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

. (page 15 of 30)
Online LibraryAmerican Technical SocietyCyclopedia 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 text (page 15 of 30)
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height by the strength of the spring we get the desired result.
For instance, if we had used a 30 pound spring (that is one which
causes the pencil of the indicator to move one inch for every 30
pounds pressure in the cylinder) the height 1* inches would
equal 30 X 1| = 45 pounds pressure.

Example. An indicator card has an area of 1.925 square
inches and is 2.2 inches long. If a 60 pound spring is used
what is the mean pressure?

The mean height equals J _ = .875 inch and
.875 X 60 = 52.5 pounds.


'Suppose the engine from which the above card was taken had
a piston 14 inches in diameter and a stroke of 24 inches. If it
were running at 150 revolutions per minnte what is its horse-
power ? Assume the mean effective pressure to be the same for
both sides of the piston.

1L p = PLAN

_ 52.5 X 2 X 153.94 X 300


=147. (about)

In most engines more work is done at one end of the cylinder
that at the other ; it is not safe then to assume the mean effective
pressure of one side the same as that of the other. Cards should
be taken from each end and calculated for mean effective pressure
separately-, then averaged. Also the area of one side of the piston
is greater than the other on account of the piston-rod. The two
ends may be figured separately or the average area of the two sides
of the piston may be used as the vahfe of A.

Another method is to find the work done at each end of the
cylinder and then add the results. This enables the engineer to
know if his valves are set so that each end does about the same
amount of work.

An engine has the following dimensions. The piston is 12
inches in diameter, the piston-rod is 21 inches in diameter and tke
length of stroke is 34 inches. While running at 92 revolutions
cards were taken. The area of the card from the head end was
5.36 square inches, that of the crank end was 5.30 square inches,
and a 40 pound spring was used. The cards were 3.72 inches
long. We wish to know what horse-power the engine developed
and which end was doing the most work.

The area of the piston is 113.097 square inches for the head
end and 113.097 3.5466 = 109.55 square inches for the crank

Then for the head end,
T p P LAN P X 34 X 113.097 X 92 RQqq p

-33,000 "llTxlipOO -

and for the crank end,

HP- PL AN _ P X 34 X 109.55 X 92 _ p

H - F- ~ 33,000 -12"X~-



The value of P is found from the cards. Since the cards
were 3.72 inches long and the area of the card from the head end

r of*

was 5.36 square inches the mean ordinate is ' . = 1.44 inches

which equals 1.44 X 40 = 57.6 pounds. The horse-power would

.8933 X 57.6 = 51.45

For the crank end the mean effective pressure is found as

J 5 ^ 3 = 1.425 and 1.425 X 40 = 57.

The horse-power would be,

.8653 X 57 = 49.32

The horse-power is evidently the sum of these two quantities
or 51.45 + 49.32 = 100.77.

The head end is doing more work than the crank end but the
difference is slight being only,

51.45 49.32 = 2.13 horse-power,
or about 2.1 per cent, of the total power.

Considerable arithmetical work is necessary when the I. JI. P.
is found from the formula,

pi A XT
I. II. P. = J -


and the chances for error are of course, great. To Safe time and
reduce the chance for error a table of engine constants has been
prepared. The number of strokes, or twice the number of revo-
lutions, multiplied by the length of stroke in feet is called the

piston speed. Then in the formula 1. H. P. ==_ , L N =


piston speed in feet per minute. In the following table, the I. H. P.
of an engine is easily computed by multiplying the constant, cor-
responding to the diameter of the piston, by the piston speed and
by the M. E. P. Or, in other words, the constants in the table
equal the horse-power for an engine with a given diameter of
piston having a piston speed of one foot per minute and a M. E. P.
of one pound.


















+ 1














+ 1






























+ *










+ 1

+ 5

















To Use the Table. If the diameter of the piston is an even
number, the constant is found in the second column ; if it contains
a fraction the constant is found by following the column horizon-
tally until the required fraction is reached. The constant multi-
plied by the piston speed in feet per minute ar.d by the M. E. P.
in pounds per square inch gives the I. H. P.

Example. An engine runs at 75 revolutions. The stroke is
4 feet; if the M. E. P,. is 48 pounds and the piston 27f inches in
diameter what is the I. H. P.

From the table the constant for a piston 27 1 inches in diam
eter is .0178355. The piston speed is 150 X 4 = 600 feet per
minute. Then the I. H. P. is,

.0178355 X 600 X 48 = 513.66

The horse-power as above calculated is called the indicated
horse-power and is usually written I. H. P. Although the above
calculation shows the amount of power the engine develops it does

Fig, 18.

not show the available power since part of the indicated horse-
power is used to run the engine itself, that is, to overcome the
friction of the parts. To determine how much power can be used
to run machinery some form of absorption dynamometer or friction
brake is attached to the engine. The power thus obtained is
called the Brake Horse Power or B. H. P. It is more satisfactory
for both the owner and builder to know the B. H. P. than to know
the I. H. P.

The Prony Brake, Fig. 18, is one of the simplest absorption
dynamometers. The two wooden blocks A and C are held together
against the rim of the pulley P by bolts . The thumb-nuts, e, , being



used to adjust the pressure. By means of the bolts the arm L is
held tp the upper block. From this arm is suspended the ball
weight, w, which by sliding along the arm counterbalances the
weight of the arm and pan at the other end. The pulley revolves
at the required speed in the direction indicated by the arrow.
The bolts are tightened until the lever remains stationary in a
horizontal position when a known weight, W, is hung at the end.

The amount of work absorbed by the brake depends upon the
weight W, the length, R, and the speed. It is independent of
the diameter of the pulley and the pressure of the block because
the moments of force about the center of the pulley are equal
when the lever L, is horizontal. Letting/ equal the co-efficient
of friction, p the pressure of the blocks and r the radius of the

fpr = WR

The work done at the face of the pulley equals the force mul-
tiplied by the distance or the pressure multiplied by the number
of feet passed through.

Let N = the number of revolutions per minute. Then the
distance passed through per minute equals 2 TT r N and the work
done equals 2 TT r N f p. Then &&fp r =' W R, the work done
at the rim of the pulley equals the left hand side of the equation
multiplied by 2 TT N, and to keep both sides equal we multiply
W R by 2 TT N. Hence the work done is expressed by the formula

2 TT N W R and,
B p 27rN WR

= .0001904 N W R

A Prony brake with an arm 4 feet long was attached to the
pulley on the fly wheel of an engine. The weight in the scale
pan was 50 pounds and the speed of the engine 300 revolutions.
Find the brake horse power.

B. H. P. = .0001904 X 300 X 50 X 4
= 11.424

The rope brake shown in Fig. 19 is easily constructed of
material at hand and being self-adjusting needs no accurate fitting.
For large powers the number of ropes may be increased. It is con-
sidered a most convenient and reliable brake. In Fig. 19 the spring




balance, B, is shown in a horizontal position. This is not at all
necessary ; if convenient the vertical position may be used. The
ropes are held to the pulley or fly-wheel face by blocks of wood, O.
The weights at W may be replaced by a spring balance if

To calculate the Brake Horse Power, subtract the pull regis-
tered -by the spring balance, B, from the load at W. The lever
arm is the radius of the pulley plus i the diameter of the rope.
The formula is,

BMP- 2?rRN (W B)

= .0001904 RN(W-B)*

Example. A rope brake is attached to a gas engine. The
average reading of the spring balance is 8 pounds. W = 80
pounds. If the radius of the brake wheel is 28 inches and the
rope 1 inch in diameter, what is the B. H. P. when the engine
makes 350 revolutions per minute ?

11 = 28 + l = 28 inches = ^ feet
B. II. P = .0001004 EN (W B)'

= .0001004 x 2 - X 72 X 350
= 11.4 Ans
If both the indicated horse-power and the brake horse-power

* NOTE: If B is greater than W, the engine is running in the opposite
direction. Use the formula B. H. P. = .0001904 R N (B W) .



are known the power used in friction is found by subtracting the
B. H. P. from the I. II. P.

The mechanical efficiency of the engine is the ratio of the
B. H. P. to the I. H. P. or,

E= JB.H.P.

I. H. P.

If an engine of 18.2 indicated horse-power develops at a trial
16.02 brake horse-power, what is its mechanical efficiency?

E= B - " P -
I. H. P.

= -l 6 ^ = .88

= 88 % Efficiency.

Brakes should be well lubricated. For small powers the
heat generated by friction between the ropes or blocks and the
rim of the wheel, will be conducted away by radiation but for
large powers some additional means is necessary. In case there
are flanges on the wheel, water can be introduced into the wheel,
the flanges keeping it from flowing out and centrifugal force keep-
ing it in contact with the rim. The. amount of water can be regu-
lated so that all may be evaporated, or a scoop can be arranged to
carry off the water. In all cases the water should flow

To Find the Area of Cards. M. E. P. or the mean effective
pressure is equal to the area of the indicator diagram divided by
the length. The length is easily found by measurement but to
find the area is more difficult since the shape is irregular. If the
figure were regular its area could be found by geometry or by
simple formulas.

The area of the indicator card can be found in two ways.
By dividing the diagram into sections and by the use of a plani-
meter. The former is only an approximate method ; the area thus
found is nearly correct if the number of divisions is great.

Tangents at each end, perpendicular to the atmospheric line
are first drawn. The horizontal distance between these tangents
is then divided into 10 or more equal parts. The horizontal length
of each section is then divided into two equal parts and lines per-
pendicular to the atmospheric line drawn through these points of



division. The sum of the lengths of all these lines is divided l>y
the number of lines to get the average. This average length or
average ordinate multiplied "by the scale of spring gives the mean
effective pressure.

Fig. 20 is the card from the crank end of an engine. The


Fig. 20.

line C L is the atmospheric line and the lines A D and E F are
drawn perpendicular to it and tangent to the extreme ends of the
diagram. The line A E is divided into 1 equal parts and lines

Fig. 21.

are drawn through points marking the centers of the divisions.
On each of these lines the length is marked. The sum of the
lengths is 15.18 and 15.18 divided by 10 is 1.518. If the scale
of spring is 40 pounds, 1.518 multiplied by 40 is the M. E. P. or
0.7 = M. E. P.




The liorizontal lengtli may be divided into any number of
?qual parts but 10 or 20 makes the computation easy. The oper-
ation of finding the M. E. P. for the head end is exactly the same.
The average M. E. P. for one revolution of the engine is the
average of the two mean effective pressures."

In case the diagram is very irregular it should be divided
into 20 equal parts instead of 10. If there is a loop in the dia-
gram as shown in Fig. 21 the area of the loop must be subtracted

from the area of the other part as it represents work done by the
piston on the steam and therefore loss.

The lengths may be marked off on a piece of paper if a good
scale is not at hand.

A more accurate result is obtained by using an instrument
called the planimeter. There are several planimeters and aver-
aging instruments in common use for determining the mean effec-
tive pressure of indicator cards. The planimeter shown in Fig.



22 is one of the most simple and is called the Amsler Polar Plani-
meter from its inventor Prof. Amsler. The cut is about one-half
the size of the instrument. It consists of two arniB free to move
about a pivot and a roller graduated in inches and tenths of
inches. A vernier is placed with the roller so the areas may be
read in hundredths of a square inch. The point A is kept sta-
tionary and the tracer B is moved once around the outline of the
diagram. The area in square inches of the diagram is read from
the roller C and the vernier E.

To Use the Planimeter. The diagram should be fastened to
some flat unglazed surface, such as a drawing board, by means of
thumb tacks, springs or pins. The point A is pressed into the
paper so that it will hold in place. The point B is set at any
point in the outline of the diagram and the roller set at zero.
Follow the outline of the diagram carefully in the direction of
the hands of a watch as indicated by the arrows in Fig. 22 until
the tracer has moved completely around the diagram. The result
is then read to hundredths of an inch from the roller. Suppose
after tracing over the outline we find that the largest figure that
has passed the zero of the vernier is 3 ; the number of graduations
(tenths) that have passed the zero to be 5 and the number
(hundredths) of the graduations in the roller that exactly coincides
with a graduation on the vernier to be 9. Then the area is 8.59
square inches.

Often at the start the roller is not adjusted so that the zeros
coincide but the reading is taken and subtracted from the final
reading. Thus if the first reading is 4.63 and the second 7.31 the
area is 7.31 4.63 = 2.68 square inches. In case the second
reading is less than the first, add 10 to the second reading then

This instrument is very valuable to an engineer who takes
indicator cards. The results obtained are very accurate, the error
being small. Ten or twelve diagrams can be measured by this
instrument in the same time that is necessary to measm-e a single
card by the method of ordinates.

It is well to run over the area three or four times and take
an average as the tracing of the diagram cannot be absolutely COP
rect at any time.




The thermal efficiency of the steam engine is found in the
same manner as that of any other heat engine. The efficiency
depends upon the limits of temperature and not upon the nature
of the working medium.
Let T l absolute temperature of the heat received by the engine.

T 2 = absolute temperature of the heat rejected by the engine.

E = efficiency of engine.

or, the efficiency equals the temperature of the heat rejected, sub-
tracted from the temperature of the heat received and the result
divided by the temperature of the heat received.

Suppose an engine is supplied with steam at 120 pounds
absolute pressure and the exhaust is atmospheric pressure. What
is the efficiency?

The absolute temperature corresponding to 120 pounds abso-
lute pressure is 341.05 -j- 461 and the temperature of atmos-
pheric pressure is212 + 461.


802.05 673

E = - _ = .16 or 16 per cent.

If the engine had been of the condensing type and the
exhaust pressure one pound above the vacuum, the efficiency
would be as follows :

The temperature of one pound absolute pressure is 101.99
+ 461.

E = W*M=_6***_ = Qr per cent>

In actual engines this efficiency cannot be obtained localise
the difference between the amount of heat received and that
rejected is not all converted into work. Part of it is lost by
radiation, conduction, leakage, etc. Also cylinder condensation
reduces the efficiency.

The Theoretical Indicator Diagram. An indicator diagram
is the result of two movements ; a horizontal movement of the



paper and a vertical movement of the pencil. The horizontal
movement exactly corresponds to the movement of the piston of
the engine and the vertical movement exactly corresponds to the
pressure of steam in the cylinder.

The shape of the indicator card depends upon the manner in
which steam is admitted to and released from the cylinder. Dif-
ferent engines give different shaped indicator cards and the cards
taken from an engine vary with the conditions. Figs. 1 and 2
show theoretical indicator cards from a non-condensing engine
without clearance ; the former being for the case that has admis-
sion during the whole stroke. The diagram of Fig. 2 shows the
cut off at ^ stroke. All practical engines have clearance and
slight compression ; so the theoretical diagram assumes the shape

shown in Fig. 23. In this card
the admission line H A is verti-
cal, the steam line A C is hori-
zontal, the expansion line C D
an hyperbolic curve, the exhaust
line D B vertical, the back pres-
sure line B F horizontal and the
compression curve an hyperbola.
The actual shape is somewhat

rig ' 23 * different from the theoretical

mainly because the valves do not open and close quickly, the ports
offer some resistance to the passage of the steam and the back
pressure is neither atmospheric in the non-condensing engine nor
absolute vacuum in the condensing engine.

The diagram shown in Fig. 24 is a practical diagram and
like those taken from engines.

The atmospheric line L M is the line drawn by the pencil of
the indicator when the connection to the engine is closed and both
sides of the piston of the indicator are open to the atmosphere.
It is the zero of the steam gage.

The admission line H A shows the rise of pressure due to the
admission of steam to the cylinder. If the steam is admitted
quickly when the engine is nearly on dead center this line will l>e
very nearly vertical.

The steam line A C is drawn while the valve admits steam



to the ' cylinder. This line is horizontal if there is no wire-

The point of cut off C, indicates the point at which the
admission of steam is stopped by the closing of the valve. This
point is rounding since the valve closes slowly. Sometimes it is
difficult to determine the exact point where cut off takes place ; it
is usually where the curve changes from concave to convex.

The expansion curve C D shows the fall in pressure as the
ssteam expands while the piston moves toward the end of the stroke.

The point of release D shows the point at which the exhaust


Fig. 24.

valve opens. The rounding is due to the slow action of the valve
when opening. Because of this slow action of the valve, release
begins a little before the end of the forward stroke.

The exhaust line D E F represents the loss in pressure which
occurs while the valve opens to exhaust at and near the end of the

The back pressure line F G shows the back pressure against
which the piston acts during the return stroke. For a condens-
ing engine this line is below the atmospheric line L M, the dis-
tance below being dependent upon the state of the vacuum in the
condenser. For cards taken from a non-condensing engine the
back pressure line is a little above the atmospheric line.

The point of exhaust closure G is the point where the valve



closes to exhaust. The exact point is not clearly denned as the
curve shows a change of pressure due to the gradual closing of the
valve. -

The compression curve G II shows the rise of pressure due
to the compression of the steam remaining in the cylinder after
the valve has closed to exhaust.

The zero line of pressure or line of absolute vacuum O X is
drawn below and parallel to the atmospheric line. The distance
between the lines O X and L M represents 14.7 pounds pressure.

The clearance line O Y is drawn perpendicular to the line of
absolute vacuum and at a distance from the end of the diagram

_l _2 ^3 4 5

Fig. 25.

equal to the same per cent, of the length of the diagram as the
clearance volume is of the piston displacement, or

L N clearance volume

L M volume of cylinder

It is readily seen that the area of an actual indicator diagram
is less than that of a theoretical card." This is because of the
round corners at cut off and exhaust, the back pressure and the
compression. Sometimes it is useful, especially in designing
engines, to draw the theoretical indicator card.



Online LibraryAmerican Technical SocietyCyclopedia 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 text (page 15 of 30)