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Rate of heat transfer/temperature change

  1. Jun 16, 2015 #1

    I have another crazy project I'm working on and I need to know if this is viable before I spend a bunch of time and money. Here is the situation:

    I have a small block of Al. 1/4" x 1/4" x 1" long. Thru the length I drill a 1/16" hole. The surface will be whatever a standard drill bit produces.

    I now cool the block to -80F and pass air thru the hole (sealed from the outside of course.)

    The air in the 1/16" x 1" hole (about .001 cubic inches) will be in contact with the Al surface of .196 sq inches for about .0005 seconds. (I can vary it between 6-9 cubic inches per second.)

    Assuming the block stays at a uniform -80F, how much temperature difference can I expect between the air at the one side of the block to what exits the 1" long block? Assume your standard air, 70-100F, 10-20% humidity, standard pressure.

    If I changed it to a block of copper what will be the difference?

    Thank you.
  2. jcsd
  3. Jun 16, 2015 #2


    Staff: Mentor

    Why can't you do your own calculations?

    Heat transfer will be proportional to the temperature difference time area times a constant. You may have to determine the value of the constant by assumption or by experiment.

    Temperature change in each medium is porportional to heat flow and the specic heat of the materials. You can look up the material properties.

    You'll have to make your own assumptions about the uniform or non uniform distribution of temperatures, or the time required to reach thermal equilibrium.
  4. Jun 16, 2015 #3


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    Given the short duration that the air is in the hole I suspect there won't be very much temperature difference between the temperature of the incoming air and the exit.
  5. Jun 16, 2015 #4
    A quick calculation shows a temperature drop of about 15oF (before ice starts forming in the hole).
    Last edited: Jun 16, 2015
  6. Jun 16, 2015 #5
    Thank you Insightful. Can you give me the formula that you used? So I can make modifications to the system? Thank you!
  7. Jun 16, 2015 #6
    If I knew the formula to use, I would not be posting here.
  8. Jun 16, 2015 #7
    U=20 Btu/(hr sqft oF)......a good "say number" (from years of designing heat exchangers)
    A=area of inside wall
    m=lb/hr air
    C=0.25 Btu/(lb oF)
    Tair=average air temp in hole. Requires trial and error calculation for exact solution, but I just used 70oF for Tair and Tin as a first approximation. Note: An exact solution is not that useful given the approximations and the very cold wall temperature.
  9. Jun 18, 2015 #8

    I calculate the change in temperature to be about 1 or 2 degrees F at most, mainly because of the short time that the air will be in the hole as CWatters said.

    The energy transfer rate is easily calculated using the Fourier conduction equation or this handy on-line calculator and it works out to 0.364 Btu , 0.107 joules/s. I used the thermal conductivity of air, 0.024 watts per meter per Kelvin since there is no insulator involved. (the air is in direct contact with the metal block)

    Based upon one Btu can raise the temp of 55 cubic feet (95,000 cubic inches) of air one degree F in 1 hour, and there is .00307 cu in of air in the hole for 0.0005 seconds, that works out to 4.3 degrees F per Btu. Therefore, 0.364 Btu = 1.6 degrees F.

    Using copper instead of aluminum will make no difference because of the direct contact with air.

    But the hole will probably ice up from dew point condensation and close very soon anyway.
  10. Jun 18, 2015 #9
    What does this mean? Units are not consistent.

    Also, the on-line calculator is for conductive heat transfer. This flow problem has conduction and convection in turbulent flow...a much higher heat transfer rate.
    Last edited: Jun 18, 2015
  11. Jun 18, 2015 #10
    That should have been Btu/hr.

    0.3632 Btu/Hr = 0.10643 watts = 0.10643 j/s
    So sorry to confuse you!

    turbulent flow in the little 1/16 inch dia hole?

    Well OK, then!

    Model it any way that you like, it is still only an approximation. Have fun!
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