Water flows out of a big tank along a tube bent at right angles; inside radius of the tube is equal to r=0.50 cm The length of the horizontal section of the  tube is equal to  $l=22cm$ The water flow rate is $Q=0.50$ litres per second. Find the moment of reaction  forces of flowing water, acting on the tube’s walls,  relative to the point O. (in Nm) 

Water flows through the tube with velocity  $v=\frac{Q}{A}.$
  At the bend, the velocity vector changes from  horizontal  $\left(\hat{i}\right)$ to vertical $(-\widehat{j})$ direction  The momentum vector of the flowing water  changes at the bend. There is a force  on the flowing water by the bend.  From Newton’s third law of motion the reaction of this  force acts on the tube at the bend. You need the  moment of this force on the tube relative to point O. The change in momentum of mass $\Delta m$  of liquid as it passes through the bend, 
 

$\Delta \overrightarrow{P}={\overrightarrow{P}}_f-{\overrightarrow{P}}_i=\Delta mv(-\widehat{j})-\Delta mv\widehat{i}=-\Delta mv\widehat{i}-\Delta mv\widehat{j}$

 Or   $\overrightarrow{F}=\frac{\Delta P}{\Delta t}=-v\frac{\Delta m}{\Delta t}\widehat{i}-v\frac{\Delta m}{\Delta t}\widehat{j}$
  $=-v\left(\rho A\frac{v\Delta t}{\Delta t}\right)\widehat{i}-v\left(\rho A\frac{v\Delta t}{\Delta t}\right)\widehat{j}=-\rho A{v}^2\widehat{i}-\rho A{v}^2\widehat{j}.$
 The reaction force $\overrightarrow{F}\text{'}$  on the tube at bend point is equal to  $=-\overrightarrow{F}$
 Therefore $\overrightarrow{F}\text{'}=\rho \pi r^2{v}^2\widehat{i}+\rho \pi r^2{v}^2\widehat{j}.$,  
 Moment of the force $\overrightarrow{F}\text{'}$  about O is $\overrightarrow{\tau }=\overrightarrow{l}\times (\overrightarrow{F}\text{'})=\left(l\widehat{i}\right)\times (\rho \pi r^2{v}^2\widehat{i}+\rho \pi r^2{v}^2\widehat{j})$

$=\rho \pi r^2{v}^{2}l\widehat{k}. \dots .\left(1\right) (v=\frac{Q}{A}=\frac{Q}{\pi r^2})$

$$\begin{gathered} \overrightarrow{\tau }=\rho \pi r^2{v}^{2}l\widehat{k} \Rightarrow \tau =\rho \pi r^{2}l{\left(\frac{Q}{\pi r^2}\right)}^2=\frac{\rho Q^{2}l}{\pi r^2}. \\ =\frac{1.0\times {10}^3\times {(0.50\times {10}^{-3})}^2\times 0.22}{3.14\times {(0.50\times {10}^{-2})}^2}=0.7Nm \end{gathered}$$