Chapter 15, Problem 62GP

(a)

To determine

The volume of air heated per day.

Answer to Problem 62GP

The volume of air heated per day is $1.774\times {10}^{10}{\text{m}}^3/\text{day}$.

Explanation of Solution

Given info:

The efficiency of the power plant is, $\eta =33\%=0.33$ .

The output of the power plant is, $W=850\text{MW}=850\text{MW}\times \frac{{10}^6\text{J}/\text{s}}{1\text{MW}}=8.50\times {10}^8\text{J}/\text{s}$ .

The rise of temperature is, $\Delta T=7°\text{C}$ .

The density of the air is, $\rho =1.2\text{kg}/{\text{m}}^3$ .

The specific heat of the air is, $c_p=1\text{kJ}/\text{kg}\cdot °\text{C}=1000\text{J}/\text{kg}\cdot °\text{C}$ .

Formula Used:

The expression to calculate the heat transferred to air is,

  $\begin{array}{c}\eta =\frac{W}{Q+W}\\ Q=W\left(\frac{1}{\eta }-1\right)\end{array}$

The expression to calculate thevolume of air heated per day is,

  $V=\frac{Q}{\rho c_p\Delta T}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}Q=\left(8.50\times {10}^8\text{J}/\text{s}\right)\left(\frac{1}{0.33}-1\right)\\ =1.72\times {10}^9\text{J}/\text{s}\\ =\left(1.72\times {10}^9\text{J}/\text{s}\right)\left(\frac{86400\text{s}}{1\text{day}}\right)\\ =1.49\times {10}^{14}\text{J}/\text{day}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}V=\frac{1.49\times {10}^{14}\text{J}/\text{day}}{\left(1.2\text{kg}/{\text{m}}^3\right)\left(1000\text{J}/\text{kg}\cdot °\text{C}\right)\left(7°\text{C}\right)}\\ =1.774\times {10}^{10}{\text{m}}^3/\text{day}\end{array}$

Conclusion:

Thus, the volume of air heated per day is $1.774\times {10}^{10}{\text{m}}^3/\text{day}$.

(b)

To determine

The area of the air layer.

Answer to Problem 62GP

The area of the air layer is $8.87\times {10}^7{\text{m}}^2$ .

Explanation of Solution

Given info:

The time for air heating is, $t=1\text{day}=24\text{h}$ .

The thickness of the air layer is, $x=200\text{m}$ .

Formula Used:

The expression to calculate thearea of the air layer is,

  $A=\frac{Vt}{x}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}A=\frac{\left(1.774\times {10}^{10}{\text{m}}^3/\text{day}\right)\left(1\text{day}\right)}{200\text{m}}\\ =8.87\times {10}^7{\text{m}}^2\end{array}$

Conclusion:

Thus, the area of the air layer is $8.87\times {10}^7{\text{m}}^2$ .