Ernest Victor Lallier.

An elementary manual of the steam engine; containing also a chapter on the theory, construction and operation of internal combustion engines for the operating engineer online

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of ten pounds per square inch, and if the piston ha$ an
area of one hundred square inches, a gain of ten pounds
per square inch will equal a total pressure of one thousand
pounds.

In the early days of engineering, when low-pressure
steam was used, it was customary to carry the pressure
during the entire length of the stroke. Modem prac-
tice, however, proves that it is more economical and
that greater efficiency is developed by using steam of
higher initial pressure. We can thus take advantage
of the expansive power of the steam by closing the inlet
valve at some early period in the stroke. This will
allow the work, during the remainder of the stroke, to
be done entirely by the expansive force of the steam.



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36 ELEMENTARY STEAM ENGINEERING.

QUESTIONS.

1. What is a mechanical horse power? (H.P.)

2. What must be known in order to calculate the horse power?

3. Give the formula for the calculation.

4. Find the H.P. of a 10' X la'' engine making 250 revolutions
per minute with a pressure of 45 lb. per sq. in.

5. Find the H.P. of a 12'' X 2$" engine making 90 revolutions
per minute with an average pressure of 50 lb. per sq. in.

6. If, in the previous question, it is desired that the engine
should develop 75 H.P., what change could be made ?

7. A 20^" X 36^" engine has 40 lb. M.E.P. (mean effective
pressure) and 60 revolutions per minute; required to find the H.P.
developed.

8. If the above engine were required to develop 175 H.P.,
what change in pressure should be made, the other conditions
being allowed to remain?

9. If the change were required to be made by increasing the
number of revolutions, how many would be necessary ?

10. If an engine were required to develop 175 H.P. at 60 revolu-
tions, M.E.P. of 40 lb., a stroke of 3 ft., what should be the diam-
eter of the cylinder?

11. In the previous question, what would be the capacity of the
cylinder if 6 per cent were allowed for clearance ?

12. If 045 lb. of steam per stroke were used, what weight of
steam per hour would be required at 200 revolutions ?

13. What is a non-condensing engine?

14. Define back pressure.

15. What is the effect of back pressure on the operation of the
J?



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CHAPTER V.

THE PISTON.

Piston speed is the distance through which the pis-
ton travels during one minute. It is not at the same
rate as the crank pin for the reason that, although me-
chanically connected together, the crank pin moves along
the circumference of a circle while the piston moves a
distance equal to the circle's diameter. Bearing in mind
that the crank pin moves along one half of the circum-
ference of the circle for each stroke, the ratio of move-
ment is

Crank-pin speed : piston speed :: ^^-^ — : i, or 1.5708: i.

2

With a mean piston speed of 1000 feet per minute
we have 1000 X 1.5708 = 1570.8 feet per minute as the
mean crank-pin speed.

As the work done on both pin and piston is the same
while the speed of the pin is 1.5708 times that of the
piston, then the mean pressure on the crank pin is

-of that on the piston.

1.5708

PISTON DISPLACEMENT, or volume swept by the pis-
ton, is equal to the area of the cylinder times the length
of the stroke, and equals the amount of steam displaced
by the piston during one stroke.

The pressure on the cross-head exerted through the
connecting rod and crank pin causes a strain on the
guides which varies throughout the stroke, and varies

37



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38 ELEMENTARY STEAM ENGmEERING.

on different engines according to the angle taken by the
connecting rod. This angularity of the connecting rod
on any individual engine depends on the length of the
rod itself, for a short rod would, evidently, at some point
produce an angle more nearly approaching a right angle
than would be possible if the rod were of two or three
times its length. The length of the connecting rod must
not be too long, however, both on account of the space
it would require and because a long rod would be flexible
and mechanically weak.

To determine the pressure or thrust on the guide we
may use the following formula:
Maximum thrust =

pressure on piston X radius of the crank in inches
length of the rod in inches

The RADIUS OF THE CRANK is measured from the
center of the crank shaft to the center of the pin.

The diameter of the circle described by the crank pin
is equal to the stroke of the engine and its radius is
sometimes called the THROW.

The portion of the total pressure on the crank pin
which turns it about the shaft center is called the TAN-
GENTIAL PRESSURE.

If the circumference of the circle. Fig. 21, represents
the path of the crank pin, then at the points a and e, or
the dead centers, no turning effort is produced. But
with the pin at c the turning effort is wholly expended in
turning the shaft, and is equal to the line cb drawn to
any convenient scale. The turning or tangential pres-
sure varies from nothing at a to maximum at c.



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THE PISTON.



39



Considering the connecting rod to be of indefinite
length and neglecting the weight of the parts, to ascer-




Fig. 21.

tain the pressure at any point as /. Through / draw the
line ff parallel to a&, resolve it into a parallelogram of
forces, as shown, and the tangential line f/will represent
the portion of the force ch used in turning the shaft,
and fr will show the amount of force pressing on the
bearing.

BRAKE TEST.

Two methods of determining the horse power of a
steam engine are illustrated in the rules given for calcu-
lating the H.P., and in the method of obtaining results
from an indicator diagram. Still another method is
often convenient: namely, by obtaining the brake H.P.



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40



ELEMENTARY STEAM ENGINEERING.



This requires the use of an apparatus called the prony
brake, and the term brake horse power "is used to
represent the power actually developed by an engine
under such a test. These pieces of apparatus vary in
design. The simple application, illustrating the prin-
ciples involved, is shown in Fig. 22. Upon the fly-wheel



l<<c^^r^b> ^-<<t^.^^^ .



?^



I





lena lh of Arm.

Fig. 22.

or pulley fastened to the main shaft is clamped a brake
consisting of two wooden pieces lined with some material
which may be saturated with water, in order to keep
the pulley and brake from over-heating. It is provided,
through the bolt a and the hand-wheel and bolt &, with
a means for readily adjusting the tightness by which the
pulley is held. The power arm of the brake c rests on
a pivot placed on a scale. Supposing the engine to be
rotating in the direction of the arrow and the hand-wheel
tightened until any further tightening would reduce the
speed below the normal running speed, then if the revo-
lutions per minute and the weight registered by the
scale be taken, the length of the brake arm being known.



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THE PISTON. • 41

we have the necessary items to determine the brake
H.P., at this speed. If, for example, the length of the
brake arm is 5 feet 3 inches, and the end resting on the
scale weighs 3 pounds before tightening the hand-wheel,
and if the engine is running at the required speed, with
the weight registering 38 pounds, there remains 35
pounds as the actual pull or force exerted by the engine.
Supposing the revolutions to be 1000 per minute, and
the brake arm were free to revolve, the extremity of
the arm would describe a circle 10 feet and 6 inches in
diameter, or 33 feet in circumference, and, as shown on
the scale, it would have a pressure or pull of 35 pounds.
Therefore, on the same principle as before, the H.P.
equals the weight times the distance divided by 33,000.
We then have the following result:

33 X 1000 X 35



33,000



= 35B.H.P.,



or the brake horse power at this particular speed.

The foot pound is the unit of mechanical
WORK. The foot pound represents the amount of work
required to lift a weight of one pound one foot high.

THE HORSE POWER IS THE UNIT OF POWER.

33,000 FOOT POUNDS EQUAL ONE HORSE POWER.

To determine the horse power we must note the
number of foot pounds of work done in a given time.



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42 ELEMENTART STEAM ENGINEERING.

QUESTIONS.

1. What is piston speed?

2. What is piston displacement?

3. What is the angularity of the connecting rod?

4. What effect has the angularity of the connecting rod on
the crank pin?

5. How may the maTimum thrust on the guides be determined ?

6. What is the crank radius?

7. What is tangential pressure?

8. What will be the mean pressure on the crank pin of an
engine 24 in. in diameter, 36 in. stroke, 45 lb. M.E.P. ?

9. In an engine having a cylinder diameter of 36 in., a stroke of
24 in. and a mean pressure of 50 lb., what is the mean pressure
on the crank pin?

10. How may the brake horse power be determined?

11. Find the B.H.P. of an engine making 250 revolutions with
a scale pressure of 100 lb. and a brake arm 3 ft long, the weight
of the arm being 5 lb.



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CHAPTER VI.

WORK DONE BT STEAM DURING FORMATION.

Let us assume a vessel in which we have placed one
pound of water at a temperature of 32'' F., and resting
upon the surface of the water is a piston and a weight
which together represent the pressure of the atmosphere.
If we apply heat to the vessel the water will become
heated and the temperature will continue to rise until
the thermometer registers 212^ F. If now the applica-
tion of the heat be continued, steam begins to form. As
it forms, the piston rises and continues to rise during the
formation of the steam. The thermometer, however,
does not indicate any change in the temperature during
this time. Finally a point is reached when the last drop
of water has been transformed into vapor. This vapor
is called SATURATED STEAM. If the application of
heat is still further continued we will produce SUPER-
HEATED STEAM, the temperature of which will rise
according to the continued application of heat and will
perform additional work by lifting the weight to a g|-eater
height. Now, if the vessel is of considerable height and
has an area of i square foot equal to 144 square inches,
as I cubic foot of water weighs 62.5 pounds, i pound of

water placed in such a vessel will stand - — or 0.016 of

62.5

a foot high. As the weight of the atmosphere at the sea

level equals 14.7 pounds per square inch, there will be

43



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44 ELEMENTARY STEAM ENGINEERING.

Upon the surface of the water in this vessel a weight
equal to 14.7 pounds per square inch or a total weight
of 14.7 times 144, equal to 21 16.8 pounds.

Starting with a temperature of 32^ and applying heat
to it as before until the thermometer registers 212^, at
which point formation of steam begins, we find that we
have added to the water the difference between the
two amounts, or 180 heat units, which amount is required
to raise the water to the point where the formation of
steam begins.

A SEAT UNIT (H.U.) is the amount of heat re-
quired to raise the temperature of one pound of water
one degree Fahrenheit, taken at its greatest density,
about 39° F.

Contintiing now the application of heat as before, until
the water has been entirely changed into steam, we find
that the piston has risen to such a point that at the
moment when the water has all become steam it occupies
the volume of 26.36 cubic feet. The total H.U. required
to produce this change were 1146, of which 180, we have
previously noted, were required to raise the water to
steaming point and the remaining 966 to effect the mo-
lecular transformation of water into steam. During the
second portion of this experiment no rise in temperature
has been observed on the thermometer and yet we know
that heat has been delivered to the water in order to
effect the transformation. The heat measured during
the first portion of the experiment is called SENSIBLE
HEAT. That absorbed during the second portion, but
not showing on the thermometer, is called LATENT
HEAT. This latent heat, 966 H.U., served two purposes.



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WORE DONE BY STEAM DUIONG FORMATION. 45



First, to change the water into steam.

Second, to do external work by lifting the weight
pressing upon the surface of the water. If the weight
or piston equals 21 16.8 pounds and we have raised it
26.36 feet| this will equal 55,799 foot pounds.

Now, as we have seen that the application of heat will
produce work, there must be some relation between heat
and work, — between a certain amount of heat and a
definite amount of work produced, which is expressed in
foot pounds.

Referring to the definition of H.U., which is the amount
of heat required to raise the temper-
ature of one pound of water i^ F.,
taken at its greatest density, about
39^ F., experiment has shown that
ONE H.U. EQUALS 778 FOOT POUNDS
OF WORK.

This being the case, we find that
to do the 55,799 foot pounds of work
previously mentioned 55,799 4- 778 or
71.72 H.U. were required. The bal-
ance of the 966 H.U. used in changing
the water into steam, or 894.28 H.n.,
was used in doing external work.

Now, if we divide these figures,
71.72, 894.28, and 180, respectively by Fig- 23-

the smaller, we obtain the ratio of i : 12.48 : 2.5, which
ratio expresses the respective proportion of the total
1 146 H.U.

To illustrate this graphically let us draw a rectangle,
Fig. 23, according to these ratios. We will note then



12.48



J



2.6



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46 ELEMENTARY STEAM ENGmEERING.

that the external work is the small shaded area at the
top. It is self-evident that the external or useful work
done by the steam is a very small part of the total work,
and as the efficiency equals the area of external work
divided by the total area we get:
-^ . area of external work i * ^ ^

^^''''''^ = iSteHJiS^ 76 °' "^^ ^-^5^-

Therefore, neglecting all other losses, which are con-
siderable, in actual practice only the heat from 125
pounds of coal per ton will produce useful work.

If, instead of raising the steam to the pressure of
14.7 pounds per square inch as in the previous instance,
we increase the pressure to 100 pounds per square inch,
on referring to a table of saturated steam we find that
one pound of steam at this pressure of 100 pounds per
square inch occupies a voltime of only 4.33 cubic feet.
Employing the same vessel as before for changing the
water to steam, one pound of water when changed to
steam will then equal 100 pounds pressure times 144
square inches times 4.33 equals 62,352 foot pounds, while
one pound of steam at atmospheric pressure we found
equals 55,799 foot pounds, thus showing a balance in
favor of the high pressure of 7553 foot pounds.

To produce the 100 pounds pressure, however, there
was required a total heat of evaporation of 1 181. 91 H.U.,
while for the steam at atmospheric pressure there were
required 1146.6 H.U., or 35.31 H.U. more for the high
than for the low pressure.

To find the work done per pound of water during the
formation of steam without expansion, we have the
following rule :



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WORK DONE BY STEAM DURING FORMATION. 47

Volume per pound times pressure per square inch
tim^s 14:4 equals foot pounds.

The volume per pound is found by referring to the
saturated steam table.

Example. — Find the external work done by one pound of
steam at 50 lb. pressure, the volume at this pressure being 8.31
cu. ft

Solution. — 8.31 x 50 x 144 = 591832 ft lb.

As ONE HORSE POWER IS EQUAL TO 33iOOO FOOT
POUNDS OF WORK IN ONE MINUTE, and m One hour
we have 60 minutes, then during one hour one horse
power will produce 33,000 times 60 or 1,980,000 foot
pounds. Then using the result obtained in the previous
example we have:

'^ ^'^^ = 33*09 pounds of water required to produce
591832

one H.P. for a period of one hour.

Reference to the table shows that very little more
heat is required to produce high-pressure than low-
pressure steam with equal weights of water; but if equal
volumes of steam are used, this will not be true, for
while 100 pounds of steam per square inch will exert
twice as much pressure on the piston as 50 pounds per
square inch, yet twice the volume of steam will be re-
quired. Therefore, in modem steam practice advantage
is taken of the expansive force of steam to produce
economical operations.



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48



ELEMENTARY STEAM ENGINEERING.



TABLE OF PROPERTIES OF SATURATED STEAM.



1


i*5
•O'g

i


.9


III


II

"SI

1


15 •

fl


id


1^ t

s


I


101.99


70.0


1043.0


III3.I


0.00299


334.5


I


2


126.27


94.4


1 026. 1


1 120.5


0.00576


173.6
1 18.5


2


3


141.62


109.8


1015.3


"^H


0.00844


3


4


153.09


121.4


1007.2


1 128.6


0.01107


90.31


4


5


162.34


130.7


1000.8


1131.5


0.01366


73.21


5


6


170.14


138.6


995.2


I 133.8


0.01622


61.67


6


7


176.90


145.4


990.5


"35.9


0.01874


53.37


I


8


182.92


151.5


986.2


1137.7


0.02125


47.06


9


188.33


156.9


982.5


1 139.4


0.02374


42.12


9


10


193.25


161.9


970.0


1 140.9


0.02621


38.15


10


147


212.00


180.9


965.7


1 146.6


0.03794


26.36


14.7


15


213.03


181.8


965.1


1 146.9


0.03826


26.14


15


20


227.95


196.9


954.6
946.0


1151.5


0.05023


19.91


20


25


240.04


209.1


1155.1


0.06199


16.13


25


30


250.27


219.4


938.9


1 158.3


0.07360
0.08508


13.59


30


35


259.19


228.4


932.6


1161.0


11.75


35


40


267.13


236.4
243.6


927.0


I 163.4


0.09644


10.37


40


45


274.29


922.0


I 165.6


0.1077
0.1 188


9.287


45


50


280.85


250.2


917.4


1 167.6


8.314


50


55


286.89


256.3


913.1


1 169.4


0.1299


7.696


55


6o


292.51


261.9


909.3


1171.2


0.1409


7.097


60


65


297.77


267.2


905.5


1 172.7


0.1519


6.583


65


70


302.71


272.2


902.1


1 174.3


0.1628


6.143


70


I^


307.38


276.9


898.8


I 175.7


0.1736


5.762


75


311.80


281.4


895.6


1 177.0


0.1843


5.426


80


85


316.02


285.8


892.5


1 178.3


0.1951


5.126


85


90


320.04


290.0




1 179.6


0.2058


4.859


90


95


323.89


294.0


886.7


1 180.7


0.2165


4.619


95


lOO


327.58


297.9


ff*-"


1181.9


0.2271


4.333


100


105


331.13


301.6


881.3


1 182.9


0.2378


4.205


105


no


334.56


305.2


878.8


1 184.0


0.2484


4.026


no


"5


337.86


308.7


876.3


1 185.0


0.2589


3.962


115


120


341.05


312.0


874-0


1 186.0


0.2695


3.7"


120


125


344.13


315.2


871.7


1 186.9


0.2800


3.571


125


130


347.12


318.4


869.4


1 187.8


0.2904


3.444


130


140


352.85


324.4


865.1


1 189.5


0.3113


3.212


140


150


358.26


330.0


861.2


1191.2


0.3321


3.011


150



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CHAPTER Vn.
THE EXPANSIVE WORKING OF STEAM.

If we inclose a volume of gas in a cylinder having a
movable piston and gage the pressure when the gas
occupies the full volume, and if we then compress the
gas to half that volume by moving the piston along the
cylinder, the pressure will be found to have increased
to double the original amount. By still further decreas-
ing the volume to one third and then to one fourth of
the original amount the pressure will be found to have
increased correspondingly, whence we have Boyle's Law:

100



1—



SO

as



I % 3 ^

Fig. 24.

The volume of a gas varies inversely as the pres-
sure, the temperature rem^aining constant.
Construct a diagram. Fig. 24, drawing first a hori-

49



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50 ELEMENTARY STEAM ENGINEERING.

zontal then a perpendicular line meeting at some point
O (the perpendicular line representing the pressure,
and the horizontal one the distances). Divide the hori-
zontal line into four parts, and mark at some suitable
point on the vertical line the number loo to represent
100 pounds. Draw a vertical line from the number i,
on the horizontal line to an equal height with the loo
mark. Joining the two we will have a rectangle, repre-
senting a volume of gas at that pressure. Increase the
volume to twice its size and the pressure will drop to
one half or 50 pounds, and a horizontal line from that
mark will meet a vertical line erected at the figure 2 on
the line of distance, producing another rectangle. Re-
peating the same process with the distances 3 and 4,
we have the resulting pressures 33.3 and 25 pounds re-
spectively. These rectangles represent areas, and the
product of the pressures and volumes indicated on their
respective lines always produces a constant number, 100
in this instance. If the comers of the parts are con-
nected, a curved line is produced which is the well-
known hyperbolic curve. This theoretical isothermal
curve is the true curve, showing the expansion of gases
when maintained at a constant temperature.

It is not, however, a true curve for steam engines, as
we are unable, for various reasons, to maintain the con-
stant temperature in the steam engine cylinder.

As another illustration, given a volume of steam at a
pressure of 100 pounds, expand it to six times its original
volume, and the curve of expansion will be as shown in
Fig. 25.

We have previously noted that the pound of water



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THE EXPANSIVE WORKING OF STEAM.



SI



converted into steam at a pressure of loo pounds per
square inch is capable of producing work equal to 62,352



SO




\^


10


.==-.


' ' ' I I
'I'll

: : : : !



Fig. 25.

foot pounds; and under the same conditions, but at
a pressure of 50 pounds per square inch, 59,832 foot
pounds of work are produced, leaving a difference in
favor of the higher pressure of 12,520 foot pounds.




Xtnglk tn Hei. i



Now, by employing a cylinder represented by the
diagram. Fig. 26, i pound of water converted into steam
at a pressure of 50 pounds per square inch occupies a



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52 ELEMENTARY STEAM ENGINEERING.

volume of slightly over 8 cubic feet. It will, therefore,
fill our cylinder, whose volume equals 4 cubic feet, twice.
If we allow the steam to enter from behind the piston,
forcing it forward, and the steam continues to enter
during the forward movement until, at the end of the
stroke, the cylinder is filled with steam at a pressure
of 50 pounds per square inch, the average pressure
throughout the stroke will be 50 pounds per square
inch, and the total work will be as follows:
144 sq. in.
50 lb. per sq. in.
7200 lb. total pressure
4 f t. length of stroke
28,800 ft. lb. per stroke

2 strokes

57,600 ft. lb. total work done.
If we transform the water into steam at a pressure of
100 pounds per square inch, while it has twice the pres-
sure, the volume will be sufficient to fill the cylinder only
once. As before, we have :
144 sq. in.
100 lb. per sq. in.
14,400 lb. total pressure
4 ft.



57,600 ft. lb. total work done.
In either case, the same amount of work has been
done, but in the one instance the work was done during
one stroke, while in the other it required two. Also, as
we have previously seen, a slightly greater amount of
coal was required to produce the higher steam pressure.
So far, the results are equal.



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THE EXPANSIVE WORKING OF STEAM. 53

Let US, however, use our steam at 100 pounds pres-
sure, but, when the piston has gone one fourth of the
stroke, close the steam valve, allowing no more steam
to enter during that stroke.

THE POINT AT WHICH THE STEAM VALVE WAS
CLOSED IS CALLED THE POINT OF CUT-OFF.

We now have in the cylinder one cubic foot of steam
at a pressure of 100 pounds per square inch. The steam
now expands until it occupies the entire volume of the
cylinder and has forced the piston to the extreme end of
its stroke.

According to the previous illustration. Fig. 24, the drop
in pressure shown by the dotted lines will vary according
to the following table :
Pressure at:
Beginning of stroke, 100 lb.; end of ist quarter, 100 lb.



" 2nd quarter, :


100 lb.; •' 2nd


>}


50 lb.


" 3rd


>i


50 lb.; " 3rd


}}


33-3 lb.


" 4th


» 33.31b.; " 4th


19


251b.


Average for


ist quarter, loo lb.








2nd


75 lb.


1 3 5 6 7 8 9 10 11 12 13 14 15 16 17

Online LibraryErnest Victor LallierAn elementary manual of the steam engine; containing also a chapter on the theory, construction and operation of internal combustion engines for the operating engineer → online text (page 3 of 17)