A 100 MW output solar thermal power plant collects enough su

A 100 MW output solar thermal power plant collects enough sunlight to operate steadily for 16 hours per day. The solar collector system (with mirrors that focus sunlight on a hot central lower) has a COP = 0.66 = energy absorbed/solar energy striking the mirrors. The power cycle has a thermal efficiency of 1/3 so the overall COP of the power plant = useful electrical output/solar energy input = 0.66 times 1/3 = 22%. Rank the following energy transfers from largest to smallest A solar energy that is absorbed by the tower receiver in MW B useful electrical output in MW C solar energy that is scattered from the mirrors or tower receiver and not absorbed m MW D waste heat transferred from the power cycle to the surrounding environment in MW

Solution

Given useful electrical output = 100 MW

From overall COP of power plant,

Solar energy input (striking the mirrors) = Electrical output / (0.22) = 100 / 0.22 = 454.54 MW

From COP of solar collector system,

Energy absorbed by the tower receiver = 0.66 * Solar energy striking the mirrors = 0.66 * 454.54 = 300 MW

Solar energy scattered and not absorbed = Solar energy input - Energy absorbed = 454.54 - 300 = 154.54 MW

Thermal energy input for power cycle = Energy absorbed = 300 MW

Waste heat transferred from power cycle = Thermal energy input - Useful energy output = 300 - 100 = 200 MW

Based on above information the ranking will be,

A > D > C > B