TITLE.PM5

192 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-5.pm5

(c)

Fig. 4.62. Constant temperature or isothermal expansion.

WpdvC

v

dv C v

e v

== =

F

HG

I

zz 1 KJ

2

1

(^22)
1
log
= p 1 v 1 loge
v
v
2
1
F
HG
I
KJ ...(4.75)
where v 1 and v 2 are the specific volumes of steam before and after expansion.
Applying first law of energy equation,
Q = ∆u + p
1
2
z. dv
= (u 2 – u 1 ) + p 1 v 1 loge
v
v
2
1
F
HG
I
KJ
= (h 2 – p 2 v 2 ) – (h 1 – p 1 v 1 ) + p 1 v 1 loge
v
v
2
1
F
HG
I
KJ
Since p 1 v 1 = p 2 v 2
∴ Q = (h 2 – h 1 ) + p 1 v 1 loge
v
v
2
1 ...(4.76)
+Example 4.54. Steam at 7 bar and dryness fraction 0.95 expands in a cylinder behind a
piston isothermally and reversibly to a pressure of 1.5 bar. The heat supplied during the process
is found to be 420 kJ/kg. Calculate per kg :
(i)The change of internal energy ; (ii)The change of enthalpy ;
(iii)The work done.
Solution. Initial pressure of steam, p 1 = 7 bar = 7 × 10^5 N/m^2
Final pressure of steam, p 2 = 1.5 bar = 1.5 ×10^5 N/m^2
Heat supplied during the process, Q = 420 kJ/kg.
The process is shown in Fig. 4.63. The saturation temperature corresponding to 7 bar is
165°C. Therefore, the steam is superheated at the state 2.