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Fluids -- Hydraulic Jump

  1. Feb 2, 2017 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    The question is stated here, though I'm happy to repost but they include a picture. I should say this is not homework, I'm doing problems for practice.
    http://web.mit.edu/2.25/www/5_pdf/5_04.pdf

    2. Relevant equations
    Conservation of momentum/mass

    3. The attempt at a solution
    Conservation of mass between stations 1 and 2 is ##h_1V_1=h_2V_2##. I'm unsure how to approach momentum since I would typically make my CV the fluid between stations 1 and 2, but there is viscous dissipation in the jump, as illustrated. Any ideas?
     
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  3. Feb 3, 2017 #2

    haruspex

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    Viscous dissipation involves internal forces, so no net affect on momentum.
    You are told to ignore horizontal stress from streambed, so no affect from that either.
     
  4. Feb 4, 2017 #3

    joshmccraney

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    Thanks! Then the momentum equation on the CV of fluid from station 1 to station 2 (assuming the hydraulic jump is stationary) is $$\partial_t \iiint \rho \vec{V} \, dV + \iint_{\partial V} \rho \vec{V} (\vec{V_{rel}}\cdot \hat{n})\, dS = \sum F\implies\\
    \iint_{\partial V} \rho \vec{V} (\vec{V_{rel}}\cdot \hat{n})\, dS = -\iint_{\partial V} P \hat{n}\, dS\implies\\
    \int_0^b \int_0^{h_1} \rho V_1 \hat{i} (V_1 \hat{i} \cdot(-\hat{i}))\, dz\,dy+\int_0^b \int_0^{h_2} \rho V_2 \hat{i} (V_2 \hat{i} \cdot(\hat{i}))\, dz\,dy = -\int_0^b \int_0^{h_1}P_1 (-\hat{i}) \, dz\,dy -\int_0^b \int_0^{h_2}P_2 \hat{i} \, dz\,dy\implies\\
    \rho (V_2^2 h_2-V_1^2 h_1) = P_1h_1-P_2h_2$$ Hydrostatic pressure is ##P=\rho g z##, so assuming this holds then average pressure along ##A_1## is $$P_1 = \frac{1}{b h_1}\int_0^b \int_0^{h_1} \rho g z \, dz \, dy = \frac{\rho g h_1}{2}$$ where I assume atmospheric pressure is 0. Then the final result for the momentum balance is
    $$V_2^2 h_2-V_1^2 h_1 = \frac{g h_1^2}{2}-\frac{g h_2^2}{2}$$ This along with continuity gives us the solution in terms of asked quantities. Is this correct?
    So viscous dissipation does not change the time derivative of momentum? Could you briefly explain why?

    So how would this balance change if we were not to ignore the horizontal streambed stress? I think the additional force would be $$-\iint_{\partial V} \mu \vec{V} \otimes \vec{V} \cdot \hat{n} \, dS$$ Would we have to take this to Navier Stokes?
     
  5. Feb 4, 2017 #4

    Nidum

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    Solutions for this problem and the previous nozzle problem can be easily found in undergraduate fluid mechanics text books and also online .
     
  6. Feb 4, 2017 #5

    joshmccraney

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    Could you list the books?
     
  7. Feb 4, 2017 #6
    This is what I got.
    The system is at steady state.
    If you wanted to include the horizontal streambed stress, you would have to get the viscous shear stress at the wall. But to do that, you would have to solve the turbulent NS equations (say using a CFD code) inside the control volume.
     
  8. Feb 4, 2017 #6
    This is what I got.
    The system is at steady state.
    If you wanted to include the horizontal streambed stress, you would have to get the viscous shear stress at the wall. But to do that, you would have to solve the turbulent NS equations (say using a CFD code) inside the control volume.
     
  9. Feb 4, 2017 #7

    Nidum

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    The one I actually have is Massey - Mechanics of Fluids 2nd Edition . There is good coverage in it of hydraulic jumps and related things like broad weirs and under water obstructions
     
  10. Feb 4, 2017 #8

    Nidum

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  11. Feb 5, 2017 #9

    joshmccraney

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    Thank you all very much!
     
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