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Circuit Breakers - tripping or not tripping?

  1. Feb 15, 2009 #1
    The problem statement:
    To save on heating costs, the owner of a greenhouse keeps 800kg of water around in barrels. During a winter day, the water is heated by the sun to 50 degrees Fahrenheit. During the night the water freezes into ice at 32 degrees Fahrenheit in 10 hours. An electrical heating system is used providing the same heating effect as the water.


    The attempt at a solution:

    P=Q/t = (542*10^6)/36000s = 15066 Watts
    I=P/V = 15066/240 = 62.77Amperes

    The minimum ampere rating that the 240 volt circuit breaker would have to be to avoid tripping is what i calculated....it turned out to be 62.77Amperes.

    The question:
    What I don't get is "does the circuit breaker trip? Why or why not?"
     
  2. jcsd
  3. Feb 16, 2009 #2

    Redbelly98

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    Weird, they would have to tell you how much current will trip the breaker in order to answer the question.

    p.s. I'm curious, how did you come up with 542 MJ for the energy?
     
  4. Feb 16, 2009 #3
    Well 4186 is spec heat of water and 2093 is spef heat of ice. I also converted the temps to kelvins. Then I input them into the equations...I did Q1= (800)(4186)(305.93)=102*10^7 and Q2=(800)(2093)(287.93)=482*10^6. Then i got Q= Q1-Q2 = 542*10^6.
     
  5. Feb 16, 2009 #4

    Redbelly98

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    That doesn't look right.

    If Q is the thermal energy change of the water, the formula is

    Q = c m ΔT​

    where c (for liquid water) is 4186 J/(kg-K), m is 800 kg, and ΔT is the change in temperature of the liquid water.

    That would be the heat removed from the liquid water as it cools from 50F to 32F. But in addition you'll also need the heat removed as the water freezes to ice, at a constant temp of 32F. Did you class talk about the heat required to freeze water or melt ice? It's an additional equation, it should be talked about in your textbook or lecture notes.
     
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