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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

    thanks
     
  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:
    [tex]u(x,t)=v(x)+w(x,t)[/tex]
    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,
    coomast
     
  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:

    100=A*10+B
    0=A*0+B

    From which you get: v=10*x

    best regards,

    coomast
     
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