Answer :
Answer:
a) The Heat Transfer to the working fluid passing through the steam generator is 3153.3 KJ/Kg
b) The Thermal Efficiency is 32.7%.
c) The heat transfer from the working fluid passing through the condenser to the cooling water is 2121 KJ/Kg
Explanation:
Given,
P1=10MPa
P2=6kPa
h1=3320 KJ/Kg and s1= 6530J/Kgk due to the properties of water at 10MPa.
h2=20100 KJ/Kg, s2=s1= 6530J/Kgk and quality, x=0.77 due to the properties at 6kPa.
h3=151 KJ/Kg, s3= 521J/Kgk and v3=0.00101[tex]m^{3} /Kg[/tex] due to the properties at P3=P2=6kPa and x=0.
h4=162 KJ/Kg, s4=s3= 521J/Kgk and v4=0.001[tex]m^{3} /Kg[/tex] due to the properties at P4=P1=10MPa.
The efficiency of turbine is 0.8.
So,( h1-h2')/(h1-h2) =0.8
By solving above equation:
h2'=2272KJ/Kg
The efficiency of pump is 0.7.
So,( h4-h3)/(h4'-h3) =0.7
By solving above equation:
h4'=166.71KJ/Kg
a) The Heat Transfer to the working fluid passing through the steam generator:
q=h1-h4'
q=3320 -166.7
q= 3153.3 KJ/Kg
b) Thermal Efficiency:
=(h1-h2')-(h4'-h3)/q
= 32.7%
c) The heat transfer from the working fluid passing through the condenser to the cooling water:
=h2'-h3
=2121KJ/Kg
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