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Heat Loss of Liquid, Solve for Time

  1. Dec 19, 2012 #1
    Hi all, i have a problem here I need to solve asap.

    1. The problem statement, all variables and given/known data

    I need to find out the Time it takes for 6900 gallons of WATER to go from:
    55 deg F. to 32 deg F.
    55 deg F. to 20 deg F.

    ambient air set at 20 deg F. if needed
    The heat loss rate is 750 Btu/Hr/degF.

    2. Relevant equations

    Q=mCdT

    3. The attempt at a solution

    Tf = 20 F
    Ti = 55 F
    m = 6900*8.35 (lbs)
    c = 1? btu/lb/degF? (was a bit confused with this)

    I have tried the formula many times but get the same answer... which I feel like is not very reasonable. Welcome any and all constructive help please =)
     
    Last edited: Dec 19, 2012
  2. jcsd
  3. Dec 19, 2012 #2

    rude man

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    It takes 8.34 BTU to heat 1 gallon of water 1 deg F.

    That is all ye need to know, as Keats once said ... the rest is high school "physics".
     
  4. Dec 19, 2012 #3
    I believe i may need to show work on how i got that number?

    Looks like thats volumetric specific heat?

    where is the time portion?
     
  5. Dec 19, 2012 #4

    haruspex

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    That /degF should refer to a temperature difference above ambient. To answer it an ambient must be specified (and it must be less than 20F).
    Note also that at 20F it will be ice.
     
  6. Dec 19, 2012 #5

    rude man

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    Why? The heat loss rate is given, makes no difference what the ambient is.
    [/quote}
    Good point, one that I didn't see.
     
  7. Dec 19, 2012 #6
    well i didnt know if it mattered or not but I think ambient air is 20 deg F. in that question.

    can somebody clarify Q in Q=mCdT for me please?
    Is it thermal energy in BTU or is it a rate? (btu/hr)
     
  8. Dec 20, 2012 #7

    haruspex

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    The heat loss rate is given as 750 Btu/Hr/degF, not 750 Btu/Hr. This makes sense because the rate of diffusion of heat is proportional to the temperature gradient, so the bigger the temp difference the higher the rate.
    The consequence is that the temp will follow a negative exponential curve, asymptotically reaching ambient. If the ambient is 20F we can get a sensible answer for reaching 32F but reaching 20F will take forever.
     
  9. Dec 20, 2012 #8
    so... no actual answers to any of my direct questions....?
    thats gg
    thanks anyway everyone.
     
  10. Dec 20, 2012 #9

    rude man

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    heat gained (or lost) = mass x specific heat x temp[erature difference
    It is in BTU, not btu/hr.

    There are 3 steps to this:
    1.water temp. is reduced from 55F to 32F
    2.Water is chanbged to ice at a constant 32F
    3.Ice is reduced in temp. from 32F to 20F.

    There is indeed a problem with this problem as haruspex pointed out. If the heat rate loss is indeed given in BTU/lb/deg F then the problem can't be solved.

    I misread the heat loss rate to be 750 BTU/hr, not 750 BTU/hr/deg T. Then the problem could be solved. As haruspex correctly pointed out, if heat loss rate is proportional to temperature difference then it matters a whole lot what the ambient temperature is. If it's 20F then yes it would take forever to reach 20F. If it's below 20F then there is a finite time for an answer. So you need to get this clarified with your instructor.
     
  11. Dec 20, 2012 #10
    I see. would it be wrong to try to find an average heat loss rate using the current information?
    can you help me set up for the process down to the 32F point?
     
  12. Dec 20, 2012 #11

    rude man

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    OK, let's assume an ambient of 20F. That can get us thru parts 1 and 2.

    Part 1: yes, you can average the heat loss down to 32F if you assume the specific heat of water is constant during this interval, which it very nearly is.

    So pick the avg temp. betw. 55F and 20F to compute the total heat loss down to 32F.
    Then, use the specific heat of water in BTU/lb/F to compute how long it took to get to 32F.
    Remember to calculate how many lbs in your water sample.

    Part 2: T is now a constant = (32F - 20F) ; use the latent heat of fusion of water to again compute the heat removed and then the time that had to take.

    You can't do part 3 unless the ambient is lower than +20F as we explained.
     
  13. Dec 20, 2012 #12

    haruspex

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    No, you can't do it by taking the average temperature difference. Cooling follows a negative exponential function of time, not a linear one.
    Q = heat content (losable part)
    C = specific heat
    m = mass
    T = temperature above ambient
    k = heat loss rate/deg temp diff
    dQ/dt = -kT
    Q = mCT
    So dT/dt = -(k/mC)T
    ln(T) = ln(T0)-kt/mC
    t = (mC/k) ln (T0/T)
     
  14. Dec 20, 2012 #13

    rude man

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    Right. I was about to edit my last post to that effect.

    I got dQ = -mcdT = k(T-TA)dt whence
    t1 = (mc/k)ln{(T1-TA)/(T0-TA)}

    where dQ = increment of heat removed (i.e. dQ > 0), t1 = time to go from 55 to 32F, T = instantaneous water temp., TA = ambient, T1 = 55F and T0 = 32F.
     
  15. Dec 20, 2012 #14

    haruspex

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    I defined T, T0 as excess above ambient, and my T0 is initial temp so corresponds to your T1, while your T0 corresponds to my T. That renders the equations the same.
     
  16. Dec 20, 2012 #15

    rude man

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    Thought so, wasn't sure. Thanks for your inputs.
     
  17. Dec 20, 2012 #16

    rude man

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    ish562, are you following all this? Sorry I misled you about being able to use average temperature change.
     
  18. Dec 28, 2012 #17
    not exactly...
    dQ = increment of heat
    = mcdT?
    i thought it was Q = mcdT
    how does the dT change in this situation??

    where is k(T-Ta) come from exactly?
     
  19. Dec 28, 2012 #18

    rude man

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    dQ is a differentially small removal of Q. Q is total heat removed = ∫dQ.

    If you have an equation with a differential on one side you must also have some kind of differential on the other side.
    If k is the heat loss per degree temperature difference per unit time, then the differential amount of heat lost (as a positive quantity) in differential amount of time dt is dQ = k(T - TA)dt because T - TA is the temperature difference between the water and the ambient.
     
  20. Dec 28, 2012 #19

    gneill

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    I don't know if the OP jsh562 is familiar with circuit theory, and in particular RC circuits, but if so it might interest him to note that each of the steps being described, namely cooling from initial temperature to freezing point, freezing the water to ice, and cooling the ice to ambient, can be modeled in the form of an equivalent electric circuit (see figure).

    In other words, we can assign the thermal parameters to analogous electrical components and then apply standard circuit analysis methods to answer the problem. Sometimes this can help to clarify what's going on...

    attachment.php?attachmentid=54310&stc=1&d=1356720674.gif

    Hint: For step 3, it would be worthwhile thinking about the engineer's ##5 \tau## rule of thumb for deciding when all the excitement is over :wink:
     

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  21. Dec 28, 2012 #20

    rude man

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    Very intersting, g, but I think most OP's have enough on their plate to assimilate the physics they're supposed to be learning ... of course, the old analog computers did something very similar via electronic gain, integrator and nonlinear blocks.
     
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