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Heat Convection

  1. Sep 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Heat is convected away from an object at the rate given by the following formula:

    ΔQ / Δt = h*A*ΔT

    Calculate the rate of convective heat loss in watts for an unclothed person standing in air at 23°C. Assume that the skin temperature is 34°C and that the body surface area is 1.5m2. Calculate for the following air speed: h = 12 w/m2 - °C ; air speed = 1m / sec.

    (h is a coefficient that depends on the shape and the orientation of the object.)


    2. Relevant equations

    ΔQ / Δt = h*A*ΔT


    3. The attempt at a solution

    Our teacher gave us this a problem to work on at home. Before solving, I tried to find all my known variables, but I don't understand where I would get some of the values of these variables from the above information. This includes the difference in time. Any help and explanation on this problem would be very helpful. I was given 3 more values to plug in for h and wind speed, so If I learn how to do this one, I'm sure I can figure out the rest of them.
     
  2. jcsd
  3. Sep 13, 2014 #2

    haruspex

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    There is no specific time period. You are asked for a rate. In fact, the left hand side of your equation should really be dQ/dt (the instantaneous rate of heat loss), not ΔQ/Δt (which would be the average rate of heat loss over a period Δt).
    There seems to be something missing in your equation on the right hand side. There's no variable for the air speed. On the other hand, h is given as Wm-2C-1. This implies h already has the affect of the air speed factored in, so I don't know why you are given it as a separate data value.
     
  4. Sep 13, 2014 #3

    CWatters

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    dQ/dt (or ΔQ/Δt) is "change in Energy"/"change in Time" which is the Power (in Watts). So the equation is really..

    Power = h*A*ΔT

    In case it's not obvious the ΔT part is the difference between air and body temperature. So you appear to have all you need.

    I agree with Haruspex that the air speed info appears to be superfluous. You can find the equation elsewhere on the web and the units given are the same (eg h is the heat transfer coefficient in W/(m2K). Perhaps check the question you posted is word for word correct.
     
  5. Sep 13, 2014 #4

    CWatters

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    Ok I just spotted this bit.

    It looks like h depends on the wind speed as he's given you four values of wind speed and the corresponding value of h.

    So run the equation four times, with each value of h. That will give you four wind speeds (supplied in the problem) and four values for the Power (that you calculate).

    For extra credit plot a graph of wind speed vs Power.
     
  6. Sep 13, 2014 #5
    In heat transfer analysis, we don't refer to it as the power. The terminology we use is what your teacher used, the rate of convective heat loss.
     
  7. Sep 13, 2014 #6
    Using P = h*A*ΔT, given:

    h = 12
    A = 1.5
    ΔT = -11, (23 - 34)

    P = -99

    However, I am confused about my units. I see that Chestermiller says Power doesn't exactly correlate to heat transfer, and when I look at my units of measure, I get the following:

    P = h*A*ΔT

    (watts) = (watts*m2*°C) / (m2 - °C)

    The terms on the left side don't match the terms on the right side? Or maybe I am looking at something wrong? If it helps, here are the other h and air speeds he gave us for the problem.

    h = 6 w/m2 - °C ; air speed = still
    h = 28 w/m2 - °C ; air speed = 5 m / sec.
     
  8. Sep 13, 2014 #7
    That's not what I said. What I said was the Power is not the conventional term that is used in practice to describe rate of heat flow. Rate of heat flow also has units of W, however.

    Chet
     
  9. Sep 13, 2014 #8
    I understand what you mean. In any case, I still don't understand the difference in terms (left side not corresponding to the right side). I'm also curious if my answer is correct.
     
  10. Sep 13, 2014 #9
    I get ##(12\frac{W}{m^2C})(1.5m^2)(34C-23C)##=198 W (rate of heat loss).

    Chet
     
  11. Sep 13, 2014 #10
    delete.
     
    Last edited: Sep 13, 2014
  12. Sep 13, 2014 #11
    I plugged in the wrong h and thinking i was getting the result for another h. Thanks again Chestermiller. Great explanation. This thread can be closed.
     
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