A central power plant, rated at 800,000 kW, operates using steam at 585 K and discards heat to a river at 295 K. If the thermal efficiency of the plant is 70% of the maximum possible value, how much heat is discarded to the river at rated power?

Question

Accepted Answer

The maximum possible thermal efficiency is given by Eq. (5.7). With [latex] T_H[/latex] as the steam-generation temperature and [latex]T_C[/latex] as the river temperature

[latex] \eta_{ ext {Carnot }}=1-\frac{T_C}{T_H} [/latex] (5.7)

[latex] \eta_{ ext {Carnot }} = 1-\frac{295}{585} = 0.4957 \quad and \quad \eta=(0.7) (0.4957)=0.3470 [/latex]

where η is the actual thermal efficiency. Combining Eq. (5.6) with the first law, written [latex]W = −Q_H − Q_C, to eliminate Q_H [/latex], yields:

[latex] \eta \equiv \frac{-W}{Q_H} [/latex] (5.6)

[latex] Q_C=\left(\frac{1 - \eta}{\eta} ight) W=\left(\frac{1 - 0.347}{0.347} ight) (-800,000)=-1,505,475 kW [/latex]

This rate of heat transfer to a modest river would cause a temperature rise of several °C.

Question 5.2: A central power plant, rated at 800,000 kW, operates using s...

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