TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 281

dharm

/M-therm/th5-4.pm5

2

T

S

pV = Cn

T 1 1

4

T 2

dS

3

=

0 287

14 1

290 6

380

0 287

0 5211

0 1363

.

.

log

.

.log

.

−.

× eeHFG IKJ+ HGF IKJ

= – 0.192 + 0.385 = 0.193 kJ/kg K (increase). (Ans.)
Example 5.35. (a) Show that approximate change of entropy during a polytropic process
equals the quantity of heat transferred divided by the mean absolute temperature.
(b) One kg of air at 290 K is compressed in a cylinder according to the polytropic law
pv1.3= constant. If the compression ratio is 16, calculate the entropy change of air during the
compression process stating whether it is an increase or decrease.
What would be the percentage error if the entropy change is calculated by dividing the
quantity of heat exchanged by the mean absolute temperature during the process?
Take γ = 1.4 and cv = 0.718 kJ/kg K.
Solution. (a) In Fig. 5.41 curve 1-2 represents the
polytropic process (pvn = c) from state point 1 to state point

2. The area under the process curve 1-2 on T-S diagram
   represents the heat transferred during the process. The slope
   of the curve 1-2 is usually small and can be considered to be
   a straight line (shotted dotted).
   Heat transferred = Area of trapezium 1-2-3-4
   = Base × mean ordinate

= dS ×

TT 12

2

F +

HG

I

KJ

= Entropy change × mean absolute

temperature during the process

or, Entropy change =
Heat transferred
Mean absolute temperature
(b) Given : m = 1 kg ; T 1 = 290 K ; pv1.3 = constant ; r = 16 ; γ = 1.4 ; cv = 0.718 kJ/kg K

For a polytropic process : T

T

v

v

n

2

1

1

2

1

=

F

HG

I

KJ

−

or, T 2 = 290 × (16)1.3–1 = 666.2 K

Now, s 2 – s 1 = c n

n

T

veT

−

−

F

HG

I

KJ

F

HG

I

KJ

γ

1

2

1

log ... per kg

= 0 718

13 14

13 1

666 2

290

.

..

.

log

−.

−

F

HG

I

KJ

F

HG

I

e KJ = – 0.199 kJ/kg K. (Ans.)

The –ve sign indicates decrease in entropy.

Heat transferred during the process is given by,

Q =

γ

γ

γ

γ

γ

γ

−

−

×=

−

−

×

−

−

=

−

−

F

HG

I

KJ

n W nRTT

n

cv n

111 1

() (^12) (T
1 – T 2 ) ...per kg
= 0 718
14 13
13 1
.
..
.
−
−
F
HG
I
KJ
(290 – 666.2) = – 90.04 kJ/kg Q c
R
v= −
F
HG
I
γ 1 KJ
Fig. 5.41