Question:(II) A nuclear power plant operates at 65% of its maximum theoretical (Carnot) efficiency between temperatures of 660°C and 330°C. If the plant produces electric energy at the rate of 1.4 GW, how much exhaust heat is discharged per hour?

The exhaust heat discharged per second gives the exhaust power of the nuclear power plant. The exhaust power is obtained as the difference of the actual power and the total power of the nuclear plant.

The actual power can be obtained by the multiplication of the total power, the maximum theoretical efficiency, and the operating efficiency.

The given data can be listed below as:

• The operating efficiency of the nuclear power plant is\(e = 65\% \left( {\frac{1}{{100}}} \right) = 0.65\).
• The temperature of the hot reservoir is\({T_{\rm{H}}} = 660^\circ {\rm{C}} = \left( {660^\circ {\rm{C}} + {\rm{273}}} \right){\rm{ K}} = 933{\rm{ K}}\).
• The temperature of the cold reservoir is\({T_{\rm{L}}} = 330^\circ {\rm{C}} = \left( {330^\circ {\rm{C}} + 273} \right){\rm{ K}} = 603{\rm{ K}}\).
• The electricity produced by the nuclear power plant is \({P_{\rm{a}}} = 1.4{\rm{ GW}}\).

The maximum theoretical efficiency of the nuclear plant can be expressed as:

\({e_{\rm{T}}} = \left( {1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}} \right)\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{e_{\rm{T}}} &= \left( {1 - \frac{{603{\rm{ K}}}}{{{\rm{933 K}}}}} \right)\\ &= 0.3537\end{aligned}\)

The total power of the nuclear power plant can be expressed as:

\(\begin{aligned}{c}{P_{\rm{a}}} &= {P_{\rm{T}}} \times {e_{\rm{T}}} \times e\\{P_{\rm{T}}} &= \frac{{{P_{\rm{a}}}}}{{{e_{\rm{T}}} \times e}}\end{aligned}\)

Here,\({P_{\rm{a}}}\)is the actual power,\({P_{\rm{T}}}\)is the total power,\({e_{\rm{T}}}\)is the maximum theoretical efficiency, and\(e\)is the operating efficiency.

Substitute the values in the above equation.

\(\begin{aligned}{c}{P_{\rm{T}}} &= \frac{{1.4{\rm{ GW}}}}{{0.3537 \times 0.65}}\\ &= 6.09{\rm{ MW}}\end{aligned}\)

The exhaust power of the nuclear power plant can be expressed as:

\({P_{\rm{e}}} = {P_{\rm{T}}} - {P_{\rm{a}}}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{P_{\rm{e}}} &= 6.09{\rm{ MW}} - 1.4{\rm{ GW}}\\ &= 4.69{\rm{ MW}}\left( {\frac{{{{10}^9}{\rm{ W}}}}{{{\rm{ MW}}}}} \right)\left( {\frac{{1{\rm{ J/s}}}}{{1{\rm{ W}}}}} \right)\left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ h}}}}} \right)\\ &= 1.69 \times {10^{13}}{\rm{ J/h}}\end{aligned}\)

Thus, the exhaust power of the nuclear plant is \(1.69 \times {10^{13}}{\rm{ J/h}}\).