wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In a power generating plant, steam enters the turbine at 150 bar, 600 C and exits at 0.5 bar. Isentropic efficiency of the turbine is 85%. Neglecting the pump work, the power generated by the plant is
Use the following data:

A
1073.55 kJ/kg
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1263.00 kJ/kg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1134.56 kJ/kg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1346.23 kJ/kg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 1073.55 kJ/kg
P(bar)l(C)h(kJ/kg)s(kJ/kgK)15060035836.67970.581.32hf=340hg=2645sf=1.091sg=7.593

At state 1 h1=3583 kJ/kg
s1=6.6797 kJ/kgK
s1=s2s=6.6797
6.6797=sf+x(sgsf)
x=6.67971.0917.5931.091
x=0.859
h2s=hf+x(hghf)
h2s=340+0.859(2645340)
h2s=2319.995 kJ/kg
Actual work =ηisen×Isentropic work
=0.85×(35832319.995)
Actual work =1073.55 kJ/kg

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon