Page 124 - Advanced thermodynamics for engineers

110 CHAPTER 5 RATIONAL EFFICIENCY OF POWER PLANT

parameters for the steam plant if (a) the isentropic efficiency of the turbine is 80%; (b) the isentropic
efficiency of the feed pump is 70%; (c) if the efficiency of the components is the combination of those
given in (a) and (b). Assume that t 0 ¼ 81.3 C.

Solution
This cycle is shown in Fig 5.5.
Conditions at 6

p 6 ¼ 0:5 bar;

t s ¼ 81:3 C; h g ¼ 2645 kJ kg; h f ¼ 340 kJ kg; s g ¼ 7:593 kJ kg K; s f ¼ 1:091 kJ kg K:

h 6 ¼ 2645 kJ=kg; s 6 ¼ 7:593 kJ=kg K:
Conditions at 5
p 5 ¼ 20 bar; s 5 ¼ s 6 ¼ 7:593 kJ kg K:

7:593 7:431

Hence; t 5 ¼ 500 þ 100 ¼ 560 C;
7:701 7:431
h 5 ¼ 3467 þ 0:6 ð3690 3467Þ¼ 3601 kJ=kg:
Work done by turbine,
w T ¼ h 5 h 6 ¼ 3601 2645 ¼ 956 kJ=kg K:

w net w T þ w P 956 2:0
Thermal efficiency of the cycle is h ¼ ¼ ¼ ¼ 0:292
th
q in h 5 h 2 3601 342:0
b 5 ¼ h 5 T 0 s 5 a 0 ¼ 3601 354:3 7:593 a 0 ¼ 910:8 a 0 kJ=kg;
b 2 ¼ h 2 T 0 s 2 a 0 ¼ 342:1 354:3 1:091 a 0 ¼ 44:4 a 0 kJ=kg:

Temperature, T 5

Q
in
3 4 W T

2
W P
1 6'
6
Q out saturated vapour
line
saturated liquid
line
Entropy, S
FIGURE 5.5
Temperature–entropy diagram for Rankine cycle with superheat.