Friley Hall is one of the largest residence halls in the United States, housing
ID: 2997415 • Letter: F
Question
Friley Hall is one of the largest residence halls in the United States, housing 1220 students. On average, each student uses 75 liters of hot water each day. Consider the addition of a trough style solar water heater to be installed on the Friley rooftop as a sustainable means to provide a portion of Friley Hall's hot water needs. Average direct beam daily solar insolation data for Ames from the previous year is provided in the table below. Water will enter the system at 20 degree and and exit the system at 50 degree. One million dollars is available for the towards the solar hot water system. The system has a total installed cost of $1000/collector. Each collector has a surface area of 2 m2 and an overall collection efficiency of 45%. The capacity factor of the system is 35%. Assume that the hot water can be stored indefinitely without heat loss. What is the average daily solar insolation (W/m2)? What is the rated system capacity of the solar hot water heating system needed to meet 100% of the hot water needs of Friley (kW)? What is the total panel area required to meet 100% of the hot water needs of Friley (m2)? Based the available budget, what quantity of hot water can be produced on an average day (L)? Based on the available budget, what fraction of Friley Hall's hot water will be provided on an average day (%)?Explanation / Answer
(a) Average dailty solar insolation = 400 * 40 + 300 * 50 + 200*65 + 150*70 + 100*80+ 75*60 / (365)
= 183.56 W/m^2
(b) hot water required = 75 * 1220 = 91500 litres
amount of power required = 1000 * 91.5 * 4200 * 30 / ( 24*60*60) = 133.4375 kW
(c) let total panel area be A
input power = 183.56 A W
efficiency = 0.45 * 0.35 = 0.1575
0.1575 = 133437.5 / 183.56 A
A = 4615.5 m^2
(d) In available budget only 1000 collectors can be purchased
Therefore A = 2000 m^2
output power = 0.1575 * 183.56 * 2000 = 57.8214 kW
let L be the quantity of hot water
1000 * L * 4200 * 30 / ( 24*60*60) = 57821.4
L = 39649 Liters of hot water can be produced
(e) fraction % = 39.649 / 91.5 = 43.33 %
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