Ii.a venturimeter is to be introduced in a 25cm dia.horizontal pipeline conveying a maximum of 150 litres per second.the pressure guge in the pipe at the point of interest is 6 m head of water.determine the minimum dia.of the threat for the venturimeter to be selected so that no negative head is informed in it. aiso calculate the discharge when u -tube difference and deflection of 20cm.?

Question

Answer

Introduction:
In this question, we are given a 25cm diameter horizontal pipeline with a maximum flow rate of 150 litres per second. We need to determine the minimum diameter of the throat for the venturimeter so that no negative head is formed in it. We also need to calculate the discharge when the U-tube difference and deflection are 20cm.

Calculating Minimum Diameter of Throat:
To calculate the minimum diameter of the throat, we can use the Bernoulli's equation. The Bernoulli's equation states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant along a streamline in a steady flow.
Mathematically, the equation can be represented as:
P1/ρg + V1^2/2g + z1 = P2/ρg + V2^2/2g + z2
Where,
P1 and P2 are the pressures at points 1 and 2 respectively,
V1 and V2 are the velocities at points 1 and 2 respectively,
z1 and z2 are the elevations at points 1 and 2 respectively,
ρ is the density of the fluid, and
g is the acceleration due to gravity.

In a venturimeter, the pressure at the throat is the lowest, so we can assume P2 = atmospheric pressure. Also, at point 1, the velocity is maximum and the pressure is 6m head of water. Therefore, the equation becomes:
(6/ρg) + (V1^2/2g) + z1 = 0 + (V2^2/2g) + z2

Since the pipeline is horizontal, z1 = z2. Also, the density of water (ρ) is constant. Therefore, the equation simplifies to:
6 + (V1^2/2g) = V2^2/2g

We know that the volume flow rate (Q) is given by:
Q = A1V1 = A2V2
Where,
A1 and A2 are the cross-sectional areas at points 1 and 2 respectively.

Since the diameter of the pipeline is given as 25cm, the area at point 1 is:
A1 = π(25/2)^2 = π(625/4) cm^2

We can rearrange the equation for volume flow rate to get:
V2 = (Q/A2)

Substituting the values in the simplified Bernoulli's equation, we get:
6 + (V1^2/2g) = (Q/A2)^2/2g

Calculating Discharge:
To calculate the discharge when the U-tube difference and deflection are 20cm, we can use the equation for volume flow rate:
Q = A1V1 = A2V2
Since the area at point 1 is known, we can rearrange the equation to get:
V1 = (Q/A1)

Substituting the given values, we get:
V1 = (150*1000)/(π(625/4))
V1 ≈ 305.19 cm/s

Since the U-tube difference is given as 20cm, we can assume the pressure difference between points 1 and 2 as:
ΔP = ρgΔh

Similar Civil Engineering (CE) Doubts

A venturimeter has its axis vertical, the inlet and throat diameter being 150 mm and 75 mm respectively. The throat is 225 mm above inlet. Petrol of specific gravity 0.78 flows up at a rate of 0.029 m3/s. The pressure difference between the inlet and throat will be (cd = 0.96 for venturimeter)

In a mercury differential manometer used for measuring pressure differences across a venturimeter in a water pipeline, if an error of 2 mm has been made in observing a differential head of 10 mm, the percentage error in pressure difference is

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