The pressure in pipeline is 140 KNomo, and the vacuum in the throat is 37.5 cm எ mercury. Four percentages of the differential head is lost between the gauges. Working from first principles find the flow rate in the pipeline assuming the venturimeter to be horizontal

Given:

$D_1=30cm=0.3m,D_2=12.5cm=.125m,p_1=140 KN⁄m^2 ,p_2$=-37.5 cm of mercury

$p_2=-\frac{37.5×13.6}{100}=-5.1 m \ of \ water;h_f=4% $of differential head.

Flow rate in pipeline, Q:

$\frac{p_1}{w}=\frac{140×10^{3}}{9810}$=14.27 m of water.

$\frac{p_2}{w}$=-5.1 m of water(calculated above)

$h_f=4%$ of differential head

=$\frac{4}{100} (\frac{p_1}{w}-\frac{p_2}{w})=\frac{4}{100} [14.27-(-5.1)]$=0.775 m of water.

Applying Bernoulli’s equation to the entrance (1) and throat (2) of the venturimeter, we get

$\frac{p_1}{w}+\frac{V_1^{2}}{2g}+z_1=\frac{p_2}{w}+\frac{V_2^{2}}{2g}+z_2+h_f$

$\frac{V_1^{2}-V_2^{2}}{2g}=(\frac{p_{2}}{w}-\frac{p_{1}}{w})+h_f$

$\frac{V_1^{2}-V_2^{2}}{2g}$=-5.1-14.27+.775=-18.59

$\frac{V_1^{2}}{2g} [1-(\frac{V_1^{2}}{V_2^{2}})^{2} ]$=-18.59

Also, $A_1 V_1=A_2 V_2$

Hence discharge, $Q=A_1 V_1=\frac{π}{4}×0.3^{2}×3.367×10^{3} l/s$

≅238 $l⁄(s$ (ANS.))