GAS POWER CYCLES 659

dharm

\M-therm\Th13-4.pm5

or

T

T

v

vr

2

1

2

1

==α ...(iii) v

v

v

v

v

v

v

v

v

vr

2

1

2

3

3

1

2

3

4

1

=×=×=

F

HG

I

KJ

α

During adiabatic expansion 4-1,

T

T

v

v

(^4) r
1
1
4
1
=F^1
HG
I
KJ

−
−
γ
()γ
T 1 =
T
r
4
()γ−^1 ...(iv)
Putting the value of T 1 in eqn. (iii), we get
T T
r r
2 =^4 − 1
()
γ .α

α
γ
T
r
(^4) ...(v)
Substituting the value of T 2 in eqn. (ii), we get
T T
(^3) r r
==^41 F
HG
I
KJ
α α − α
γ
γ
γ
(). T 4
Finally putting the values of T 1 , T 2 and T 3 in eqn. (i), we get
η = 1 – γ
T
r
T
r
T
r
T
r
r
4
1
4
44
1
γγ
γγγ
α
α
γ α
α
− −
−FHG IKJ
F
H
G
G
G
G
I
K
J
J
J
J
=− −
−
F
HG
I
KJ
()
.
Hence, air standard efficiency = 1 – γ.
r
r
−
−
F
HG
I
KJ
α
γγα
...(13.14)
Example 13.31. A perfect gas undergoes a cycle which consists of the following processes
taken in order :
(a) Heat rejection at constant pressure.
(b) Adiabatic compression from 1 bar and 27°C to 4 bar.
(c) Heat addition at constant volume to a final pressure of 16 bar.
(d) Adiabatic expansion to 1 bar.
Calculate : (i) Work done/kg of gas.
(ii) Efficiency of the cycle.
Take : cp = 0.92, cv = 0.75.
Solution. Refer Fig. 13.31.
Pressure, p 2 = p 1 = 1 bar
Temperature, T 2 = 27 + 273 = 300 K
Pressure after adiabatic compression, p 3 = 4 bar
Final pressure after heat addition, p 4 = 16 bar
For adiabatic compression 2-3,
T
T
p
p
3
2
3
2
(^1) 122 1
4 122
1

F
HG
I
KJ
=F
HG
I
KJ
γ− −
γ
.

. = 1.284 Lγ= = =
N

M

O

Q

P

c

c

p

v

092

075

. 122
.
.