Question
Download Solution PDFIt is proposed that solar energy be used to warm a large collector plate. The energy would in turn, be transferred as heat to a fluid within a heat engine, and the engine would reject energy as heat to the atmosphere having assumed temperature of 20°C. Experiments indicate that about 1880 KJ/m2h of energy can be collected when the plate is operating at 90°C. The minimum collector area that would be required for plant producing 1 KW of useful shaft power will be approximately
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The collector absorbs solar energy to operate a heat engine. The maximum efficiency of the engine is given by Carnot's principle. The useful power is calculated by multiplying collected solar power with Carnot efficiency.
Calculation:
- Source temperature: \( T_h = 90^\circ C = 363~K \)
- Sink temperature: \( T_c = 20^\circ C = 293~K \)
- Collected solar energy: \( 1880~kJ/m^2 \cdot h = \frac{1880000}{3600} = 522.22~W/m^2 \)
Carnot efficiency:
\( \eta = 1 - \frac{T_c}{T_h} = 1 - \frac{293}{363} = 0.1939 \)
Useful power per m²:
\( 522.22 \times 0.1939 = 101.28~W/m^2 \)
Required power output: 1 kW = 1000 W
\( \text{Area} = \frac{1000}{101.28} \approx 9.87 \approx 10~m^2 \)
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