VAPOUR POWER CYCLES 555

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\M-therm\Th12-1.pm5

From steam tables :

At 35 bar : h 1 = hg 1 = 2802 kJ/kg, sg 1 = 6.1228 kJ/kg K

At 0.26 bar : hf = 251.5 kJ/kg, hfg = 2358.4 kJ/kg,

vf = 0.001017 m^3 /kg, sf = 0.8321 kJ/kg K, sfg = 7.0773 kJ/kg K.

(i) The pump work :

Pump work = (p 4 – p 3 ) vf = (35 – 0.2) × 10^5 × 0.001017 J or 3.54 kJ/kg

Also Pump work 3.

3. kJ / kg

hh

h

f f

f

43

4

54

251.5 54 255.04

−= =

∴=+=

L

N

M

O

Q

P

Now power required to drive the pump

= 9.5 × 3.54 kJ/s or 33.63 kW. (Ans.)

(ii) The turbine work :

s 1 = s 2 = sf 2 + x 2 × sfg 2

6.1228 = 0.8321 + x 2 × 7.0773

∴ x 2 =

61228 0 8321

7 0773

..

.

−

= 0.747

∴ h 2 = hf 2 + x 2 hfg 2 = 251.5 + 0.747 × 2358.4 = 2013 kJ/kg

∴ Turbine work = m& (h 1 – h 2 ) = 9.5 (2802 – 2013) = 7495.5 kW. (Ans.)

It may be noted that pump work (33.63 kW) is very small as compared to the turbine work

(7495.5 kW).

(iii)The Rankine efficiency :

ηrankine =

hh

hhf

12

(^12)
−
−
= 2802 2013
2802 251.5
−
−
=^789
2550.5
= 0.3093 or 30.93%. (Ans.)
(iv)The condenser heat flow :
The condenser heat flow = m& (h 2 – hf 3 ) = 9.5 (2013 – 251.5) = 16734.25 kW. (Ans.)
(v)The dryness at the end of expansion, x 2 :
The dryness at the end of expansion,
x 2 = 0.747 or 74.7%. (Ans.)
Example 12.7. The adiabatic enthalpy drop across the primemover of the Rankine cycle is
840 kJ/kg. The enthalpy of steam supplied is 2940 kJ/kg. If the back pressure is 0.1 bar, find the
specific steam consumption and thermal efficiency.
Solution. Adiabatic enthalpy drop, h 1 – h 2 = 840 kJ/kg
Enthalpy of steam supplied, h 1 = 2940 kJ/kg
Back pressure, p 2 = 0.1 bar
From steam tables, corresponding to 0.1 bar : hf = 191.8 kJ/kg
Now, ηrankine = hh
hhf
12
(^12)
−
−

840
2940 191.8− = 0.3056 = 30.56%. (Ans.)
Useful work done per kg of steam = 840 kJ/kg
∴ Specific steam consumption =^1
840
kg/s =^1
840
× 3600 = 4.286 kg/kWh. (Ans.)
+Example 12.8. A 35 kW (I.P.) system engines consumes 284 kg/h at 15 bar and 250°C. If
condenser pressure is 0.14 bar, determine :