Exam. on Fluid Mechanics - No.2
Instruction
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Answer ALL Questions.
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Electronic calculators may be used.
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Use following values if necessary. Density of water: w=103kg/m3, Acceleration of gravity: g=9.81m/s2
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Use proper units.
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Define all the variables clearly. Show sketches as much as possible.
Question 1
A venturi meter has a main diameter of 65mm at section A and a throat diameter of 26mm at section B. A differential
mercury U-tube pressure gauge was connected between section A and B. When measuring the flow of a liquid of mass
density 898kg/m3 the reading of a mercury differential pressure gauge was 71mm. The coefficient of the meter was 0.97
(Appeared in 1996,2000,2002)
and the specific gravity of mercury is 13.6.
(1) Submit a neat and well-labeled sketch of the venturi meter. (10marks)
(2) Apply Bernoulli’s theorem to section A and B.
3
(3) Determine the actual discharge in m /hour.
(20marks)
(10marks)
* “coefficient of the meter” and “coefficient of discharge” are equivalent.
Question 2
A large tank has a circular sharp edged orifice 930mm2 in area at a depth of 2.7m below constant water level. The jet
issues horizontally and in a horizontal distance of 2.34m it falls 0.54m. The measured discharge is 4.2x10 -3m3/s.
Submit a neat and well-labeled sketch (10marks) and calculate the coefficients of velocity, contraction and discharge
(Appeared in 2002)
(20marks).
Tip: You can refer to this example.
Problem: A jet of liquid issues horizontally from a small orifice in the vertical side of the tank. Derive an
expression for the actual velocity of the jet v if the jet falls a distance y vertically in a horizontal distance x
measured from the vena contracta. If the head of liquid above the orifice is h, what will be the coefficient of
velocity?
Solution: If t is the time for a liquid to travel from the vena contracta to a point (x,y), horizontal distance is x=vt,
vertical distance is y=1/2gt2. Eliminating t, we have v=(gx2/2/y)1/2. Since theoretical velocity of at a small orifice is
v’ =(2gh)1/2, coefficient of velocity is Cv=v/v’=(x2/4/y/h)1/2.
Question 3
(1) Develop a formula for the theoretical discharge over a
notch shown if approaching velocity can be neglected.
(20marks)
(2) What is the actual discharge of this notch if H=0.15m,
L=0.2m, =30o, and coefficient of discharge is 0.75?
(10marks)
H

L