Answer to Question #219322 in Mechanics | Relativity for EMMANUEL

Question

 A venturi meter of 15 cm inlet diameter and 10 cm throat is load horizontally in a pipe to

measure the flow of oil of 0.9 specific gravity. The reading of a mercury manometer is 20 cm.

Calculate the discharge in m3/s.

Answer 1

Gives

Area

$A_1=\frac{\pi d_1^2}{4}=\frac{\pi \times15^2}{4}=176.7cm^2$

$A_2=\frac{\pi d_2^2}{4}=\frac{\pi \times10^2}{4}=78.54cm^2$

$Q=\frac{A_1A_2 \sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}$

$$H=h[\frac{r_m}{r}-1]=20\times[\frac{13.6}{0.9}-1]=282.2 cm oil$$

$$Q=\frac{176.7\times78.542 \sqrt{2\times981\times282.2}}{\sqrt{176.7^2-78.54^2}}=6538.2cm^3/sec$$

$Q=0.065m^3/sec$