Question

A car wash needs 1000 gallons of warm water a day heated from 40 degrees F to 110 degrees F. How large a solar collector would be needed to do this? (Assume the incident solar energy is 1800 Btu per square foot each day, and the collector efficiency is 60 percent. (Computer your answer in square feet, but do not enter units. If your answer is 17 square feet, please enter 17.0)

Answer 1

Volume of water = 1000 gallons = 3785.41 L

Hence, Mass of water = 3785.41 kg

Change in temperature = 38.9^oC

Therefore, Heat required for water to heat up, H = mcΔT

= (3785.41 kg) x (4186 J/kg^oC) x (38.9^oC)

= 616218110 J

= 584062.1 Btu

Hence, Heat collected from solar collector = (584062.1 Btu) / (0.60) = 973436.8 Btu

Hence, Area of collector required = (973436.8 Btu) / (1800 Btu/sq.ft) = 540.8 sq.ft

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