TITLE.PM5

584 ENGINEERING THERMODYNAMICS

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\M-therm\Th12-3.pm5

T

ss

5

5 ¢

0.1 bar

40 bar 0.1 bar

150 bar

150 bar

1

331

(^22)
4
4
h
600°C
x = 0.896
Fig. 12.33 Fig. 12.34
Moisture contents in exit from L.P. turbine = 10.4%
x 4 = 1 – 0.104 = 0.896
(i)Reheat pressure : From the Mollier diagram, the reheat pressure is 40 bar.
(Ans.)
(ii)Thermal efficiency, ηth :
Turbine work = (h 1 – h 2 ) + (h 3 – h 4 )
= (3580 – 3140) + (3675 – 2335) = 1780 kJ/kg.
Assuming specific volume of water = 10–3 m^3 /kg, the pump work = 10–3 (150 – 0.1)
= 0.15 kJ/kg, i.e., may be neglected in computing of ηth, h 5 = h 4 = 191.8 kJ/kg, (hf at 0.1 bar) from
steam tables,
Qinput = (h 1 – h 5 ) + (h 3 – h 2 )
= (3580 – 191.8) + (3675 – 3140) = 3923.2 kJ/kg
%ηth =^1780
3923 2.
× 100 = 45.37%. (Ans.)
(iii)Specific steam consumption :
Steam consumption =
15 10
1780
×^3
= 8.427 kg/s
Specific steam consumption =
8 427 3600
15 10^3

. ×
×
= 2.0225 kg/kWh. (Ans.)

(iv) Rate of pump work :

Rate of pump work = 8.427 × 0.15 = 1.26 kW. (Ans.)

12.6. Binary Vapour Cycle

Carnot cycle gives the highest thermal efficiency which is given by TT

T

12

1

−. To approach

this cycle in an actual engine it is necessary that whole of heat must be supplied at constant
temperature T 1 and rejected at T 2. This can be achieved only by using a vapour in the wet field but
not in the superheated. The efficiency depends on temperature T 1 since T 2 is fixed by the natural
sink to which heat is rejected. This means that T 1 should be as large as possible, consistent with
the vapour being saturated.