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Comparing two functions and determining coefficients. I get 1/1.

  1. Nov 30, 2011 #1
    The objective is to find some coefficients by comparing two equations.

    T_2 \cdot \frac{R_1}{R_1+R_2} + T_1 \cdot \frac{R_2}{R_1+R_2} [/itex]


    [itex]T_2 \cdot k_1 + T_1 \cdot k_2[/itex]

    I compare and set

    [itex]k_1 = \frac{R_1}{R_1+R_2} (1)[/itex]

    [itex]k_2 = \frac{R_2}{R_1+R_2} (2)[/itex]

    I expand the equations and throw them around

    [itex](1) -> R_1 = R_2 \cdot \frac{k_1}{1-k_1} (3)[/itex]
    [itex](2) -> R_2 = R_1 \cdot \frac{k_2}{1-k_2} (4)[/itex]

    I put in (3) in (4) and get

    [itex]R_2 = R_2 \cdot \frac{k_1}{1-k_1} \cdot \frac{k_2}{1-k_2} -> 1 = \frac{k_1}{1-k_1} \cdot \frac{k_2}{1-k_2}[/itex]

    So i dont know how to get the values of [itex]R_1[/itex] and [itex]R_2 [/itex] as a function of [itex] k_1 [/itex] and [itex]k_2.[/itex]

    The solution for this is given and is correct when I put it in the equations above and it fulfills everything but i just don't know how he has gotten it.

    [itex]k_x[/itex] is a long function based on parameters that i just rebrand for simplicity.
  2. jcsd
  3. Nov 30, 2011 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    If those are two equations then where are the "=" signs? Do you mean "functions" instead of "equations"? Are the two functions assumed to be equal?
  4. Dec 1, 2011 #3
    No they have it all expressed as

    T_m = T_2 \underbrace{\left( \frac{\displaystyle r_2^2}{\displaystyle r_2^2-r_1^2}-\frac{\displaystyle 1}{ \displaystyle 2 \ln{\left( \displaystyle r_2 / r_1 \right) }} \right)}_{k_1} + T_1

    \underbrace{\left( \frac{ \displaystyle 1}{\displaystyle 2 \ln{\left(\displaystyle r_2 / r_1 \right) }}-\frac{\displaystyle r_1^2}{\displaystyle r_2^2-r_1^2} \right)}_{k_2}

    and then they have the equation as i have given above

    [itex]T_m = T_2 \cdot \frac{\displaystyle R_1}{\displaystyle R_1+R_2} + T_1 \cdot \frac{\displaystyle R_2}{\displaystyle R_1+R_2}[/itex]

    And then they say "Comparing the coefficients of [itex]T_1[/itex] and [itex]T_2[/itex] in the two equations we get"

    R_1 = \frac{1}{\displaystyle 4\pi kL} \left( \frac{\displaystyle 2r_2^2\ln{\left(r_2/r_1\right)} }{\displaystyle r_2^2-r_1^2}-1 \right)

    R_2 = \frac{1}{\displaystyle 4\pi kL} \left( 1- \frac{\displaystyle 2r_1^2\ln{\left(r_2/r_1\right)} }{\displaystyle r_2^2-r_1^2} \right)


    What i wanna know is how they arrive at the values for [itex]R_1 [/itex] and [itex]R_2[/itex]
  5. Dec 1, 2011 #4
    There must be something else you are not telling because the equations for R1 and R2 contain 1/(4.pi.k.L) ....which is nowhere in the coefficients k1 and k2....
  6. Dec 1, 2011 #5
    Those are for thermal conductivity and and the length, when you put them in the equation they are cancelled out.

    Tm is a mean temperature in an lumped parameter cylinder with no internal heat generation. And that's how they arrive at the solution. No further explaining done. And i wanna know how they arrive at the solution rather than just look at the solution. Not holding back anything.
  7. Dec 1, 2011 #6
    Then, maybe you need to reveal the entire problem at hand....we are driving blind here...without knowing the context...maybe if we knew more...for example, you just revealed that we are talking about a cylinder without internal heat generation...what can you tell about T1 and T2, where are they supposed to be?

    Sparing us from the details is not working...tell us everything, now.
  8. Dec 1, 2011 #7
    lol ok, the r2 is outer radius at temperature T2 and r1 is inner with T1. L is the length.

    And the resistance is set so that


    If we use Ohms law here, then Tm = T2+Q*R2

    We know that Q = (T1-T2)/(R1+R2) -> Tm = T2 + R2*(T1-T2)/(R1+R2)

    Tm = T2*R1/(R1+R2) + T1*R2/(R1+R2)

    Thats the equation form the begining and then he has the other equation that he compares with which i have already given. And then he magically arrives at that solution.

    That specific equation he derives form a cylinder with the equation

    d^2T/dr^2 + 1/r* dT/dr + g/k = 0 but here g = 0 as there is no internal heat generation.
  9. Dec 1, 2011 #8

    So, if R1 and R2 add up to the total thermal resistance in the radial direction of the cylinder (tube, rather)...then

    R1/(R1+R2) is a fraction of it (a number less than 1.0), and

    R2/(R1+R2) is another fraction ( <1.0) and complementary...

    meaning, together, they add up to 1, correct?

    R1/(R1+R2) + R2/(R1+R2) = (R1+R2)/(R1+R2) = 1.0

    that means that also k1+k2=1 ...the knowledge of this may help you in your calculations

    Also, those formulas about the thermal resistance look too complicated...why do you have r2 in there? Make sure that when calculating the thermal resistance you do not mix what the cross-sectional area available for heat flow is...in other words, if the heat is flowing from the inside wall of the cylinder to the outside wall...the area for flow at a given radius is 2∏rL ...this appears in the denominator of the formula for resistance:

    dR = dr/(2∏rLκ)

    when you integrate this over a given range, you do not have r2, you just get a difference of ln:


    Now, if the heat is flowing in the axial direction of the cylinder, well, then the area available for heat flow does include r2...so, just make sure you have all your ducks in a row....
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