Question:-
Heat supplied ( kJ kg-1) to the cycle is
Option (A)
2372
Option (B)
2576
Option(C)
2863
Option(D)
3092
Correct Option:
(C)
Question Solution:

Given data

Rankine cycle on T-S diagram

P3=4MPa

T3=350°c

ht =0.9

H3=3092.5 KJ/kg

S3 =S4 =6.5821 KJ/kg k

H1=225.94 KJ/kg=hf

Hg=2599.1 KJ/kg

Hfg=hg-hf=2373.16 KJ/kg

S1=Sf=0.7549 KJ/kg k

Sf=8.0085 KJ/kg k

Sfg=8.0085-0.7549 = 7.2536 KJ/kg k

S4=S1+x Sfg=0.7549+x7.2536=6.5821
x=0.8

h4=h1+x hfg=225.94 + (0.8x2373.16)

h4=2132.4 KJ/kg


Net work output of the cycle = ht.
(h3-h4) =0.9(3092.5-2132.4) =864 KJ/kg 860 KJ/kg

Heat supplied to the cycle = h3-h2

We don’t have information regarding h2

Therefore neglect compressor work and take h2=h1

Therefore Qs=3092.5-225.94 =2866 KJ/kg 2863 KJ/kg
question-answer-faq-2383