Example Calculation of Solar Collector Area
Problem Statement - Part 1: Consider a 2000 sq. ft. house in Albuquerque, NM. It has been determined using the methods from the articles mentioned above, that the heat loss rate from the house is 5.6 Btu/oF-day-ft2 and that there are an average of 955 oF-day heating degree days in January in Albuquerque. How many Btu/day are needed on the average to heat this house in the month of January?
Solution: The total heat needed for the month of January is: (5.6 Btu/oF-day-ft2)(2000 ft2)(955 oF-days) = 10,696,000 Btu. Dividing by the 31 days in January gives: 345,032 Btu/day as the average needed for heat in January.
Problem Statement - Part 2: Approximately how many square feet of south facing, flat plate solar collector, at a tilt angle of 50o from the horizontal, would be needed to provide the 345,032 Btu/day of heat to the home in Albuquerque, NM.
Solution: From the first reference below (discussed in some detail in the "Estimating Solar Radiation Rate ..." article mentioned above), it is noted that the latitude of Albuquerque, NM, is 35.05 oN, so the given 50o tilt angle is approximately equal to the latitude plus 15o. Also, from the same reference, it can be found that the solar radiation rate to a south facing, flat plate solar collector tilted at latitude plus 15o from the horizontal is: 5.8 kWh/day/m2 (1840 Btu/day/ft2). The 1840 Btu/day/ft2, however, is the rate of solar radiation striking the collector. The fraction of that figure that will be delivered as useful heat to the house is the efficiency of the solar collector. From the second reference below, typical efficiency for a flat plate solar collector could range from 30 to 70%, depending on many factors. Assuming 50% collector efficiency, the collector area needed is: (345,032 Btu/day)/[(0.50)(1840) Btu/day/ft2] = 375 ft2.