Question

A 25 mm diameter horizontal pipe has a neck of 18 mm diameter. water flows in the broader point of pipe at 1.8 m/s. What is the speed of water in the neck? What is the 1. volume 2. mass flow rate?
$(3.5m/s1.88\times {10}^{-4}{m}^3/s\mathrm{2.0.88}kg/s)$

Answer 1

By equation of continuity flow

$AV$ = constant

$A_1{V}_1=A_2{V}_2$

$\Rightarrow \left(\pi r_1^2\right)V_1=\left(\pi r_2^2\right)V_2$

$\Rightarrow (25{)}^2\times 1.8m/sec=(18{)}^2\times V_2$

$\Rightarrow V_2=\frac{625}{18\times 18}\times \frac{1.8}{10}$

$=\frac{125}{36}$ m/sec

$V_2=3.5$ m/sec

Volume flow rate $=A_1{V}_1$

$=\pi (25\times {10}^{-3}{)}^2\times 1.8$

$=0.0035m^2/sec$

Volume state $=\left(\pi R_1^2\right)V_1$

$V=3.14\times {(\frac{25}{2}\times {10}^{-3})}^2\times 1.8$

$=88\times {10}^{-4}{m}^3/sec$