Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Steady state temperature of insulated rod

  1. Feb 3, 2009 #1
    Problem: Find the steady state temperature of a laterally insulated rod subject to the following conditions. Length of the rod 10, , left end kept at 0 and right end at 100

    Solution: i got this equation from a book i dont know if this applied to this problem but i dont know what to do.
    u(x,t) = v(x) + w(x,t) , v is the steady solution and w the transient putting this is the PDE of Heat , we get two equations , one of which is a function of v and this can be easily solved ( but i dont see what the equations are) .

    From this point i dont know what to do

  2. jcsd
  3. Feb 4, 2009 #2
    Hello jc2009,

    The basic equation of heat conduction in 1 dimension is given as:
    [tex]\frac{\partial u}{\partial t}=k \cdot \frac{\partial ^2 u}{\partial x^2}[/tex]
    Now if it can be assumed that the equation can be written as the sum of a transient and steady state part as follows:
    with v(x) the steady-state and w(x,t) the transient solution and u(x,t) the total solution, you can obtain two new equations by substituting this into the original PDE. This is thus:
    [tex]\frac{\partial v}{\partial t}+\frac{\partial w}{\partial t}=k \cdot \left[\frac{\partial ^2 v}{\partial x^2}+ \frac{\partial ^2 w}{\partial x^2}\right][/tex]
    Now the first term is zero because the steady-state is independent of time. This gives two new equations as follows:
    [tex]\frac{\partial w}{\partial t}=k \cdot \frac{\partial ^2 w}{\partial x^2}[/tex]
    [tex]0=\frac{\partial ^2 v}{\partial x^2}[/tex]
    The second one gives you the steady-state solution, the first one the transient one. In your case the solution for the transient can be written as:
    [tex]v(x)=A\cdot x + B[/tex]
    Using the boundary conditions, you get:
    [tex]v(x)=10\cdot x[/tex]

    Hope this helps,
  4. Feb 4, 2009 #3
    excellent answer, very detailed, thank you coomast
  5. Feb 4, 2009 #4
    In this part :
    I think you meant to say "THe solution for the steady state,(not the transient one) can be written as:
    [tex]v(x)=A\cdot x + B[/tex]
    Using the boundary conditions, you get:
    [tex]v(x)=10\cdot x[/tex]

    because you if you integrate 2 times the v(x) part using the boundary condition you get
    u'' = 0 , Integrate [ u '' ] = u' = A
    Integrate [u ' ] = Ax+B ----> v(x) = 10x , or this is the transient one ?
    Also how exactly you got 10x? , is this because of the length of the rod ? , how do you know A = 10 , and B=0 in Ax+B ?
    Last edited: Feb 4, 2009
  6. Feb 5, 2009 #5
    Hello jc2009,

    Indeed I meant the steady-state solution :-)

    The boundary conditions are:
    v=0 at x=0
    v=100 at x=10

    Putting this into the solution v=A*x+B gives you the system to solve:


    From which you get: v=10*x

    best regards,

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook