TITLE.PM5

56 ENGINEERING THERMODYNAMICS

dharm
M-therm/th2-2.pm5

REVERSIBLE WORK

Example 2.21. A fluid at a pressure of 3 bar, and with specific volume of 0.18 m^3 /kg,

contained in a cylinder behind a piston exapnds reversibly to a pressure of 0.6 bar according to a

law, p =

C

v^2 where C is a constant. Calculate the work done by the fluid on the piston.

Solution. Refer Fig. 2.38.

Fig. 2.38

p 1 = 3 bar = 3 × 10^5 N/m^2

v 1 = 0.18 m^3 /kg

Work done = Shaded area = pdv

1

2

z

i.e., Work done, W =

C

v

dv C dv

v

C v

v

v

1 2

2

1 2

(^221)
21
1
2
zz==−+
−+
,
= L−
NM
O
QP
=−L
NM
O
QP
=−L
N
M
O
Q
C v− C v C vvP
v
v
v
v
1
1 12
2
1
1112
...(i)
Also C = pv^2 = p 1 v 12 = 3 × 0.18^2 = 0.0972 bar (m^3 /kg)^2
and v 2 =
C
p 2
0 0972
= 06
.

. = 0.402 m

(^3) /kg
Substituting the values of C, v 1 and v 2 in eqn. (i), we get
Work done, W = 0.0972 × 10^5
1
0.18
1
0.402
L −
NM
O
QP
Nm/kg
= 29840 Nm/kg. (Ans.)