Textbook Problem

Calculating Efficiencies

A solar power tower has 50 heliostats, each of which has an active area of $\text{2}{\text{.7 m}}^{\text{2}}$. The tower produces 29.7 kW of power. By assuming that the solar power falling on the active area is ${\text{1000 W/m}}^{\text{2}}$, what is the percent efficiency of the tower?

Interpretation Introduction

Interpretation:

The percent efficiency of the tower is to be calculated.

Concept Introduction:

Hundreds of sun-tracking mirrors, called heliostats, form a tower.

The percent efficiency of heliostats is given by:

$\%\text{ efficiency}=\frac{\text{Output power}}{\text{Input power}}\times 100\%$

Answer to Problem 10.1YT

Solution: $22\%$.

Explanation of Solution

Since the total power produced is $29.7\text{ kW}$, the output power is as follows:

The conversion of kW into W is as follows:

$29.7\text{ kW}=29.7\times {10}^3\text{ W}$.

Since the total number of heliostats is $50$, each of which has an area of $2.7{\text{ m}}^2$, the total active area is as follows:

$\begin{array}{c}\text{A}=50\times 2.7{\text{ m}}^2\\ =135{\text{ m}}^2\end{array}$

Hence, the active area is $135{\text{ m}}^2$.

The solar energy falls on the active area at the rate of $1000\text{ W}/{\text{m}}^2$. So, the input power is calculated as follows:

$\begin{array}{c}\text{Input power}=135\cancel{{\text{m}}^2}\times \frac{1000\text{ W}}{\cancel{{\text{m}}^2}}\\ =135\times {10}^3\text{ W}\end{array}$

Therefore, the input power is $135\times {10}^3\text{ W}$.

The percent efficiency of heliostats is calculated as follows:

$\%\text{ efficiency}=\frac{\text{Output power}}{\text{Input power}}\times 100\%$

Substitute $135\times {10}^3\text{ W}$ for input power and $29.7\times {10}^3\text{ W}$ for output power in the above equation as:

$\begin{array}{c}\%\text{ efficiency}=\frac{29.7\times {10}^3\cancel{\text{W}}}{135\times {10}^3\cancel{\text{W}}}\times 100\%\\ =0.22\times 100\%\\ =22\%\end{array}$

Conclusion

The tower gives a percent efficiency of $22\%$.