TITLE.PM5

272 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-3.pm5

Characteristics gas constant,

R =

R

M

(^0) =^8314
44
= 189 Nm/kg K
To find T 2 , using the relation,
p 2 V 2 = mRT 2
∴ T 2 =
pV
mR
(^22) = 9 10 0 003
0 04 189
××^5
×
.

. = 357 K

Now sA – s 2 = R loge

p

p

2

1

=

189

103

loge 9

1

F

HG

I

KJ

= 0.4153 kJ/kg K

Also at constant pressure from 1 to A

sA – s 1 = cp loge

T

T

2

1

= 0.88 loge 357

293

F

HG

I

KJ

= 0.1738 kJ/kg K

Then (s 1 – s 2 ) = (sA – s 2 ) – (sA – s 1 )

= 0.4153 – 0.1738 = 0.2415 kJ/kg K

Hence for 0.04 kg of carbon dioxide decrease in entropy,

S 1 – S 2 = m(s 1 – s 2 ) = 0.04 × 0.2415

= 0.00966 kJ/K. (Ans.)

Note. In short, the change of entropy can be found by using the following relation :

(s 2 – s 1 ) = cp loge T

T

2

1

• R loge p
  p

2

1

= 0.88 loge 357

293

189

103

F

HG

I

KJ

− loge

9

1

F

HG

I
KJ
= 0.1738 – 0.4153 = – 0.2415 kJ/kg K
∴ S 2 – S 1 = m(s 2 – s 1 ) = 0.04 × (– 0.2415)
= – 0.00966 kJ/K
(– ve sign means decrease in entropy)
or S 1 – S 2 = 0.00966 kJ/K.

Example 5.27. Calculate the change of entropy of 1 kg of air expanding polytropically in a
cylinder behind a piston from 7 bar and 600°C to 1.05 bar. The index of expansion is 1.25.
Solution. The process is shown on a T-s diagram in Fig. 5.35.
Initial pressure, p 1 = 7 bar = 7 × 10^5 N/m^2
Initial temperature, T 1 = 600 + 273 = 873 K
Final pressure, p 2 = 1.05 bar = 1.05 × 10^5 N/m^2
Index of expansion, n = 1.25
Mass of air = 1 kg
To find T 2 , using the relation,

T

T

2

1

= p

p

n

2 n

1

1

F

HG

I

KJ

−

∴ T^2

873

= 1.05

7

125 1

F 125

HG

I

KJ

. −
.