in a vertical plane. The axis at inlet is horizontal and the centre of the outlet section is 1.5 m below the centre of the inlet section. Total volume of water in the bend is 0.9 m'. Neglecting friction, calculate the magnitude and direction of the force exerted on the bend by water flowing through it at 250 litress and when inlet pressure is 0.15 N/mm”.

Given: angle of bend θ=$120^{\circ}$

Diameter of pipe at inlet $D_{1}$=30cm=0.30m

Diameter of pipe at outlet $D_{2}$=15cm=0.15m

Area at inlet , $A_{1}=\frac{π}{4} D1^{2}=0.0706m^2$

Area at outlet, $A_{2}=\frac{π}{4} D_2^{2}=0.01767m^2$

Pressure at inlet =$0.15N⁄mm^{2} =0.14×10^6 N⁄m^2 $

Q=250litres/sec=$0.25m^{3}sec$

$z_{2}$=-1.5m

$V_1=\frac{Q}{A_1} =\frac{.25}{.0706}$=3.541m/s

$V_2=\frac{Q}{A_2} =\frac{.25}{.01767}=14.1483m/s$

Applying Bernoulli’s equation at section 1 and 2

$\frac{p_1}{ρg}+\frac{V_1^{2}}{2g}+z_1=\frac{p_2}{ρg}+\frac{V_2^{2}}{2g}+z_2$

$z_2=-5$

$\frac{0.14×10^{6}}{1000×9.81}+\frac{3.541^{2}}{2×9.81}=\frac{p_2}{1000×9.81}+\frac{14.1483^2}{2×9.81}$-1.5

$p_2=60897.14N/m^2$

Forces on the bend in x and y directions are given by

$F_x=ρQ[V_1-V_2 cosθ]+p_1 A_1-p_2 A_2 cosθ$

=$1000×.25[3.541-14.1483cos120]+.14×〖10〗^6×.0706-60897.4×.01767cos120 =13075.82N$

And

$F_y=ρQ(-V_2 sinθ)-p_2 A_2 sinθ$

=$1000×.25(-14.1483sin120)-60897.14×.01767sin120$

=-3995.08N

-ve sign indicates the $F_y$ is acting in the downward direction.

Therefore resultant force, $F_R=2\sqrt{F_x^{2}+F_y^{2}}=2\sqrt{13075.82^{2}+-3995.08^{2}}$

$F_R=13672.52N$ ….(ANS.)

The angle made by resultant force with x-axis given by,

tanθ=$\frac{F_y}{F_x} =\frac{3995.08}{13075.82}$=0.3055

$θ=16.98^0$ …..(ANS)