TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 283

dharm

/M-therm/th5-4.pm5

∴ Overall change of entropy
= (S 2 – S 1 ) + (S 3 – S 2 )
= 0.9771 + (– 0.6938) = 0.2833 kJ/K. (Ans.)
Example 5.37. A closed system contains air at a pressure 1 bar, temperature 300 K and
volume 0.018 m^3. This system undergoes a thermodynamic cycle consisting of the following three
processes in series : (i) Constant volume heat addition till pressure becomes 5 bar, (ii) Constant
pressure cooling, and (iii) Isothermal heating to initial state.
Represent the cycle on T-S and p-V plots and evaluate the change in entropy for each
process.
Take cp = 0.718 kJ/kg K and R = 0.287 kJ/kg K.
Solution. Given : p 1 = 1 bar ; T 1 = 300 K ; V 1 = 0.018 m^3 ; p 2 = 5 bar ;
cv = 0.718 kJ/kg K ; R = 0.287 kJ/kg K.

2

T

S

V=C

p=C

3 1

T=C

2

p

V

V=C

p=C

1

3

T=C

Fig. 5.43. T-S and p-V diagrams.

Mass of air, m =

pV

RT

11

1

1 10^5 0 018

0 287 1000 300

= ××

××

().

(. )

= 0.0209 kg

Refer to Fig. 5.43.

lConstant volume process 1-2 :

p

T

p

T

1

1

2

2

= or T 2 = T 1 ×= ×p

p

2

1

300

5

1

= 1500 K

∴ Change in entropy, S 2 – S 1 = mcv loge

T

T

2

1

F

HG

I

KJ

= 0.0209 × 0.718 × loge

1500

300

F

HG

I

KJ = 0.0241 kJ/K. (Ans.)

lConstant pressure process 2-3 :

T 3 = T 1 = 300 K

Now, change in entropy,

S 3 – S 2 = mcp loge

T

T

3

2

F

HG

I

KJ

= m(cv + R) loge

T

T

3

2

F

HG

I

KJ