Nuclear Powerplant Efficiency

[ZephyR1]

[SOLVED] Nuclear Powerplant Efficiency

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?

 

[ZephyR1]

nevermind i had it right all along doh!

 

[Moetasim]

I would like to provide a response to the solution attempted for the given problem of nuclear power plant efficiency. The equations used in the solution are correct, however, there seems to be a miscalculation in the final step. The correct value for the exhaust heat discharged per hour would be 1.855*10^13 J/h, which is equivalent to 18.55 TJ/h (terajoules per hour). This value can also be verified by using the first equation, e=(Qh-Qc)/Qh, where Qh is the heat input and Qc is the heat output. Substituting the values, we get e=(3.87*10^11-3.091*10^11)/3.87*10^11 = 0.2 or 20%, which is the same as the calculated efficiency of 78% (0.78).

In conclusion, the correct value for the exhaust heat discharged per hour by the nuclear power plant operating at 78% efficiency would be 1.855*10^13 J/h or 18.55 TJ/h. It is important to double-check calculations and units to ensure accurate results in scientific problem-solving.