TITLE.PM5

VAPOUR POWER CYCLES 575

dharm

\M-therm\Th12-3.pm5

h

s

(^12)
3
4
5
6
3 ¢
5 ¢
6 ¢
100 bar
25 bar
22 bar
6 bar
0.05 ba
r
95 bar
s
1
2
4
3
5
6
3 ¢
5 ¢
6 ¢
100 bar
25 bar
22 bar
6 bar
0.05 bar
95 bar
T
( )aT-sdiagram ( )bh-sdiagram
Fig. 12.22
Pump Work. Take specific volume of water as 0.001 m^3 /kg.
(Wpump)L.P. = (1 – m 1 – m 2 )(6 – 0.05) × 0.001 × 10^2
= (1 – 0.13566 – 0.1371) × 5.95 × 0.1 = 0.4327 kJ/kg.
(Wpump)H.P. = 1 × (100 – 6) × 0.001 × 10^2 = 9.4 kJ/kg
Total pump work (actual) =

4327 4

9

+

= 10.925 kJ/kg

Turbine output (indicated) = (h 2 – h 3 ) + (1 – m 1 )(h 4 – h 5 ) + (1 – m 1 – m 2 )(h 5 – h 6 )

= (3460 – 3111.5) + (1 – 0.13566)(3585 – 3207)

+ (1 – 0.13566 – 0.1371)(3207 – 2466)

= 1214.105 kJ/kg

Net electrical output = (Indicated work – Pump work) × ηmech. × ηgen.

= (1214.105 – 10.925) × 0.9 × 0.96 = 1039.55 kJ/kg

[Note. All the above calculations are for 1 kg of main (boiler) flow.]

∴ Main steam flow rate = 120 10^3600

1039.55

××^3
= 4.155 × 10^5 kJ/h.
Amounts of bled off are :
(a) Surface (high pressure) heater,
= 0.13566 kg/kg of boiler flow
or = 0.13566 × 4.155 × 10^5
i.e., = 5.6367 × 104 kg/h. (Ans.)
(b) Direct contact (low pressure) heater
= 0.1371 kg/kg of boiler flow
or = 0.1371 × 4.155 × 10^5
i.e., = 5.697 × 104 kg/h. (Ans.)
(iii) Overall thermal efficiency, ηoverall :

Heat input in boiler =

hh 110 −

ηboiler

=

3460 962

09

−

.

= 2775.6 kJ/kg of boiler flow.

Heat input in reheater = hh^43 −

ηboiler

= 3585 3111 5

09

−.

.

= 526.1 kJ/kg of boiler flow