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Deriving an expression for the heat flux as a function of radius (HARD!)? watch

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    Hi guys, I have the answer but I don't know know the thought process behind it, can you help me?
    The cylindrical coordinates are:
    1/r (r dT/dr) + g/λ = 0

    the answer is in the attachment. But how do I get from the above equation to it?
    (I accidentally cut abit off. The rest reads: heat flux = -λ dt/dr = g0r/2 ).

    Many Thanks
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    (Original post by Mr Mondai)
    Hi guys, I have the answer but I don't know know the thought process behind it, can you help me?
    The cylindrical coordinates are:
    1/r (r dT/dr) + g/λ = 0

    the answer is in the attachment. But how do I get from the above equation to it?
    (I accidentally cut abit off. The rest reads: heat flux = -λ dt/dr = g0r/2 ).

    Many Thanks
    I'm confused what you're stuck with. You have the equation you mentioned in the post, which I bolded. You re-arrange, integrate w.r.t r, then use the conditions that when r = 0, \, \dfrac{dT}{dr} = 0, thus you can find the constant of integration.

    Are you familiar with how to integrate?
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    (Original post by Phichi)
    I'm confused what you're stuck with. You have the equation you mentioned in the post, which I bolded. You re-arrange, integrate w.r.t r, then use the conditions that when r = 0, \, \dfrac{dT}{dr} = 0, thus you can find the constant of integration.

    Are you familiar with how to integrate?
    I'm sorry if I sound abit thick, but: aren't all the r's on the same side? and when you integrate 1/r you get ln (r) ?, then integrating r dT/dr = r ?
    Again, sorry but I'm having a very hard time with this
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    ahhhhh my mistake, I have attached a pic of the question (b), now
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    (Original post by Mr Mondai)
    ahhhhh my mistake, I have attached a pic of the question (b), now
    The equation you initially gave is not what they used. They times through by r, then just integrate, where are you stuck?

    Spoiler:
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    \dfrac{1}{r}\dfrac{d}{dr}\left( r\dfrac{dT}{dr}\right) = \dfrac{-g_0}{\lambda}

    \dfrac{d}{dr}\left( r\dfrac{dT}{dr}\right) = \dfrac{-g_0r}{\lambda}

    Integrate w.r.t r

    Given \dfrac{dT}{dR} = 0 when r = 0 you can find that C = 0


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    (Original post by phichi)
    the equation you initially gave is not what they used. They times through by r, then just integrate, where are you stuck?

    Spoiler:
    Show


    \dfrac{1}{r}\dfrac{d}{dr}\left( r\dfrac{dt}{dr}\right) = \dfrac{-g_0}{\lambda}

    \dfrac{d}{dr}\big( r\dfrac{dt}{dr}\right) = \dfrac{-g_0r}{\lambda}

    integrating w.r.t r

    </span></font><font color="#505050"><span style="font-family: arial">r\dfrac{dt}{dr} = </span></font><font color="#505050"><span style="font-family: arial">\dfrac{-g_0r^2}{2\lambda} + c

    given \dfrac{dt}{dr} = 0 when r = 0 you can find that c = 0


    ahhhhhhhhh thank you!!!! I understand it now! Thank you for your help :d
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    (Original post by Mr Mondai)
    ahhhhhhhhh thank you!!!! I understand it now! Thank you for your help :d
    I changed it slightly as LaTeX broke on me, but you're purely misreading the equation I assume. No worries
 
 
 
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