Question

There is a uniform tube of cross-section area . A Liquid of density $\rho$ is flowing in the tube flowing with uniform speed $v$. Wat will be the force applied by liquid at the bend ?$\theta {=}^o$

Answer 1

toward x-dircetion, Angle bisect into half, as show in fgure

Inlet velocity $V_i$ = outlet velocity $V_o$ = $V$

Change in velocity in y-direction, $\Delta V_y=V_o\sin {30}^o-V_i\sin {30}^o=0$

Change in velocity in x-direction, $\Delta V_x=V_o\cos {30}^o-(-V_o\cos {30}^o)=2V_o\frac{\sqrt{3}}{2}=\sqrt{3}{V}_o$

Rate of fluid flow, $\stackrel{\bullet }{m}=\rho AV$

$Force=rateofchangeofmomentum=\stackrel{\bullet }{m}\Delta V$

$Force=\rho AV.\sqrt{3}V=\sqrt{3}\rho A{V}^2$

Hence, force on tube is $\sqrt{3}\rho A{V}^2$