TITLE.PM5

612 ENGINEERING THERMODYNAMICS

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\M-therm\Th13-1.pm5

Example 13.6. An ideal engine operates on the Carnot cycle using a perfects gas as the
working fluid. The ratio of the greatest to the least volume is fixed and is x : 1, the lower tempera-
ture of the cycle is also fixed, but the volume compression ratio ‘r’ of the reversible adiabatic
compression is variable. The ratio of the specific heats is γ.
Show that if the work done in the cycle is a maximum then,

(γ – 1) loge

x

r

+

1

rγ−^1

• 1 = 0.
  Solution. Refer Fig. 13.1.
  V
  V

3

1

= x ;

V

V

4

1

= r

During isotherms, since compression ratio = expansion ratio

∴

V

V

3

4

=

V

V

2

1

Also

V

V

3

4

=

V

V

3

1

×

V

V

1

4

= x ×

1

r

=

x

r

Work done per kg of the gas

= Heat supplied – Heat rejected = RT 1 loge x

r

• RT 2 loge
  x
  r

= R(T 1 – T 2 ) loge

x

r = RT^2

T

T

1

2

− 1

F

HG

I

KJ

loge

x

r

But

T

T

1

2

=

V

V

4

1

F^1

HG

I

KJ

−γ

= ()rγ−^1

∴ Work done per kg of the gas,

W = RT 2 ()rγ−^1 − 1 loge x

r

Differentiating W w.r.t. ‘r’ and equating to zero

dW

dr

= RT 2 ()()lor r g {( ) }

x

xr x

r

γγ− −×−RS − e γ r −

T

U

V

W

+−

L

N

M

O

Q

P

(^12112) = 0
or ()r
r
γ− −−F
HG
I
KJ
(^111) + (γ – 1) × rγ− (^2) log
e^
x
r
= 0
or – r
r
γγ−− (^22) ++^1 r (γ – 1) loge
x
r
= 0
or r
rr
x
er
γ
γ γ
−
−+ − + −
R
S
|
T|
U
V
|
W|
2
(^12)
(^11)
.
()log = 0
or – 1 +
1
rr. γ−^2 + (γ – 1) loge^
x
r
= 0
(γ – 1) loge
x
r

• 1
  rγ−^1 – 1 = 0. ...... Proved.