eg Estimate th he collector surface area needed to heat 100 gallons of water a day from 50 F to 120 F when the daily insolation is 1000 Btu/ft. Assume an efficiency of 50% Solution Recall that I Btu will heat one pond of water by 1 F, and a gallon of water weigh about 8 pounds Then Btu heat energy needed=100 gal、81b1B1 ×70F=56.000 aay Because the efficiency is 50% there must be twice the needed amount, or 112, 000 Btu, of solar energy incident on the collector each day. The area needed will be 1120008/y=112f2 1000 Btu/ft .da 复旦大学环境科学与工程系 Department of Environmental Science and Engineering, Fudan Universityeg. Estimate the collector surface area needed to heat 100 gallons of water a day from 50 ̊F to 120 ̊F when the daily insolation is 1000 Btu/ft2. Assume an efficiency of 50% day Btu F F lb Btu gal lb day gal 1 70 56,000 1 8 heat energyneeded 100 × = ⋅ = × ×   Solution: Recall that 1 Btu will heat one pond of water by 1 ̊F, and a gallon of water weigh about 8 pounds. Then Because the efficiency is 50% there must be twice the needed amount, or 112,000 Btu, of solar energy incident on the collector each day. The area needed will be: 2 2 112ft 1000 ft 112,000 = Btu ⋅day Btu day