ETC.
Power is distinguishable from force or pressure in that the term power
means force or pressure in motion, and since this motion cannot occur
without the expenditure of the force or pressure, power may, with
propriety, be termed the expenditure of force or pressure.
If we suppose a piston to stand in a vertical cylinder sustaining a
weight upon its surface and compressing the air within the cylinder, so
long as there is no motion no work is done, as the term "work" is
understood in a mechanical sense, and the weight merely produces a
pressure. If, however, the weight be removed, the compressed air will
force the piston upward, performing a certain quantity of work which may
best be measured by the amount of power exercised or expended.
The mechanical value of a given amount of power cannot be either
increased, diminished or destroyed by means of any mechanical device or
appliance whatsoever through which it may be transmitted.
It may be concentrated, as it were, by decreasing the amount of its
motion. It may be distended, as it were, by increasing the distance
through which it moves, or it may be expended in giving or producing
motion, but in either case the amount of duty or work done is the exact
equivalent of the amount of power applied.
A gain or increase in speed is not, therefore, a loss of power, but
merely a variation in the mode of using or utilizing such power.
For instance, 1 lb. moving through a distance of 12 inches in a given
time represents an amount of power which may be employed either as 1 lb.
moving a foot, 2 lbs. moving six inches, or 1/2 lb. moving through 24
inches, in the same space of time, the amount of the power or duty
remaining the same in each case, the method of utilization merely having
differed.
It is an inexorable law of nature that power is concentrated in
proportion as the amount of its motion is diminished, or distended in
precise proportion as such motion is increased.
[Illustration: Fig. 3351.]
Suppose, for example, that in Fig. 3351 L is a lever having its fulcrum
at F, which is 4 inches from end A, and 8 inches from end B, and
(leaving the weight of the lever out of the question) if we place an 8
lb. weight on a it will just balance 4 lbs. at B.
If the lever is moved, the amount of motion will be twice as much at end
B as it is at end A.
If we apply the power at A, the lever has become a means of converting 8
lbs. moving a certain distance into 4 lbs. moving twice that distance,
and nothing has been either gained or lost.
If we apply the power at B, the lever has merely been used as a means of
converting 4 lbs. moving a certain distance into 8 lbs. moving one half
that distance, and nothing has been gained or lost.
Suppose that end A was moved an inch, and the power at that end will be
8 inch pounds or 8 lbs. moving an inch, whereas at the end B the power
is 4 lbs. moving 2 inches; we have, therefore, reduced the weight in the
same proportion that we have increased the distance moved through.
Suppose now that the lever is moved to the position denoted by the
dotted line M M, and the leverages will be altered; that at end A
becoming that denoted by the distance from F to the vertical C, and that
for end B being denoted by the distance from F to the vertical D.
This occurs because we are dealing with gravity, which always acts in a
vertical line.
[Illustration: Fig. 3352.]
A crow bar is an excellent example of the application of the lever. In
Fig. 3352, for example, we have a 1 lb. weight on the long end of the
lever, and as we are dealing with a weight, the effective length of the
long end of the lever is from the fulcrum _f_ to _w_, which is divided
into 10 equal divisions. The short end of the lever is from _f_ to _p_,
which is equal to one division, hence the 1 lb. is balanced by the 10
lbs.
[Illustration: Fig. 3353.]
A simple method of distending power is by means of pulleys or gear
wheels. Suppose, for example, that in Fig. 3353, we have a weight of 12
lbs. suspended from a shaft or drum, whose radius _a_ is 10 inches, and
that on the same shaft there is a pulley, whose radius _b_ is 20 inches,
and the two weights will balance each other.
In this case the falling of either weight would not effect the leverage,
because the distance of both weights would remain the same from the
centre of the shaft. The leverage of the 12 lbs. is denoted by the line
_a_, and that of the 6 lbs. by _b_.
So far as the transmission of power is concerned, therefore, pulleys are
in effect revolving levers, which may be employed to concentrate or to
distend power, but do not vary its amount.
[Illustration: Fig. 3354.]
Suppose we have two shafts, on the first of which are two pulleys, B and
C, Fig. 3354, while upon the second there are two pulleys D and E. A
belt H, connecting C to D. Let the pulleys have the following
dimensions:
If we take the first pair of wheels B and C, we have that the velocity
will vary in the same ratio or degree as their diameters vary,
notwithstanding that their revolutions are equal.
Radius. Diameter. Circumference.
B = 5-1/8 inches. 10-1/4 inches. 32.2 inches.
C = 10-1/4 " 20-1/2 " 64.4 "
D = 7-5/8 " 15-1/4 " 47.9 "
E = 15-1/4 " 30-1/2 " 95.8 "
The velocity is the space moved through in a unit of time, and as it is
the circumference of the pulley that is considered, the velocity of the
circumference is that taken; thus, if we make a mark on the
circumferences of the two pulleys, B and C, Fig. 3354, the velocity of
that on C will be twice that upon B, or in the same proportion as the
diameters.
Let there be suspended from the circumference of B 10 lbs. weight, and
let us see the degree to which this power will be distended by this
arrangement of pulleys, supposing the weight to rotate B, and making no
allowance for the friction of the shaft.
Suppose the weight to have fallen 32.2 inches, and we have 10 lbs.
moving through 32.2 inches, this power it will have transmitted to
pulley B.
To find what this becomes at the perimeter of C, we must reduce the
number of lbs. in the same proportion that the perimeter of C moves
faster than does that of B; hence we divide the circumference of one
into the other, and with the sum so obtained divide the amount of the
weight; thus, 64.4 (circumference of C) ÷ 32.2 (circumference of B), =
2; and 10 lbs. ÷ 2 = 5 lbs., which, as the circumference of C is twice
that of B, will move twice as fast as the 10 lbs. at B, hence for C we
have 5 lbs. moving through 64.4 inches.
Now C communicates this to D by means of the belt H, hence we have at D
the same 5 lbs. moving through 64.4 inches.
Now E moves twice as fast as D, because its circumference is twice as
great, and both are fast upon the same shaft, hence the 5 lbs. at D
becomes 2-1/2 lbs. at E, but moves through a distance equal to twice
64.4, which is 128.8 inches. To recapitulate, then, we have as follows:
The weight gives 10 lbs. moving through 32.2 inches.
Pulley B " 10 " " " 32.2 "
" C " 5 " " " 64.4 "
" D " 5 " " " 64.4 "
" E " 2-1/2 " " " 128.8 "
That the amount of power is equal in each case, may be shown as follows:
For C, 5 lbs. moving through 64.4 inches is an equal amount of power to
10 lbs. moving through 32.2 inches, because if we suppose the first pair
of pulleys to be revolving levers, whose fulcrum is the centre of the
shaft, it will be plain that one end of the lever being twice as long as
the other, its motion will be twice as great, and the 5 at 10-1/4 inches
just balances 10, at 5-1/8 inches from the fulcrum, as in the common
lever.
In the case of D we have the same figures both for weight and motion as
we have at C, because D simply receives the weight or force and the
motion of C. In the case of E, we have the motion of the weight
multiplied four times; for the distance E moves is 128.8 inches, which,
divided by 4, gives 32.2 inches, which is the amount of motion of the
weight, hence the 10 lbs. of the weight is decreased four times, thus 10
lbs. ÷ 4 = 2-1/2 lbs., hence the 2-1/2 lbs. moving through 128.8 inches
is the same amount of power as 10 lbs. moving 32.2 inches, and we may
concentrate or convert the one into the other, by dividing 128.8 by 4,
and multiplying the 2-1/2 lbs. by 4, giving 10 lbs. moving 32.2 inches.
If, therefore, we make no allowance for friction, nothing has been lost
and nothing gained.
Thus far, we have taken no account of the time in which the work was
done, more than as one wheel is caused to move by the other, and all of
them by the motion of the weight, they must all have begun and also have
to move at the same time. Suppose, then, that the time occupied by the
weight in falling the 32.2 inches was one minute, and the amount of
power obtained may be found by multiplying the lbs. of the weight by the
distance it moved through in the minute, thus 10 lbs. moving 32.2 inches
in a minute gives 32.2 inch lbs. per minute, being the amount of power
developed by the 10 lb. weight in falling the 32.2 inches.
We may now convert the power at each pulley perimeter or circumference
into inch pounds by multiplying the respective lbs. by the distance
moved through in inches, as per the following table:
Distance moved.
Lbs. Inches. Inch lbs. of power.
Weight at B 10 × 32.2 = 322
" " C 5 × 64.4 = 322
" " D 5 × 64.4 = 322
" " E 2-1/2 × 128.8 = 322
If we require to find the power in foot lbs. per minute, we divide by 12
(because there are 12 inches in a foot), thus 322 inch lbs. ÷ 12 = 26.83
foot lbs. per minute.
Now suppose that B was moved by a belt, with a pull of 10 lbs. at its
perimeter, and made 100 revolutions in a minute instead of one, then the
pull at the perimeters of C, D, and E would remain the same, but the
motion would be 100 times as great, and the work done would therefore be
increased one hundred fold. It will be apparent, then, that the time is
as important an element as the weight.
The velocity and power of gear wheels are calculated at the pitch
circle.
Now suppose the gear A in Fig. 3355 has 30, gear B 60, gear C 10 and
gear E 80 teeth, and that 5 lbs. be applied at the pitch circle of A; to
find what this 5 lbs. would become at the pitch circle of E, we multiply
it by the number of teeth in B and divide it by the number of teeth in
C, thus:
Lbs.
At pitch of circle A 5
Number of teeth in B 60
---
Number of teeth in C 10 ) 300
---
30
--
Answer, 30 lbs. at the pitch circle of E.
Now suppose that on the shaft of A there is a pulley 20 inches in
diameter, and that on this pulley there is a belt exerting a pull of 5
lbs., while on the shaft of E there is a pulley 16 inches in diameter,
and to find how much this latter pulley would pull its belt, we proceed
as follows:
2 ) 20 = Diameter of pulley on A.
--
10 = Radius of pulley on A.
5 = Pull on pulley A.
--
Number of teeth in A = 30 ) 50 = Pull at centre of shaft
----- of A.
1.666
60 = Number of teeth on B.
------
Number of teeth on C = 10 ) 99.960 = Pull at axis of shaft
------ of B.
9.996
80 = Number of teeth on E.
-------
Radius of pulley on shaft of E 8 ) 799.680 Pull at axis of shaft E.
-------
99.96 Pull at perimeter of last
pulley.
We have in this case treated each pulley as a lever whose length
equalled the radius of the pulley, while in the case of the wheel we
have multiplied by the number of teeth when the power was transmitted
from the circumference to the shaft, and divided by the number of teeth
(the number of teeth representing the circumference) when the power was
transmitted from the shaft to the teeth.
We thus find that power is composed of three things, first, the amount
of impelling force; second, the distance that force moves through; and
third, the time it takes to move that distance.
If we take a number of pulleys, say four, and arrange them one after
another so that they drive by the friction of their circumferences, then
the amount of power transmitted by each will be equal and the velocities
will be equal, whereas, if we arrange them as in Fig. 3354, the power
will be equal for each, but the velocities or space moved through in a
given time will vary.
What is known as the unit of power is the foot lb., being the amount of
power exerted in raising or lifting one lb. one foot, and from what has
already been said, it will be perceived that this is the same amount of
power as 12 lbs. moving a distance of one inch.
Watt determined that the power of a horse was equal to that necessary to
raise 33,000 lbs. one foot high in a minute, and this is accepted, in
English speaking countries, as being a horse power.
An engine or machine has as much horse power as it has capacity to lift
33,000 lbs. a foot high in a minute.
CALCULATING THE HORSE POWER OF AN ENGINE.
The horse-power of an engine may be calculated as follows:
_Rule._--Multiply the area of the piston by the average steam pressure
upon the piston throughout the stroke, and by the length of the stroke
in inches, which gives the number of inch pounds received by the piston
from the steam during one stroke.
As there are two piston strokes to one revolution of the engine, we
multiply by two, and thus get the number of inch pounds received by the
piston in one revolution.
By multiplying this by the number of revolutions the engine makes in a
minute, we get the number of inch pounds of power received by the piston
in a minute.
By dividing this by 12, we get the number of foot pounds the piston
receives per minute, and dividing this by 33,000 lbs. we get the
horse-power of the engine.
It has already been stated that Watt determined that a horse was capable
of exerting a power equal to the raising of 33,000 lbs. one foot high in
a minute, hence, having foot pounds of the engine per minute, dividing
them by 33,000 gives the horse power.
This gives the amount of power received by the piston, but it is evident
that the engine cannot exert so much power, because part of it is
expended in overcoming the friction of the moving parts of the engine.
The amount of the piston power expended in overcoming the friction
depends upon the fit of the parts, upon the lubrication and the amount
of the load.
Thus, the friction of the cross head guides, of the cross head pin, of
the crank pin and of the crank shaft bearings will increase with the
amount of resistance offered to the piston motion.
The average pressure on the piston is a difficult thing to find,
however, for several reasons.
First, because the pressure in the cylinder may, during the live steam
period, vary from that in the steam chest because of the ports being too
small or from the passages being choked from a defective casting.
Second, because the steam is wire drawn during the time that the slide
valve is closing the port to effect the cut off.
Third, because the live steam in the port and passage at the time the
cut off occurs gives out some power during the period of expansion.
Fourth, because there is some condensation of the steam in the cylinder
after the point of cut off, and there is no means of finding by
calculation how much loss there may be from this cause.
During the live steam period there is also loss from condensation in the
cylinder, but this is made up for by steam from the steam chest.
Fifth, the loss from condensation after the cut off has occurred will
vary with the speed of the engine, and is greater in proportion as the
piston speed is less, because there is more time for the condensation to
occur in.
Sixth, there is some pressure on the piston between the time that the
exhaust begins and the piston ends its stroke.
Seventh, because the compression absorbs some of the piston power.
Assuming the average pressure on the piston to be known, however, we may
calculate the horse power as follows:
_Example._--What is the horse power of an engine whose piston is 20
inches in diameter, and stroke 30, the revolutions per minute being 120,
and the average pressure on the piston 60 lbs. per square inch?
Diameter of piston 20
Diameter of piston 20
---
Diameter of piston squared 400
---
.7854
400
------------
Area of piston = 314.1600 ([<--] these two ciphers neglected.)
60 average steam pressure.
--------
lbs. pressure on piston 18849.60 ([<--] this cipher neglected.)
30 length of stroke in inches.
--------
565488.0 inch lbs. per stroke.
2 two piston strokes per revolution.
-------
12 ) 1130976 inch lbs. per revolution.
-------
94248 foot lbs. per revolution.
120 revolutions per minute.
-------
1884960
94248
--------
11309760 foot lbs. per minute.
33000 ) 11309760 ( 342.72 = horse power of engine.
99000
140976
132000
------
89760
66000
-----
237600
231000
------
66000
66000
-----
In working out the calculation, the ciphers that are decimals and are on
the right hand are neglected or taken no account of, because they
represent no value and may therefore be discarded.
[Illustration: Fig. 3355.]
Thus the area of the piston is 314.1600 inches, the two right hand
ciphers having no value. Again the lbs. pressure on the piston is
18849.60 lbs., and the right hand cipher, having no value, is discarded.
The inch lbs. per stroke is 565488.0, and the decimal cipher,
representing nothing, is discarded when multiplying by the 2.
We have in this case taken no account of the fact that the piston rod
prevents the steam from acting against a part of the piston area during
one stroke; hence for correct results we must subtract from the area of
the piston one half the area of the piston rod.
The horse power thus obtained is that which the engine receives from the
steam, and is more than the engine is capable of exerting to drive
machinery, because a part of this power is consumed in overcoming the
friction of the working parts of the engine.
TESTING THE HORSE POWER OF AN ENGINE.
[Illustration: Fig. 3356.]
The useful horse power of a stationary engine may be readily and
accurately obtained by means of a pair of scales, and a brake, as shown
in Fig. 3356, which is constructed and used as follows:
On the crank shaft of the engine is a pulley enveloped by a friction
brake, which consists of an iron band, to which wooden blocks are
fastened.
The ends of the iron band do not meet, but are secured together by a
bolt as shown.
By screwing up the bolt the wood blocks are brought to press against the
circumference of the wheel.
This forms a friction brake that would revolve with the wheel, were it
not for two arms that are secured to the brake, and rest at the other
end upon a block placed upon a pair of scales.
The principle of action of this device is that the amount of friction
between the brake and the wheel is weighed upon the scales, and this
amount, multiplied by the velocity of the wheel at its circumference and
divided by 33,000, is the horse-power of the engine.
It is necessary, in arranging this brake, to have its end rest upon the
scale at the same height from the floor as the centre of the crank
shaft, so that the line marked 5' 3" (5 feet 3 inches), which represents
the length of the lever, shall stand parallel with the surface of the
platform of the scale.
To test the horse-power, we proceed as follows:
Suppose the pressure of the end of the lever on the scale is found by
the weight on the scale beam to be 540 lbs., the diameter on which the
brake blocks act being 3 feet, the length of the leverage being 5 feet 3
inches, as marked, and the engine making 150 revolutions per minute, and
the calculation is as follows:
540 lbs. on scale.
5.25 leverage in feet.
-----
2700
1080
2700
------
radius of pulley in feet 1.5 ) 2835.00 ( 1890 lbs. at pulley perimeter.
15
---
133
120
---
135
135
----
...0
====
Then
3.1416
3 diameter of pulley in feet.
------
9.4248 circumference of pulley in feet.
150 revolutions per minute.
-------
4712400
94248
---------
1413.7200 velocity of pulley perimeter.
1890 pounds at pulley perimeter.
---------
12723480
1130976
141372
----------
2671930.80 foot lbs. per minute.
==========
Then
33000 ) 2671930.80 ( 80.9
264000
------
319308
297000
------
223080
======
Answer, 80-9/10 horse power.
In this calculation we have nothing to do with the size of the cylinder
or the steam pressure, because the scale beam tells us how many lbs. the
brake exerts on the scale, and we treat the brake and brake pulley as
levers. Thus by multiplying the lbs. on the scale by the leverage of the
brake arm we get the number of lbs. exerted at the centre of the crank
shaft, and by dividing this by the radius of the brake pulley we get the
number of lbs. on the circumference, or, what is the same thing, the
perimeter of the brake pulley.
By multiplying the circumference of the pulley in feet by the
revolutions per minute, we get the speed at which the pounds travel, and
by multiplying this speed by the number of lbs. we get the foot lbs.
per minute, which, divided by 33,000, gives us the effective horse power
of the engine.
This effective horse power is correct, because in loading the engine by
the brake the crank pin, the cross head guides, etc., are all placed
under the same friction as they would be if it was a circular saw, or
some other piece of machinery or machine that the engine was driving.
SAFETY VALVE CALCULATIONS.
Among the most frequent questions asked in an engineer's examination are
those relating to the safety valves of boilers.
These questions may be easily answered from a study of the following:
The safety valve is a device for relieving the boiler of steam after it
has reached a certain pressure.
This it accomplishes by letting the steam escape after it has reached
the required pressure.
At what pressure the safety valve will blow off depends upon the
position of the weight on the safety valve lever.
The calculations referring to this part of the subject are, finding how
much weight will be required to be placed at a given point on the lever,
in order, with a given sized valve, to blow off at a given pressure.
Finding the position on the lever of a given amount of weight, in order
to blow off at a certain pressure.
Finding, with a given sized valve and a given weight, how to mark off
the lever and where the notches must be cut for given pressures.
In each of these calculations there are three elements: first, the area
of the valve and the steam pressure, which constitute the effect of
acting to lift the valve; second, the amount of the weight and its
position upon the lever, which acts to keep the valve closed; and third,
the weight of the lever and of the valve, which act to keep the valve
closed.
[Illustration: Fig. 3357.]
In Fig. 3357 we have a drawing of a safety valve shown in section, and
if there was no weight upon the lever, the pressure of steam the valve
would hold in the boiler would be that due to the weight of the valve
and of the lever upon the valve.
To find out how much this would be, we would have to put the valve
itself and the pin _a_ on a pair of scales and weigh them.
Then put a piece of string through the hole at _a_ in the lever, and see
how much it weighed when suspended from that point.
Suppose the valve and pin to weigh 2 lbs. and the lever (suspended by
the string) 10, and the total will be 12 lbs.
Next we find the area of the valve, and suppose this to be 8 square
inches; then we may find how much pressure the valve would keep in the
boiler, by dividing the area of the valve into the weight holding the
valve down, thus:
Lbs.
Weight of valve and pin, 2
" " lever, 10
--
Area of valve, 8 ) 12
---
Pressure the valve would hold, 1.5 lbs.
The area of the valve is that part of its face receiving the steam
pressure when the valve is seated, so that if the smallest part of the
valve diameter is equal to the diameter of the seat bore, the diameter
from which the valve area is to be calculated will be that denoted by D
in the figure, and cannot in any case be less than this. But if the
smallest end of the valve cone is of larger diameter than the smallest
end of the seat cone (which should not, but might be the case), then it
is the smallest diameter of valve cone that must be taken in calculating
the area, because that is the area the steam will press against.
Now suppose we rest a 20 lb. weight on the top of the valve that is on
the point denoted by I, and there will be 32 lbs. holding the valve
down, thus, weight of valve 2 lbs., of lever 10, and weight added, 20
lbs., and to find how much pressure this would hold in the boiler, we
divide it by the valve area, thus:
Weight on valve.
Valve area = 8 ) 32
--
4 = pressure valve will hold.
But suppose we put the weight on the lever, in the position shown in the
figure, which is six times as far from the fulcrum F of the lever as the
valve is, and its effect on the valve will be six times as great as it
would if placed directly upon the valve, so that leaving the weight of
the valve and of the lever out of the question (as is commonly done in
engineers' examinations), we may find out what pressure the valve will
hold, as follows:
_Rule._--Divide the length of the lever by the distance from the centre
of the valve to the centre of the fulcrum. Multiply by the amount of the
weight in lbs., and divide by the area of the valve.
_Example._--The area of the valve is 8 inches, the distance from the
centre of the fulcrum to the centre of the valve is 4 inches, and the
distance from the fulcrum to the point of suspension of the weight 24
inches, the weight is 40 lbs., what pressure will the valve hold?
Length of lever.
From fulcrum to valve, 4 ) 24
--
6
40 amount of weight.
---
Area of valve, 8 ) 240
---
30
Lbs. per square inch the valve will hold = 30.
The philosophy of this is clear enough when we consider that as the
weight is six times as far from the fulcrum as the valve is, and each 1
lb. of weight will press with a force of 6 lbs. on the valve, hence the
40 lbs. will press 240 lbs. on the valve, and as the valve has 8 square
inches, the 240 becomes 30 lbs. for each inch of area.
_Example._--The area of a safety valve is 8 inches, the distance from
the fulcrum to the valve is 4 inches, and the weight is 40 lbs., how far
must the weight be from the fulcrum to hold in the boiler a pressure of
30 lbs. per square inch?
In lbs.
From fulcrum to valve, 4 ) 40 amount of weight.
--
10
Area of valve, 8 square inches.
Pressure required, 30
---
10 ) 240
---
24
Answer = 24 inches from the fulcrum.
_Example._--The diameter of a safety valve is 4 inches, the distance
from the centre of the fulcrum to the valve is 3 inches, a 50 lb. weight
is 30 inches from the fulcrum, what pressure will the valve hold?
3 diameter of valve.
3
--
9
.7854
-----
36
45
72
63
------
7.0686 = area of valve.
======
3 ) 30
--
10
50 = weight
10 = leverage of weight.
-------
Area of valve, 7.068 ) 500.000 ( 70.7 = lbs. pressure per sq. in.
49476
-------
52400
49476
-----
2924
HEAT.
The heat unit, or the unit whereby heat is measured, is the quantity of
heat that is necessary to raise 1 lb. of water from its freezing
temperature (which is 32° Fahrenheit) 1°, and this unit is sometimes
termed a _thermal unit_.
The reason that some specific temperature, as 32° Fahrenheit, is taken,
is because the quantity of heat required to heat a given quantity of
water 1° increases with the temperature of the water; thus, it takes
more heat to raise 1 lb. of water from 240° to 245° than it does to
raise it from 235 to 240, although the temperature has been raised 5° in
each case.
The whole quantity of heat in water or steam is not, however, sensible
to the thermometer, or, in other words, is not shown by that instrument.
The heat not so shown or indicated is termed _latent_ heat.
Water obtains latent heat while passing from a solid to a liquid state,
as from ice into water, and while passing from a liquid to a gaseous
state, as while passing from water into steam, and the existence of
latent heat in steam may be shown as follows:
If we take a body of water at a temperature above freezing, and insert
therein a thermometer, the decrease in the temperature as the water
becomes frozen will be shown by the thermometer. If, then, its
temperature being say at zero, heat be continuously imparted to the ice,
the thermometer will mark the rise in temperature until the ice begins
to melt, when it will remain stationary at 32° so long as any ice
remains unmelted, and it is obvious that all the heat that entered the
water from the time the ice began to melt until it was all melted became
latent, and neither sensible to the sense of feeling nor to the
thermometer. Similarly, if the water, after the ice is all melted, be
heated in the open air, the thermometer will mark the rise of
temperature until the water boils, after which it will show no further
rise of temperature, although the water still receives heat. The heat
that enters the water from boiling until it is evaporated away is the
latent heat of steam. The latent heat of water is 143° Fahrenheit, and
that of steam when exposed to the pressure of the atmosphere, or under
an atmospheric pressure of 15 lbs. (nearly), is 960°, which may be shown
as follows:
If a given quantity of water, as say 1 lb., has imparted to it a
continuously uniform degree of heat sufficient to cause it to boil in
one hour, then it will take about 5-1/3 more hours to evaporate it all
away, hence we find the latent heat by taking the difference in the
amount of heat received by the water, and that shown by the thermometer
thus:
Degrees.
Temperature by thermometer at boiling point 212
Less the temperature of the water at first 32
---
Heat that entered the water in the first hour 180
Hours that the water was subsequently heated 5-1/3
-------
900
One-third of 180 = 60
---
Heat that entered the water during the 5-1/3 hours 960 degrees.
This, however, is not quite correct, as it would take slightly more than
5-1/3 hours to boil the water away, and the heat that entered the water
after it commenced to boil would be about 966 degrees.
If the steam that arose from the water while it was boiling were
preserved without increasing the pressure under which it boiled, and
without losing any of its heat, it will have a temperature the same as
that of the water from which it was boiled, which is a temperature of
212°, so that neither the steam nor the water account, by the
thermometer, for the 966° of heat that entered the water after it
boiled, hence the 966° became latent, constituting the latent heat of
the steam when boiled from and at a temperature of 212°.
The total heat of steam is the sensible heat, or that shown by the
thermometer, added to the latent heat; hence the heat necessary to
evaporate water into steam at a temperature of 212° (which corresponds
to a pressure of 14.7 lbs. per square inch) is 212° + 966°, which is
1178°, and these, therefore, are the number of degrees that must be
imparted by the coal to the water, in order to form steam at a
temperature of 212°.
WATER.
Water is at its greatest density when at a temperature of 39.1°
Fahrenheit, that is to say, it occupies its least space and weighs the
most per given quantity (as per cubic inch) when at that temperature.
At a lower temperature water expands, its freezing point being 32°
Fahrenheit, below which it forms ice. The weight of a cubic foot of
water when at its maximum density (39.1) is 62.382 lbs. Water also
expands as its temperature is increased above 39.1°; thus, while it is
heated from 39.1 to 212°, its volume increases from 1 to 1.04332. The
expansion for each degree of heat added to its temperature increases
from 0 at 40° Fahrenheit to .0043 at 212°.
The rate of expansion of water at a temperature above 212° is unknown.
STEAM.
At every temperature above freezing point water passes from the liquid
into a gaseous state, the gas being termed steam. While water is below
its boiling point its evaporation occurs at its surface only; but when
its mass is heated to boiling point, and additional heat is imparted to
it, evaporation occurs from the water lying against the surface from
which it receives the heat, and an ebullition is caused by the vaporized
water passing through the mass, the ebullition being what is known as
boiling.
The temperature at which water boils depends upon the pressure acting
upon its surface, the boiling point being at a lower temperature in
proportion as the pressure is reduced; thus water at the top of a
mountain, where the pressure of the atmosphere is less than at the sea
level, would boil at a lower temperature than 212°, which is the boiling
point when the atmospheric pressure is 14.7 lbs., which it is assumed to
be at the sea level. Conversely, the boiling point is raised in
proportion as the pressure upon its surface is raised, whether that
pressure consists of air or of steam. As, however, the pressure is
increased, the boiling point is at a higher temperature. So long as the
steam is in contact with the water both are at the same temperature, as
denoted by the thermometer (although they do not contain the same
quantity of heat, as will be show presently), and the steam is termed
_saturated_ steam.
The pressure of saturated steam cannot be either increased or diminished
without either increasing or diminishing its temperature, hence there is
a definite relation of pressure to temperature, which enables the
pressure to be known from the temperature, or conversely, the
temperature to be known from the pressure. But if the steam be separated
from the water and heated, it may be what is termed _superheated_, which
is that it may be surcharged with heat or contain more heat than
saturated steam at the same pressure. Such additional heat, however, is
latent.
The pressure of steam is the lbs. of force it exerts upon a given area,
as upon a square inch. In non-condensing engines the effective pressure
of the steam is its pressure above that of the atmosphere, because the
exhaust side of the piston being exposed to the atmosphere receives the
atmospheric pressure, which must be overcome by a corresponding
pressure of steam on the steam side of the piston, and this pressure is
not, therefore, available for producing work or power in the engine.
In condensing engines, however, the exhaust side of the piston is (as
nearly as practicable), relieved of the atmospheric pressure, and
assuming a perfect vacuum to be formed, the whole of the steam pressure
is exerted to propel the piston, in which case the steam pressure is
termed the _absolute_ pressure.
In considering the weight or density or the expansion of steam, its
_absolute_ and not its effective pressure must obviously be taken.
What is termed dry steam is _saturated_ steam that does not contain what
may be termed entrained water, which is water held in suspension in the
steam, which may be caused by the surface of the water through which the
steam is allowed to rise being too small in proportion to the volume of
steam formed, in which case the rapid passage of the steam through the
water causes it to carry up water with it and hold it in suspension,
this action being termed _foaming_ or _priming_.
Suppose, for example, that a boiler be filled with water up to the
bottom of the steam dome, then all the steam formed would require to
find exit from the water within the area of the dome, and the violence
of the ebullition would cause foaming. Obviously, then, to obtain dry
steam there must be provided a sufficient area of water surface for the
steam to pass through.
But water so entrained is evaporated into steam, if the steam is wire
drawn, that is, allowed to expand and reduce in pressure.
THE EXPANSION OF STEAM.
A cubic inch of water, when evaporated into steam at a pressure of 14.7
lbs. per square inch, occupies as steam a space or volume of 1644 cubic
inches, and its weight will be equal to that of the water from which it
was evaporated.
If additional heat be imparted (after its evaporation into steam), such
additional heat becomes latent and does not cause an increase of
sensible temperature or of pressure.
The weight of a given volume of steam, therefore, bears a definite and
constant relation to the pressure and sensible temperature of the steam,
so that the pressure or the sensible temperature being known, the weight
of a given volume, as say a cubic foot, may be known therefrom. Or the
weight of a cubic foot of steam being known, its sensible temperature
and pressure may be known therefrom.
This would not be the case if steam expanded by heat. Suppose, for
example, we have a cubic foot of steam at any absolute pressure, as say
15 pounds per square inch, a cubic foot weighing .0387 of a lb., and its
sensible temperature will be 213°. Now it is evident that the weight
will remain the same whatever the amount of heat that may be imparted to
the steam. Now if the steam were maintained within the cubic foot of
space, and was capable of expansion by the absorption of additional
heat, its pressure would increase and its weight remaining the same,
there would be no definite relation between the weight and the
temperature and pressure.
But if the cubic foot of steam were allowed to expand so as to occupy
more space, then additional heat is necessary to prevent its
condensation.
The relation between the temperature, pressure, and weight of steam is
not quite proportional to the volume, because steam is not a perfect
gas, and does not, therefore, strictly follow Mariotte's law.
A perfect gas is one that during expansion or compression follows the
law laid down by Boyle and Mariotte, this law being that, if maintained
at a constant temperature, the volume is inversely proportional to the
pressure.
For example, the quantity of gas that, if confined in a cubic inch of
space, would give a pressure of 80 lbs. per square inch, would give a
pressure twice as great (or 160 lbs. per inch of area), if confined in
one-half the space, that is, if compressed into one-half of a cubic
inch. Conversely, if the cubic inch was allowed to expand until its
pressure was 40 lbs., it would occupy 2 cubic inches of space, assuming,
of course, that the temperature remains the same. Since, however, if a
gas be compressed, its temperature is increased by reason of the
friction of the particles moving one upon the other, the law of Mariotte
may be better explained as follows:
Suppose we have three vessels, A, B, and C, filled with a fluid which is
a perfect gas, the temperatures being equal. Let the pressure be: A 40,
B 80, and C 160 lbs. per square inch, then 2 cubic inches of the fluid
in B will weigh the same as 4 cubic inches in A, because that in B is at
twice the pressure of that in A, and the 2 cubic inches in B will weigh
the same as 1 cubic inch in C, because its pressure is one-half that of
C, or, what is the same thing, whatever number of cubic inches of the
fluid in C it takes to weigh a pound, it will take twice as many in B,
and four times as many in A to weigh one pound.
But steam is not a perfect gas, as is evidenced by the fact that its
volume does not increase in a ratio inverse to its pressure. For
example, if a cubic inch of water be evaporated into steam at a pressure
of 14.7 lbs. per square inch, its volume will be 1644 cubic inches, and
its temperature 212° Fahrenheit.
But if the cubic inch of water be evaporated into steam at twice the
pressure, which is 29.4 lbs per square inch, its volume will be 838
inches.
The volume then is not inversely as the pressure, although the actual
quantity and weight remain the same, as is proven by the fact that if
the steam at either pressure were condensed it would pass back into the
cubic inch of water from which it was generated.
This may be accounted for in the difference in the boiling point of the
water in the two pressures, or in other words, by the difference in the
temperatures; thus the boiling point of the water at a pressure of 14.7
lbs. is 212°, while that for the pressure of 29.4 is increased about
38.4 degrees, and the steam is at the higher pressure expanded by these
38.4 degrees of heat, which adds to its pressure, although not affecting
its actual quantity or weight.
The amount of this expansion may be estimated as follows:
Taking the 1644 cubic inches, and supposing the steam to be a perfect
gas, we divide it by 2 to obtain half the volume, 1644 ÷ 2 = 822.
If then we subtract this 822, which is the volume of the steam if it
acted as a perfect gas from the 838 it actually occupies, we get 16
(838-822 = 16), which is the number of cubic inches of expansion due to
the increase in the boiling temperature.
THE CONVERSION OF HEAT INTO WORK.
When steam performs work a certain portion of the heat it contains is
converted into work, the steam simply being a medium of conveying the
heat into the cylinder in which the motion of the piston converts this
proportion of heat into work. It has been proven that a given quantity
of heat will pass into a given quantity of work, and conversely that a
given quantity of work is convertible into a given quantity of heat, and
it has also been proven that so much heat is convertible into so much
work, independent of the temperature of the heat during its conversion
into work, power, or energy, all three of these words being used to
imply pressure, force, or weight in motion.
The accepted measurement of the conversion of heat into work is known as
_Joule's equivalent_; _Joule_ having determined that the amount of power
exerted in raising 772 lbs. one foot is the equivalent of the amount of
heat that is required to raise the temperature of 1 lb. of water when at
or near its freezing point (that is, at a temperature of 32°) one
degree.
This is called the _mechanical equivalent of heat_, being merely the
quantity of heat necessary to do a certain amount of work, but having no
relation to the time in which that work was done.
The conversion of heat into work and of work into heat may be
demonstrated as follows: Suppose a cylinder to be so situated that heat
can neither be transferred to it or from it, and that saturated steam be
admitted under the piston so as to fill one-half of the cylinder at a
pressure of 50 lbs.
Suppose then that we raise the piston from an independent application of
power, the steam simply expanding to fill the space given by the piston,
but not exerting its force to move the piston.
Now suppose the experiment is repeated, permitting the force of the
steam to lift the piston, and the temperature of the steam will be less
in the second than it was in the first, proving that in the second
experiment a certain portion of the heat in the steam was converted into
the work of raising the piston.
If we desire to reconvert the work into heat, we may force the piston
back again to its original position, and its temperature will be
restored to what it was before we allowed it to raise the piston. It is
here, of course, assumed that there is no friction in moving the piston
in the cylinder.
The apparent or external work performed by steam in expanding and moving
a piston against a given resistance is measurable by multiplying the
amount of the resistance against which the piston moved by the distance
it moved through, thus:
Suppose a piston weighs 100 lbs. and had resting upon it a weight of 50
lbs., and that it be raised by the expansive action of steam a distance
of a foot, then, since the total resistance it moved against would be
(supposing it to move frictionless in the cylinder) 150 lbs., and since
the amount of motion was 1 foot, the external or apparent amount of work
performed by the steam will be 150 foot lbs., or 150 lbs. moved 1 foot.
But in expanding, the steam has performed a certain amount of what is
called _internal_ work, that is to say, its particles or atoms have done
work in expanding, and this work has been done at the expense of some of
the heat in the steam, so that the loss of heat due to the motion of the
piston is the amount of heat converted into work in moving the piston
against the piston resistance, added to that converted into the internal
work due to the expansion of the steam.
It is because of this internal work that the steam in expanding does not
strictly follow Mariotte's law.
The mechanical theory of heat is, that the atoms of which bodies are
composed are at absolute rest when at a temperature of 461.2° below the
zero of Fahrenheit, which is supposed to be absolute cold, and at any
degree of temperature above this the atoms are in motion; the extent and
force of their motion determines what we know as the temperature of the
body.
Atoms are capable of transmitting their motion to adjoining atoms of the
same or of other bodies, losing, of course, the amount of motion they
transmit, and it is in this way that heat is conveyed from one to
another part of the same body, or from one body to another, this being
known as the heat of _conduction_.
But heat may be conveyed by means of what is known as _radiation_, and
also by _convection_.
Thus, the air surrounding a heated body becomes heated, and by reason of
its expansion it then becomes lighter and rises, a fresh supply of
cooler air taking its place, becoming in turn heated, and again giving
place to cooler air; the heat thus conveyed away by the fluid or air is
conveyed by what is termed _convection_.
Heat also passes from a body in straight lines or rays, which do not
heat the air through which they pass to their own temperature, but do
impart that temperature to a solid body, as iron or water; the heat that
passes from a body in this manner is termed radiant heat, or the heat of
_radiation_.
In the cylinder of a steam engine, therefore, the heat contained in the
steam is disposed of as follows:
A certain portion of it is converted into work through the medium of the
piston.
Another portion is conveyed away by the walls of the cylinder, this
portion including the heat of convection and that of radiation.
Yet another portion is converted into internal work. Referring to the
latter, suppose that steam is permitted to expand and its atoms will be
in motion, which motion has been derived at the expense of or from the
conversion of a certain quantity of heat.
The amount of the heat so converted obviously depends upon the amount of
the motion. Suppose, for example, that steam is generated in a closed
vessel as in a steam boiler, and that a certain pressure having been
attained, the steam is permitted to pass off as fast as it is formed
from the boiler, then the amount of atomic motion will remain constant,
because the pressure remains constant; but suppose instead of the steam
passing off, it be confined within the boiler, then the pressure will
increase and there will be a greater resistance to the motion of the
atoms, hence their motion will be less, and less of their heat will
therefore be converted into atomic motion, and, as a consequence, more
of it will exist in the form of sensible heat; hence while the pressure
of steam continues to increase, its heat is increased, not only by
reason of the heat it receives from the furnace, but also by reason of
that abandoned by the steam, because it is prevented by the pressure,
from expending it in atomic motion.
CHAPTER XL.--THE INDICATOR.
The indicator is an instrument which marks or draws a figure, or diagram
as it is called, which shows the pressure there is in the cylinder at
every point in the piston stroke, while it also shows the resistance
offered by the same body of steam to the piston on its return stroke.
From the form of this figure or diagram, the engineer is enabled to
discover whether those parts of the engine whose operation regulates the
admission of the steam to and its exhaust from the cylinder are
correctly adjusted.
From the diagram the engineer may find the average or mean effective
pressure of steam on the piston throughout the stroke, for use in
calculating the power of the engine.
He may also locate the point of cut off, of release, the amount of back
pressure, the degree of perfection of the vacuum in a condensing engine,
and the amount of compression.
From the area of the diagram the engineer may also estimate the quantity
of steam that is used, and supposing it to be dry steam, he may
calculate the amount of water used to make the steam, and assuming one
pound of coal to evaporate so much water, he may calculate the amount of
coal used to produce the steam.
The indicators commonly used upon steam cylinders contain two principal
mechanical movements; first, a drum revolving the piece of paper upon
which the diagram is to be marked, and second, a piston and parallel
motion for moving the pencil to mark the diagram upon the revolving
paper.
The drum is given a motion that, to insure a correct diagram, is exactly
timed with the piston motion.
The pencil is given a vertical movement; this movement must bear a
constant and uniform relation to the pressure of the steam in the engine
cylinder.
[Illustration: Fig. 3358.]
An indicator may be attached to each end of the cylinder or in the
middle, with a pipe passing to each end of the cylinder, as in Fig.
3358, but an indicator of the usual construction and such as here
referred to, can take a diagram, or _card_ as it is sometimes called,
from but one end of the cylinder at a time. The stop valves A and B are
used, so that the communication between the indicator and one end of the
cylinder may be shut off while a diagram is being taken from the other
end, while both ends may be shut off when the indicator is not being
used.
In the figure a piece of paper (or card, as it is commonly called) is
shown in place upon the drum with a diagram upon it.
[Illustration: Fig. 3359.]
[Illustration: Fig. 3360.]
The Thompson Indicator is shown in Fig. 3359, and in section in Fig.
3360.
The Tabor Indicator is shown in Fig. 3361, and in section in Fig. 3362.
Both are made with the piston and parallel motion as light as possible,
in order to enable the taking of diagrams at as high a speed of engine
revolution as possible.
Each consists of a cylinder and piston, the bottom surface of the latter
being in communication with the bore of the engine cylinder, so as to
receive whatever steam pressure there may be in the cylinder.
This indicator piston receives, on its upper surface, the pressure of a
spiral spring, which acts to resist the steam pressure.
The indicator piston rod actuates an arm or line on the end of which is
a pencil, which, by means of a parallel motion, is caused to move in a
straight line.
The paper or _card_ being in place upon the drum, and steam let into
the indicator, the pencil lever is moved until the pencil touches the
paper as lightly as possible, and as a result of the combined movements
of the pencil and drum, the diagram is marked, its form being
illustrated in Fig. 3363, which represents a diagram placed above a
cylinder, and the engine piston in three positions; first at the
beginning of the stroke; second, at the point of cut off (which is
supposed to be at one-third of the stroke); and third, at the point of
release where the valve first opens the port for the exhaust. For
convenience, the diagram is shown as long as the cylinder, but the
actual diagram usually measures about 2-1/2 inches high and 4-1/4 inches
long.
[Illustration: Fig. 3361.]
[Illustration: Fig. 3362.]
[Illustration: Fig. 3363.]
Supposing the cylinder to be filled with air, and the engine piston in
position 1, and the indicator piston would be at the corner A of the
diagram; but if steam were admitted, the pencil would rise vertically,
marking the line from A to B, which is therefore called the _admission
line_, or by some, the _induction line_.
If on reaching B the pressure was enough to move the engine piston, that
piston and the indicator drum would move simultaneously, and as long as
live steam was admitted the line from B to C would be drawn, hence this
is called the _steam line_, its length denoting the live steam period.
The cut off occurs when the engine piston is in position 2, and the
indicator pencil at C.
From this point the pencil will fall, in proportion as the steam
pressure falls from expansion until the exhaust begins, the piston then
being in position 3, and the pencil at D.
The line from C to D is therefore called the _expansion line_ or
_expansion curve_, and the point D the _point of release_ or _point of
exhaust_.
We have now to explain that in reality the whole of the remainder of the
line of the diagram is, in reality, the exhaust line, yet there is a
difference between the part of the line from point D to the end E of the
diagram, and that part from E to A, inasmuch as that during the period
of exhaust from D to E, the pressure is helping to propel the piston,
while after E is reached, whatever steam pressure there may remain in
the cylinder acts to retard the piston.
The line from D to E is therefore the exhaust line, and that from E to A
is the _back pressure line_ or _counter pressure line_.
In this example it has been supposed that while the piston was moving
from position 3 to the end of its stroke, and the pencil from D to E,
the indicator piston would have a steam pressure on it equal to
atmospheric pressure, hence the line from E to A, in this case,
represents the atmospheric line, and also the back pressure line.
The atmospheric line is a line drawn when there is no steam admitted to
the indicator, and represents a pressure above a perfect vacuum equal to
the pressure of the atmosphere. Its use is to show the amount of back
pressure, and in a condensing engine to show the degree of vacuum
obtained.
It also forms a line wherefrom the line of perfect vacuum, or that of
full boiler pressure, may be marked.
The steam pressure at any point in the stroke is denoted by the height
of the diagram above the atmospheric line, but the steam pressure thus
taken is obviously above atmospheric, and is thus the same as the
pressure of a steam gauge, which is also above the atmospheric pressure,
and therefore represents the pressure that produces useful effect in a
non-condensing engine.
This is what may be called a theoretical diagram, because, first, it
supposes the steam not to be admitted to the cylinder until the piston
was at the end of its stroke, and to attain its full pressure in the
cylinder before the piston lead begins to move, whereas, in order to
attain a full steam pressure at the beginning of the stroke, the valve
must have lead.
Second, it supposes the cut off to be effected simultaneously, whereas
the valve must have time to move and close the port, and during this
time the steam pressure will fall, and the curve C of the diagram will
therefore be rounded more or less according to the rapidity with which
the valve closed.
Third, it supposes the steam to have exhausted down to atmospheric
pressure by the time the piston had reached the end of the stroke,
whereas the piston will have moved some part of the back or return
stroke before the steam will have had time to exhaust down to
atmospheric pressure; and,
Fourth, it supposes the steam to remain at atmospheric pressure until
the piston arrives at the end of its return stroke, whereas the valve
will begin to close the port and cause the steam remaining in the
cylinder to compress before the piston has completed its return stroke.
In practice the diagram will, under favorable conditions, accord nearer
to the shape shown in the lower part of Fig. 3363, in which the closure
of the port for the cut off is shown by the curve at F. At the point
denoted by _g_ the valve began to close, and at the point denoted by _h_
the cut off was completely effected, and the expansion curve began.
The curve beginning at D is caused by the gradual opening of the exhaust
port.
The height of the line of back pressure above the atmospheric line shows
the amount of back pressure.
At the point _m_, where the back pressure line rises into a curve, the
valve had closed, shutting in the cylinder a portion of the exhaust
steam, which is afterwards compressed by the piston.
This curve is therefore called the _compression line_ or _compression
curve_. The point at which it begins cannot be clearly seen when the
exhaust port is closed slowly.
The compression curve ends at _p_, where it merges into the admission
line, but the exact point where the compression ends and the admission
begins cannot always be located, this being the case when the port is
opened slowly or the compression extends through a large portion of the
stroke.
The admission line is, however, in most cases nearly vertical when the
valve has lead, because the valve opens the port quickly while the
engine piston is moving at its slowest.
A diagram as drawn by the indicator does not account for all the steam
that is used in the cylinder, however, as will be seen from Fig. 3363,
because, as the paper drum of the indicator receives its motion from the
engine cross head, its length represents the length of the piston
stroke, whereas, there is a part of the cylinder bore between the piston
(when it is at the end of the stroke) and the cylinder cover that is
filled with steam as is also the steam passage.
This steam performs no useful work during the live steam period, but
obviously expands during the expansion period, and therefore affects the
expansion curve, and must be taken account of in calculating the
consumption of steam, of water, or of coal from the diagram, or in
marking in the true expansion curve.
In calculating the horse power, however, it may be neglected, as it does
not enter into that subject.
But in any calculation involving the amount of steam used, the clearance
must be marked in by a line at a right angle to the admission line and
distant from the nearest point of the admission line to an amount that
bears the same proportion to the whole length of the diagram as the
clearance does to the whole contents of the cylinder.
The clearance line is shown at L, L´, in Fig. 3363, its distance from
the admission line representing the amount of clearance which includes
the contents of the steam port and passage, as well as that of the
cylinder bore that is between the cylinder cover and the piston, when
the latter is at the end of the stroke.
A method of measuring the amount of clearance has already been given
with reference to stationary steam engines.
[Illustration: Fig. 3364.]
A diagram for a condensing engine is shown in Fig. 3364, which
corresponds to Fig. 3363, except that the line of perfect vacuum or no
pressure is marked in.
It represents a perfect vacuum, and must be marked on all diagrams from
which the consumption of steam is to be calculated, because the quantity
of steam used obviously includes that which is used in counter balancing
the pressure of the atmosphere.
Learners often get confused on this point, hence it may be more fully
explained as follows:
Suppose the engine piston to be blocked, in the middle of the cylinder,
and has on one side of it a pressure of 20 lbs. of steam by steam gauge,
and on the other the pressure of the atmosphere, and we might pump out
the steam, thus leaving the cylinder empty on that side of the piston.
The atmosphere would then exert a pressure of about 14-1/2 lbs. per
square inch on one side of the piston, and if we slowly admitted steam
again, it would have to get up a pressure of 14-1/2 lbs. per square inch
before the atmospheric pressure would be counterbalanced and the piston
be in equilibrium.
But the steam gauge would at this time stand at zero, and not show that
there was any steam in the cylinder, because the zero of the steam gauge
is atmospheric pressure.
When, therefore, the steam gauge showed a pressure of 20 lbs. of steam
in the cylinder, there would actually be a pressure of 34-1/2 lbs. of
steam per square inch.
The clearance line and the vacuum line must both, therefore, be marked
on the diagram when the quantity of steam used is to be computed from
the diagram, and also when the proper or theoretical expansion curve is
to be marked on the diagram.
This is clear, because in finding the expansion curve for a given volume
of steam the whole of its volume must be taken into account, and this
whole volume is represented by the area inclosed within the clearance
line, the steam line, the expansion curve, the exhaust line, and the
line of perfect vacuum, or line of no pressure.
The atmospheric line should be drawn after the diagram has been taken,
and while the indicator is hot, as the expansion of the indicator
affects the position of this line. It is drawn with the steam shut
entirely off from the indicator, whose piston therefore has atmospheric
pressure on both sides of it.
Whether the engine is condensing or non-condensing, the same amount of
steam (all other things being equal) is used, the only difference being
that in a condensing engine a greater portion of the steam is available
for driving the piston.
If the condenser produced a perfect vacuum, the whole of the steam would
be utilized in propelling the piston.
The "line of no pressure," or of perfect vacuum, is marked as far below
the atmospheric line as will represent the pressure of the atmosphere,
which is, at the sea level, about 14.7 lbs. per square inch when the
barometer stands at 29.99 inches.
THE BAROMETER.
A barometer is an instrument for denoting the pressure or weight of the
atmosphere, which it does by means of a column of mercury inclosed in a
tube, in which there is a vacuum, which may be produced as follows:
A tube having a parallel bore and closed at one end is filled with
mercury and while the finger is placed over the open end of the tube, it
is turned upside down and inverted in a cup of mercury that is open to
receive the pressure of the atmosphere.
The finger is then removed from the end of the tube and the mercury will
fall, leaving a vacuum at its upper end.
The pressure of the atmosphere on the surface of the mercury in the cup
forces the mercury up the tube, because the surface of the mercury in
the tube has no atmospheric pressure on it, the action being the same as
that already described with reference to the principles of action of a
pump.
The weight of the atmosphere is equal to the weight of that part of the
column of mercury that is above the surface of the mercury in the cup,
hence lines may be drawn at different heights representing the weight of
the atmosphere, or of any other gas, when the column of mercury stands
at the heights denoted by the respective lines.
But as mercury expands by heat, a definite degree of temperature must be
taken in marking a column, to represent the weight, this temperature
being 32° Fahrenheit.
Similarly, as the weight of the atmosphere varies, according to the
height at which it is taken from the surface of the earth, a definite
height must be taken.
The sea level is that usually taken, the mean or average atmosphere (at
that level) being 14.7 lbs. per square inch.
For higher altitudes, the mean atmospheric pressure in lbs. per square
inch may be found by multiplying the altitude or height above sea level
by .00053, and subtracting the product from 14.7.
Each pound on the square inch is represented by a height of 2.036 inches
of mercury, hence the height of a column of mercury at a temperature of
32° that will balance the mean weight of the atmosphere is 29.92 inches,
and to avoid fractions, it is usual (for purposes not requiring to be
very exact) to say that the atmospheric pressure at sea level is
represented by 30 inches of mercury.
The atmospheric pressure is also, to avoid using fractions, taken
roughly at 15 lbs. per square inch at sea level.
Each 2 inches of mercury will, under these conditions, represent 1 lb.
of pressure.
Vacuum gauges are based upon the same principles and subject to the same
variations as to altitude as mercury gauges or the barometer.
To find the absolute pressure, or pressure above zero, or a perfect
vacuum, we may add the pressure of the boiler steam gauge to that shown
by the mercury gauge or barometer.
In Fig. 3364 the line of no pressure is marked at 15 lbs. per square
inch below the atmospheric line of the diagram, the atmospheric pressure
being for convenience taken as 15 lbs. above a perfect vacuum.
The line of no pressure serves as a guide in showing the effectiveness
of the condenser, as well as for computing the volume of steam used, but
is not necessary in computing the horse power of a non-condensing
engine, because the gauge pressure has its zero marked to correspond
with the atmospheric pressure.
In computing the consumption of steam or water from the diagram,
therefore, both the clearance line and the line of no pressure must be
marked on the diagram, and lines of the diagram extended so as to
include them, thus accounting for all the steam that leaves the steam
chest from the piston stroke.
Indicator springs are varied in strength to suit the pressure of steam
they are to be used for.
The scale of the spring is the number of lbs. pressure per square inch
represented by a vertical motion of the pencil; thus, a 40 lb. spring is
one in which a pressure of steam of 40 lbs. per square inch would cause
the piston to rise an inch above the atmospheric line of the diagram.
The strength or tension of the spring is so adjusted as to cause the
diagram to be about 2-1/2 inches high, let the steam pressure be what it
may. The following are the scales of springs of the Thompson and Tabor
indicator.
THOMPSON INDICATOR.
Scale of Used for pressure above atmosphere
spring. if not more than
15 lbs. 21 lbs. per square inch.
20 " 38 " " " "
30 " 94 " " " "
40 " 90 " " " "
60 " 143 " " " "
TABOR INDICATOR.
10 lbs. 14 " " " "
12 " 20 " " " "
16 " 30 " " " "
20 " 40 " " " "
24 " 48 " " " "
30 " 60 " " " "
32 " 64 " " " "
40 " 80 " " " "
48 " 96 " " " "
50 " 100 " " " "
60 " 120 " " " "
64 " 128 " " " "
80 " 160 " " " "
A spring that is strong enough for a given pressure may be used for any
less pressure.
The height of the diagram will, however, be less, and accuracy is best
secured by having the diagram up to the limit of about 2-1/2 inches,
using a spring that is light enough to secure this result.
Diagrams of high speed engines, however, will have their lines
more regular in proportion as a stronger spring is used.
This occurs because the spring, being under more tension, is less liable
to vibration.
An indicator requires careful cleaning and oiling with the best of oil,
as the slightest undue friction seriously impairs the working of the
instrument.
Instructions upon the care of the instrument, and how to take it apart,
etc., are usually given by the makers of the indicator.
There are various methods of giving to the paper drum of the indicator a
motion coincident with that of the engine piston, but few of them give
correct results.
Reducing levers, such as shown in Fig. 3365, are constructed as follows:
[Illustration: Fig. 3365.]
Fig. 3365 represents a reducing lever with the indicators attached. A C
is a strip of pine board three or four inches wide and about one and
one-half times as long as the stroke of the engine. It is hung by a
screw or small bolt to a wooden frame attached overhead. A link C
one-third as long as the stroke is attached at one end to the lever, and
at the other end to a stud screwed into the cross head, or to an iron
clamped to the cross head by one of the nuts that adjust the gibs, or to
any part of the cross head that may be conveniently used. The lever
should stand in a vertical position when the piston is at the middle of
the stroke. The connecting link C, when at that point, should be as far
below a horizontal position as it is above it at either end of the
stroke. The cords which drive the paper drums may be attached to a screw
inserted in the lever near the point of suspension; but a better plan is
to provide a segment, A, B, the centre of which coincides with the point
of suspension, and allow the cord to pass around the circular edge. The
distance from edge to centre should bear the same proportion to the
length of the reducing lever as the desired length of diagram bears to
the length of the stroke. On an engine having a stroke of 48 inches, the
lever should be 72 inches, and the link C 16 inches in length, in which
case, to obtain a diagram 4 inches long, the radius of the segment would
be 6 inches. It is immaterial what the actual length of the diagram is,
except as it suits the operator's fancy, but 4 inches is a length that
is usually satisfactory. It may be reduced to advantage to 3 inches at
very high speeds. The cords should leave the segment in a line parallel
with the axis of the engine cylinder.
The pulleys over which they pass should incline from a vertical plane
and point to the indicators wherever they may be located. If the
indicators and the reducing lever can be placed so as to be in line with
each other, the pulleys may be dispensed with and the cords carried
directly from the segment to the instruments, a longer arc being
provided for this purpose. The arm which holds the carrier pulleys on
each indicator should be adjusted so as to point in the direction in
which the cord is received.
In all arrangements of this kind the reduced motion is not
mathematically exact, because the leverage is not constant at all points
of the stroke.
Pantagraph motions have been devised for overcoming these defects. Two
forms have been successfully used, which, if well made, well cared for,
and properly handled, reproduce the motion on the reduced scale with
perfect accuracy. They are shown in working position in Figs. 3366 and
3367.
[Illustration: Fig. 3366.]
Fig. 3366 represents the manner of attaching the pantagraph motion, or
_lazy tongs_, as it is sometimes called, when the indicators are applied
to the side of the cylinder. It works in a horizontal plane, the pivot
end being supported by a post B erected in front of the guides, and the
working end receiving motion from an iron attached to the cross head.
By adjusting the post to the proper height and at a proper distance in
front of the cross head, the cords may be carried from the cord pin C to
the indicators, without the intervention of carrier pulleys.
[Illustration: Fig. 3367.]
If the indicators are attached to the side of the cylinder, the simplest
form of pantagraph shown in Fig. 3367 may be used. The working end A
receives motion from the cross head, and the front piece B is attached
to the floor. The cord pin D is fixed in line between the pivot and the
working end, and the pulleys E, attached to the block C, guide the cords
to the indicators.
The indicator rigging that gives the best results at high speeds is a
plain reducing lever like that first described, provided at the lower
end with a slot that receives a stud, screwed into the cross head. The
length of the lever should be one and one-half times the engine stroke,
as given on the preceding page.
Whatever plan is followed, it is desirable to avoid the use of long
stretches of cord. If the motion must be carried a long distance, strips
of wood may often be arranged in their place and operated with direct
connections. Braided linen cord, a little in excess of one-sixteenth of
an inch in diameter, is a suitable material for indicator work.
To take a diagram, a blank card is stretched smoothly upon the paper
drum, the ends being held by the spring clips. The driving cord is
attached and so adjusted that the motion of the drum is central.
For convenience two diagrams, one from each end of the cylinder, may be
made on the same card, as shown in Fig. 3368.
[Illustration: Fig. 3368.]
TESTING THE EXPANSION CURVE.
The usual manner of testing the expansion curve of a diagram is to
compare it with a curve representing Mariotte's law for the expansion of
a perfect gas.
A theoretic expansion curve that will accord with Mariotte's law may be
constructed on the diagram by the following method:
The diagram, as drawn by the indicator, will have the atmospheric line
upon it, and from this as a basis we may mark in the line of no pressure
or line of perfect vacuum.
To do this we draw, beneath the atmospheric line, a line as far beneath
it as will represent the atmospheric line, on the same scale as the
spring used, in the indicator, to draw the diagram.
Suppose, for example, that a 30 lb. spring was used, and assuming the
atmospheric pressure to be 15 lbs. per inch, then the line of no
pressure would be drawn half an inch below the atmospheric line, because
15 lbs. pull on the spring would cause it to distend half an inch.
The clearance line must then be drawn in, according to directions that
have already been given.
The next thing to do is to divide the length of the diagram into any
convenient number of equal parts, by vertical lines parallel to, and
beginning at, the clearance line, as shown in Fig. 3369. These lines are
numbered as shown, ten of them being used because that is a convenient
number, but any other number would do.
We next decide at which part of the diagram its expansion curve and the
test curve shall touch, and in this example we have chosen that it shall
be at line 10.
We have now to find what pressure the length of line 10 represents on
the scale of the indicator spring, which in this case we will suppose to
be 25 lbs., the line measuring 25/30 of an inch, and a 30 lb. spring
having been used to draw the diagram. Next multiply the pressure (25
lbs.) by the number of the line (10), and divide the product (250) by
the number of each of the other lines in succession, and the quotient
will be the pressures to be represented by the lines.
For example, for line 9 we have that 250 divided by 9 gives 27.7, hence
line 9 must be long or high enough to represent a pressure of 27.7 lbs.
above a perfect vacuum, or in this case 27.7/30 of an inch.
For line 8 we have that 250 divided by 8 gives 31.25 lbs., hence line 8
must be high enough to represent a pressure of 31.25 lbs. above a
perfect vacuum.
The atmospheric line is, in this case, of no other service than to form
a guide wherefrom to mark in the line of no pressure, or of perfect
vacuum.
[Illustration: Fig. 3369.]
Now take the case of line 5, and 250 divided by 5 gives 50, hence the
height of line 5 must represent a pressure above vacuum of 50 lbs.
Having carried this out for all the lines from line 10 to line 1, we
draw in the true expansion curve, which will touch the tops of all the
lines.
[Illustration: Fig. 3370.]
Another method of drawing this curve is shown in Fig. 3370. Having drawn
the clearance line B C, and vacuum line D C, as before and chosen where
the curves shall touch (as at _a_), then draw from _a_ a perpendicular
_a_ A.
Draw line A B, parallel to the vacuum line, and at any convenient height
above or near the top of the diagram.
From A draw A C, and from _a_ draw _a_ _b_ parallel to D C, then from
its intersection with A C, erect the perpendicular _b_ _c_, locating on
A B, the theoretical point (_c_) of cut-off.
From a number of points on A B (which may be located without regard to
equally spacing them), such as E, F, G and H, draw lines to C, and also
drop perpendicular lines, as E _e_, F _f_, G _g_, H _h_.
From the intersection of E C with _b_ _c_, draw a horizontal line to
_e_. From the intersection of F C with _b_ _c_, draw a horizontal line,
and so on; and where these horizontals cut the verticals (as at _e_,
_f_, _g_, _h_) are points in the curve, which begins at _c_, and passes
through _e_, _f_, _g_, _h_, to _a_.
But this curve does not correctly represent the expansion of steam. It
would do so if the steam remained or was maintained at a uniform
temperature; hence it is called the isothermal curve, or curve of same
temperature. But in fact steam and all other elastic fluids fall in
temperature during their expansion, and rise during compression, and
this change of temperature slightly affects the pressure.
A curve in which the combined effects of volume and resulting
temperatures is represented is called the _adiabatic_ curve, or curve of
no transmission; since, if no heat is transmitted to or from the fluid
during change of volume, its sensible temperature will change according
to a fixed ratio, which will be the same for the same fluid in all
cases.
A sufficiently close approximation to the adiabatic curve to enable the
non-professional engineer to form an idea of the difference between the
two may be produced by the following process:
Taking a similar diagram to that used for the foregoing illustrations,
as in Fig. 3371. Fix on a point A near the terminal, where the total
pressure is 25 pounds. As before, this point is chosen in order that the
two curves may coincide there.
Any other point might have been chosen for the point of coincidence; but
a point in that vicinity is generally chosen, so that the result will
show the amount of power that should be obtained from the existing
terminal. This point is 3.3 inches from the clearance line, and the
volume of 25 pounds 996, that is, steam of that pressure has 996 times
the bulk of water.
Now if we divide the distance of A from the clearance line by 996, and
multiply the quotient by each of the volumes of the other pressures
indicated by similar lines, the products will be the respective lengths
of the lines measured from the clearance line; the desired curve passing
through their other ends. Thus, the quotient of the first or 25 lb.
pressure line divided by 996 is .003313; this, multiplied by 726, the
volume of 35 lbs. pressure, gives 2.4, the length of the 35 lb. pressure
line; and so on for all the rest.
The application of either of the above curves will show that some
diagrams are much more accurate than others, even though taken from
engines of the same design and quality of workmanship.
As a general rule, those from large engines will be more correct than
from small ones, and those from high more correct than from low speeds,
and in either case efficiently covering the steam pipes and jacketing
the cylinder, to prevent condensation, will improve the diagram.
The character of the imperfection in the expansion curve, shown by the
application of a test curve, is generally too high a terminal pressure
for the point of cut off, the first part of the curve being generally
the most correct, and nearly all the inaccuracy appearing in the last
half.
The usual explanation of this is, that the steam admitted during the
live steam period condenses because of having to heat the cylinder, and
that this water of condensation re-evaporates during the latter part of
the stroke when this water of condensation is at a higher temperature
than the expanded steam, and thus increases the pressure.
A leaky admission valve may generally, however, be looked for (or else
wet steam), if the expansion curve rises much during its lower half.
TO CALCULATE THE HORSE POWER FROM A DIAGRAM.
In calculating the horse power of an engine, the only assistance given
by the indicator is, that it provides a means of obtaining the average
pressure of the steam throughout the piston stroke.
There are two methods of doing this, one by means of a planimeter or
averaging instrument, and the other by means of lines called
_ordinates_.
The ordinates or lines are drawn at a right angle to the atmospheric
line, as shown in Fig. 3372, and each line is taken to represent the
average height or length of one-half of the space between itself and the
next lines.
[Illustration: Fig. 3371.]
Suppose, for example, that we require to get the area of that part of
the diagram that lies between the dotted lines in the figure, and it is
clear that the average height of this part of the diagram is represented
by the height of the full line between them.
Any number of ordinates may be used, and the greater their number the
greater the accuracy obtained. It is, however, usual to draw 10.
[Illustration: Fig. 3372.]
The end ordinates A and D, in the figure, should be only half the
distance from the ends of the diagram that they are from the next
ordinate, as will be seen when it is considered that the ordinate is in
the middle of the space it represents.
The ordinates being drawn their lengths, are added together, and the sum
so obtained is divided by the number of ordinates, which gives the
average height of the ordinates.
Suppose, then, that the average height of the ordinate is two inches,
and that the scale of the spring of the indicator that took the diagram
was 30 lbs., then the average pressure, shown by the diagram, will be 60
lbs. per square inch. Or in other words, each inch in the height of the
ordinate represents 30 lbs. pressure per square inch.
The mean effective pressure having been found, the indicated horse power
(or I. H. P. as it is given in brief) is found by multiplying together
the area of the piston (minus half the area of the piston rod when great
accuracy is required) and the travel of the piston in feet per minute,
and dividing the product by 33,000, an example having been already
explained.
It is to be observed, however, that when great accuracy is required a
diagram should be taken from each end of the cylinder, as the mean
effective pressure at one end of the cylinder may vary considerably from
that at the other.
This will be the case when a single valve is used with equal lap,
because, in this case, the point of cut off will vary on one stroke as
compared with the other, which occurs by reason of the angularity of the
connecting rod.
When cut off valves or two admission valves are used, it may occur from
improper adjustment of the valves. It occurs in all engines, because on
one side of the piston the piston rod excludes the steam from the
piston face, unless, indeed, the piston rod passes through both covers,
in which case the rod area must be subtracted from the piston area.
If the expansion curve in a diagram from a non-condensing engine should
pass below the atmospheric line, then the mean effective pressure of
that part of the card that is below the atmospheric line must be
subtracted from the mean effective pressure of that part that is above
the atmospheric line, because the part below represents back pressure or
pressure resisting the piston motion.
The planimeter affords a much quicker and more accurate method of
obtaining the average steam pressure from a diagram.
[Illustration: Fig. 3373.]
Coffin's averaging instrument or planimeter is shown in Fig. 3373. The
diagram is traced by the point O, and the register wheel gives the area
of the diagram.
A quick method of approximating the mean effective pressure (or M. E. P.
as it is called) of a diagram is to draw a line _a_ _b_, in Fig. 3374,
touching the expansion curve at _a_, and so inclined that the space _e_
is, as near as the eye can judge, equal to the space _d_. Then the line
_f_ drawn in the middle of the diagram, and measured on the scale of the
spring that was used to take the diagram, represents the mean effective
pressure, or M. E. P. of the diagram.
CALCULATING THE AMOUNT OF STEAM OR WATER USED.
The amount of water evaporated in the boiler is not accounted for by an
indicator diagram or card, and the full reasons for this are not known.
It is obvious, however, that the loss, from the steam being unduly wet
or containing water held in suspension, is not shown by the diagram, and
this amount of loss will vary with the conditions.
Thus the loss from this cause will be less in proportion as the point of
cut off occurs earlier in the stroke, because, as the water is at the
same temperature as the steam, it will, as the temperature of the steam
reduces from the expansion, evaporate more during the expansion period,
doing so to a greater extent in proportion as the cut off is early, on
account of there being a wider variation between the temperature of the
steam at the point of cut off and at the end of the stroke. On the other
hand, however, in proportion as the cut off is earlier, the
proportionate loss from condensation during the live steam period is
greater, because a greater length of the cylinder bore is cooled during
the expansion period, and it has more time to cool in.
Whatever steam is saved by the compression, from the exhaust, must be
credited to the engine in calculating the water consumption from the
indicator card or diagram, since it fills, or partly fills, the
clearance space.
In engines which vary the point of cut off, by varying the travel of the
induction or admission valve, the amount of compression is variable with
the point of cut off, and increases in proportion as the live steam
period diminishes; hence to find the actual water or steam consumption
per horse power per hour, diagrams would require to be taken
continuously from both ends of the cylinder during the hour; assuming,
however, that the point of cut off remains the same, that the amount of
compression is constant, that the steam is saturated, and neither wet
nor superheated, steam and the water consumption may be computed from
the diagram as follows:
WATER CONSUMPTION CALCULATIONS.--An engine driven by water instead of
steam, at a pressure of 1 lb. per square inch, would require 859.375
lbs. per horse power per hour; the water being of such temperature and
density that 1 cubic foot would weigh 62-1/2 lbs. If the mean pressure
were more than 1 lb., the consumption would be proportionately less;
and, if steam were used, the consumption would be as much less as the
volume of steam used was greater than an equal weight of water. Hence,
if we divide the number 859.375 by the mean effective pressure and by
the volume of the terminal pressure, the result will be the theoretical
rate of water consumption in pounds per I. H. P. per hour.
[Illustration: Fig. 3374.]
For the terminal pressure we may take the pressure at any convenient
point in the expansion curve near the terminal, as at A, Fig. 3375, in
which case the result found must be diminished in the proportion that
the portion of stroke remaining to be made, A _a_, bears to the whole
length of the stroke _a_ _b_; and it may also be diminished by the
proportion of stroke remaining to be made after the pressure at A has
been reached in the compression curve at B. In other words, A B is the
portion of the stroke A B, during which steam at the pressure at A is
being consumed. Hence the result obtained by the above rule is
multiplied by A B, and the product divided by _a_ _b_.
To illustrate, suppose the mean effective pressure of the diagram to be
37.6 lbs., and the pressure at A, 25 lbs., of which the volume is 996.
Then 859.375/(37.6 × 996) = 22.94 pounds water per I. H. P. per hour,
the rate that would be due to using an entire cylinder full of steam at
25 pounds pressure every stroke. But as the period of consumption is
represented by B A (_b_ _a_ being the stroke), the following correction
is required:
(22.94 × 3.03)/3.45" = 20.15; 3.03 inches being the portion B A, and
3.45 inches being the whole length _b_, _a_. This correction allows for
the effects of clearance as well as compression, since, if more
clearance had existed, the pressure at A would not have been reached
till later in the stroke, and the consumption line B A would have been
longer.
[Illustration: Fig. 3375.]
But such a rate can never be realized in practice. Under the best
attainable conditions, such as about the load indicated on the diagram,
or more on a large engine with steam tight valves and piston, and well
protected cylinder and pipes, the unindicated loss will seldom be less
than 10 per cent., and it will be increased by departure from any of the
above conditions to almost any extent. It will increase at an
accelerating ratio as the load is diminished, so that such calculations
applied to light load diagrams would be deceptive and misleading; in
fact, they have but little practical value, except when made for
comparison with tests of actual consumption for the purpose of
determining the amount of loss under certain given conditions.
DEFECTIVE DIAGRAMS.
In seeking the causes that may produce a defective diagram, the
following points should be remembered:
The indicator must be kept in perfect order, thoroughly clean and well
lubricated, so that its parts will move freely. It should always be
cleaned throughout after using.
The motion of the indicator drum should be an exact copy, on a reduced
scale, of that of the piston at every point in the stroke.
The steam pipes from the cylinder to the indicator, if any are used,
must be large enough to give a free and full pressure of steam, and care
must be taken that the water of condensation does not obstruct them or
enter the indicator.
The cord should be as strong as possible, or if long, fine wire should
be substituted.
The pencil should be held to the card with just sufficient force to make
a fine line with a sharp pencil.
The diagram should be as long as the atmospheric line, any difference in
this respect showing unequal tension of the cord, probably from unequal
pressure of the pencil to the paper or card.
A fall in the steam line could arise from too small a steam pipe, and
this could be tested by a diagram taken from the steam chest. It could
also occur from too small a steam port or an obstructed steam passage as
well as from a leaky piston.
An expansion curve that is higher than it should be may arise from a
leaky valve, letting in steam after the cut off had occurred, or if at
the later point of expansion curve, it may be caused by the steam being
wet or containing water, which evaporates as the temperature falls from
the expansion.
An expansion curve that is lower than it should be may be caused by a
leaky piston, by a valve that leaks on the exhaust side but not on the
steam side, or if the exhaust valve is separate from the steam valve, it
may leak while the steam valve is tight.
It may also be caused by the cylinder being unduly cooled, as from water
accumulating in a steam jacket.
There are many defects in the adjustment of the valve gear, or of
improper proportion in the parts, that may be clearly shown by a
diagram, while there are defects which might exist and that would not be
shown on the diagram.
It is possible, for example, that a steam valve and the engine piston
may both leak to the same amount, and as a result the expansion curve
may appear correct and not show the leak.
Insufficient valve lead would be shown by the piston moving a certain
portion of its stroke before the steam line attained its greatest height
in Fig. 3376, in which from A upwards, the admission line, instead of
rising vertically, is at an angle to the right, showing that the piston
had moved a certain portion of its stroke before full pressure of steam
was admitted.
That too small a steam port or steam pipe did not cause this defect may
be known from the following reasoning:
The port opened when the pencil was at A, which shows that the valve had
lead. At this time the piston was near the dead centre and moving slower
than it was when the pressure reached its highest point on the diagram,
and since the steam line is fairly parallel with the atmospheric line,
it shows that the port was large enough to maintain the pressure when
the piston was travelling fast, and therefore ample when the piston was
moving slow.
The remedy in this case is to set the eccentric back.
With less compression the point A would be lower.
Excessive lead is shown in Fig. 3377 by the loop at A, where the
compression curve extends up to the steam line, and the lead carries the
admission line above it, because of the piston moving against the
incoming steam.
To mark in the theoretical compression curve, the vacuum line and the
clearance line must be drawn in as in the figure, and ordinates must be
drawn.
According to the diagram, in Fig. 3377, the compression is clearly
defined to have begun at C, and at that time the space filled by steam
is represented by the distance from C to the clearance line. The
pressure above vacuum (or total pressure) of the steam in the cylinder
when the compression began is represented by the length or height of the
dotted line 1.
Now suppose the piston to have moved from the point C, where compression
began, to line 2 (which is midway between line 1 and the clearance
line), and as the compressed steam occupies one-half the space it did
when the piston was at C, therefore the steam pressure will be doubled,
and line 2 may be drawn making it twice as high as line 1.
[Illustration: Fig. 3376.]
[Illustration: Fig. 3377.]
Line 2 is now the starting point for getting the next ordinate, and line
3, must be marked midway between line 2 and the clearance line, and
twice as high as line 2, because at line 3 the steam will occupy half
the space it did at line 2. Line 4 is obviously midway between line 3
and the clearance line.
Through the tops of these lines we may draw the theoretical compression
curve, which is shown dotted in.
To find the amount of steam actually saved by the compression, we have
to consider the compression curve only, beginning at the point of the
diagram where it is considered that the compression actually began, and
ending where the compression line joins the admission line, and the
horizontal distance between these two points represents the length of
the cylinder bore filled by the compression.
To find the average amount to which the steam is compressed, we must
draw within this length of the diagram, and within the boundaries of the
compression curve, and the line of no pressure ordinates corresponds to
those given for finding the average shown pressure of a diagram, as
explained with reference to that subject, taking care to have the end
ordinates spaced half as wide as the intermediate ones, as explained
with reference to Fig. 3372.
CHAPTER XLI.--AUTOMATIC CUT OFF ENGINES.
An automatic cut off engine is one in which the valve gear is so acted
upon by the governor as to keep the speed of the engine uniform under
variations of the load the engine drives, and notwithstanding variations
in the boiler pressure. This it accomplishes by varying the point in the
piston stroke at which the live steam is cut off. This is economical
because it enables the engine to use the steam more expansively than is
possible with engines having throttling governors, which govern the
engine speed by wire drawing the steam.
There are two principal forms of automatic cut off engines, first, those
in which the steam valve spindle or rod is released from the parts that
move it to open for admission, while dash pots, weights, or springs
close the valve to effect the cut off; and second, those in which the
travel of the valve is varied so as to alter the point of cut off.
The first usually employ fly ball governors which actuate cams or stops
to trip the valves for the steam cylinders. The second usually employ
wheel governors or speed regulators, as they are sometimes termed.
The distinctive features in the action of the first, of which the
Corliss engine is the most important, is that as two admission and two
exhaust valves are used, therefore the amount of the valve lead, the
point of exhaust and amount of the compression remain the same at
whatever point in the piston stroke the cut off may occur; whereas in
the second, the lead increases, the cut off occurs earlier, and the
compression increases in proportion as the cut off occurs earlier in the
piston stroke. In this class of engine the steam valve travels as
quickly when opening the steam port for a short and early period of cut
off as it does for a late one, hence the amount of steam port opening is
as full, with reference to the piston motion, for an early as it is for
a late point of cut off. In other words, there is the same amount of
steam port opening for the first, second, third, and fourth inch of
piston motion, let the point of cut off occur at whatever point in the
piston motion it may. In engines which vary the point of cut off by
reducing the travel of the slide valve, this is accomplished by using
double ported valves or griddle valves.
[Illustration: Fig. 3378.]
Fig. 3378 represents the arrangement of the valves in a Corliss engine,
V and V^{1} being the steam valves and V^{2} and V^{3} the exhaust
valves. These valves, it will be seen, extend crossways of the cylinder
and are circular. In the figure the valves are shown in the position
they would occupy when the piston was at the crank end of the cylinder,
as in the figure.
The principles of a Corliss valve gear will be understood from the
following, which is derived from a book by the author of this work, and
entitled _Modern Steam Engines_.
In 3379 and 3380 the valve gear (which is the distinctive feature of the
engine) is represented with the parts in the position they occupy when
the cut off occurs at half stroke, the piston having moved from the head
end of the cylinder. In Figs. 3381 and 3382 the parts are shown in
position with the crank on the dead centre and the piston at the crank
end of the cylinder, valve _v_ having opened its port to the amount of
the lead.
Referring to Fig. 3379, motion from the eccentric is imparted by the rod
M to the wrist plate Y, to which are connected the rods C, C´, for
operating the admission valves, whose shafts are seen at S, S´, and the
rods F, F´, for operating the exhaust valves, whose shafts are seen at
T, T´.
The mechanism for the steam or admission valves may be divided into
three elements: first, that for operating the valve to open the port for
admission; second, that for closing the valve to effect the cut off; and
third, that which determines the point in the stroke at which the cut
off shall occur.
The first consists of the rod M, wrist plate Y, and the rods C and C´,
which operate the bell cranks _r_ _r_, _r´_ _r´_ which are fast on the
valve shafts S, S´. Upon the ends of bell cranks _r_ _r_, _r´_ _r´_, are
pivoted latch links _u_, _u´_, which have in them a recess for the latch
blocks, of which one is seen at _e_ (the rod R´ and its connection with
the valve stem being shown broken away to expose _e_ to view). During
the admission the latch block abuts against the end _y_ of the recess
_w_ and is tripped therefrom by the cam _n´_. The ends of arms _g_ of
the latch links abut against the hub of the arms _d_, _d´_ upon which
are cams _n_, _n´_, and at _a_, _a´_ are springs for keeping the ends
_g_ of latch links _u_, _u´_ against the hubs and cams of _d_, _d´_.
Referring now to the valve mechanism at the head end only, suppose the
piston to be at the head end of the cylinder, and latch block _e_ will
be seated in the recess provided in _a_ to receive it, and as the bell
crank moves, the latch block will be raised by the latch link, which is
carried by a crank arm corresponding to that seen at _x_ at the crank
end of the cylinder, and as this crank arm is fast upon the valve
spindle, the lifting of _e_ will open the valve for admission. As soon,
however, as the end _g_ of the latch link meets the cam _n´_, the latch
link will be moved so that the end _y_ of its recess will leave contact
with the latch block _e_ and the dash pot will cause rod R´ to descend
instantaneously and close the valve, thus effecting the cut off.
The period of admission, therefore, is determined by the amount of
motion the latch link _u´_ is permitted to have before its end _g_ meets
the cam _n´_, which trips the latch link, and therefore frees _e_ from
the latch link recess.
[Illustration: Fig. 3379.]
The point at which the cut off will occur, therefore, is determined by
the position of the cam _n´_, because if _n´_ is out of the way, the end
_g_ of the latch link will not meet it, the latch link will not
disengage from the latch block _e_, and the cut off would be effected by
the lap of the valve, and independently of the dash pot. As in Fig. 3379
the parts are shown in the positions they occupy at the instant the cut
off is to occur, therefore the cam _n´_ has just tripped the latch link,
and the end of _e_ has just left contact with the end _y_ of the recess
_w_ in the latch link _u´_.
The point in the stroke at which the tripping of _u´_ from _e_ will
occur and effect the cut off is determined by the governor, because _d´_
is connected to the governor through the rod G´. In proportion as the
governor balls rise, _d´_ is moved from left to right, and the end of
cam _n´_ meets _g_ earlier, or, vice versa, in proportion as the
governor balls fall, the arm _d´_ is moved to the left, _g_ will meet
the end of cam _n´_ later, and the point of cut off will be prolonged.
[Illustration: _VOL. II._ =THE CORLISS VALVE GEAR.= _PLATE XXXIV._
Fig. 3381.]
We now come to the means employed to close the valve quickly and without
shock when the latch block is released from the latch link. Referring
then to the crank end of the cylinder, the latch block for that valve is
carried upon arm _x_, to which is attached the rod R from the dash pot
piston (the arm corresponding to _x_, but at the head end being shown
removed to expose the latch block to view). We may now turn again to the
head end of the cylinder, rod R´ corresponding to rod R at the other
end, and it is seen that R´ connects to a dash pot piston _p´_ having a
stepped diameter, the lower half fitting into bore H´, and the upper
half fitting into a bore H. The piston _p´_ fits the bore H´ and fills
it when the rod R´ is at the bottom of the stroke, hence as _p´_ is
raised there is a vacuum in H that acts to cause _p´_, and therefore R´
and _x_, to fall quickly and close the valve the instant the latch block
is released from the latch link. To prevent the descent of rod R´ and
piston _p´_ from ending in a blow, a cushion of air is given in H by the
following construction:
[Illustration: Fig. 3380.]
At S and S´ are valves, threaded to screw and unscrew, the ends forming
a valve for a seat entering H.
As the rod R´ and its piston _p´_ descend, the air in H finds exit
through a hole at _h_ until that hole is closed by the piston _p´_
covering it, after which the remaining air in H can only find exit
through the opening left by the end of the valve S´, and this amount of
opening is so regulated by the adjustment of S´ that a certain amount of
air cushion is given, which prevents _p´_ from coming to rest with a
blow. The head of valve S´ is milled or knurled, and a spring _t´_ fits,
at its end, into the milled indentation, thus holding it in its adjusted
position. The under surface of the upper part of _p´_ is covered by a
leather disc, while the part that fits in H´ is kept air-tight by a
leather-cupped packing.
The connection of the cam arms _d_ and _d´_ with the governor is shown
in Figs. 3381 and 3382, in which the parts are shown in the position
they would occupy when the crank is on the dead centre and the piston at
the crank end of the cylinder. The rod G´ connects the cam arm _d´_
with the upper end of lever A, which is connected to the governor and
vibrates on its centre as the governor acts upon it.
[Illustration: Fig. 3382.]
Now suppose the speed to begin to diminish, and the governor balls to
fall, and the direction in which A will move will be for its lower end
to move to the right, thus moving _d_ to the right and carrying its cam
away from the end of the latch link, which will therefore continue to
open the port for a longer period of admission. Or, referring to Fig.
3381, it is plain that, if the governor balls were to lower from a
reduced governor speed, G´ would move to the left and cam _n´_ would be
moved away from contact with the end _g_ of the catch link, which, not
being tripped, the admission would continue. On the other hand, suppose
the governor balls to rise from an increase of governor speed, and _d´_
(Fig. 3379) would be moved to the right, and the cam _n´_ meeting _g_
earlier, correspondingly hastening the cut off.
The governor is driven by a belt from a pulley on the crank shaft to the
pulley W, Fig. 3381, whose shaft conveys motion to the governor spindle
through the medium of a pair of bevel pinions in which _v_ represents
(referring again to Fig. 3378) the steam or admission valve for the
crank end port, and _v_^{1} that for the head end port, while _v_^{2} is
the exhaust valve for the crank end, and _v_^{2} that for the head end
of the cylinder. All four valves are shown in the positions they would
occupy when the crank was on the dead centre and the piston at the crank
end of the cylinder, hence the valve positions shown correspond to the
positions the parts of the valve motion occupy in Fig. 3381.
The faces of the valves are obviously arcs of circles of which the axes
of the shafts _s_, _s´_ are the respective centres. Valve _v_ has opened
its port to the amount of the lead, which in this class of engine varies
usually from 1/32 to about 1/16 inch. As separate exhaust valves are
employed, the point of release, and (as the same valve edge that effects
the release also effects the compression) therefore that of the
compression, may be regulated at will by adjusting the lengths of the
rods F, F´, Fig. 3379, which have at one end a right and at the other a
left hand screw, so that by turning back the check nuts and then
revolving the rods their lengths will be altered.
Similarly the amount of admission lead may be adjusted by an adjustment
of the lengths of rods C, C´, which also have right and left hand
screws. Referring now to the admission valve _v_, it is seen that its
operating rod C is at a right angle to bell crank _r_, _r_, hence the
amount of valve motion will not be diminished to any appreciable extent
by reason of the wrist plate end of rod C moving in an arc of a circle,
and the point of attachment of rod C to the wrist plate is such that,
during the admission, the valve practically gives as quick an opening as
though rod C continued at a right angle to _r_. But, if we turn to valve
_v´_, which has closed its port and covers it to the amount of the lap,
we find that bell crank _r´_ and its operating rod C´ are in such
positions with relation to the wrist plate, that the motion of the
latter will have but little effect in moving the bell crank _r´_. This
is an especial feature of the Corliss valve motion and is of importance
for the following reasons:
The lap of the valve (which corresponds to the lap of a plain D slide
valve) is usually, in this class of engine, such as to cut off the steam
at about 7/8 stroke, but the adjustment of the cam position is usually
so made that, from the action of the governor, the latest point of cut
off will occur when the piston has made 5/8 of its stroke, the range of
cut off being from this to an admission equal to the amount of the lead.
As the eccentric is fixed upon the shaft, the speed at which the valve
opens the port for the admission is the same for all corresponding
piston positions. Thus suppose the piston has moved an inch from the end
of the stroke, and the valve speed will be the same, whether the cut off
in that stroke is to occur at quarter stroke or half stroke, and as the
valve continues to open the port until it is tripped, therefore, at the
moment it is tripped, the direction of valve motion must be suddenly
reversed.
As the duty of its reversal falls upon the dash pot, it is desirable to
make this duty as light as possible, which is accomplished by the wrist
motion, which acts to reduce the valve motion after the port is opened a
certain amount for the admission.
We have, therefore, that during the earlier part of the admission, the
port opening is quick because of the eccentric throw being a maximum,
while during the later part of the port opening, this rapid motion is
offset or modified by the wrist motion, thus lessening the duty of the
dash pot and enabling it to promptly close the valve.
The range of governor action, so far as the governor itself is
concerned, is obviously a constant amount, because a certain amount of
rise and fall of the governor balls will move the cams a given amount.
But the range of cut off may be varied as follows: At Z, Z´, are
adjustment nuts, by means of which the lengths of rods G, G´ may be
varied.
Lengthening rod G obviously moves arm _d_ and its cam _n_ further from
the end of the latch link _u_, and therefore prolongs the admission
period.
Shortening the rod G´ causes cam _n´_ to move around and away from the
leg _g_ of the latch link, and prolongs the admission.
The adjustment of the lengths of G and G´ may therefore be employed for
two purposes; first, to prolong the point of cut off, and maintain the
speed when the engine is overloaded, or to hasten the point of cut off
for a given engine speed, and thus adjust the engine for a lighter load.
HIGH SPEED AUTOMATIC CUT OFF ENGINES.
What are termed high speed engines are those whose pistons run at a
velocity of more than about 600 feet per minute, some making as high as
800 or 900 feet in regular work. High speed engines are usually provided
with an automatic cut off, and a majority of them vary the point of cut
off, by means of shifting the eccentric across the shaft, so as to
reduce the eccentric throw, and therefore the valve travel. This causes
the valve to cut off the steam earlier.
The eccentric, instead of being fixed upon the crank shaft, has an
elongated bore, and is hung on an arm that is pivoted at its other end
after the manner of a pendulum. This arm is called the eccentric hanger.
A wheel governor is usually employed to shift the eccentric across the
shaft. In some cases, however, two valves are employed, one effecting
the admission, the release, and the compression, and the other the cut
off.
When two valves are employed, the lead, the point of cut off, the point
of release, and the point of compression may be maintained equal for all
points of cut off; whereas, when a single valve is employed, the lead,
the point of release, and the compression will vary with the point of
cut off, or, in other words, will be different for every different point
of cut off.