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Heat Transfer

  1. Apr 2, 2013 #1
    Hello Gentlemen,

    Long time reader first time poster.

    I have a problem that I am trying to solve below.

    Problem 1

    A metal copper rod say 20mm diameter has a constant heat applied to it of 150°C on one end. Ambient temp of Air is 25°C.

    What length rod do I need so the other end is at 50°C one equilibrium is reached?

    Problem 2

    Also another one that I am looking at is. If the electric stove hot plate is at a constant 100°C area of 100mmX100mm what would be the temperature of the air 200mm above the hot plate using ambient air temp of 25°C again?

    They seem like straight forward problems but for the life of me I cant get my head around it. Any help would be appreciate it.

    Thank you in advance.
     
  2. jcsd
  3. Apr 2, 2013 #2
    I sense these are homework problems. Are they? Here are my thoughts on the problems. I don't have them all worked out, but these are some things to consider:

    Problem 1: It's a steady-state problem, so the heat flow in must equal the heat flow out. The heat flow due to conduction along the rod must be convected away by the air, so [itex]kA_{xsec}\frac{\partial T}{\partial x} = hA_{surf}(T_{wall}-T_∞)[/itex]. You didn't mention any moving air, so it's a free convection. I'm assuming there's a linear heat gradient across the rod, so [itex]\frac{\partial T}{\partial x} = \frac{dT}{dx} = \frac{T_h-T_c}{L}[/itex]. If you don't have your h for convection, then you have to solve for it using nondimensional numbers like Grashof, Prandtl, Nusselt and such.

    Problem 2: This one's a free convection problem with heated surfaces radiating to air. To determine the heat flow out of the plate, we can use an equation derived in Heat Transfer by Holman (p.346) for heated horizontal surfaces facing upward for free convection to air: [itex]h = 1.32\left ( \frac{T_{plate}-T_{air}}{P/A} \right )^{1/4}[/itex], where P is the perimeter of the plate, A is the surface area. Then you can use [itex]q = hA_{surf} (T_{plate}-T_{air})[/itex] to get the heat flow out of the plate. The problem with this kind of analysis is that it's assumed that the temperature above the thermal boundary layer of the plate is the free-stream temperature T (in this case, 25 deg C). Since the plate is radiating upwards, eventually that temperature will change depending on the conduction of heat through the air. You might need to do this numerically to get something exact.
     
  4. Apr 2, 2013 #3
    In problem 1, you are dealing with a "cooling fin" situation. Timethereaper is on the right track, but the differential equation is really:

    [tex]kA_{xsec}\frac{d^2T}{dx^2}-hP(T-T_{amb})=0[/tex]

    where P is the perimeter of the rod. The boundary conditions are T = 150 at x = 0, and zero flux at x = L.
     
  5. Apr 3, 2013 #4
    Absolutely Not. I am a CNC Machinist by trade and Learning Physics / Match / 3d cad in my spare time. I am one of those that finds TV boring.

    Problem 1 came about when heat shrinking inference fit couplers together, so I was curious if there is a simple equation that explain heat dissipation over distance. I just used copper as an example cos its a good heat conductor.

    Problem 2 was just my curiosity i thought they would be straight forward to answer but apparently i need to study some advanced topics before I can tackle this.
     
  6. Apr 3, 2013 #5
    Yeah, that sounds right. I had a feeling that my equation wasn't the right one.

    Sorry, my bad. The questions just had that textbook feel to them. +1 for learning in your spare time instead of just TV.
     
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