TITLE.PM5

VAPOUR POWER CYCLES 565

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\M-therm\Th15-2.pm5

Energy balance for the feed heater is written as :

mh 2 + (1 – m) hf 5 = 1 × hf 2

m × 2100 + (1 – m) × 125 = 1 × 570.9

2100 m + 125 – 125 m = 570.9

1975 m = 570.9 – 125

∴ m = 0.226 kg per kg of steam supplied to the turbine

∴ Steam supplied to the turbine per hour

=^11200

0226.

= 49557.5 kg/h

Net work developed per kg of steam

= (h 1 – h 2 ) + (1 – m) (h 2 – h 3 )

= (3100 – 2500) + (1 – 0.226) (2500 – 2100)

= 600 + 309.6 = 909.6 kJ/kg

∴ Power developed by the turbine

= 909.6 ×

49557 5

3600

.
kJ/s
= 12521.5 kW. (Ans.) (Q 1 kJ/s = 1 kW)
+Example 12.13. In a single-heater regenerative cycle the steam enters the turbine at
30 bar, 400°C and the exhaust pressure is 0.10 bar. The feed water heater is a direct contact type
which operates at 5 bar. Find :
(i)The efficiency and the steam rate of the cycle.
(ii)The increase in mean temperature of heat addition, efficiency and steam rate as com-
pared to the Rankine cycle (without regeneration).
Pump work may be neglected.
Solution. Fig. 12.17 shows the flow, T-s and h-s diagrams.

Turbine

Boiler

Power output

1

2

3

4

5

6

7

1 kg

1 kg

(a)

Pump

Pump

(1 – m)kg

Heater

5 bar m kg

(1 – m) kg

0.1 bar

30 bar, 400 Cº

Condenser