TITLE.PM5

566 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th15-2.pm5

1

2

4 3

5

6

7

30 bar

1 kg

5 bar

m kg

0.1 bar

(1 – m) kg

(1 – m)kg

1 kg

(400 C)º

T

s

(b)

h

s

1

2

3

4

(^56)
7
(400 C)º
1 kg
(1–m) kg
30 bar
5 bar
0.1 bar
m kg
(1–m) kg
(c)
1 kg
Fig. 12.17
From steam tables :
At 30 bar, 400°C : h 1 = 3230.9 kJ/kg, s 1 = 6.921 kJ/kg K = s 2 = s 3 ,
At 5 bar : sf = 1.8604, sg = 6.8192 kJ/kg K, hf = 640.1 kJ/kg
Since s 2 > sg, the state 2 must lie in the superheated region. From the table for superheated
steam t 2 = 172°C, h 2 = 2796 kJ/kg.
At 0.1 bar : sf = 0.649, sfg = 7.501, hf = 191.8, hfg = 2392.8
Now, s 2 = s 3
i.e., 6.921 = sf 3 + x 3 sfg 3 = 0.649 + x 3 × 7.501
∴ x 3 =
6 921 0 649
7 501
..
.
−
= 0.836
∴ h 3 = hf 3 + x 3 hfg 3 = 191.8 + 0.836 × 2392.8 = 2192.2 kJ/kg
Since pump work is neglected
hf 4 = 191.8 kJ/kg = hf 5
hf 6 = 640.1 kJ/kg (at 5 bar) = hf 7
Energy balance for heater gives
m (h 2 – hf 6 ) = (1 – m) (hf 6 – hf 5 )
m (2796 – 640.1) = (1 – m) (640.1 – 191.8) = 448.3 (1 – m)
2155.9 m = 448.3 – 448.3 m
∴ m = 0.172 kg
∴ Turbine work, WT = (h 1 – h 2 ) + (1 – m) (h 2 – h 3 )
= (3230.9 – 2796) + (1 – 0.172) (2796 – 2192.2)
= 434.9 + 499.9 = 934.8 kJ/kg
Heat supplied, Q 1 = h 1 – hf 6 = 3230.9 – 640.1 = 2590.8 kJ/kg.