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Heat loss due to viscosity

  1. Dec 7, 2012 #1
    Hello,

    1. The problem statement, all variables and given/known data

    I am trying to find an expression for the energy transferred into heat due to friction between layers of viscid fluid. I am provided with the following diagram (see attachment), and am instructed to first find the force at both ends of the cylinder exerted on the surface area of the cylinder of thickness dr, then multiply the force by the velocity, then integrate.

    2. Relevant equations


    3. The attempt at a solution

    F2 = P2*A = 2P2*pi*L*dr; F1 = P1*A = 2P1*pi*L*dr
    I believe the integration should hence be:
    Int[ΔP(R2-r2)/(4ηL)*2pi*L*dr] between 0 and R.

    Would that be correct? I am not sure.
     

    Attached Files:

    Last edited: Dec 7, 2012
  2. jcsd
  3. Dec 8, 2012 #2
    Odd, I was certain someone would have replied by now. Is there anything amiss with my formulation?
     
  4. Dec 8, 2012 #3

    haruspex

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    It doesn't look right to me. π(R2-r2) would be the area of the annulus between radii r and R. I would expect to see the area of the annulus for the cylindrical element only, πrdr, prior to integration. Does your formula give the right dimensionality? How do you get the velocity?
    Maybe if you explain your steps I can help more.
     
  5. Dec 8, 2012 #4
    Hi haruspex,
    V(r) = (ΔP/4ηL)(R2-r2)
    Does it make more sense now?
     
  6. Dec 8, 2012 #5

    haruspex

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    It's making more sense. It would help if you were to quote all the standard equations you're relying on (like you're supposed to).
    That's computing velocity in the sense of L/T, right? You then multiply by 2pi*L*dr (representing what?) to get L3/T. But you want power, ML2/T3, yes? Seems to me you need to invoke the pressure (again) and area of the end of the cylindrical element (2pi r dr) to get a force.
     
  7. Dec 8, 2012 #6
    Sorry about that.
    As I wrote, I was instructed, by the question itself, to first find the force at both ends of the cylinder exerted on the surface area of a cylinder of thickness dr, then multiply the force by the velocity, then integrate for the entire surface area.
    Wouldn't 2pi*L*dr be that surface area?
     
  8. Dec 8, 2012 #7

    haruspex

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    There's no L in the diagram or description, so I assumed that was the length of the cylinder.
    The pressures act on the ends of the cylinder, so to get the force you have to multiply by the areas they act on, namely, at the ends of the cylinder. Remember we're computing power as velocity * force. This is, strictly speaking, a vector dot product, so the velocity and force must be measured in the same direction. An area element can be treated as a vector, normal to the element. So you need the area to be normal to the velocity.
     
  9. Dec 8, 2012 #8
    L is indeed given as the length of the cylinder.
    Should the area element hence be: 2*pi*r*dr?
    Ought the integration then to be: int[v(r)*ΔP*A] between 0 and R, where R is the radius of the cylinder?
     
  10. Dec 8, 2012 #9

    haruspex

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    where A = 2*pi*r*dr? Yes, that looks right to me. (At least the dimensions are right for power!)
     
  11. Dec 9, 2012 #10
    Hi haruspex,
    This integration yielded QΔP, where Q = dV/dt = pi*ΔP*R^4/8ηL.
    I can see that the units match, yet is the expression itself correct? May you please confirm? Moreover, should the integration indeed be done thus (seems a bit peculiar to me, as if something is incorrect)?
     
  12. Dec 9, 2012 #11

    haruspex

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    Looks ok to me. Is it the ΔP2 that bothers you? Think of it like voltage in V2/R.
     
  13. Dec 9, 2012 #12
    Another way of solving this problem, which should give you the same answer, is to integrate the local viscous heating rate over the volume of the pipe. The local viscous heating rate is given by ηγ2/2, where η is the viscosity and γ is the local shear rate:

    γ = dv/dr

    with v being given by the parabolic velocity profile. [itex]v = 2\bar{v}(1-(r/R)^2)[/itex]

    where [itex]\bar{v}[/itex] is the cross sectional average velocity. Consider trying this to see what you get.
     
  14. Dec 9, 2012 #13
    Thank you!! :-)
     
  15. Dec 10, 2012 #14
    Oops. The local rate of viscous heat generation per unit volume for this problem should be ηγ2, not ηγ2/2. Sorry. My mistake.

    Chet
     
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