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## Q. 6.P.4

The rate of flow of water in a 150 mm diameter pipe is measured by means of a venturi meter with a 50 mm diameter throat. When the drop in head over the converging section is 100 mm of water, the flowrate is 2.7 kg/s. What is the coefficient for the converging cone of the meter at that flowrate and what is the head lost due to friction? If the total loss of head over the meter is 15 mm water, what is the coefficient for the diverging cone?

## Verified Solution

The equation relating the mass flowrate G and the head loss across a venturi meter is given by:

$G=\frac{C_D A_0}{v} \sqrt{\frac{2 v\left(P_1-P_2\right)}{1-\left(A_0 / A_1\right)^2}}$             (equation 6.19)

$G=C_D \rho \frac{A_1 A_2}{\sqrt{\left(A_1^2-A_1^2\right)}} \sqrt{\left(2 v\left(P_1-P_2\right)\right)}$             (equation 6.32)

$G=C_D \rho C^{\prime} \sqrt{ }\left(2 g h_v\right)$             (equation 6.33)

where $C^{\prime}$ is a constant for the meter and $h_v$ is the loss in head over the converging cone expressed as height of fluid.

$A_1=(\pi / 4)(0.15)^2=0.0176 m ^2$

$A_2=(\pi / 4)(0.05)^2=0.00196 m ^2$

$C^{\prime}=\left(0.0176 \times 0.00196 / \sqrt{\left.\left(0.0176^2-0.00196^2\right)\right)}=0.00197 m ^2\right.$

$h_v=0.1 m$

∴        $2.7=\left(C_D \times 1000 \times 0.00197\right) \sqrt{ }(2 \times 9.81 \times 0.10)$ and $\underline{\underline{C_D=0.978}}$

In equation 6.33, if there were no losses, the coefficient of discharge of the meter would be unity, and for a flowrate G the loss in head would be $\left(h_v-h_f\right)$ where $h_f$ is the head loss due to friction.

Thus:                     $G=\rho C^{\prime} \sqrt{\left[2 g\left(h_v-h_f\right)\right]}$

Dividing this equation by equation 6.33 and squaring gives:

$1-\left(h_f / h_v\right)=C_D^2$ and $h_f=h_v\left(1-C_D^2\right)$

∴                         $h_f=100\left(1-0.978^2\right)=\underline{\underline{4.35 mm }}$

If the head recovered over the diverging cone is $h_v^{\prime}$ and the coefficient of discharge for the converging cone is $C_D^{\prime}$, then $G=C_D^{\prime} \rho C^{\prime} \sqrt{\left(2 g h_v^{\prime}\right)}$
If the whole of the excess kinetic energy is recovered as pressure energy, the coefficient $C_D^{\prime}$ will equal unity and G will be obtained with a recovery of head equal to h0v plus some quantity $h_f^{\prime}$, $G=\rho C^{\prime} \sqrt{\left[2 g\left(h_v^{\prime}+h_f^{\prime}\right)\right.}$

Equating these two equations and squaring gives:

$C_D^{\prime 2}=1+\left(h_f^{\prime} / h_v^{\prime}\right)$ and $h_f^{\prime}=h_v^{\prime}\left(C_D^{\prime 2}-1\right)$

Thus the coefficient of the diverging cone is greater than unity and the total loss of head $=h_f+h_f^{\prime}$.
Head loss over diverging cone $=(15.0-4.35)=10.65$ mm

The coefficient of the diverging cone $C_D^{\prime}$ is given by:

$G=C_D^{\prime} \rho C^{\prime} \sqrt{\left(2 g h_v^{\prime}\right)}$

$h_v^{\prime}=(100-15)=85 mm$

and:          $2.7=\left(C_D^{\prime} \times 1000 \times 0.00197\right) \sqrt{(2 \times 9.81 \times 0.085)}$ or $\underline{\underline{C_D^{\prime}=1.06}}$