1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Heat Equation Question

  1. Oct 6, 2014 #1

    joshmccraney

    User Avatar
    Gold Member

    Hi PF!

    Given: ##u_t = u_{xx} +1## (heat equation) with the following B.C.: ##u_x(0,t)=1, u_x(L,t)= B, u(x,0)=f(x)##. My professor then continued by stating that in equilibrium, we have ##0 = u_{xx} +1 \implies u = -x^2/2 + C_1 x + C_2##. So far I'm on board, although by "equilibrium" does he mean steady state (after a long period of time)?

    Next, we show that the equilibrium solution must satisfy ##-L + 1 = B## (via the B.C., after showing that ##C_1 = 1##).

    He goes on to say that this condition is imposed from the source of heat and the two flux B.C. By source, does he mean the ##1## on the R.H.S of the initial equation?

    Now here's where it get's tricky. He then says to perform an energy balance, thus:

    $$\frac{d}{dt} \int_0^L c \rho u dx = -u_x(0) + u_x(L) + \int_0^L Q_0 dx = -1 +B +L=0$$

    Now I understand the L.H.S completely, but where is he getting the heat source on the R.H.S, namely the ##Q_0## term? Also, shouldn't the thermal constant ##k## show up prefixing the flux terms (the ##u_x## terms) or is this only true if we applied the divergence theorem and fouriers law of heat? Also, why is there a minus sign in front of the ##u_x(0)## term and a positive sign in front of the ##u_x(L)##? Isn't it energy in minus energy out?

    He then proceeds by stating that initial energy equals final energy in equilibrium, thus we must have $$\int_0^L f(x) dx = \int_0^L u(x) dx$$ where i know ##f(x)## is initial but is ##u(x)## final since it does not depend on ##t##?

    thanks a ton!!!!
     
  2. jcsd
  3. Oct 6, 2014 #2

    joshmccraney

    User Avatar
    Gold Member

    Sorry guys, I'm good now on everything except the flux question (why aren't those derivatives reversed, namely, ##u_x(0)-u_x(L)##)?
     
  4. Oct 6, 2014 #3
    If heat is coming in at x = 0, the temperature has to be decreasing with distance from the boundary. So -ux(0) describes the rate at which heat is entering through the boundary (i.e., the heat flux at the boundary).

    Chet
     
  5. Oct 7, 2014 #4

    joshmccraney

    User Avatar
    Gold Member

    Hey Chet!

    Ahhh I see, you're looking at those derivatives from their definition. This makes total sense; I can't believe I missed this. Thanks a ton!
     
  6. Oct 7, 2014 #5

    joshmccraney

    User Avatar
    Gold Member

    So to clarify that I'm understanding the flux correctly, if we are in 1-D at the left edge and given ##u'(0) > 0 ## then we see that ##(u(a)-u(0))/(a-0) : a>0 \implies u(a)-u(0) > 0 \implies u(a) > u(0)## which tells us that temperature is greater to the right of zero (for some neighborhood), and thus heat should be moving out (left) at ##x=0##? Thus, we place the minus sign in front of ##u_x(0)##, which was done.

    At the right point where ##x=L## we see that ##u'(L)>0 \implies u(L) > u(a) : L>a## and thus since the temperature is slightly bigger at ##L## than at ##a## heat should be flowing into the rod (to the left), making me think there should be a negative in front of the ##u'(L)## when balancing our energy. Can you help me out here?

    Thanks again!
     
    Last edited: Oct 7, 2014
  7. Oct 7, 2014 #6
    Putting the minus sign in means you are referring the the rate at which heat is coming in at x = 0. In the energy balance, this contributes to the rate at which temperature (internal energy) is rising.
    There shouldn't be a minus sign. u'(L)>0 means the heat is coming into the rod. In the energy balance, this also contributes to the rate at which temperature (internal energy) is rising.

    Chet
     
  8. Oct 7, 2014 #7

    joshmccraney

    User Avatar
    Gold Member

    Gotcha! So it seems like positive flux always means heat (or whatever) is entering, right?

    So how is the logic that I wrote wrong? I know it's not right, but why?
     
  9. Oct 7, 2014 #8
    No. Heat flux is a vector quantity. Its direction in opposite to that of the temperature gradient. Have you had vector calculus yet? Do you know what the Del (aka Nabla) gradient vector operator is?
    I'll get to that after I hear back from you.

    Chet
     
  10. Oct 7, 2014 #9

    joshmccraney

    User Avatar
    Gold Member

    Yes, I've had be of calculus and am very familiar with del, used for curl, divergence, and gradient.

    And I totally spaced that heat flux is the double partial, right? It would be u_xx. Then why arent they balancing the time rate of change with the flux? Isnt it correct to say time rate of change of heat equals net heat flux plus source?
     
  11. Oct 7, 2014 #10
    Heat flux is not second partial; it is determined by the first partials of temperature T. The heat flux vector is given by:
    [tex]\vec{h}=-k\vec{\triangledown} T[/tex]
    Thus, the heat flux vector is directed opposite to the temperature gradient vector.
    For your rod problem, this equation reduces to:
    [tex]\vec{h}=-k\frac{\partial T}{\partial x}\vec{i}_x[/tex]
    where ##\vec{i}_x## is the unit vector in the x direction. To get the rate of heat flow into the rod at either x = 0 or x = L, you dot the heat flux vector with an inwardly directed normal to the rod at that location, and then multiply by the cross sectional area A. At x = 0, the inwardly directed normal to the rod is ##+\vec{i}_x##, while, at x = L, the inwardly directed normal to the rod is ##-\vec{i}_x##. So the rate of heat flow into the rod at x = 0 is:
    [tex]Q(0)=-kA\frac{\partial T}{\partial x}\vec{i}_x \centerdot \vec{i}_x=-kA\frac{\partial T}{\partial x}[/tex]
    Similarly, the rate of heat flow into the rod at x = L is given by:
    [tex]Q(L)=-kA\frac{\partial T}{\partial x}\vec{i}_x \centerdot (-\vec{i}_x)=+kA\frac{\partial T}{\partial x}[/tex]
    Hope this makes sense.

    Chet
     
  12. Oct 9, 2014 #11

    joshmccraney

    User Avatar
    Gold Member

    Honestly, why can't you teach at my university and i just learn from you?!?! This makes SO MUCH SENSE! (Sorry, I know we're suppose to keep this site formal, but dang. Mind blowing!) You're the man!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Heat Equation Question
  1. A question of heat (Replies: 2)

Loading...