TITLE.PM5

VAPOUR POWER CYCLES 569

dharm

\M-therm\Th15-2.pm5

= 1(3231 – 2700) + (1 – 0.1672) (2700 – 2085)

= 1043.17 kJ/kg

Heat supplied per kg of steam = h 0 – 1 × 4.186 × 130

= 3231 – 544.18 = 2686.82 kJ/kg.

ηThermal =

Work done

Heat supplied

=

1043.17

2686.82

= 0.3882 or 38.82%. (Ans.)

Example 12.15. Steam is supplied to a turbine at 30 bar and 350°C. The turbine exhaust
pressure is 0.08 bar. The main condensate is heated regeneratively in two stages by steam bled
from the turbine at 5 bar and 1.0 bar respectively. Calculate masses of steam bled off at each
pressure per kg of steam entering the turbine and the theoretical thermal efficiency of the cycle.
Solution. Refer Fig. 12.19.

h

s

1

2

3

0

30 bar, 350 C

º

5 bar1.0 bar

0.08 bar

T u r b i n e

Heater-1 m 1 mm 12 +

(H.P.) Heater-2

(L.P.)

Drain cooler

1 kg hf 2

hf 1

hf 4

1– m – m 12

hf 3

hf 3

1 kg

m, h (^11) m, h
22
1 – m 1
1 kg
1– m – m 12
(^123)
Condenser
0
Fig. 12.19
The following assumptions are made :

1. The condensate is heated to the saturation temperature in each heater.


2. The drain water from H.P. (high pressure) heater passes into the steam space of the L.P.
   (low pressure) heater without loss of heat.


3. The combined drains from the L.P. heater are cooled in a drain cooler to the condenser
   temperature.


4. The expansion of the steam in the turbine is adiabatic and frictionless.
   Enthalpy at 30 bar, 350°C, h 0 = 3115.3 kJ/kg.
   After adiabatic expansion (from Mollier chart)
   Enthalpy at 5 bar, h 1 = 2720 kJ/kg
   Enthalpy at 1.0 bar, h 2 = 2450 kJ/kg
   Enthalpy at 0.08 bar, h 3 = 2120 kJ/kg
   From steam tables : hf 1 = 640.1 kJ/kg (at 5.0 bar)
   hf 2 = 417.5 kJ/kg (at 1.0 bar)
   hf 3 = 173.9 kJ/kg (at 0.08 bar)