VAPOUR POWER CYCLES 561
dharm
\M-therm\Th15-2.pm5
2
3
p
V
2 ′
Loss of work due
to incomplete
expansion
1
p
(10 bar)
400 C
1
º
p
(0-9 bar)
2
p
(0.3 bar)
3
10 bar
0.9 bar
0.3 bar
T
Constant volume line s
3
2
2 ′
1
Fig. 12.14. p-V and T-s diagrams.
or tsup 2 = 382 – 273 = 109 °C. (Ans.)
∴ h 2 = hg 2 + cps (Tsup 2 – Ts 2 )
= 2670.9 + 2.1 (382 – 366.5) = 2703.4 kJ/kg.
(ii) Quality of steam at the end of constant volume operation, x 3 :
For calculating v 2 using the relation
v
T
g
s
2
2
=
v
Tsup
2
2
(Approximately)
1.869
369.7
=
v 2
382
or v 2 = 1.869 382
369 7
×
.
= 1.931 m^3 /kg
Also v 2 = v 3 = x 3 vg 3
1.931 = x 3 × 5.229
or x 3 =
931
229
= 0.37. (Ans.)
(iii)Power developed, P :
Work done = (h 1 – h 2 ) + (p 2 – p 3 ) v 2
= (3263.9 – 2703.4) +
(0.75 0.3) 10 1.931
1000
−××^5
= 560.5 + 86.9 = 647.4 kJ/kg
∴ Power developed = Steam flow rate × work done (per kg)
= 1 × 647.4 = 647.4 kW. (Ans.)
(iv)Specific steam consumption, ssc :
ssc =^3600
Power
= 1 3600
647 4
×
.
= 5.56 kg/kWh. (Ans.)