Answer:
17060700 btu/h
Step-by-step explanation:
Power output from power plant = 150 kW.
This power output is at 3% efficiency, this means power available in pond is 0.30P
Now, 150 kW = 0.03P
P = 150/0.03 = 5000 kW of power in the pond.
From basic conversion,
1 kW = 3412.14 btu/h
5000 kW = 5000 x 3412.14
average value of the required solar energy collection rate equal
17060700 btu/h