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Insolation, efficiency, and area

  1. Oct 31, 2014 #1
    1. The problem statement, all variables and given/known data
    What size FPC is needed to supply a family's domestic hot water needs in March in Denver, CO? Assume 80 gallons per day are needed (1gal=8.3 lb.) for the water and that the collector-heat exchanger system has an average efficiency of 40%. The collector tilt angle is equal to latitude.

    2. Relevant equations
    Example given in chapter:
    Calculate the size collector needed to heat 80 gallons of water per day from 50 deg to 130 deg F in March in Los Angeles. Assume the insolation is 1700 Btu/d/ft^2, and the FPC efficiency is 50%.
    Q=mc(change of Temp)
    Q= 80gal x 8.3 lb/gal x 1 Btu/lb/deg F x (130-50)deg F = 53,100 Btu/d
    The heat available from the FPC will be Q = insolation x area x efficiency
    53,100 Btu/d = 1700 Btu/d/ft^2 x area x 0.5
    Therefore, the collector area = 62 ft^2

    3. The attempt at a solution
    Following the example I did:
    Insolation in March for Denver, CO is 2060 Btu/ft^2/d (provided in the text book index)
    80 gallons x 8.3 lbs/gal = 664 lbs
    mean temp in March in Denver, CO = 45 deg F (provided in index of text)
    efficiency = 40%
    664 lbs x 1 Btu/lb/deg F x (130 - 45) deg F = 56,440 Btu/d
    56,440 Btu/d = 2060 Btu/ft^2/d x area x 0.4 = 69 ft^2

    My professor marked it wrong and said it's 56 ft^2, where did I go wrong?
     
  2. jcsd
  3. Oct 31, 2014 #2

    haruspex

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    I see nothing wrong with your calculation. Are you sure the target temperature is still 130F? But my guess is the prof used 50% efficiency by mistake.
     
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