Question:(II) A certain power plant puts out 580 MW of electric power. Estimate the heat discharged per second, assuming that the plant has an efficiency of 32%.

The power developed by the power plant is the work done by the plant per unit time. The heat discharged by the power plant per unit time depends on the plant’s efficiency and the power developed by the plant.The unit of power developed by the plant can be expressed in terms of watts or megawatts.

The given data can be listed below as:

• The efficiency of the power plant is\(e = 32\% \left( {\frac{1}{{100}}} \right) = 0.32\).
• The power developed by the power plant is\(P = 580{\rm{ MW}}\).
• The time taken by the plant to discharge heat is \(t = 1{\rm{ s}}\).

The work done by the power plant can be expressed as:

\(\begin{aligned}{c}W &= {Q_{\rm{H}}} - {Q_{\rm{L}}}\\{Q_{\rm{H}}} &= W + {Q_{\rm{L}}}\end{aligned}\)

Here,\({Q_{\rm{H}}}\)is the heat added or supplied to the power plant and\({Q_{\rm{L}}}\)is the heat discharged by the power plant.

The efficiency of the power plant can be expressed as:

\(\begin{aligned}{c}e &= \left( {\frac{W}{{{Q_{\rm{H}}}}}} \right)\\e &= \left( {\frac{W}{{W + {Q_{\rm{L}}}}}} \right)\\\frac{{W + {Q_{\rm{L}}}}}{W} &= \frac{1}{e}\\1 + \frac{{{Q_{\rm{L}}}}}{W} &= \frac{1}{e}\end{aligned}\)

This can be solved as:

Here, the work (W) per unit time is the power (P) and the heat discharged per unit time (one second) is denoted by .

Substitute the values in the above equation.

Thus, the heat discharged per second by the power plant is 1232.5 MW.