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Heat Transfer Problem - Radiation missing Emissivity value

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Two students, A and B, are discussing a heat transfer problem regarding what the surface temperature of a cartridge electrical heater would be when exposed to a water flow (CASE 1) and an air flow (CASE 2).

    Case 1:

    hw = 5000 W / (m2 * K); Tsurr = 20ºC

    Case 2:

    ha = 50 W / (m2 * K); Tsurr = 20ºC

    Student A insists that the surface temperature of the cartridge will be larger in CASE 2 than in CASE 1, while student B states the opposite.
    The cartridge is shaped as a solid cylinder of length L=200 mm and diameter D=20 mm, which is known to dissipate 2 kW irrespective of the case. Find the surface temperature of the cartridge in each case to find out if student A is right or wrong.


    2. Relevant equations

    qtot = qconv + qrad

    qtot = A * q''conv + A * q''rad

    I assumed that since [itex]\alpha[/itex] wasn't mentioned, [itex]\epsilon[/itex] = [itex]\alpha[/itex] and I could use this form:

    qtot = 2[itex]\pi[/itex]RL * h(Ts-Tsurr) + 2[itex]\pi[/itex]RL * [itex]\epsilon[/itex] [itex]\sigma[/itex] (Ts4 - Tsurr4)

    3. The attempt at a solution

    So really, given the above equations, this should be a "plug and chug" problem by saying that qtot = 2 kW and solving the equation for Ts. However, I cannot do that because the surface emissivity (or absorption) is not given. So I started to think whether or not the cartridge had any radiation at all. I looked through my notes and the textbook and could not find a way to quantify [itex]\epsilon[/itex] using the given information. That's when I came here. Am I wrong to assume that there is any radiation? Or am I doing something completely wrong?

    Thanks for any help!
     
  2. jcsd
  3. Sep 15, 2011 #2
    There will be radiation heat transfer. However, it may (or may not) be small compared to the convection. First, assume ϵ =1 and see if radiation heat transfer is significant.

    If it is, then you can do a practical treatment of the emissivity as follows. Most cartridge heaters have an interior heated element insulated by magnesium oxide. Then the outside surface sheath is usually either 304 or 316 stainless steel or Incoloy for temperatures above about 450 F. The outside sheath material will apply to your emissivity. You can look up the emissivity for these materials. State your reasoning in your solution.
     
  4. Sep 15, 2011 #3
    But how can I see if the radiation heat transfer is significant if I don't know a surface temperature?
     
  5. Sep 15, 2011 #4
    You had the idea in your original post: "qtot = 2 kW and solving the equation for Ts"

    You will now need an numerical or trial and error solution.
     
  6. Sep 16, 2011 #5
    Alright, so I solved this equation (using [itex]\epsilon[/itex] = 1):

    qtot = [itex]\epsilon[/itex][itex]\sigma[/itex](Ts4 - Tsurr4)*2[itex]\pi[/itex]RL

    using 2 kW for qtot and I got Ts = 1294.3 K. So... what does that mean?

    I guess I'm confused on where you can draw the line of whether or not there is significant heat transfer due to radiation or not. When I solve the problem for just convection, I get Ts = 325 K, which seems a whole lot more reasonable than 1294.3 K.

    What do you think?
     
  7. Sep 16, 2011 #6
    Jake222,

    You want to do both terms at the same time as you had in your first post:

    qtot = qconv + qrad = 2 kW

    qconv = 2 Pi R L * h(Ts-Tsurr)
    qrad = 2 Pi R L * ϵ σ (Ts^4 - Tsurr^4)

    You can pick a value of Ts, compute qconv, compute qrad, then add together and see if they equal 2 kW. When they do, figure qrad / qconv.

    The 1293 K surface temperature you figured would only apply in a vacuum (no convection).
     
  8. Sep 16, 2011 #7
    Oh, okay. I see now.

    Thanks so much!
     
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