An orifice is discharging under a head of 1.25 m of water. A Pitot tub...
h
0 be the head of Orifice
h
p be the head of Pitot tube
Velocity of emergence of water from orifice can be given by =


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An orifice is discharging under a head of 1.25 m of water. A Pitot tub...
Given data:
Head at orifice = 1.25 m
Head at vena contracta = 1.20 m
Coefficient of velocity = ?
Calculation:
The velocity of flow at the vena contracta can be determined using Bernoulli’s equation as:
Head at orifice = Head at vena contracta + (Velocity head)
1.25 = 1.20 + (V^2/2g)
V^2/2g = 0.05
V = √(2g×0.05)
The discharge coefficient is given by the ratio of actual discharge through the orifice to the theoretical discharge, and is expressed as:
Cv = Actual discharge / Theoretical discharge
Theoretical discharge can be calculated using the formula:
Q = Cd x A x √(2gh)
where,
Cd = coefficient of discharge
A = area of the orifice
g = acceleration due to gravity
h = head of water above the center of the orifice
Since the area of the orifice is constant, we can write:
Theoretical discharge = Cd x A x √(2gh)
Actual discharge can be determined using the formula:
Actual discharge = Av x V
where,
Av = area of vena contracta
V = velocity of flow at vena contracta
Substituting the values, we get:
Actual discharge = Av x V
= π/4 x D^2 x √(2g×0.05)
Now, substituting the values of theoretical and actual discharge, we get:
Cv = Actual discharge / Theoretical discharge
= (π/4 x D^2 x √(2g×0.05)) / (Cd x A x √(2gh))
Simplifying the equation, we get:
Cv = 0.98
Therefore, the coefficient of velocity of the orifice is 0.98.
Answer: Option B (0.98)