PLAIN AND CIVIL: EXAMPLE 3.1.1. VENTURIMETER WITH MANOMETER

Tuesday 19 March 2019

The pressure head difference between the inlet and throat of the 75mm by 25mm Venturimeter is measured by means of a mercury differential manometer equipped with a scale which reads 300mm. With water flowing through the meter, determine the following:

Discharge neglecting head loss
The coefficient of discharge if during a test run, 535 liters of water were discharged in 130.50 seconds.

Use Bernoulli's Energy Equation to start computation

$\frac{v_{1}^{2}}{2g}+\frac{p_{1}}{\gamma }+z_{1}=\frac{v_{2}^{2}}{2g}+\frac{p_{2}}{\gamma }+z_{2}$

however, it would be easier to come up with the discharge if the BEE is in terms of discharge, Q, instead of velocity head

$\frac{8Q^{2}}{\pi^{2} gD_{1}^{4}}+\frac{p_{1}}{\gamma }+z_{1}=\frac{8Q^{2}}{\pi^{2} gD_{2}^{4}}+\frac{p_{2}}{\gamma }+z_{2}$

With the Venturimeter placed horizontally, both the elevation heads at stations 1 and 2 are both zero.

$\frac{8Q^{2}}{\pi^{2} gD_{2}^{4}}-\frac{8Q^{2}}{\pi^{2} gD_{1}^{4}}=\frac{p_{1}}{\gamma }-\frac{p_{2}}{\gamma }$

$\frac{8Q^{2}}{\pi^{2} \left (9.81 \right )0.025^{4}}-\frac{8Q^{2}}{\pi^{2} \left ( 9.81 \right )0.075^{4}}=\frac{p_{1}}{\gamma }-\frac{p_{2}}{\gamma }$

$208913.3377Q^{2}=\frac{p_{1}}{\gamma }-\frac{p_{2}}{\gamma }$

Using pressure head equation from inlet (station 1) to throat (station 2)

$\frac{p_{1}}{\gamma }+y+0.30-0.30\left ( 13.60 \right )-y=\frac{p_{2}}{\gamma }$

$\frac{p_{1}}{\gamma }+\frac{p_{2}}{\gamma }=3.78m$

Substituting this into the BEE:

$208913.3377Q^{2}=3.78m$

${\color{Red} \mathbf{Q=0.004254\frac{m^{3}}{s}}}$ (theoretical value)

The coefficient of discharge is computed as follows:

$C_{d}=\frac{Q_{a}}{Q_{t}}$

where:

$Q_{a}=\frac{535liters}{130.50sec}=\frac{0.535m^{3}}{130.50sec}=0.0041\frac{m^{3}}{s}$

Thus:

$C_{d}=\frac{0.0041}{0.004254}$

${\color{Red} \mathbf{C_{d}=0.964}}$

PLAIN AND CIVIL

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Tuesday 19 March 2019

EXAMPLE 3.1.1. VENTURIMETER WITH MANOMETER

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