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Steady state temperature

  1. Jan 31, 2009 #1
    Problem 1 : Find the steady state temperature of a laterally insulated rod subject to the following conditions. Length of the rod 20, left end insulated and right end held at 100

    I am not sure if i fully understand this problem, is this asking me to list the Boundary value conditions ? like , u(x,0) = 10 , u(x,20) = 100, u_x (0) = 0 ?
    or a function ?

    Problem 2 : The vertical displacements of a membrane were found to be given by the function u(x,y,t) = cos(15t) sin(3x) cos(4y) , What is the corresponding wave's speed c ?

    for this problem first i need to take the derivative u'(x,y,t) but with respect to what?
     
  2. jcsd
  3. Feb 1, 2009 #2
    You need to assume that the solution can be written as the sum of the steady state and the transient solution. This can be written as:
    [tex]u(x,t)=v(x)+w(x,t)[/tex]
    In which v the steady state solution and w the transient is. Putting this in the PDE of heat you get two equations, one of which is a function of v and this can be easily solved. The equation is now:
    [tex]\frac{d^2v}{dx^2}=0[/tex]
    Using the boundary conditions gives you for the steady state solution:
    [tex]v(x)=100[/tex]
    A constant for the whole bar, which is correct because no heat can be added or extracted due to the insulation. A constant must arrise.

    Substitute the solution in the wave equation:
    [tex]\frac{\partial^2u}{\partial t^2}=c^2 \cdot
    \left[\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}
    \right][/tex]
    You can extract the velocity c from this.

    coomast
     
  4. Feb 4, 2009 #3
    My question is how come v(x) = 100, if you integrate v'' , you'll have to get a constant times x + constant , why just 100 ?
     
  5. Feb 5, 2009 #4
    Hello jc2009,

    Indeed the solution is v(x)=A*x+B. The two boundary conditions are on the left side (x=0)
    [tex]\frac{dv}{dx}=0[/tex]
    and on the right hand side (x=20)
    [tex]v=100[/tex]

    This gives you v=100 for the complete bar after applying these conditions.

    best regards,

    coomast
     
  6. Feb 5, 2009 #5
    if you solve B=0 then
    v(20)=A20+b =100
    A=5

    then v(x) = 5x

    how did you get 100?
     
  7. Feb 5, 2009 #6
    Hello jc2009,

    You are not applying the boundary conditions right. The left hand side is insulated, thus the derivative is 0. This is:
    [tex]\frac{dv}{dx}=0=\frac{d}{dx}(Ax+B)=A[/tex]
    Thus A=0 and the other boundary condition gives you
    [tex]v=100=B[/tex]

    best regards,

    coomast
     
  8. Feb 5, 2009 #7
    [tex]\frac{dv}{dx}=0=[/tex] , is [tex]u_{x}[/tex] right
    so whenever an end is insulated [tex]u_{x}[/tex] is the correct format of that boundary
    THis has always caused me problems in this PDE class im taking, in this problem [tex]u_{x}[/tex] is used because it is insulated , but in other problems where insulation is not present they anyways use [tex]u_{x}(0,t)=0[/tex] for a vertical side , so besides using [tex]u_{x}[/tex] in a rod what does exactly [tex]u_{x}[/tex] means for other problems , for example : a 10 X 20 plate ? why do they use [tex]u_{x}[/tex] for a side of a plate even when there is no insulation? why not just use u(x,t)
     
  9. Feb 6, 2009 #8
    Yes, [tex]u_x=\frac{\partial u}{\partial x}[/tex]. Now stating that this is zero is indeed what is required for an insulated boundary. This comes from the law of fourier of conductivity. What does this means? Consider a wall with a thickness d, area A, conductivity k and a temperature T1 and T2 at the two sides. Now the heat flux going through the wall is descibed by the law of fourier as:
    [tex]f=-k\cdot A \cdot \frac{T_1-T_2}{d}[/tex]
    or for a very thin wall, which can then be used to integrate for different situations:
    [tex]f=-k\cdot A \cdot \frac{\partial T}{\partial x}[/tex]
    In which x the direction of heat flow, thus perpendicular to the wall sides. Looking at this equation you immediately see the influences of the different parameters. The larger the area A, the larger the flux and the thicker the wall, the smaller the flux. Also the higher the conductivity, the higher the flux as is with a higher temperature difference.

    If the flux f is zero due to insulation you get:
    [tex]0=f=-k\cdot A \cdot \frac{\partial T}{\partial x}=\frac{\partial T}{\partial x}[/tex]
    This is the boundary condition you need to use in your problems.

    More info at: http://en.wikipedia.org/wiki/Heat_conduction

    hope this clears some problems. Best regards,

    coomast
     
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