[Basic Hydraulics] The Pitot Tube/Manometers

  • Thread starter MarleyDH
  • Start date
  • #1
26
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Homework Statement



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


Homework Equations



Bernoulli's Equation.

Equation of Continuity. (Q1 = Q2)

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?
 

Answers and Replies

  • #2
1,198
5
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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