American Technical Society.

# Cyclopedia of engineering : a general reference work on steam boilers, pumps, engines, and turbines, gas and oil engines, automobiles, marine and locomotive work, heating and ventilating, compressed air, refrigeration, dynamos motors, electric wiring, electric lighting, elevators, etc. (Volume 2) online

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Font size To Draw the Theoretical Card. To draw an ideal diagram
draw P X equal to the length of stroke and O P equal to the
clearance shown in Fig. 25. Draw O Y and P A perpendicular
to O X and draw Y S parallel to O X and at a height correspond-
ing to the boiler pressure.

The line of initial pressure A C is then drawn parallel to Y S
and is usually taken as from 90 to 95 per cent, of the boiler pres-
sure if there is no special cause for loss. Then take A C as the

P X

portion of the stroke at which steam is admitted so that =

.A. \j

the ratio of expansion. The expansion line is considered a hyper-
bolic curve with O Y and O X as asymptotes. To draw the
"hyperbolic curve. First draw the line A C B parallel to the
atmospheric line and F D B and R C perpendicular to it. Then
make points 1, 2, 3, 4, etc., on C B and connect them with the
point O. At the points 1', 2', 3', 4', etc., where these lines inter-
sect the line R C draw parallels to C B until they meet perpen-
diculars from 1, 2, 3, 4. The point of intersection of these lines
are points in the hyperbolic curve C D, as shown in Fig. 25.
Any number of points may be used ; but there must be enough to
determine the curve.

The area A C D F N H is the theoretical card, with a given
boiler pressure, ratio of expansion and an assumed back pressure.
The actual card for the same data would be more nearly like the
shaded area which lies mostly within the outline of the theoretical
card. In designing engines it is well to know the ratio of the
actual to the ideal card for all types of engines.

This ratk> varies with the speed, type of engine, and kind of
valves and has the following values.

For ordinary plain slide valve engines, .8 to .9

For engines with high speed, about, .75

For compound engines, .75 to .80

For compound engines, high speed, .65 to .75

For triple expansion engines, .60 to .70

For triple expansion engines, high speed, .50 to .60

223

40

INDICATORS.

CARDS FOR COMPOUND ENGINES.

In Fig. 26 the ideal cards of a tandem compound engine of
the Woolf type are shown. The diagram A C D E F G H is
from the high pressure cylinder and the diagram L M N P Q II
from the low pressure. S T is the atmospheric line.

The line H A is the admission line, A C the steam line, and
C D the expansion line. These lines are similar to those of a
single cylinder engine diagram. At D, the point of release, the
pressure drops slightly as steam is admitted to the low pressure

Fig. 26.

cylinder. The back pressure line E F, of the high pressure cylinder
is parallel to the steam line L M, of the low pressure cylinder.
They would coincide if there were no resistance to the flow of
steam and no heat loss. The flow of steam from the high to the
low pressure cylinder is cut off at F and the steam remaining in
the high pressure cylinder is compressed in the cylinder and in the
pipes. The exhaust closes at G and steam is compressed in the
small cylinder from G to H.

224

INDICATORS.

41

Cut off occurs in the low pressure cylinder at M, and the
steam in the cylinder expands, the curve M N being an equilateral
hyperbola. Release takes place at N and the pressure falls to
that of the condenser. The back pressure line P Q and the com-
pression curve Q R are like those of a single engine.

The diagrams shown in Fig. 27 are from a Woolf engine.
The lines are similar to those of the theoretical diagram shown in
Fig. 26.

With compound engines of the Woolf type the steam passes

Fig. 27.

directly from the high pressure cylinder to the low pressure. In
the cross compound, the cranks being at 90, the piston in one
cylinder is at the end of the stroke when the other is at the middle
of the stroke. Therefore, a receiver must be used. The ideal
cards from a cross compound engine are shown in Fig. 28. The
dimensions are as follows :

Volume of high pressure cylinder ^ 5 cu. ft.

Volume of receiver = 5 cu. ft.

Volume of low pressure cylinder 15 cu. ft.

Cut off of high pressure cylinder = ^ stroke.

Cut off of low pressure cylinder = - stroke.

Initial pressure (absolute) =120 Ibs.

Back pressure (absolute) = 3 Ibs.

825

42

INDICATORS.

Clearance and compression are not considered. The stroke
of the high pressure cylinder begins at A. Steam is cut off at ^
stroke at C.

Since steam is cut off at | stroke, the ratio of expansion in
the high pressure cylinder is 4. Then the volume of steam at D
is 4 times that at C and since p v = p l v l , the* pressure at D is

120 X 1

_ = 30 pounds. Steam is released at D and passes to the

receiver. The pressure at release, L, of the low pressure cylinder
is found from the equation p v = p l v. The volume of steam
in the high pressure cylinder at cut off, is | cubic feet and the
I20r C, ,A

Fig. 28.

volume at L is 15 cubic feet; then the total ratio of expansion is

120 Y 1

15 -f- 2 = 12. Then the pressure atL is -^ = 10 pounds.

12

Since the low pressure cylinder cuts off at -| the stroke the
volume at G is 1 that at L.

Then asp v = p l v^ p X \$ = 10 X 1, or p = 20 pounds.

Hence the pressure at G is 20 pounds.

We can now find the pressure at E. The pressure in the
receiver when the low pressure cylinder cut off, was 20 pounds
because the low pressure cylinder was in communication with it.

INDICATORS. 43

Then as tlie piston in the small cylinder is at the end of the stroke
when the steam is cut off in the large cylinder, steam at 30
pounds pressure rushes from the high pressure cylinder into the
receiver and mixes with steam at 20 pounds pressure. When
steam is cut off in the low pressure cylinder the high pressure
cylinder and the receiver are in communication with each other ;
the total volume being the volume of the high pressure cylinder
plus that of the receiver or 5 -f- 5 = 10 cubic feet. What we
wish to find is the pressure of the steam of this volume. The
formula is,

Since there are 10 cubic feet in the receiver and high pressure
cylinder, the value of V is 10. The volume of steam in the
high pressure cylinder is 5 cubic feet and its pressure is 30 pounds.
The volume of steam in the receiver is 5 cubic feet and its pressure
is 20 pounds, then,

10 P = 5 X 30 + 5 X 20

250
P = _ - = 25 pounds.

Then the pressure at E 25 pounds.

While the piston of the high pressure cylinder is on the return
stroke, steam in that cylinder is compressed from E to F ; the
volume being 10 cubic feet. When the piston has completed i
the return stroke, the volume at F is equal to |- the volume of the
high pressure plus the volume of the receiver or |- -}- 5 = 7-| cubic
feet.

Then with the formula^ v = p 1 v-^ we obtain,

F = 25X10 = 3333

or the pressure at F is 33.33 pounds.

Then as the cranks are 90 apart, and the high pressure piston
is at the middle of the stroke, the low pressure piston must be at
the beginning of its stroke ; or the pressure in the low cylinder is
the same as that in the high and in the receiver, or 33.33 pounds.
Since the stroke is beginning in the low pressure cylinder, steam
pressure in the large cylinder falls as the volume increases, which

227

I -I

INDICATORS.

js shown by the line F G. Cut-off occurs in the low pressure
cylinder at G, and the steam expands until the piston reaches L ;
the curve being an equilateral hyperbola. At L the release occurs
and the pressure drops to that in the condenser ; in this case about
3 pounds. The back pressure line is of course the pressure in the
condenser.

The cards of Fig. 29 are from a cross compound engine. The
rise of pressure in the middle of the back pressure line of the

Fig. 29

diagram from the small cylinder is due to the fluctuation of pres-

COflBINED DIAGRAHS.

The indicator cards of multi-cylinder engines may be com-
bined so that the pressures and volumes are shown in their proper
relations. To do this, the cards are reduced to the same scale of
pressure and the same scale of volume. To make the combined
diagram of convenient size, the length of the low-pressure card is

228

INDICATORS.

45

left as it is and the length of the high-pressure diagram is shortened
in the ratio of the cylinder volumes.

Perhaps the best way to show the method is by an illustra-
tion. . The cards shown in Fig. 30 were taken from a compound
condensing engine.

Ratio of cylinder volumes =1:3
Initial pressure = 138 pounds

Spring, high-pressure card = 60 pounds
Spring, low-pressure card = 30 pounds

Fig. 30.

First draw the line of zero pressure A B and the line of zero
volume E F (see Fig. 31). The line O P is drawn parallel to E F
and at a distance proportional to the clearance of the low-pressure
cylinder. Similarly the distance between M N and E F represents
the clearance of the high-pressure cylinder. Now draw the atmos-
pheric line C D. In this case it will be .49 inch above A B_

because a 30-pound spring is used, and

14.7
30

:.49. Reproduce the

low-pressure card without change, as shown.

Divide the high-pressure card with 10 (or more) ordinates,
and reproduce it with volumes and pressures of the same scale as
the low-pressure card. Since a 60-pound spring was used, each

AH
ordinate will be twice as long (because . = 2). The distance

30

229

46

INDICATORS.

between the ordinates will be -|- as great because the high-pressure
cylinder is ^ the volume of the low. Draw the ordinates as shown.
The distance between ordinates L and M will be ^ of the length
of the high-pressure card. The points c and e are on the fifth
ordinate, and are twice as far above the atmospheric line as they
are in the original high-pressure card.

After locating all the points draw the curve through them.
Now draw the theoretical expansion curve R S through the point
of cut-off of the high-pressure cylinder by the method explained

Fig. 81.

on page 39. The difference between the area included between
the theoretical curve and the lines of no pressure and no volume,
and the sum of the actual areas, represents approximately the
losses.

This method is not strictly accurate, because all the steam
used in the high-pressure cylinder does not pass to the low-pressure
cylinder: a small portion is left for compression. By thus com-
bining the cards the action of the valves may be discussed, provided
such data as size, type and speed are known.

The combined diagram of a multi-expansion engine drawn

230

INDICATORS.

47

according to the above method is shown in Fig. 32. The volumes
arid pressures are first reduced to the scale of the low-pressure
card. The cards are then redrawn at the proper distances from
the lines of no pressure and no volume ; the clearance in each
cylinder being considered.

HORSE-POWER OF COMPOUND ENGINES.

The I. H. P. of multi-cylinder engines may be found by add-
ing the I. II. P. of the several cylinders. Another method is to
reduce all the pressures to the area of the low-pressure cylinder.
This is done by dividing the M. E. P. of each cylinder by the
inverse ratio of hs volume to that of the low-pressure cylinder.

Suppose the M. E. P. of the high-pressure cylinder of a com-
pound engine is 78 pounds as found from the indicator card. It
the volume of the high-pressure cylinder is 1 that of the low, the

7 R

M. E. P. of the high referred to the low would be . = 26 pounds.

3

Then if the card from the large cylinder shows a M. E. P. of 30
pounds, the total M. E. P. is 30 -f- 26 56 pounds, and the
I. H. P. is found by inserting 56 as the value of P and using the
area of the low pressure as A, in the formula for I. H P.

48 INDICATORS.

INDICATOR CARDS.

The indicator reveals defects in the steam distribution. That
is, if the valve or valves are set so that the events of the stroke
are too early or too late or if more work is done at one end of the
cylinder than at the other, the engineer finds it out by examining
the indicator card.

Sometimes an engine appears to run well and the owner is
perfectly satisfied with it ; but from the indicator diagrams it is
found that considerable saving might be effected by correcting the
defects in the valve setting.

On looking at a diagram one might say it was a faulty card

Fig. 33.

and yet for that type, size and speed it is perhaps the best that
could be obtained from the engine. The same form of diagram is
not possible from different types of engines. The diagram from a
Corliss engine, having four valves, is different from that of the
plain slide valve engine ; also the diagrams from high speed engines
differ from those of low speed.

The most common faults in the distribution of steam in the
cvlinder can be divided into four classes.

a = Admission too early or too late.
I = Cutoff too early or too late.
c = Release too early or too late.
(I = Compression too early or too late.
In the following figures,

A is the point of admission,
C is the point of cut off,
B is the point of release,
Gr is the point of compression.

232

INDICATORS.

49

The diagram shown in Fig. 33 shows too early admission.

perpendicular to the atmospheric line as it is in Fig. 24. The
diagram also shows cut off, release and compression early. When
the valve is of the plain slide valve type all the events are likely
to be too early if one of them is. The reason for the event being

Fig. 34.

too early is that the eccentric has too much angular advance,
Hence the remedy is to decrease the angular advance.

Fig. 34 shows a diagram having the opposite defects to those
of Fig. 33. The events are too late. The admission line curves
forward and the line shows that admission does not take place
until after the stroke is well begun. Release occurs at the end of

Fig. 35.

the stroke. In this case the angular advance of the eccentric
should l>e increased until the admission line is perpendicular to the
atmospheric line.

Fig. 35 shows a card having too much back pressure. This
may be due to a small exhaust port or pipes, or the passage of steam

50

INDICATORS.

through coils of pipe for heating. The card shows the other events
to be good. A diagram having too late cut off is shown in Fig.
36. The pressure at release is high. When the engine is running
under this condition much of the benefit from expansion is lost.

Fig. 3G.

A diagram from a condensing engine is shown in Fig. 37.
These oscillations are caused by the vibration of the indicator
piston and spring. To avoid these vibrations never use a very
light spring for high speed.

The diagram of Fig. 21 shows too early cut off. In this case
the cut off is so early that the expansion line extends below the

atmospheric line making a loop. The area of the loop must Iwi
mltracted from the area of the card as explained on page 33.

Fig. 38 shows a pair of diagrams from a plain sjide valve
engine. The admission lines are good. The sloping steam lines

234

VERTICAL CROSS COMPOUND BLOWING ENGINE.
Buckeye Engine Company.

INDICATORS.

51

show wire-drawing due to the slow action of the valve or too
small ports or pipes. This wire-drawing decreases the area of the
card which indicates loss. The greatest fault is the inequality of
area of the diagram for the ends. The late cut off and conse-
quent late compression of one end causes more area than the too
early cut off and too early compression of the other end. These

Fig. 38.

eccentric and the length of the valve rod.

The diagram of Fig. 39 indicates too early compression. The
compression curve extends above the initial pressure line. The
area of the loop must lie subtracted from the card area, when corn-

Fig. 39.

pucing the I. H. P. If the cut off is kept the same and the com-
pression made what it should be, the gain in area would be the
area included between the full line and the dotted line plus the
area of the loop. The remedy for this case is to decrease
the* inside lap.

235

52 INDICATORS.

The amount of compression varies with the speed and type of
engine. Slow speed engines require less compression or cushion-
ing than high speed. The exhaust steam should never be com-
pressed higher than the boiler pressure.

For high speed engines the compression should extend to
about .9 the initial pressure. For medium speeds alwut .5 and for
low speed .2 to .4.

In the case of a slide valve engine it is not always possible to set
the valve so that the card may have all the events as they should
be. Sometimes the laps of the valves should be altered. For too
much compression decreases the inside lap. For too early cut off
decrease the outside lap.

If the valve travel is increased, compression is retarded, that
is, decreased ; release occurs sooner.

STEAM CONSUMPTION.

The amount of steam used by an engine is called its steam
consumption and for comparison, it is customary to state the
amount of steam consumed per indicated horse-power per hour.
By means of the indicator diagram the steam consumption can be
computed approximately.

To find the Steam Consumption from the Diagram. The

diagram shown in Fig. 40 is from a 20 X 36 engine, running at a
speed of 80 revolutions per minute. A 40-pound spring was used.

By measuring the card, we find the mean ordinate to be .91
inch and the M. E. P. = .91 X 40 = 36.4 pounds.

I. H. P. = Engine Constant X M. E. P. X piston speed.
= .00952 X 36.4 X 480.
= 166.33.

In Fig. 40 L M is the atmospheric line and O R the line
of zero pressure drawn so that O L 14.7 pounds. O N is the
clearance volume = 8 per cent of the piston displacement. The
line P Q is drawn from O R to some point on the compression

INDICATORS.

53

curve. From D, a point on the expansion curve before release,
the line D F is drawn perpendicular to O R.
Then from the diagram,

O R = 3.24 inches.
O F = 3.00 inches.
O P = .345 inch.
D F = .795 inch.
P Q = .795 inch.

The length of stroke is 36 inches or 3 feet, and the length of
the diagram 3 inches. Then 1 inch of the length of the card cor-
responds to 1 foot of the stroke. The scale of spring used is 40.
Therefore we can easily reduce the above dimensions to pounds
pressure and to feet.

O R = 3.24 feet.
O F = 3.00 feet.
O P = .345 feet.
D F = 31.80 pounds.
P Q = 31.80 pounds.
The area of the piston (head end) is,
wi 8.1410 X 400 .

Fig. 40.

We can now find the volume of steam at any point of the
stroke.

When the piston is at D, the volume is,

2.18166 X 3 = 6.54498 cubic feet.

237

54 INDICATORS.

When the piston is at Q, the volume is,

2.18166 X -345 .75267 cubic foot.

From the steam tables we can find the weight of a cubic foot
of steam at a given pressure.

The weight of 1 cubic foot at 31.8 pounds absolute pressure
is .07773 pound. Then the weight of steam present when the
piston is at D is,

G.54498 X .07773 = .50887 pound.
The weight of steam present when the piston is at Q is

.75267 X .07773 = .05852 pound.

The weight of steam in the cylinder is .50887 pound and the
weight of steam kept for compression is .05852 pound. The
weight exhausted per stroke is,

.50887 .05852 = .45035 pound.
The amount used per I. H. P. per hour is,
.45035 X 2 X 80 X 60 =

166.33

This may be stated in words as follows :

Measure the pressure from the vacuum line to some point in
the expansion curve before release and from the steam tables find
the weight of a cubic foot at that pressure. Multiply the volume
(in cubic feet) of the cylinder, including clearance to that point,
by the weight per cubic foot. The result is the weight of steam
in the cylinder. As this weight includes the steam used for com-
pression it must be corrected to obtain the weight used per stroke.
Take some point on the compression curve and measure its
absolute pressure. Then compute the weight of steam to this
point. Subtract this weight from the weight of steam to the
point in the expansion curve and the result is the weight of steam
used per stroke.

Multiply the weight of steam used per stroke by the number
of strokes and divide by the indicated horse-power as found from
the card. The final result is the number of pounds of steam con-
sumed per horse-power per hour.

We may also calculate the steam consumption by taking the
point of the expansion curve near the cut-off.

O B = 1.23 inches = 1.23 feet of the stroke.
B C = 1.8 inches = 72 pounds.

INDICATORS. 55

The weight of 1 cubic foot of steam at 72 pounds absolute
pressure is .1671 pounds. Then the volume of steam at C is,

2.18166 X 1.23 2.68344 cubic feet.
The weight at C is,

2.68344 X .1671 = .44840 pound.

The weight kept for compression is the same as previously
found, i. e., .05852 pound.

Then the weight of steam used per stroke is,

.44840 .05852 = .38988 pound.
The steam consumption per I. H. P. per hour is,

88988 X|X 80X60 = 22J .

If the valve doesn't leak, the amount of steam just after cut
off should equal the amount just before release, but our calculation
shows that there is .45035 .38988 = .06047 pound more at
release than at cut off. This shows that at entrance .06047 pound
was condensed before the piston reached C and was re-evaporated
before release. This calculation gives an idea of the amount of
cylinder condensation.

If there is considerable compression as in Pig. 40 the above
method may be simplified by taking the two points I) and Q
at the same height above the vacuum line. The pressures will
then be the same.

Let W =. weight of steam used per I. II. P. per hour.

w = weight of one cubic foot of steam at the absolute

pressure D F.

L = length of the diagram, N It.
I = distance between Q and D.
P = mean effective pressure.

Then,

W = 18 ' 75Q X w X I
P L

Example. A card was taken from a 12 X 14 engine. Length
of card L = 3.5 inches, I = 2.875 inches. Absolute pressure at

239

56 INDICATORS.

D = 30 pounds. M. E. P. 30.75 pounds. What is the steam
consumption per I. H. P. per hour ?

13,750 w X I

W =

P X L
13.750 X .0736 x 2.875

30.75 X 3.5
= 27.03 pounds.

This method gives only an approximate steam consunr don
On account of the leakage of valves and the initial condensa
tion of steam in the cylinder, the actual consumption is somewhat
greater. The excess varies considerably and makes all results so
obtained of little value. In our calculation we used one diagram
only, that for the head end; we assumed the diagram from the
crank end to be the same. The indicated horse-power for each end
should be computed, as it is subject to considerable variation. In
the above formula the average mean effective pressure should be
used. The results of calculations of steam consumption from
indicator cards are so unreliable that engineers place little depend-
ence upon them. The results may be used as a basis for estimates,
but for accurate knowledge, an engine test must be resorted to.

EXAMPLES FOR PRACTICE.

1. Given the following data to find the M. E. P. Area of
card 2.79 square inches, length of card 3.1 inches, scale of spring
50. Ans. 45 pounds.

2. Steam enters a cylinder at a pressure of 210 pounds by
gage and leaves at 3 pounds pressure (absolute). What is the
thermal efficiency? Ans. 29 per cent.

3. An engine has a B. H. P. of 78.96. The I. H. P. is
86.73. What is the mechanical efficiency. Ans. 91 per cent.

4. An engine develops 149.97 I. H. P. If the piston
diameter is 1 7| inches, the stroke 30 inches, and the M. E. P. 40
pounds, what is the speed ? Ans. 100 revolutions per minute.

5. The theoretical M. E. P. of a triple expansion marine
engine is 4=8 pounds. What is the probable actual M. E. P.?

6. The cylinders of a triple expansion are 12, 30, and 75
inches in diameter respectively. The stroke is 24 inches. The

340

INDICATORS.

57

mean effective pressures from the cards were 118.4, 51.6 and
19.34 pounds respectively. What is the I. H. P. when the speed
is 125 revolutions per minute? Ans. 2,050 I. H. P.

7. Find the steam consumption from the following card.
The engine was running at 75 revolutions per minute, and devel-
oping 230 I. H. P. when the card was taken. A 60-pound spring
was used, and the M. E. P. for this end was 46 pounds. Assume
crank-end card to .be the same as the head end.