TITLE.PM5

626 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th13-2.pm5

(b) Determine the air-standard efficiency of the cycle when the cycle develops maximum

work with the temperature limits of 310 K and 1220 K and working fluid is air. What will be the

percentage change in efficiency if helium is used as working fluid instead of air? The cycle

operates between the same temperature limits for maximum work development.

Consider that all conditions are ideal.

Solution. Refer Fig. 13.13.

v 2 1 vv

p

3

2

4

1

Fig. 13.13

(a) The work done per kg of fluid in the cycle is given by

W = Qs – Qr = cv (T 3 – T 2 ) – cv (T 4 – T 1 )

But

T

T

2

1

= v

v

1

2

F^1

HG

I

KJ

−γ

= ()rγ−^1

∴ T 2 = T 1. ()rγ−^1 ...(i)

Similarly, T 3 = T 4. ()rγ−^1 ...(ii)

∴ W = cv TTr T

r

31 −−+^131 T 1

L

N

M

O

Q

.( )− − P

()

γ
γ ...(iii)
This expression is a function of r when T 3 and T 1 are fixed. The value of W will be maximum
when,
dW
dr

= 0.

∴

dW

dr

= – T 1. (γ – 1) ()rγ−^2 – T 3 (1 – γ) ()r−γ = 0

or T 3 ()r−γ = T 1 ()rγ−^2

or

T

T

3

1

= ()r^21 ()γ−

∴ r =

T

T

3

1

F^121

HG

I

KJ

/(γ−)

. ......Proved.