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Flow rate through a submerged orifice

  1. Jan 5, 2017 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations



    3. The attempt at a solution

    I attempted to apply the Bernoulli equation, making point 1 the top surface of the left hand side and point 2 the instant at the sharp edge.
    However I ended up with; pA+ρgz=p+½ρU2

    Normally when it is not submerged, the formula can narrow down to U=√2gH, where H=z in this case, but I'm not sure how to eliminate the pressure in the liquid.

    Is liquid also trying to move from the right to the left?
  2. jcsd
  3. Jan 5, 2017 #2


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    No. It goes with the flow :smile:

    Sorry for being corny. What are pA, z and p ? How many unknowns are there in your attempt equation ?

    "The problem statement, all variables and given/known data" is rather lacking from my point of view...
  4. Jan 5, 2017 #3
    pA would be atmospheric pressure, z is the distance from the contraction point to the top of the left hand side surface, so 1.2m, and then p would be the pressure at the contraction part where it leaves.

    The previous part to the problem, in part a) asks me to write down the Bernoulli equation, in b) it tells me to apply the equation to the flow through the sharp edged orifice shown in the picture and sketch the streamline you select to analyse the flow.
    I chose to analyse it from the surface and the contraction point.

    I've solved questions where the right hand side is air, so the pA's would normally cancel out, but I'm not sure with this one.

    Would it be possible to make point 1 a point of very small distance from the surface so that the two p's are equal?
  5. Jan 5, 2017 #4


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    Point 1? That sounds like the start of the flow. Did you mean at the end?
    What would the pressure be at the right of the orifice if it were blocked up?
  6. Jan 5, 2017 #5
    I mean point 1 as in right at the top of the 1.8m surface.

    And what would it be with the liquid there too or if the liquid hadn't have been there? The pressure would be pA I think if it was blocked up?
  7. Jan 5, 2017 #6


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    I don't see how that changes anything. Aren't you already taking the start of the flow as being there?
    What the pressure would be with the liquid levels as shown but the hole blocked.
  8. Jan 5, 2017 #7
    I see. The pressure would be ρg(1.2) on the left and ρg(0.6) on the right?
  9. Jan 5, 2017 #8


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    That would seem reasonable.
  10. Jan 6, 2017 #9
    Thank you! I continued and equated the difference in pressure with the hole blocked with p1-p2 and that gave me the correct answer for U. :)
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