278 ENGINEERING THERMODYNAMICS

dharm
/M-therm/th5-4.pm5

For constant volume process
= 20 × 0.7432 loge 2 = 10.3 kJ/K. (Ans.)
Example 5.31. 5 m^3 of air at 2 bar, 27°C is compressed up to 6 bar pressure following
pv1.3 = constant. It is subsequently expanded adiabatically to 2 bar. Considering the two processes
to be reversible, determine the net work. Also plot the processes on T-S diagram.
Solution. Refer Fig. 5.39.
Given : V 1 = 5 m^3 ; p 1 = 2 bar ; T 1 = 27 + 273 = 300 K ; p 2 = 6 bar ; p 3 = 2 bar
Net work :
p

v

2

1

3

pv1.3= C

pv = Cg

T(K)

s

2

1

3

pv1.3= C

pv = Cg

Fig. 5.39. p-V diagram. Fig. 5.40. T-s diagram.

Mass of air, m = pV

RT

11

1

210 5^5

287 300

= ××

×

= 11.61 kg.

Considering polytropic compression process 1-2, we have

T

T

p

p

n

2 n

1

2

1

1

=

F

HG

I

KJ

−

or T^2

13 1

13

300

6

2

=F

HG

I

KJ

−

or T 2 = 386.5 K.

Considering isentropic process 2-3, we get

T

T

p

p

p

p

2

3

2

3

1

2

1

(^1) 14 1
6 1
2

F
HG
I
KJ

F
HG
I
KJ
=F
HG
I
KJ
γ−−−
γ
γ
γ .4 = 1.369 (
Q p 3 = p 1 )
∴ T 3 = T^2
1 369
386 5

. 1 369

.

.

= = 282.3 K

Now, work done during polytropic compression 1-2,

W1–2 =

mR T T

n

()..( 12 .)

1

11 61 0 287 300 386 5

13 1

−

−

= ×−

−

= – 960.7 kJ

and, work done during adiabatic expansion 2-3,

W2–3 =

mR T()..(..)T

.

23

1

11 61 0 287 386 5 282 3

14 1

−

−

=

×−

γ −

= 868 kJ

∴ Net work done = W1–2 + W2–3 = – 960.7 + 868 = – 92.7 kJ

Hence net work done on the air = 92.7 kJ. (Ans.)

The process plotted on T-s diagram is shown in Fig. 5.40.