Find the force exerted by water on the bend if the intensity of pressure at inlet to bend is 8,829 $N/cm^2$ and rate of flow of water is 600 lit/s.

Given:

Angle of bend, $\theta = 45˚$

Dia of inlet, $D_1 = 600 mm$

$= 0.6 m$

Area $A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} \times (0.6)^2$

= $0.2827m^2$

Dia. of outlet, $D_2 = 300mm$

$= 0.3m$Area $A_2 = \frac{\pi}{4} D_2^2 = \frac{\pi}{4} (0.3)^2$

$= 0.07068m^2$

Pressure at inlet, $P_1 = 8.829 N/cm^2$

$= 8.829 \times 10^4 N/m^2$

$Q = 600 lit/S$

$= 0.6 m^3/s$

$v_1 = \frac{Q}{A} = \frac{0.6}{0.2827} = 2.122 m/s$

$v_1 = \frac{Q}{A_2} = \frac{0.6}{0.07068} = 8.488m/s$

Applying Bernoulli's equation at section (1) and (2),

$\frac{p_1}{pg} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{pg} + \frac{v_2^2}{2g} + z_2$

$(z_1 = z_2)$

$\therefore$ $\frac{p_1}{pg} + \frac{v_1^2}{2g} = \frac{p_2}{pg} + \frac{v_2^2}{2g}$

$\frac{8.829 \times 10^4}{1000 \times 9.81} + \frac{2.122^2}{2 \times 9.81} = \frac{p_2}{pg} + \frac{8.488^2}{2 \times 9.81}$

$9 + 0.2295 = \frac{p_2}{pg} + 3.672$

$\frac{p_2}{pg}$ $= 9.2295 - 3.672$

$\frac{p_2}{pg}$ $= 5.5575m$ of water.

$\therefore$ $p_2 = 5.5575 \times 1000 \times 9.81$

$p_2 = 5.45 \times 10^4 N/m^2$

Forces on the bend in x and y direction are given by equation.

$F_x = PQ [ v_1 - v_2 cos \theta] + p_1 A_1 - p_2 A_2 cos \theta$

$1000 \times 0.6 [2.122 - 8.488 cos 45˚] + 8.829 \times 10^4 \times 0.2827 - 5.45 \times 10^4 \times 0.07068 cos 45˚$

$= [ -2327.9 + 24959.6] - [2720.3]$

$F_x = 19911.4 N$

Now,

$F_y = pq [ -v_2sin \theta] - p_2 A_2 sin \theta$

$= 1000 \times 0.6 [ -8.488 sin 45˚] - 5.45 \times 10^4 \times 0.07068 \times sin 45˚$

$F_y = -3601.1 - 2721.1$

$= - 6322.2 N$

Negative sign means fy is acting in downward direction.

$F_R = \sqrt{fx^2 + fy^2}$

$= \sqrt{(19911.4)^2 + (-6322.2)^2}$

$F_R = 20890.9 N$

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

$tan \theta = \frac{fy}{fx}$

$= \frac{6322.2}{19911.4}$

$tan \theta = 0.3175$

$\theta = tan^-1 (0.3175)$

$\theta = 17˚ 36'$