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Convection heat transfer coefficients ?

  1. Sep 30, 2011 #1
    hi,

    if i have a cylindrical rod of copper and i heat is at one end face while the other surface areas are open to the atmosphere then besides the conductivity of copper i will also need the convective heat transfer coefficient to analyse the setup.

    my question is : how do can we estimate this convective heat transfer coefficient (W/m2K) ; will it depend on the solid material ie. copper, aluminium, ploymer, etc or will it only depend on the fluid is. atmospheric air, water etc ?

    i want to approximate value of the coefficient in case its a free convection scenario...
    any and all help is appreciated.

    thanks..
    dipan
     
  2. jcsd
  3. Sep 30, 2011 #2
    I recommend the Churchill and Chu correlation. A simplier one that is pretty good is by Morgan.
     
  4. Sep 30, 2011 #3
    Hhhhmmm....how about this for a setup...

    Instead of heating up only one end of the copper rod and letting the heat flow from there to the other end and sides (flowing axially and radially)...

    ...how about you actually impose a direct current (DC) through the copper rod and let such current produce heat (I2R)? This way you know how much total heat (Q=I2R or Q=VI) you are injecting into the system. R, above, is the electric resistance of the copper rod which you can estimate given the geometry of the rod and known values of copper resistivity and temperature coefficient (you know, to calculate the resistance of the rod at its final temperature)...or you can simply also measure the voltage across the rod and your losses would simply be Q=VI

    so, if you could place wires (to apply V,I) on the ends of the rod and thermally "plug" those ends, somehow...like holding them via big insulation blocks...then you know that once the copper rod reaches some steady final temperature (you need to monitor this temperature, too, somehow), you record that temperature, the voltage across the rod and the current through it and the temperature of the surrounding air....

    Assuming all the heat into the system is being lost to the surrounding air, then

    ΔT=QRsurf

    where

    ΔT=Trod - Tair

    Q=VI

    Rsurf=1/hA,

    R is, basically, the thermal resistance to heat from the copper surface to the surrounding air

    A is the surface of the cylinder without the ends...pi x D x length

    h is the heat transfer coefficient you are looking for and the only unknown...

    It is important to wait until the system reaches steady state and the copper has absorbed all the heat that is going to absorb (due to its once thermal capacitance) so that you know that at that time the VI that you are providing is truly simply going into the air so that the equation above applies.

    is that kind of along the lines of what you are trying to achieve?
     
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