GAS POWER CYCLES 627

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\M-therm\Th13-2.pm5

(b) Change in efficiency :

For air γ = 1.4

∴ r =

T

T

3

1

F 12 14 1

HG

I

KJ

/(.−)

=

1220

310

F^108

HG

I

KJ

/.

= 5.54

The air-standard efficiency is given by

ηotto = 1 –

1

()rγ−^1 = 1 –

1

(. ) 554 14 1. − = 0.495 or 49.5%. (Ans.)

If helium is used, then the values of

cp = 5.22 kJ/kg K and cv = 3.13 kJ/kg K

∴γ =

c

c

p

v

=^522

313

.

.

= 1.67

The compression ratio for maximum work for the temperature limits T 1 and T 3 is given by

r = T

T

3

1

(^121) 12 167 1
1220
310
277
F
HG
I
KJ
=F
HG
I
KJ

/(− ) /(.−)
.
γ
The air-standard efficiency is given by
ηotto = 1 –^111
()rγ−^11 (. ) 277.^67 −^1
=− = 0.495 or 49.5%.
Hence change in efficiency is nil. (Ans.)
Example 13.15. (a) An engine working on Otto cycle, in which the salient points are 1, 2,
3 and 4, has upper and lower temperature limits T 3 and T 1. If the maximum work per kg of air is
to be done, show that the intermediate temperature is given by
T 2 = T 4 = TT 13.
(b) If an engine works on Otto cycle between temperature limits 1450 K and 310 K, find the
maximum power developed by the engine assuming the circulation of air per minute as 0.38 kg.
Solution. (a) Refer Fig. 13.13 (Example 13.14).
Using the equation (iii) of example 13.14.
W = cv TTr
T
r
31 −−+^131 T 1
L
N
M
O
Q
.( )− − P
()
γ
γ
and differentiating W w.r.t. r and equating to zero
r =
T
T
3
1
F^121
HG
I
KJ
/(γ−)
T 2 = T 1 ()rγ−^1 and T 4 = T 3 /()rγ−^1
Substituting the value of r in the above equation, we have
T 2 = T 1
T
T
T T
T
(^3) TT
1
12 1 1
1 3
1
12
13
F
HG
I
KJ
L
N
M
M
O
Q
P
P

F
HG
I
KJ

/(γ−)γ− /
Similarly, T 4 =
T
T
T
T
T
T
(^3) TT
3
1
12 1 1
3
3
1
12 31
F
HG
I
KJ
L
N
M
M
O
Q
P
P

F
HG
I
KJ

/(γ−)γ− /