TITLE.PM5

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.)