280 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-4.pm5

Example 5.34. 1 kg of air initially at 8 bar pressure and 380 K expands polytropically
(pv1-2 = constant) until the pressure is reduced to one-fifth value. Calculate :
(i)Final specific volume and temperature.
(ii)Change of internal energy, work done and heat interaction.
(iii)Change in entropy.
Take : R = 0.287 kJ/kg K and γ = 1.4.
Solution. Given : m = 1 kg ; p 1 = 8 bar ; T 1 = 380 K ; Law of expansion : pv1.2 = constant ;

p 2 = p^1

5

8

5

= = 1.6 bar ; R = 0.287 kJ/kg K ; γ = 0.4.

(i)Final specific volume and temperature, v 2 , T 2 :

p 1 v 1 = RT 1

or, v 1 = RT

p

1

1

3

5

0 287 10 380

810

= ××

×

(. ) = 0.1363 m (^3) /kg.
Also, p 1 v 1 n = p 2 v 2 n or v 2 = v 1 ×
p
p
1 n
2
1
F
HG
I
KJ
∴ v 2 = 0.1363 ×^8
16
1
12
.
F.
HG
I
KJ
= 0.5211 m^3 /kg. (Ans.)
Again, T
T
p
p
n
2 n
1
2
1
1

F
HG
I
KJ
−
or T^2
12 1
12
380
1
5
=F
HG
I
KJ
.−
.
∴ T 2 = 290.6 K
Alternatively : KT pv
(^2) R
22
1 6 10^5 0 5211
287
==×× =290 6
L
N
M
M
O
Q
P
P
...
(ii)Change of internal energy, work done and heat interaction :
Change of internal energy
u 2 – u 1 = cv(T 2 – T 1 ) =
R
TT
γ−
−
1 21
()

0 287
14 1

. 290 6 380
.
(. )
−
− = – 64.14 kJ/kg (Decrease). (Ans.)

Work done, W1-2 = pv pv

n

TT

n

11 2 2 2

11

−

−

= −

−

R( 1 )

=

0 287 380 290 6

12 1

.( .)

.

−
− = 128.29 kJ/kg (Work done by air). (Ans.)
Heat interaction, Q1–2 = (u 2 – u 1 ) + W = – 64.14 + 128.29 = 64.15 kJ/kg (Heat
received). (Ans.)

Alternatively Q: n W ..

.

= − ..

−

×=−

−

L ×=

N

M

O

Q

P

γ

γ 1

14 12

14 1

128 29 64 15 kJ/kg

(iii)Change in entropy, (s 2 – s 1 ) :

s 2 – s 1 = cv loge

T

T

R

v

ev

2

1

2

1

+ log