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Thermodynamic (Solar Hot Water System) Problems?

  1. Dec 21, 2013 #1
    1. The problem statement, all variables and given/known data

    A solar hot water system is designed to supply 100% of the hot water demand of 0.6 m3/day when the isolation on the tilted collector is 22,000 kJ/m2 with a collector efficiency of 50%. The tap water temperature is 8ºC and the desired hot water temperature is 58ºC. Determine the required collector area.

    2. Relevant equations

    P0 = W0/t
    W0 = C m ΔT
    P0 = ε P

    3. The attempt at a solution
    I tried to solve this problem, but how can I use the 0.6 m3/day, somehow I was used it by replacing m3 by kg because density of water is 1 cal/gm*0C, but some illogical value raise as a result like Ac = 1.32*10-7 m2
     
    Last edited: Dec 21, 2013
  2. jcsd
  3. Dec 21, 2013 #2

    SteamKing

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    Since when is density measured in units of cal/gm-C?
     
  4. Dec 21, 2013 #3
    Thermodynamic (Solar Hot Water System) Problems?miss print

    I'm sorry its miss print.
    I mean that the density of water is 1 kg/m3.
    So according to the equation Ï= m/v when Ï=1 m=v
    Hence we can replace 0.6 m3 by 0.6 kg
     
  5. Dec 21, 2013 #4

    SteamKing

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    Is the density of water equal to 1 kg/m^3? That seems to be an awfully small number. Take a moment to visualize how large a volume a cube measuring 1 m on each side is.
     
  6. Dec 22, 2013 #5
    Tring to solve the problem

    Any mistake or faults you see in the following solve?
    P0=W0/t
    W0=C*m*ΔT
    P0=ε*P
    P=1/ε *(C.m.ΔT) ------ (1)
    ε is 50% = 0.50
    C is 1 cal/gm *0C
    since Ï= m/V and Ïwater= 1 ∴ m=V , m=600 gm
    ΔT=Tm-Ta = 58-8=50 0C
    by replacing the values above in equation (1)
    P=1/0.50 * (1 * 600 * 50)
    P= 60,000 cal = 60,000 * 4.186 = 251160 J = 251,16 kJ
    Ac=P/Q
    Ac=251.16 / 22,000 = 0.0114 m2 Collector area
    .
    .
    .
    is this result is illogical or a step mistake was happened.
    Thanks in advance.
     
  7. Dec 22, 2013 #6

    SteamKing

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    Hello! You're not listening! Water does not have a density of 1 kg per cubic meter! That is an absurdly low density! You can barely make a cup of coffee with 600 g of hot water!
     
  8. Dec 22, 2013 #7
    They have interesting sizes of coffee cups where you live. Starbucks' largest is Trenta, at 920 ml, but it is for iced drinks only, I hear. Their largest, which is actually used to serve hot beverages, is Venti, 590 ml.

    The customary kitchen "cup" unit is 250 ml (except in the US, where it is a tad less).
     
  9. Dec 22, 2013 #8

    SteamKing

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    Yeah, but when the water only has a density of 1 kg/m^3 like the poster thinks, it's hard to get that full-bodied flavor you want in a premium cuppa Joe.
     
  10. Dec 22, 2013 #9
    This forum make me bored meanwhile two of the big members have fighting without giving any respect to the new members, nevertheless I not get any interest.
     
  11. Dec 22, 2013 #10

    SteamKing

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    You're still not paying attention to what is being said about major errors in your calculation.


    WATER DOES NOT HAVE A DENSITY OF 1 KG / M[itex]^{3}[/itex]!
     
  12. Dec 24, 2013 #11
    It's gotten bro, you know it's passed, our questions not on this point, and no need to this red bold things, I'm not haha. Anyway very thanks for your helps. Bye Bye this forum.
     
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