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## Q. 6.P.5

A venturi meter with a 50 mm throat is used to measure a flow of slightly salt water in a pipe of inside diameter 100 mm. The meter is checked by adding 20 cm³/s of normal sodium chloride solution above the meter and analysing a sample of water downstream from the meter. Before addition of the salt, 1000 cm³ of water requires 10 cm³ of 0.1 M silver nitrate solution in a titration. 1000 cm³ of the downstream sample required 23.5 cm³ of 0.1 M silver nitrate. If a mercury-under-water manometer connected to the meter gives a reading of 221 mm, what is the discharge coefficient of the meter? Assume that the density of the liquid is not appreciably affected by the salt.

## Verified Solution

If the flow of the solution is x m³/s, then a mass balance in terms of sodium chloride gives:

$(x \times 0.0585)+\left(20 \times 10^{-6} \times 58.5\right)=0.1375\left(20 \times 10^{-6}+x\right)$

and:                                                                           $x=0.0148 m ^3 / s$

or, assuming the density of the solution is 1000 kg/m³, the mass flowrate is:

$(0.0148 \times 1000)=14.8 kg / s$

For the venturi meter, the area of the throat is given by:

$A_1=(\pi / 4)(50 / 1000)^2=0.00196 m ^2$

and the area of the pipe is:

$A_2=(\pi / 4)(100 / 1000)^2=0.00785 m ^2$

From equations 6.32 and 6.33:

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

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

$C^{\prime}=A_1 A_2 / \sqrt{\left(A_1^2-A_2^2\right)}=0.00204 m ^2$

$h = 221 mm Hg _{\text {under-water }} =0.221(13500-1000) / 1000=2.78$ m water

and hence: $14.8=\left(C_D \times 1000 \times 0.00204\right) \sqrt{(2 \times 9.81 \times 2.78)}$

and:               $C_D=\underline{\underline{0.982}}$