600 ENGINEERING THERMODYNAMICS
dharm
\M-therm\Th12-4.pm5
h 5 = 2210 kJ/kg
h 4 ′ – h 5 ′ = 0.83(h 4 ′ – h 5 ) = 0.83(3072.5 – 2210) ~− 715.9 kJ/kg
∴ h 5 ′ = h 4 ′ – 715.9 = 3072.5 – 715.9 = 2356.6 kJ/kg
From steam tables, we have :
hf 6 = 163.4 kJ/kg ; hf 8 = 762.6 kJ/kg
h 1 – h 2 ′ = 0.785(h 1 – h 2 ) = 0.785(3285 – 2980) = 239.4 kJ/kg
∴ h 2 ′ = h 1 – 239.4 = 3285 – 239.4 = 3045.6 kJ/kg
( ) Schematic arrangement of the planta
1
H.P.
üýþ
3- supplyf
1kg
Superheater
I.P.
Reheater
2 1kg 3 25 bar,420°C
1kg
70 bar,
450°C
1kg
L.P.
9
8 7 6
4 ¢ 1kg 5 ¢
mkg (1–m) kg
Feed water
heater
(open type)
P-2 (Pump) P-1 (Pump)
(1–m) kg
(1–m) kg
Condenser
0.07 bar
s
T
6
7
8
9
( )bT-sdiagram
70 bar
1kg
1 (470°C) 3 (420°C)
2 ¢
4 ¢
4
55 ¢
25 bar 2
mkg
10 bar(1–m) kg
0.07 bar
s
h
( )ch-sdiagram (Pump work not shown)
1
70 bar
25 bar
10 bar
0.07 bar
5 ¢
5
4 4 ¢
3
2 2 ¢
4 ¢
Fig. 12.42
(ii) The quantity of steam bled per kg of flow at the turbine inlet, m :
Considering energy balance for the feed water heater, we have :
m × h 4 ′ + (1 – m) hf 7 = 1 × hf 8
m × 3072.5 + (1 – m) × 163.4 = 1 × 762.6 (Q hf 7 = hf 6 )