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Heat equilibrium between two different rods

  1. Nov 15, 2007 #1
    [SOLVED] Heat equilibrium between two different rods

    1. The problem statement, all variables and given/known data
    A rod 1.3 m long consists of a 0.8 m length of Aluminium joined end-to-end to a 0.5 m length of brass. The free end of the aluminium section is maintained at 150 oC and the free end of the brass piece is maintained at 20 oC. No heat is lost through the sides of the rod. In the steady state, what is the temperature of the point where the two metals are joined ? (Thermal conductivity of brass k_b = 109 W/mK and Thermal conductivity of Aluminum k_a = 237 W/mK).

    2. Relevant equations

    [tex]H=-kA[/tex] [tex]\frac{\Delta T}{\Delta x}[/tex]

    3. The attempt at a solution

    Well, you know in steady state that there is no net heat flow so H_aluminum = H_brass. So you have...(I give up for tying to figure this formatting out for now so please forgive me)

    [tex](-k_{a} A (T_{h} - T_{e})) / l_{a} = (-k_{b} A (T_{e} - T_{c})) / l_{b}[/tex] , with [tex]T_{e}[/tex] being the point in the middle and [tex]T_{h}[/tex] and [tex]T_{c}[/tex] being the hold and cold ends respectively.

    Since the cross sectional area of the two are the same, those will cancel and youre left with...

    [tex](-k_{a} T_{h} + k_{a} T_{e}) / l_{a} = (-k_{b} T_{e} + k_{b} T_{c}) / l_{b}[/tex]

    solving for [tex]T_{e}[/tex] you get...

    [tex]k_{a} T_{h} l_{b} + k_{b} T_{c} l_{a} = k_{b} T_{e} l_{a} + k_{a} T_{e} l_{b}[/tex]
    Which then goes to...
    [tex]T_{e} = (k_{a} T_{h} l_{b} + k_{b} T_{c} l_{a}) / (k_{b} l_{a} + k_{a} l_{b})[/tex]

    However, when I put in the values I have, I get an answer HIGHER than the original [tex]T_{h}[/tex]!

    I dont get what Im doing wrong here.

    I have a problem after this where I have to solve for [tex]T_{e}[/tex] when 3 rods come together but I clearly cant do that one properly if I cant solve this one first!
  2. jcsd
  3. Nov 16, 2007 #2
    Little bump before bed. Maybe someone can figure out what Im doing incorrectly :(
  4. Nov 16, 2007 #3

    Shooting Star

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    You are doing incorrect arithmetics, that's all. Recheck you calc. The ans should be 95 C.
  5. Nov 16, 2007 #4
    Well is the algebra at least correct? Because Ive gone over that time and time again and cant find anything wrong. Maybe Im losing the forest in the threes though...
  6. Nov 16, 2007 #5

    Shooting Star

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    The algebra is correct. Do the arithmetic very carefully once.
  7. Nov 16, 2007 #6

    Shooting Star

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    I forgot to mention this. You had written "Well, you know in steady state that there is no net heat flow..." That is not correct. In a steady state, heat does flow, but the temp at every point remains constant.
  8. Nov 16, 2007 #7
    There is one thing I want to add here:

    Your first equation is wrong in terms of + and -.

    When your x goes up, your T will drop. This means you should say:

    -ka.A.(-(Th-Te))/la = - kb.A.(-(Te-Tc))/lb
  9. Nov 16, 2007 #8
    Firstly, thank you for the assistance. I found out where the wrong step was. I figured it was something I was doing wrong algebraically.

    Secondly, Im sorry if I worded my statement improperly. I was trying to imply the fact that at the middle point, there is no NET heat flow that will cause a temperature change. It has come to equilibrium. I apologize for any confusion this may have caused.
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