Chapter 15, Problem 67GP

To determine

The amount of electrical power saved due to shading the windows.

Answer to Problem 67GP

The electrical power saved is $179.59\text{W}$ .

Explanation of Solution

Given info:

The room temperature is, $T_1=21°\text{C}=\left(21°\text{C}+273\right)\text{K}=294\text{K}$ .

The outside temperature is, $T_2=32°\text{C}=\left(32°\text{C}+273\right)\text{K}=305\text{K}$ .

The power entered in room through the windows is, $P_w=5.3\text{kW}=5300\text{W}$

The reduced power due to shading of windows is, $P_r=500\text{W}$ .

Formula Used:

The expression to calculate the electrical power saved is,

  $E=\left(P_w-P_r\right)\left(\frac{T_2}{T_1}-1\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}E=\left(5300\text{W}-500\text{W}\right)\left(\frac{305\text{K}}{294\text{K}}-1\right)\\ =179.59\text{W}\end{array}$

Conclusion:

Thus, the electrical power saved is $179.59\text{W}$ .