TITLE.PM5

186 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-5.pm5

Steam at 80 bar and 350°C is in a superheated state, and the specific volume from tables is
0.02995 m^3 /kg. Hence the mass of steam in the cylinder is given by

m=0 02995.0 028. = 0.935 kg

Internal energy at state 1, (80 bar, 350°C),

u 1 = h 1 – p 1 v 1

= 2987.3 – 80 10^002995

10

5

3

××. or u

1 = 2747.7 kJ/kg.

(i)State of steam after cooling :

At state 2, p 2 = 50 bar and v 2 = 0.02995 m^3 /kg, therefore, steam is wet, and dryness fraction

is given by,

x v

(^2) vg
2
2
0
0 0394
==.02995

. = 0.76.
(ii)Heat rejected by the steam :
Internal energy at state 2 (50 bar),
u 2 = (1 – x 2 ) uxuf 22 + 2 g
= (1 – 0.76) × 1149 + 0.76 × 2597 = 2249.48 kJ/kg
At constant volume, Q = U 2 – U 1 = m(u 2 – u 1 )
= 0.935(2249.48 – 2747.7) = – 465.5 kJ
i.e., Heat rejected = 465.5 kJ. (Ans.)
Fig. 4.58 shows the process drawn on T-s diagram, the shaded area representing the heat
rejected by the system.

1

2

80 bar

50 bar

273

T

s (kJ/kg K)

Fig. 4.58