Chapter 24, Problem 59A

To determine

To show: The calculations for the given maximum power input as revealed in the answer key.

Explanation of Solution

Given:

A power plant is generating $420\text{ MW}$ of electricity by working in at $540°\text{C}$ source temperature and $30°\text{C}$ sink temperature.

Formula used:

The efficiency of the power plant is,

  ${\eta }_{\text{Ideal}}=\frac{{\text{T}}_{\text{source}}{\text{-T}}_{\text{sink}}}{{\text{T}}_{\text{source}}}=\frac{\text{Work Done}}{{\text{Q}}_{\text{source}}}$

Calculation:

Temperature in kelvins is,

  $\begin{array}{c}540°\text{C}=\text{(540+273) K}\\ =\text{813 K}\end{array}$

  $\begin{array}{c}30°\text{C}=\text{(30+273) K}\\ =\text{303 K}\end{array}$

The ideal efficiency of the power plant is

  $\begin{array}{l}{\eta }_{\text{Ideal}}=\frac{{\text{T}}_{\text{source}}{\text{-T}}_{\text{sink}}}{{\text{T}}_{\text{source}}}=\frac{813-303}{813}\\ {\eta }_{\text{Ideal}}=\frac{510}{813}=0.627\end{array}$

Heat given to the power plant is

  $\begin{array}{l}{\eta }_{\text{Ideal}}=\frac{\text{Work Done}}{{\text{Q}}_{\text{source}}}\Rightarrow {\text{Q}}_{\text{source}}=\frac{\text{Work Done}}{{\eta }_{\text{Ideal}}}\\ {\text{Q}}_{\text{source}}=\frac{420\text{ MJ}}{0.627}=670\text{ MJ}\end{array}$

Conclusion:

Thus, results obtained by calculation and those revealed by the answer key matches which is $670\text{ MJ per second}=\text{670,000 KJ per second}$ .