Answer:
a)75.8%
b)2.517KW/K
Explanation:
Hello!
To solve this problem follow the steps below
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
1. use thermodynamic tables to find the following variables.
a.enthalpy and entropy at the turbine entrance
h1=Enthalpy(Water;T=500;P=6000)
=3422KJ/kg
s1=Entropy(Water;T=500;P=6000)
=6.881KJ/kgK
b. enthalpy and ideal entropy at the turbine outlet
h2i=Enthalpy(Water;s=6.881;P=20)
=2267KJ/kg
s2i=s1=6.881KJ/kgK
2. uses the output power and the first law of thermodynamics to find the real enthalpy at the turbine's output
W=m(h1-h2)
h2=h1-W/m
h2r=3422-2626/3=2546.6KJ/kg
3.
find efficiency with the following equation
[tex]eficiency=\frac{h1-h2r}{h1-h2i}[/tex]
[tex]\frac{h1-h2r}{h1-h2i}=\frac{3422-2546.6}{3422-2267} =0.758=75.8%[/tex]
4.
find the real entropy at the turbine exit
s2=Entropy(Water;h=2546,6;P=20)=7.72KJ/kgK
5.Finally find the entropy generated, using the following equation
ΔS=m(s2-s1)=(3kg/s)(7.72 KJ/kgK-6.881 KJ/kgK)=2.517KW/K
