194 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-5.pm5

(iii)Work done :
From non-flow energy equation,
Q = (u 2 – u 1 ) + W
∴ W = Q – (u 2 – u 1 ) = 420 – 123.65 = 296.35 kJ/kg
i.e., Work done by the steam = 296.35 kJ/kg. (Ans.)

Note. The work done is also given by the area on the Fig. 4.60 pdv

v

v

1

2

z

F

HG

I
KJ
, this can only be evaluated
graphically.
Example 4.55. In a steam engine cylinder the steam expands from 5.5 bar to 0.75 bar
according to a hyperbolic law, pv = constant. If the steam is initially dry and saturated, calculate
per kg of steam :
(i)Work done ;
(ii)Heat flow to or from the cylinder walls.
Solution. Initial pressure of steam, p 1 = 5.5 bar = 5.5 × 10^5 N/m^2
Initial condition of steam, x 1 = 1
Final pressure of steam, p 2 = 0.75 bar = 0.75 bar × 10^5 N/m^2
At 5.5 bar, v 1 = vg = 0.3427 m^3 /kg
Also p 1 v 1 = p 2 v 2

∴ v 2 =

pv

p

11

2

5 0 3427

075

= .5×.

. = 2.513 m

(^3) /kg
At 0.75 bar, vg = 2.217 m^3 /kg.
Since v 2 > vg (at 0.75 bar), therefore, the steam is superheated at state 2.
Interpolating from superheat tables at 0.75 bar, we have
u 2 = 2510 + 2.
2.
513 2 271
588 2 271
−
−
F
HG
I
KJ
.
.
(2585 – 2510)
= 2510 +
0
0
.242
.317
× 75 = 2567.25 kJ/kg.
For dry saturated steam at 5.5 bar
u 1 = ug = 2565 kJ/kg
Hence, gain in internal energy
= u 2 – u 1 = 2567.25 – 2565 = 2.25 kJ/kg
The process is shown on a p-v diagram in Fig. 4.64, the shaded area representing the work
done.
Now, Wpdvv
v
=z
1
2
=z FHconstantv IKdv
v
v
1
2
Q pv p
v
L ==
NM
O
QP
constant, and constant

L
N
M
M
O
Q
P
P
constant loge
v
v
v
1
2