TITLE.PM5

BASIC CONCEPTS OF THERMODYNAMICS 57

dharm
M-therm/th2-2.pm5

+Example 2.22. A cylinder contains 1 kg of a certain fluid at an initial pressure of 20 bar.

The fluid is allowed to expand reversibly behind a piston according to a law pV^2 = constant until

the volume is doubled. The fluid is then cooled reversibly at constant pressure until the piston

regains its original position ; heat is then supplied reversibly with the piston firmly locked in

position until the pressure rises to the original value of 20 bar. Calculate the net work done by the

fluid, for an initial volume of 0.05 m^3.

Solution. Refer Fig. 2.39.

Fig. 2.39

Mass of fluid, m = 1 kg

p 1 = 20 bar = 20 × 10^5 N/m^2

V 1 = 0.05 m^3

Considering the process 1-2

p 1 V 12 = p 2 V 22

∴ p 2 = p 1

V

V

1

2

2

F

HG

I

KJ = 20

V

V

1

(^21)
2
F
HG
I
KJ [Q V^2 = 2V^1 (given)]
=^204 = 5 bar
Work done by the fluid from 1 to 2 = Area 12 ML1 = pdV
1
2
z
i.e., W1–2 =
C
V
dV
v
v
1 2
2
z , where C = p 1 V 12 = 20 × 0.05^2 bar m^6
∴ W1–2 = 10^5 × 20 × 0.0025 −
L
NM
O
QP
1
005
01
V.
.