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Conceptual fluid mechanics question

  1. Nov 13, 2005 #1
    Suppose we have a tube whose cross-sectional area narrows from [itex]A_1[/itex] to [itex]A_2[/itex]. We attach this to a tank which supplies fluid a constant flow rate [itex]Q[/itex]. Assuming the fluid is nonviscous, we can apply Bernoulli's equation to the narrow and wide ends to calculate the pressure drop [itex]\Delta p = p_1 - p_2[/itex].

    [itex]p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2[/itex]
    [itex]\Delta p = \frac{1}{2}\rho(v_2^2 - v_1^2)[/itex].

    where v_1 and v_2 can be determined using the continuity equation.

    Now assume that the fluid is viscous. Could you please point out any holes in the following reasoning?

    1. By the continuity equation, the fluid speeds in the wide and narrow sections of the tube are the same as in the nonviscous case: ie [itex]v_1[/itex] and [itex]v_2[/itex] respectively.

    2. Since the fluid is viscous, energy is lost from the fluid as it moves. To keep track of the lost energy, I will write

    [itex]p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2 + [\textrm{viscous head}][/itex]

    [itex]\Delta p = \frac{1}{2}\rho(v_2^2 - v_1^2) + [\textrm{viscous head}][/itex]

    which means that the pressure drop is even larger when viscosity is included. This contradicts the answer to an old exam question I attempted recently. See

    https://www.physicsforums.com/showthread.php?t=99620

    Thanks in advance.

    James
     
    Last edited: Nov 13, 2005
  2. jcsd
  3. Nov 14, 2005 #2
    Bump.

    Sorry I'm desperate (exam tomorrow).

    Thanks.

    James
     
  4. Nov 14, 2005 #3

    Physics Monkey

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    Well, I'll take a shot at this. I don't understand the reasoning on the solution guide. They seem to suggest that the speed [tex] v_2 [/tex] will be less than it would be in a frictionless situation, but the whole point of the pump is that it provides a constant flow rate. The continuity equation clearly tells you that [tex] v_2 [/tex] is same as in the case with no viscosity. The pump simply has to work harder to keep that same flow rate going (just like you have to suck more on your straw to drink a milkshake at the same rate that you can drink water through a straw). If this argument is correct then I don't see how the pressure drop couldn't be greater in the viscous situation.

    I hate to contradict another physicist, but I don't see anyway around it. Maybe it's just late and I'm missing something simple.
     
  5. Nov 14, 2005 #4
    Your reasoning is equivalent to mine, I think they must have gotten it wrong.
     
  6. Nov 14, 2005 #5
    By the way, if you have a chance, could you please take a look at my other thread on Lenz's Law? Thanks a million!

    James.
     
  7. Nov 14, 2005 #6

    Physics Monkey

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    Yes, I feel confident now that you have it right. I thought of a simpler example that shows the pressure has to decrease more. Consider a uniform horizontal pipe (one that does not contract) and place regular vertical tubes to measure the pressure. If the fluid wasn't viscous then the head in each tube would remain at the same height, but a viscous fluid will lose head even though the speed of the fluid is the same along the pipe (again a consequence of continuity) because the pressure is dropping. So the pressure drops more than it did in the non-viscous case.

    Edit: Here is a resource I just found that you might find useful http://www.du.edu/~jcalvert/tech/fluids/bernoul.htm
     
    Last edited: Nov 14, 2005
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