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ZephyR1
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[SOLVED] Nuclear Powerplant Efficiency
A nuclear power plant operates at 78% of its maximum theoretical (Carnot) efficiency between temperatures of 540°C and 330°C. If the plant produces electric energy at the rate of 1.3 GW, how much exhaust heat is discharged per hour?
1)e=(Qh-Qc)/Qh
2)e=W/Qh
3)e=1-(Tc/Th)
e=(Qh-Qc)/Qh, so...
e=[(540+273)-(330+273)]/(540+273)
e=.258302583
e=.258302583(.78) since its operating at 78% efficiency
e=.2014760148
e=W/Qh, so...
.2014760148=(1.3*10^9)(60))/Qh
Qh=3.87*10^11
e=1-(Qc/Qh), so...
Qc/3.87*10^11=(1-.2014760148)
Qc=3.091*10^11
3.091*10^11*60
Qc=1.855*10^13 J/h
I don't know where I went wrong?
Homework Statement
A nuclear power plant operates at 78% of its maximum theoretical (Carnot) efficiency between temperatures of 540°C and 330°C. If the plant produces electric energy at the rate of 1.3 GW, how much exhaust heat is discharged per hour?
Homework Equations
1)e=(Qh-Qc)/Qh
2)e=W/Qh
3)e=1-(Tc/Th)
The Attempt at a Solution
e=(Qh-Qc)/Qh, so...
e=[(540+273)-(330+273)]/(540+273)
e=.258302583
e=.258302583(.78) since its operating at 78% efficiency
e=.2014760148
e=W/Qh, so...
.2014760148=(1.3*10^9)(60))/Qh
Qh=3.87*10^11
e=1-(Qc/Qh), so...
Qc/3.87*10^11=(1-.2014760148)
Qc=3.091*10^11
3.091*10^11*60
Qc=1.855*10^13 J/h
I don't know where I went wrong?
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