TITLE.PM5

FIRST LAW OF THERMODYNAMICS 195

dharm

/M-therm/Th4-5.pm5

Fig. 4.64

The constant is either p 1 v 1 or p 2 v 2

i.e., W = 5.5 × 10^5 × 0.3427 × loge

p

p

1

2

Q pv 11 pv2 2 vv^2 pp

1

1

2

L =

NM

O

QP

or =

= 5.5 × 10^5 × 0.3427 × loge

5.

.75

5

0

F

HG

I

KJ = 375543 N-m/kg.

Using non-flow energy equation, we get

Q = (u 2 – u 1 ) + W

= 2.25 +
375543
103 = 378 kJ/kg
i.e., Heat supplied = 378 kJ/kg. (Ans.)

Example 4.56. Dry saturated steam at 100 bar expands isothermally and reversibly to a
pressure of 10 bar. Calculate per kg of steam :
(i)The heat supplied ;
(ii)The work done.
Solution. Initial pressure of steam, p 1 = 100 bar
Final pressure of steam, p 2 = 10 bar
The process is shown in Fig. 4.65, the shaded area representing the heat supplied.
At 100 bar, dry saturated : From steam tables,
s 1 = sg = 5.619 kJ/kg K and ts 1 = 311°C
At 10 bar and 311°C the steam is superheated, hence interpolating

s 2 = 7.124 +

311 300

350 300

−

−

F

H

I

K (7.301 – 7.124) or s^2 = 7.163 kJ/kg K.