TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 275

dharm

/M-therm/th5-4.pm5

∴ T 2 = 733 × 1.

0.4

012 1.4

7

F

HG

I

KJ

= 733 × (0.1446)0.286 = 421.6 K
= 421.6 – 273 = 148.6°C.
But the actual temperature is 160°C at the pressure of 1.012 bar, hence the process is
irreversible. Proved.
(ii)Change of entropy per kg of air :
The change of entropy s 2 ′ – s 1 , can be found by considering a reversible constant pressure
process between 2 and 2′.

∴ s 2 ′ – s 2 = cp loge T

T

2

2

′ = 1.005 log

e^

433

421.6

= 0.02681 kJ/kg K

i.e., Increase of entropy, s 2 ′′′′′ – s 1 = 0.02681 kJ/kg K. (Ans.)

+Example 5.29. A fluid undergoes a reversible adiabatic compression from 4 bar, 0.3 m^3

to 0.08 m^3 according to the law, pv1.25 = constant.

Determine :(i) Change in enthalpy ; (ii) Change in internal energy ;

(iii)Change in entropy ; (iv)Heat transfer ;

(v)Work transfer.

Solution. Refer Fig. 5.38.

Fig. 5.38

Initial volume, V 1 = 0.3 m^3

Initial pressure, p 1 = 4 bar = 4 × 10^5 N/m^2

Final volume, V 2 = 0.08 m^3

Law of compression : pv1.25 = constant.

For reversible adiabatic process,

p 1 V 1 n = p 2 V 2 n

or

p

p

2

1 =

V

V

n

1

2

F

HG

I

KJ

∴ p 2 = p 1 ×

V

V

n

1

2

F

HG

I

KJ =^4

3

08

25

×FHG IKJ

0.

0.

1.

= 20.87 bar.