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[Basic Hydraulics] The Pitot Tube/Manometers

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine the flow rate, Q.

    Ignore energy losses.

    Given:

    D1 = 100 mm
    D2 = 75 mm
    Fluid in the pipe is water.
    Fluid in the metre is mercury.
    S.G. of Mercury is ~13.6.
    The difference in height of the mecury columns is 80mm.

    Diagram is below.

    t8mfz4.jpg

    2. Relevant equations

    Bernoulli's Equation.

    Equation of Continuity. (Q1 = Q2)

    3. The attempt at a solution

    I approached this as a manometer question, but with that orifice in the flow I was unsure on how to proceed. I chose a datum as the lower of the two levels of mercury. The expressions I got were:

    Px-x = Pressure due to the flow + Pressure due to the height of the water above the mercury + the pressure due to the mercury above the datum. This was for the left hand side.
    Px-x = P1 + ρwgh + ρmgh', where h' = 80 mm and h is an unknown height which cancels out when the left side and right side are equated.

    For the right hand side I said

    Px-x = Pressure due to the flow + Pressure due to the height of the water.
    Px-x = P2 + ρwg(h+h')

    For this second equation I feel there is something missing, surely that opening in the flow causes an additional term to be added?
     
  2. jcsd
  3. Nov 7, 2011 #2
    The tube on the right is facing the flow so it is measuring total pressure. You should include the V^2/(2*g) term. The water must come to a stop at the mouth of the tube. Pressure generated is V^2/2g.
     
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