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Answer :
Final Answer:
The exposed surface area of the solar panel required to yield a thermal efficiency of 20% and a power output of 3.6 kW for the heat engine is 51.48 m^2.
Option b) is the correct answer.
Explanation:
1. Calculate the total solar power absorbed by the panel:
Total solar power = Solar radiation fl-u-x × Absorption efficiency
Total solar power = 900 W/m² × 0.85 = 765 W/m²
2. Determine the total power input to the heat engine:
Total power input = Total solar power × Exposed surface area
3.6 kW = 765 W/m² × Exposed surface area
Exposed surface area = 3.6 kW / 765 W/m² = 0.004706 m²
3. Calculate the required exposed surface area:
Exposed surface area = Power output / (Solar radiation fl-u-x × Absorption efficiency × Thermal efficiency)
Exposed surface area = 3.6 kW / (900 W/m² × 0.85 × 0.20) = 51.48 m^2
Option b) 51.48 m^2 is the correct answer.
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