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Homework Help: Trouble with Galilean transform problem heat equation

  1. Jun 2, 2017 #1
    1. The problem statement, all variables and given/known data
    The common form of the heat-diffusion equation governing the temperature distribution
    $$\rho C_p \frac{\partial T}{\partial t}=k\nabla^2T$$

    Is this equation valid in any inertial frame of reference? (i.e. does it have the property of Galilean invariance?) If not, can it be “fixed” so that it is invariant?
    After cold milk is added to a cup of hot coffee, it is traditionally stirred. One effect of this stirring is to
    increase the temperature gradients in the mixture, thereby increasing heat diffusion and causing a uniform
    temperature to be reached more quickly.
    Assuming that the temperature diffuses according to the “fixed” equation you derive in Problem 1, where c p and k are constant, which of the kinematic components (convection, rotation, strain rate) set up by the stirring is responsible for increasing the magnitude of temperature gradients in the cup and thereby equilibrate the temperature faster?

    2. Relevant equations

    3. The attempt at a solution
    They did not give me a convection term in the equation, so it is throwing me off. They also did not give me the cylindrical version of the heat transfer equation... so using it for the cup coffee stir problem also is throwing me off.
    transforms the equation to :
    $$\rho C_p(\frac{\partial T}{\partial t'}-V\frac{\partial T}{\partial x'}) = K\frac{\partial^2 T}{\partial x'^2}$$

    So, I THINK For this to be invariant, V would have to be very small or dT/dx' would have to be 0.
    What throws me off for part 2, it says if the temperature diffuses according to the "fixed" equation I derive in problem 1... When they say fix it ,they say make it invariant... So doesn't the V term have to dissapear?

    But just using my common sense, if stirring the liquid is increasing temperature gradients... then that would mean the temperature gradients of adding milk.... would have to be different at different radius of the cup (weirdly, instead of just a splash at the center). Rotating it would make it look like a higher temperature concentration per second, and thus look like an increased gradient. So I would say rotation is responsible.

    But my problem is... I don't see how to connect problem 1 and problem 2. A cartesian coordinate system problem without a convection term, fixed to be invariant, used on a cup being stirred. Maybe I am confused about the usage of the term invariant... because even if it was cylindrical, there would be (I think) :
    $$\theta'=\theta + (w_z x r)t$$
    As a Galilean transform, but to make it invariant, wouldn't the w_z term have to disappear? So how can rotation be the factor increasing the gradient.
    Where am I going wrong here? Its probably in my conception of the transform, but yeah please help
  2. jcsd
  3. Jun 3, 2017 #2


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    No. You need to add the convection term to the equation to make it invariant!

    I am sorry, but this is impossible to parse... the point they are making is that the convection currents you induce when stirring mixes hot and cold fluid and you end up with a situation where the typical distance from hot to cold is shorter - thus a stronger gradient and more rapid equilibration.
  4. Jun 4, 2017 #3
    Hi, thank you for your help,
    Maybe I was reading the problem wrong? I thought it wanted me to fix the transformed equation to make it invariant but it wants me to fix the original equation to make the transformed version invariant...?

    Trying out what you said and adding convection to "fix" the original equation :
    $$\rho C_p \frac{\partial T}{\partial t}=K\frac{\partial^2T}{\partial x^2}+V\frac{\partial T}{\partial x}\\\\
    \rho C_p [\frac{\partial T}{\partial t'} -V \frac{\partial T}{\partial x'} ]=K\frac{\partial^2T}{\partial x'^2}+V\frac{\partial T}{\partial x'}\\\\
    \rho C_p \frac{\partial T}{\partial t'} =K\frac{\partial^2T}{\partial x'^2}+[1+\rho C_p]V\frac{\partial T}{\partial x'}

    But... this is still not invariant right? I don't think the following is valid but it's all I could come up with:
    $$\rho C_p \frac{\partial T}{\partial t}=K\frac{\partial^2T}{\partial x^2}+\alpha V\frac{\partial T}{\partial x}\\\\
    \rho C_p \frac{\partial T}{\partial t'} =K\frac{\partial^2T}{\partial x'^2}+[\alpha +\rho C_p]V\frac{\partial T}{\partial x'}\\\\
    \rho C_p \frac{\partial T}{\partial t'} =K\frac{\partial^2T}{\partial x'^2}+\alpha 'V\frac{\partial T}{\partial x'}

    Or am I just completely off for what I think Galilean invariance of an equation is? I am confused.
  5. Jun 4, 2017 #4


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    The velocity is not invariant under the Galilean transform ...
  6. Jun 4, 2017 #5
    Still lost.
  7. Jun 4, 2017 #6


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    You are mixing velocities. One velocity is the velocity ##\vec v## of the medium, the other is the velocity ##\vec u## of the Galilean transform. The velocity of the medium changes ##\vec v \to \vec v' = \vec v - \vec u## under the Galilean transform. You also have dimensional errors in your equations that you should fix.
  8. Jun 4, 2017 #7
    Adding convection to "fix" the problem gives:
    $$\rho C_p \left(\frac{\partial T}{\partial t}+V\frac{\partial T}{\partial x}\right)=K\frac{\partial^2T}{\partial x^2}$$This will transform properly.
  9. Jun 5, 2017 #8
    With the 2nd frame moving at velocity U, and a particle in that frame moving at a velocity V':
    $$\rho C_p \left(\frac{\partial T}{\partial t}+V\frac{\partial T}{\partial x}\right)=K\frac{\partial^2T}{\partial x^2}\\\\
    \rho C_p \left(\frac{\partial T}{\partial t'}-U\frac{\partial T}{\partial x'}+(V'+U)\frac{\partial T}{\partial x'}\right)=K\frac{\partial^2T}{\partial x^2}\\\\
    \rho C_p \left(\frac{\partial T}{\partial t'}+V'\frac{\partial T}{\partial x'}\right)=K\frac{\partial^2T}{\partial x^2}$$

    Thanks both of you for your help. I was messing up with the velocity.
  10. Jun 5, 2017 #9
    Nice job.
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