Question:-
Common Data Questions: 48 & 49

In a steam power plant operating on the Rankine cycle, steam enters the turbine at 4MPa, 350ºC and exits at a pressure of 15kPa. Then it enters the condenser and exits as saturated water. Next, a pump feeds back the water to the boiler. The adiabatic efficiency of the turbine is 90%. The thermodynamic states of water and steam are given in the table.





State h(kJ kg-1) s(kJ kg-1k-1) v(m3kg-1)
Steam: 4MPa, 350°C 3092.5 6.5821 0.06645
Water: 15kPa hf hg sf sg vf vg
225.94 2599.1 0.7549 8.0085 0.001014 10.02


h is specific enthalpy, s is specific entropy and v the specific volume; subscripts f and g denote saturated liquid state and saturated vapour state.


Q. The net work output ( kJ kg-1) of the cycle is
Option (A)
498
Option (B)
775
Option(C)
860
Option(D)
957
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-2382