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Heating a metallic rod from one end by continuous heat flux

  1. Feb 3, 2015 #1
    What is the solution of heat transfer equation when there is continuous heating ( flux of energy) to one end of a metal rod? what will be the temperature profile before reaching the steady state? when I try the boundary condition where one end (x=0) is kept at T1, I got the general solution of the space dependent component X(x)= C1 sin(ux)+C2cos(ux). substituting x=o leads to C2=T1, then I stopped !! How to retrieve the other constant and the general solution too? will the rod reach steady state given the other end is exposed to room temperature.
  2. jcsd
  3. Feb 3, 2015 #2
    Please specify where the heat flux is applied. What is the boundary condition at the other end of the rod? What is the initial condition? What is the steady state solution to the heat equation for the prescribed boundary conditions?

  4. Feb 4, 2015 #3
    The rod of length L, is put at the x-axis with its left end at the origin and its right end is at x=L. The boundary conditions (BCs): u(0,t)=T1, u(L,t)=T2. where T1>T2. Inital condition u(x,0)=T2. The steady state is expected at t approach infinity: U(x,t)= T1 - (T1-T2/L)x.

    One point which I could not figure out and it might affect BCs, is that if the left end is in contact with hot reservoir at temperature T1, then that end should not approach T1 because if it does there would not be continuous flux to it which is not the case when steady state is reached! so what is the maximum temperature the end at x=0 could reach if it is in continous contact with a reservoir at T1.

    Also, very important, what equation describes u(x,t) before reaching the steady state?
  5. Feb 4, 2015 #4
    The temperature at the boundary jumps from T2 to T1 at time zero, and it stays at that temperature for all time. The heat flux at the boundary jumps from zero to an infinite value initially, and then decreases monotonically to the steady state value at long times.

    The way to solve a problem like this is to represent the temperature in the rod at some arbitrary time as:

    u(x,t) = T1 - (T1-T2/L)x + v(x,t)

    You then solve for v(x,t), which is much easier to obtain. Boundary conditions on v are v = 0 at x = 0 and v = 0 at x = L. The initial condition on v is

    v(x,0) = -[T1 - (T1-T2/L)x]

    Show that, as with u(x,t), v(x,t) satisfies the transient heat conduction equation.

    For v, you now have a transient heat conduction problem in which there is an initial temperature profile in the rod, but the ends are set to v = 0 at t = 0.

  6. Feb 4, 2015 #5
    But v(x,0) at ends (x=0) and (x=L) does not satisfy u(x,0) because u(x,0) at all location is T2 not 0.
    the heat flux q= -k du/dx, so if the end at x=0 rise to T1 which is equal to the reservoir temperature, du/dx=0 and so does q? Also how about the flux at x=L at different times?
    Last edited: Feb 4, 2015
  7. Feb 4, 2015 #6
    Ooops. Correction: v(x,0) = T2 - [T1 - (T1-T2/L)x]
    No. The reservoir is assumed to have infinite conductivity, so it can supply whatever flux is necessary. The temperature gradient within the rod in the vicinity of x = 0 is very high at short times, and this corresponds to a very high heat flux.

    In the transient behavior of the rod, at short times, the temperature will decrease very rapidly with distance from x = 0. Beyond a certain distance, both the temperature change and the heat flux will be very close to zero. So the boundary layer region near x = 0 will grow with time until it penetrates the entire rod. At short times, the heat flux at x = L will be very close to zero. As time progresses, it will rise to the final steady state value.

  8. Feb 5, 2015 #7
    Thank you, that was very helpful.

    Technically speaking, what if v(x,0) is an even function, for example what if we choose u(x,0) so as v(x,0)=cos x. Then how the coefficient, bn of sin(π nx/L) of v(x,0), which is suppose to be odd function, be calculated?
  9. Feb 5, 2015 #8
    You have to take v as an odd function in order to satisfy the boundary condition on v at x = 0. A half-wave sine series is involved in satisfying the initial condition on v.

  10. Feb 5, 2015 #9
    Hereby I show as you said that the flux is infinite at x=o at t=0 and then reduces to a finite value of steady state as time goes by to infinity. However, at x=L and t=0, the heat flux is T1-T2/L ???

    the solution to that problem is u(x,t)= T1-(T1-T2/L)x + v(x,t)
    = T1-(T1-T2/L)x + -2(T1-T2)/π Σ(1/n) sin (nπx/L) Exp (-n2π2t/L2

    and from Fourier law, the heat flux, φ= -K∂u/∂x

    so φ(x,t)= T1-(T1-T2/L) + 2(T1-T2)/L (Exp (-n2π2t/L2) Σ cos (nπx/L)

    so φ((0,0) = ∞ exactly as you kindly said.
    φ((L,0) = (T1-T2)/L which is the finite final value which is not zero
    φ((0,∞) = (T1-T2)/L
    and finally, φ((L,∞) = (T1-T2)/L
  11. Feb 5, 2015 #10
    No, at x = L and t = 0, the heat flux is zero.
    No. We know that at t = 0 , φ = 0 at x = L because T = T2 for all x < L. So, you must have made a mistake somewhere. For one thing, the first term in your equation for φ, T1, should not be there. Secondly, what happened to the summation term in your equation for the flux at t = 0?

  12. Feb 5, 2015 #11
    yes sorry, the true flux is φ(x,t)= (T1-T2/L) + 2(T1-T2)/L (Exp (-n2π2t/L2) Σ cos (nπx/L).

    At t=0, the exponential term =1 but all cos (nπx/L) at x=L goes to zero, because cos(nπ)=(-1)n. So even-n will cancel out odd-n which leads Σ cos (nπx/L) to be zero at x=L. However, still (T1-T2/L) is not zero which mean φ(L,0) is not zero too.
  13. Feb 5, 2015 #12
    Your Fourier coefficients are incorrect. There should be π's and n's in them.

  14. Feb 5, 2015 #13
    n&π appear in the equation of v(x,t) as well as u(x,t), but as we differentiate u(x,t) to x to get the flux from the equation of φ= - K ∂u(x,t)/∂x, n&π will be cancelled out.

    u(x,t)= T1-[x.(T1-T2)/L] + (-2)(T1-T2)/π Σ (1/n) sin nπx/L. Exp (n2π2/L2. t

    so true, the Fourier coefficient bn=-2(T1-T2)/nπ

    But, taking the first differential to x, ∂u(x,t)/∂x= -(T1-T2)/L + (-2)(T1-T2)/π Σ nπ/L (1/n) cos nπx/L. Exp (n2π2/L2. t

    so π will be cancelled as well as n.
    Last edited: Feb 5, 2015
  15. Feb 5, 2015 #14
    Your Fourier coefficients can't be correct. Show us your derivation of the Fourier coefficients please.

  16. Feb 5, 2015 #15
    I checked your Fourier coefficients and they are correct. I can see what's happening at x = L. Even though the solution for the temperature converges correctly at x = L, the solution for the flux does not converge at this location at time zero. The slope fluctuates between -(T1-T2)/L and +(T1-T2)/L as n increments by 1. The average as n varies is zero.

  17. Feb 6, 2015 #16
    For me, I can see what fluctuating is the second term of -∂u/∂x= (T1-T2)/L + 2(T1-T2)/L Σ cos nπx/L. Exp-(n2π2/L2. t)
    so at t=0 and x=L, the average of the second term is 0 leaving ∂u(L,0)/∂x = (T1-T2)/L which is not zero!
    Last edited: Feb 6, 2015
  18. Feb 6, 2015 #17
    What do you get if you only include the term n=1 in the summation? What do you get if you include only the terms up to n=2? n=3?
  19. Feb 6, 2015 #18
    for n=1, -∂u/∂x= (T1-T2)/L + 2(T1-T2)/L (cos π) = (T1-T2)/L + 2(T1-T2)/L(-1) = -(T1-T2)/L
    for n=1&2, -∂u/∂x= (T1-T2)/L + 2(T1-T2)/L (cos π+cos 2π) = (T1-T2)/L +2(T1-T2)/L(-1+1) = (T1-T2)/L
    for n=1->3 -∂u/∂x= (T1-T2)/L + 2(T1-T2)/L (cos π+cos 2π+cos 3π) = (T1-T2)/L + 2(T1-T2)/L(-1+1-1) = -(T1-T2)/L
    Last edited: Feb 6, 2015
  20. Feb 6, 2015 #19
    So now you see what I'm saying, right?

  21. Feb 6, 2015 #20
    Yes thank you for your help. The issue of flux is over now.

    Now, if the initial condition of v(x,t) is an even, then no Fourier coefficients could be calculated right? This means the initial condition could not be chosen arbitrarily? This also mean that the initial condition is dependent on the boundary conditions which I don't understand.
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