CHAPTER XIII
EFFECT OF LOSS OF
HEAT, GENERATED DURING
COMPRESSION, ON THE ULTIMATE USEFUL ENERGY RESIDING
IN A GIVEN QUANTITY OF COMPRESSED AIR
117.
By an accepted law of
thermodynamics, work and heat are mutually convertible at the ratio
of about 778 ft.lb. of work for every B.T.U.
In Article 41a it was stated that the work expended in
compressing air is all converted into heat. According to the law
quoted, we should expect the compressed, and therefore heated, air
to be capable of performing useful work, equal to the amount
expended in compressing it. Neglecting friction in the air engine,
this would actually be the case, if the compressed air could be used
immediately after compression and before it has lost any of its
heat.
If, on the other band, the compressed air be allowed
to cool down to the temperature which it possessed before
compression, as happens in all compressed air installations, it
would seem logical, by applying the same law quoted above, to reason
as follows:
Since the work of compression is all converted into
heat, the ability for doing useful work must have disappeared after
all this heat has been abstracted.
In the following articles it will be shown:
a. That the work of compression is all converted
into heat.
b. That, after all the heat of compression has
been abstracted, there still remains in the compressed air a certain
amount of energy for doing useful work.
c. That this is due to the energy residing in the air
before compression.
a. Referring to Fig. 17, the total work of
compressing adiabatically a volume
V_{1}
cubic feet of free air from an absolute pressure
P_{1} to an
absolute pressure P_{2}
is represented by the area
MABR. Expressed in footpounds, it is equal to 144 times
the numerical value of this area.
In
Article 44 we found: 
Area
MABR 
= 
P_{2}V_{2}
– P_{1}V_{1}

(1) 
n –
1 







therefore, total work of
compression 
W_{1} 
= 
144 
P_{2}V_{2} –
P_{1}V_{1} 
footpounds.

(2) 

n –
1 
Let 
P_{1}

= 
14.7 lb. absolute pressure per
square inch. 


P_{2} 
= 
89.7 lb. absolute pressure per
square inch. 



= 
75 lb. gage. 


V_{1} 
= 
13.09 cu. ft. which is the
volume of 1 lb. of free air at sea level and at 60^{○}
Fahr. 


n 
= 
1.406. 















1 

From equation (7),
Article 41, deduce: 
V_{2} 
= 
( 
P_{1} 
) 
n 

V_{1} 
P_{2} 

















1 





_{
0.71} 





whence 
V_{2} 
= 
V_{1} 
( 
P_{1} 
) 
n 
= 
13.09 
( 
14.7 
) 

= 
3.62 cu.
ft. 


(3) 
P_{2} 

89.7 





Substituting values in
equation (2) we get: 
W_{1} 
= 
144 
89.7 
X 
3.62 
— 
14.7 
X 
13.09 
= 
47,000
ft.lb. 
(4) 


0.406 
















After the air has been compressed adiabatically to an
absolute pressure P 2 its absolute temperature will be according to
equation (11), Article 41:







n1 





_{
0.29} 






T_{2} 
= 
T_{1} 
( 
P_{2} 
) 
n 
= 
60 + 461 
( 
89.7 
) 

= 
880°
absolute 


(5) 
P_{1} 

14.7 

















= 
419° Fahr. 



After compression, the original pound of air occupies
a volume V_{2}
= 3.62 cu. ft. and has a temperature of 419° Fahr. which is (41960)
= 359 degrees more than its initial temperature.
Now, we can imagine a volume
V_{2}
of air weighing 1 lb. to have a temperature of 60° Fahr. If we raise
the temperature of this air by (T_{2}
 T_{1}) = (880561) = 359 degrees without
changing its volume, we heat under constant volume. The specific
heat C_{v} of
air in this case is 0.168 and the amount of heat put into this pound
of air, expressed in
B.T.U.'s. is 
C_{v
}(T_{2} – T_{1}) = 0.168
X 359 = 60.3 B.T.U.’s. 
Expressed in footpounds it is: 
K_{v}
(T_{2} – T_{1}) = 131.6 X 359
= 47,000 ft.lb. 
(6) 
A comparison of equation
(6) with (4) shows that the mechanical equivalent of the heat
required to raise the temperature of 1 lb. of air from an absolute
temperature T_{1}
to an absolute temperature T_{2}
is identical with the mechanical energy expended in
compressing adiabatically 1 lb. of atmospheric air having an
absolute temperature T_{1}
to a pressure which raises the temperature of the air to an absolute
temperature T_{2}.
In other words, the mechanical work of compressing air adiabatically
is all converted into heat energy.
b. If we now allow this volume
V_{2} = 3.62 cu. ft. of compressed air, having a
temperature of 419° Fahr., to cool down to initial temperature of
60° Fahr. under constant volume, its pressure will decrease to a
pressure P_{3},
which we find
from
the formula: 
P_{3} 
= 
P_{2} 
T_{3} 
= 
89.7 
X 
521 
= 
53.2 lb.
absolute 


T^{2} 
880 













The energy residing in
this volume V_{3}
= 3.62 cu. ft. of air for doing useful work in expanding
adiabatically down from an absolute pressure of 53.2 lb. to
atmospheric pressure is represented by the area
BCGF in the diagram,
Fig. 18, and expressed in footpounds it is 144 times the numerical
value of this area.



From article 110 we deduce: 
Area
BCGF

= 
P_{3}
V_{2} – P_{1} V_{1}




n –
1 







Hence
energy 
W 
= 
144 
P_{3}
V_{2} – P_{1} V_{1}













n –
1 








Applying it to the case in hand:
P_{3}
= 53.2 lb. absolute per sq. in.
V_{2}
= 3.62 cu. ft.
P_{1}
= 14.7 lb. per sq. in.







1 





_{
0.71} 






V_{1} 
= 
V_{2} 
( 
P_{3} 
) 
n 
= 
3.62 
( 
53.2 
) 

= 
9.02 cu.
ft. 
(From
equation 13, Article 41.) 
P_{1} 

14.7 




















n = 1.406.



Hence 
W 
= 
144 
53.2 
X 
3.62 
— 
14.7 
X 
9.02 
= 
21,300
ft.lb. 




0.406 


















Comparing this with the work of
compression, we have:



Substituting values in
equation (2) we get: 
21,300 
= 
0.45 
= 
45 per
cent. 



47,000 











That is, theoretically, after cooling
down to initial temperature, there still remains in the compressed
air energy for doing expansive work to the amount of 45 per cent of
the energy expended in compressing it.
Referring to the diagram in Fig. 17, it will be noted
that part of the total work of compression represented by the area
MABR is performed by
the atmospheric air rushing into the cylinder behind the piston
during the compression stroke and not by energy furnished by the
compressor. This work is represented by the area
MAFR.
In practice, the air, after being compressed, is
delivered into the receiver. The work of delivery is jointly
performed by the compressor and by the atmospheric air. The
compressor's work is represented by the area
FBCD and the work of the
atmosphere by the area RFDO.
The net work of compression and delivery done by the air compressor
alone is represented by the area
ABCD. The compressor's share of delivery work is always
available for doing useful work in the air engine because in forcing
a volume of compressed air from the aircylinder into the receiver,
an equal volume of air is displaced therein, and this displacement
process is extended into the pipe line and finally into the air
engine, where, in making room for itself, this volume of compressed
air drives the piston forward, and thus does useful work.
It may be asked: What becomes of the energy
contributed by the atmospheric air toward compression and delivery
which is represented by the area
MADO in Fig. 17?
This energy is actually stored up in the compressed
air when the latter leaves the compressor. It could do useful work
if it were practicable to exhaust the air from the engines into a
vacuum. But since we must exhaust against atmospheric pressure, the
energy is consumed in the process of exhaustion and is therefore not
available for useful work. It is not included in the formulas
expressing power to be furnished by the compressor because it is
furnished gratis by the atmosphere; and it is not included in the
formulas expressing the useful work which a volume of compressed air
can perform, because it is not available for such work.
The following example shows the effect of heat loss
upon the total power stored up in a mass of air by the compressor.
Example.  To
compress adiabatically in one stage 100 cu. ft. of free air per
minute at sea level to 60 lb. gage and deliver it into the receiver,
requires (theoretically) 13.40 h.p. (from column 4 Table V).
If the temperature of the free air was 60° before
compression, after compression it will be 375° Fahr. (column 6 Table
V) and the volume of the compressed air will be 31.44 cu. ft.
(column 5 Table III)
If used immediately after compression, before having
lost any heat, it could do work (theoretically) to the amount of
13.40 h.p. by expanding adiabatically down to atmospheric pressure.
But if allowed to cool, before use, to initial
temperature under constant volume, the pressure will decrease to a
pressure P_{3}
which we find from the following formula:




P_{3} 
= 
P_{2} 
T_{3} 
= 
(60 + 14.7) 
60 + 461 
= 
46.6 lb.
absolute 


T^{2} 
375 + 461 












A volume of 31.44 cu.
ft. of air per minute at 46.6 lb. absolute, if allowed to expand
adiabatically down to atmospheric pressure could perform
(theoretically) an amount of work found from equation (1) Article
111:










n –
1 





Horsepower 
= 
144 n
P_{2}V_{2} 

1 – 
( 
P_{a} 
) 
n 





33,000 (n1) 

P_{2} 






























_{0.29} 






= 
144 x 1.406
x 46.6 x 31.44 

1 – 
( 
14.7 
) 


= 
6.30 h.p. 


33,000 x
0.406 

46.6 


















which is about 47 percent of
the power expended in compression and delivery.
When friction and other
imperfections are taken into account, this percentage decreases
materially.
Adding 15 percent to the power of production we get
15.43 h.p
Subtracting 15 percent from the available theoretical
energy we get 5.35 h.p., and the comparative value shrinks to 35
percent. This is further diminished by losses during transmission
which are pointed out under Articles 9394 and 97105.
c. The answer to the question,
why energy still remains in the compressed air after all the heat of
compression has been dissipated, is that a certain capacity for work
resides in the air which is due to the latter's ability to expand
when the proper conditions prevail.
Such
conditions could be brought about by confining a volume of
atmospheric air in a cylinder under a piston and then create a
partial vacuum on the other side of the piston; the atmospheric air
in the cylinder would expand and push out the piston, that is,
perform work. But creating a vacuum requires extra work, and is
therefore not of practical application in air engines.
As a
matter of fact, after all the heat generated during compression of a
volume of air has been dissipated, the compressed air possesses no
more energy than it did before compression, but
the energy which it did possess has,
by mechanical compression, been made available for doing
useful work.
To
do work, however, the air requires energy in the form of heat and
while expanding, it consumes heat that was contained in its mass
before compression. As a consequence the temperature of the
expanded air falls below that of the surrounding atmosphere. The
amount of heat consumed is equivalent to the amount of work
performed and equal to the amount of heat that would be generated in
compressing this air from the pressure at which it exhausts from the
air engine to the pressure at which it enters the same.
The
consumption of heat from the mass of the expanding air is manifested
by the cold created in and around the cylinders of an engine using
air expansively. Theoretically this is exactly the reverse of the
generation of heat in the air cylinders of a compressor.
117a.
Determination of the Value of “n"
used in adiabatic compression and expansion formulas:
From equation (6), Article 117, we have:
Work of adiabatic compression of 1 lb.
of free air: W =
K_{v}
(T_{2} – T_{1}) footpounds
(1)
in which K_{v}
= specific heat of air at constant volume, expressed in footpounds.
T_{2}
= final absolute temperature of air after being compressed to an
absolute pressure P_{2}.
T_{1}
= initial absolute temperature of air at an absolute pressure
P_{1}.
In the
diagram, Fig. 17, the area MABR
represents the mechanical work of compressing a volume
V_{1} of air
from an absolute pressure P_{1}
to an absolute pressure P_{2},
the volume of compressed air being
V_{2}.
From
equation (1)
Article 117: 
Area
MABR 
= 
P_{2}V_{2}
– P_{1}V_{1}




(2) 
n –
1 
Let
P_{1} and P_{2}
be the absolute pressures in pounds per square foot; then the work
performed,
corresponding to area
MABR: 
W 
= 
P_{2}V_{2}
– P_{1}V_{1}

footpounds 


(3) 
n – 1 
Let,
furthermore, V_{1}
and V_{2}
represent volumes occupied by l lb. of air when under an absolute
pressure of P_{1}
or P_{2}
respectively; then from equation (5) Article 20:

P_{1}V_{1}
= RT_{1} 


and 
P_{2}V_{2}
= RT_{2} 


Substituting these values in equation (3) we have: 
W 
= 
RT_{2}
– RT_{1} 
= 
R (T_{2}
– T_{1}) 

(4) 
n –
1 
n –
1 








From
equation (7) Article 20 we have: 
R =
K_{p}
– K_{v} 

Substituting in equation (4) we get: 
W 
= 
(K_{p}
– K_{v})(T_{2}
– T_{1}) 



(5) 
n –
1 
This work is equal to the work
expressed by equation (1), therefore:









K_{v}
(T_{2} – T_{1})

= 
(K_{p}
– K_{v})(T_{2}
– T_{1}) 



(5) 

n – 1 








or 
nK_{v}
–
K_{v} 
= 
K_{p}
– K_{v} 












whence 
n 
= 
K_{p} 




(6) 
K_{v} 




as first
stated under Article 40.
