VAPOUR POWER CYCLES 567

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\M-therm\Th15-2.pm5

(i)Efficiency of cycle, ηηηηηcycle :

ηcycle =

W

Q

T

1

=

934 8

2590 8

.

.

= 0.3608 or 36.08%. (Ans.)

Steam rate =

3600

934 8.

= 3.85 kg/kWh. (Ans.)

(ii) Tm 1 =

hh

ss

1 f

17

− 7

−

=

2590.8

6. 921 −1. 8604 = 511.9 K = 238.9°C.
   Tm 1 (without regeneration)

=

hh

ss

1 f

14

− 4

−

=

3230 9 191 8

6 921 0 649

..

..

−

−

= 3039.1

2726.

= 484.5 K = 211.5°C.

Increase in Tm 1 due to regeneration

= 238.9 – 211.5 = 27.4°C. (Ans.)

WT (without regeneration)

= h 1 – h 3 = 3230.9 – 2192.2 = 1038.7 kJ/kg

Steam rate without regeneration

=

3600

1038 7.

= 3.46 kg/kWh

∴ Increase in steam rate due to regeneration

= 3.85 – 3.46 = 0.39 kg/kWh. (Ans.)

ηcycle (without regeneration) =

hh

hhf

13

(^14)
−
−

1038.7
3230.9 191.8− = 0.3418 or 34.18%. (Ans.)
Increase in cycle efficiency due to regeneration
= 36.08 – 34.18 = 1.9%. (Ans.)
Example 12.14. Steam is supplied to a turbine at a pressure of 30 bar and a temperature
of 400°C and is expanded adiabatically to a pressure of 0.04 bar. At a stage of turbine where the
pressure is 3 bar a connection is made to a surface heater in which the feed water is heated by
bled steam to a temperature of 130°C. The condensed steam from the feed heater is cooled in a
drain cooler to 27°C. The feed water passes through the drain cooler before entering the feed
heater. The cooled drain water combines with the condensate in the well of the condenser.
Assuming no heat losses in the steam, calculate the following :
(i)Mass of steam used for feed heating per kg of steam entering the turbine ;
(ii)Thermal efficiency of the cycle.
Solution. Refer Fig. 12.18.
From steam tables :
At 3 bar : ts = 133.5°C, hf = 561.4 kJ/kg.
At 0.04 bar : ts = 29°C, hf = 121.5 kJ/kg.
From Mollier chart :
h 0 = 3231 kJ/kg (at 30 bar, 400°C)
h 1 = 2700 kJ/kg (at 3 bar)
h 2 = 2085 kJ/kg (at 0.04 bar).