily take part in combustion and are therefore spoken of as combustibles. They are carbon, hydrogen and sulphur. The sulphur content is generally
277
278
**STEAM POWER**
very small, and the carbon and hydrogen are therefore the most important constituents.
The combustion of each of these elements will be considered in detail in the following section, but before this can be done two other ideas must be developed.
The smallest particle of an element which can be conceived of as entering into combination to form a compound is known as an atom of that element. It has been found that the atoms of each element have an invariable and characteristic weight. The lightest atom is that of hydrogen, and its weight is considered unity. The atom of carbon is therefore said to have an atomic weight equal to twelve. Similarly the atomic weight of nitrogen is fourteen and that of oxygen is sixteen.
The simplest particle which can be formed by the combination of two atoms is a molecule. Like or unlike atoms may combine to form molecules. Thus two hydrogen atoms combine to form a molecule of hydrogen, and hydrogen gas as it ordinarily exists may be pictured as made up of a collection of such molecules. Similarly, gaseous oxygen and gaseous nitrogen may be pictured as collections of molecules which are made up of two like atoms.
When unlike atoms combine to form a molecule, they form a molecule of a compound. Obviously a molecule of any compound is the smallest particle of that compound which can exist.
For convenience, the different elements are represented by abbreviations; thus oxygen is represented by O, nitrogen by N, hydrogen by H, carbon by C and sulphur by S. When these abbreviations are written in chemical equations, such as will be given later, they stand for an atom of the substance. Hence O in a chemical equation would mean one atom of oxygen. The symbol O₂ is used to mean two atoms of oxygen in combination, hence, one molecule
![]() A diagram showing the structure of a water molecule (H₂O).
COMBUSTION
279
of oxygen. The symbol 2O₂ means two groups of two oxygen atoms in combination, hence two molecules of oxygen.
The simplicity and elegance of this system will become apparent as the chemical equations which follow are developed.
121. Combustion of Carbon. Carbon can unite with oxygen or burn to form two different compounds—carbon monoxide (CO) and carbon dioxide (CO₂). The monoxide is formed by the combination of one atom of oxygen with one atom of carbon; the dioxide, by the combination of two atoms of oxygen with one atom of carbon. Thus, therefore, containing twice as much oxygen as does the monoxide.
Carbon burned to carbon monoxide has not combined with the largest possible quantity of oxygen, and combustion is therefore said to be incomplete in such cases. When, however, carbon dioxide is formed, the carbon has combined with as much oxygen as possible and combustion is said to be complete.
It will be shown later that much more heat is liberated when the dioxide is formed than when carbon burns to the monoxide. Hence, when liberation of heat is the object of combustion, the process should be so conducted as to result in the formation of the maximum quantity of dioxide and the minimum quantity of monoxide.
122. Combustion to CO. The combustion of carbon and oxygen to form the monoxide can be represented by the equation
C+O=CO,...(80)
or by the equation
2C+O₂=2CO,...(81)
The former is the simpler and will be considered first, but the latter is the more perfect and indicates more to the trained eye than does the simpler form.
The simple equation states that one atom of carbon combined with one atom of oxygen to form one molecule
280
**STEAM POWER**
of carbon monoxide. It can, however, be so interpreted as to show much more than this. The carbon atom is twelve times as heavy as the hydrogen atom, while the oxygen atom is sixteen times as heavy as that of hydrogen. The equation
$$C + O = CO,$$
therefore, shows that an atom which is twelve times heavier than the hydrogen atom, unites with one which is sixteen times heavier than the hydrogen atom to form a molecule which is $28 - 12 - 16$ times heavier than the hydrogen atom.
In other words, the weights of carbon and oxygen combining are in the ratio of $\frac{12}{3} : \frac{1}{1}$ or $4 : 1$. If a sufficient number of carbon atoms to weigh one pound be used, a quantity of oxygen weighing $1\frac{1}{4}$ lbs. will be required to combine with them to form carbon monoxide. The resultant carbon monoxide will contain the pound of carbon and the $1\frac{1}{4}$ lbs. of oxygen and will therefore weigh $2\frac{1}{4}$ lbs.
The same weight relations would hold irrespective of the weight of carbon used, and the simpler equation may therefore be put
$$1 \text{ weight of } C + 1\frac{1}{4} \text{ weights of } O = 2\frac{1}{4} \text{ weights of } CO.$$ (82)
**ILLUSTRATIVE PROBLEM**
To illustrate the use of this equation, assume that 9 lbs. of carbon are burned to carbon monoxide and that it is desired to find the weight of oxygen used, and the weight of the product. The weight of carbon used is 9 lbs., that is, $1\frac{1}{4} \times 9 = 12$ lbs. The weight of the product must be $2\frac{1}{4}$ times the weight of the carbon, that is $2\frac{1}{4} \times 9 = 21$ lbs.; or, it must be the weight of the carbon burned plus the weight of the oxygen used, that is, $9 + 12 = 21$ lbs.
In general, the oxygen used for combustion is obtained from the atmosphere, which may be considered as a mechanism.
COMBUSTION
281
ical mixture of oxygen and nitrogen in unvarying proportions. These proportions are roughly, 0.23 of oxygen to 0.77 of nitrogen by weight, or 0.21 of oxygen to 0.79 of nitrogen by volume, as shown in Table VII. The weight of air that contains one pound of oxygen is therefore
$$\frac{0.23}{0.23} = 4.35 \text{ lbs., and this weight of air contains}$$
$$4.35 - 1 = 3.35 \text{ lbs. of nitrogen.}$$
In the problem previously considered it was found that 12 lbs. of oxygen would be required to burn 9 lbs. of carbon to CO. The total weight of air required to obtain this amount of CO is $$12 \times 4.35 = 52.2 \text{ lbs. and it will contain}$$
$$52.2 - 12 = 40.2 \text{ lbs. of nitrogen.}$$
By simple arithmetical calculations of the type just given all the weight relations involved in the combustion of C to CO can be determined. The volume of air required in any given case can be found by multiplying the weight of air at the temperature and pressure given in Table VII. Thus, in the illustrative problem already considered, it was found that 32.2 lbs. of air would be required to burn 9 lbs. of C to CO. The volume of this air at 62° F, would be
$$52.2 \times 13.14 = 685 \text{ cu.ft.}$$
It is found that a quantity of heat equal to about
4800 B.L.U., is evolved for each pound of carbon burned to CO; that is the calorific value of C burned to CO is 4500 B.L.U.
Returning now to Eq. (81), which was said to be more useful than the simpler form given as Eq. (80), it will be necessary to consider a rather simple law of gases. It has been shown experimentally that equal volumes of all gases contain equal numbers of molecules when at the same temperature and pressure. This statement is known as Avogadro's Law.. It has also been shown that the molecules of gaseous oxygen contain two atoms.
The equation in question,
$$2C + O_2 = 2CO$$
282
STEAM POWER
can therefore be read, two atoms of carbon combine with one molecule of oxygen to form two molecules of carbon monoxide. But, if every molecule of O yields two molecules of CO it follows from Avogadro's law that the CO formed will occupy twice the volume of the oxygen it replaced at the same temperature and pressure. If the equation be imagined as containing a numeral 1 before the O₂, it will be obvious that the coefficients of the terms representing gas molecules give volume relations directly. This equation therefore gives both volume and weight relations.
TABLE VIII
Properties of Air
Considering it to consist only of nitrogen and oxygen. |
|
Relative Proportions. |
Ratio of N to O. |
Ratio of Air to O. |
|
Exact. |
Approx. |
|
Exact. |
Approx. |
|
Exact. |
Approx. |
| By weight. |
0.766 N |
0.77 N |
|
0.243 O |
0.23 O |
|
3.27 |
3.35 |
4.27 |
4.35 |
|
|
|
|
|
|
|
|
|
|
|
| By volume. |
0.701 N |
0.79 N |
|
0.209 O |
0.21 O |
|
3.75 |
3.76 |
4.76 |
4.76 |
| Spec. wt. at Atmos. Press.
(Lbs. per Cu.ft.) | Spec. Vol. at Atmos. Press.
(Cu.ft. per Lb.) |
| | | At 32° F. | | At 62° F. | | At 32° F. | | At 62° F. | |
|---|
| | | 0.08072 | | 0.07009 | | 12.39 | | 13.14 | |
|---|
Weight of air containing one pound of oxygen, is approximately 435 lbs.
123. Combustion to CO₂. The combination of carbon and oxygen to form the dioxide is represented by the equation
C+4O₂=CO₂, . . . . . . (83)
COMBUSTION - 283
which shows that one atom of carbon (twelve times heavier than hydrogen) combines with two atoms of oxygen (each sixteen times heavier than hydrogen) to form a molecule of CO$_2$, which is forty-four times heavier than an atom of hydrogen. Therefore the weight of carbon and oxygen combining are as $12 : 16 = 8 : 2$ so that $2^2$ lbs. of oxygen are required to burn a pound of carbon to carbon dioxide. Writing this in the form of an equation, gives
1 weight of C+$2^2$ weights of O=$3^2$ weights of CO$_2$. . . . (84)
The weight of air required can readily be found by multiplying the required oxygen by the number 4.35, previously shown to be the number of pounds of air containing one pound of oxygen. Thus, the required air is $2^2 \times 4.35 = 11.57$ pounds per pound of C burned to CO$_2$. This number is commonly rounded out to 12 in engineering literature.
The equation given shows volume relations directly. It is evident, therefore, that one molecule of O yields one molecule of CO$_2$, and hence that the volume of the product is exactly equal to the volume of the oxygen used in forming it if measured at the same temperature and pressure. This is a very important relation, and is often made use of in engineering calculations.
Experiment shows that when carbon burns to the dioxide, 14,000 B.t.u. are liberated per pound of carbon burned, that is, the calorific value of C burned to CO$_2$ in 14,000.
124. Combustion of CO to CO$_2$. Since carbon which has burned to carbon monoxide has not combined with the greatest possible quantity of oxygen, the monoxide can take up more oxygen by burning to the dioxide. This process is represented by the formula
$$2\text{CO} + \text{O}_2 = 2\text{CO}_2$$ . . . . . . . . . . (85)
284
STEAM POWER
which shows that two molecules of monoxide combine with one molecule of oxygen to form two molecules of the dioxide.
The volume of CO₂ formed is therefore equal to that of the CO burned.
So far as the ultimate result is concerned, it makes no difference whether carbon is burned directly to CO₂ or is first burned to CO and then the CO is burned to CO₂. The total oxygen used per pound of carbon burned to CO₂ and the total heat liberated per pound of carbon burned to CO₂ are the same.
Thus, for the oxygen, one pound of C burned to CO₂ requires 23 lbs. of oxygen; but one pound of C burned to CO requires 13 lbs. of oxygen, and 13 lbs. more will be required when this CO is burned to CO₂. The result is therefore the same.
For heat liberated, one pound of C burned to CO₂ liberates about 14,600 B.t.u.; but one pound of C burned to CO liberates about 4500 B.t.u. and 10,100 B.t.u. are liberated when this CO is burned to CO₂. Since the sum of 4500 and 10,100 is equal to 14,600 the result is again the same.
Data on the combustion of C to CO and CO₂ and the combustion of CO to CO₂ are collected in convenient form in Table IX.
125. Conditions Determining Formation of CO and CO₂.
Excluding certain complicated considerations which are not of great importance in steam-power engineering, it may be said that when carbon is being burned at a certain rate (pounds per hour) and all the oxygen brought into contact with the carbon burns out, that carbon burns to CO or to CO₂. If enough or more than enough oxygen to burn the carbon to CO is brought into contact, that oxide will be formed. If there is not enough to burn all the carbon to the dioxide, both oxides are formed in certain proportions, which can be calculated.
Since combustion to CO yields only 4500 B.t.u. per
COMBUSTION 285
Pound of C and combustion to CO₂ yields 14,600 B.t.u.
per pound of C, the importance of supplying sufficient oxygen to burn all carbon to the dioxide in cases where the liberation of the maximum quantity of heat is desirable is obvious. In actual practice the oxygen is furnished by supplying air and it is found necessary in most cases to supply more than the amount of air theoretically required under other conditions, because a large part of the carbon to the dioxide. This comes from the great difficulty met in obtaining contact between the oxygen of the air and the carbon which is to be burned, that is, in bringing all the oxygen of the air into intimate contact with the fuel being burned in real apparatus.
**TABLE IX**
**COMBUSTION DATA FOR CARBON**
(Per pound of carbon.)
| Product. |
Oxygen Required. |
Nitrogen Accompanying Oxygen. |
| Pounds. |
Cu.ft. at 62° F. and 147 Lbs. |
Pounds. |
Cu.ft. at 62° F. and 147 Lbs. |
|
|
|
|
| CO |
1.333 |
16.0 |
3.46 |
00.1 |
| CO₂ from C... |
2.667 |
32.0 |
9.92 |
120.2 |
| CO₂ from CO... |
1.333 |
16.0 |
4.46 |
00.1 |
| Product. |
Air Required. |
Quantity of Product (Not included). |
Heat Liberated. |
| Pounds. |
Cu.ft. at 62° F. and 147 Lbs. |
Pounds. |
Cu.ft. at 62° F. and 147 Lbs. |
|
|
|
|
| CO |
5.79 |
76.1 |
32.0 |
1,500 |
| CO₂ from C... |
11.38 |
152.2 |
3.67 |
1,400 |
| CO₂ from CO... |
5.79 |
76.1 |
3.67 |
| | | |
|---|
| Cu.ft. | Cu.ft. | Cu.ft. | Cu.ft. |
|---|
| at 62° F. | at 62° F. | at 62° F. | at 62° F. |
|---|
| and 147 Lbs. | and 147 Lbs. | and 147 Lbs. | and 147 Lbs. |
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