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Chemistry In Focus

6th Edition
Tro + 3 others
Publisher: Cengage Learning
ISBN: 9781305084476

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BuyFindarrow_forward

Chemistry In Focus

6th Edition
Tro + 3 others
Publisher: Cengage Learning
ISBN: 9781305084476
Chapter 10, Problem 10.1YT
Textbook Problem
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Calculating Efficiencies

A solar power tower has 50 heliostats, each of which has an active area of 2 .7 m 2 . The tower produces 29.7 kW of power. By assuming that the solar power falling on the active area is 1000 W/m 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:

% efficiency=Output powerInput power×100%

Answer to Problem 10.1YT

Solution: 22%.

Explanation of Solution

Since the total power produced is 29.7 kW, the output power is as follows:

The conversion of kW into W is as follows:

29.7 kW=29.7×103 W.

Since the total number of heliostats is 50, each of which has an area of 2.7 m2, the total active area is as follows:

A=50×2.7 m2=135 m2

Hence, the active area is 135 m2.

The solar energy falls on the active area at the rate of 1000 W/m2. So, the input power is calculated as follows:

Input power=135 m2×1000 Wm2=135×103 W

Therefore, the input power is 135×103 W.

The percent efficiency of heliostats is calculated as follows:

% efficiency=Output powerInput power×100%

Substitute 135×103 W for input power and 29.7×103 W for output power in the above equation as:

% efficiency=29.7×103 W135×103 W×100%=0.22×100%=22%

Conclusion

The tower gives a percent efficiency of 22%.

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Chemistry In Focus
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