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) …