A pipe of 300 mm diameter conveying $0.30 m^3/s$ of water has a right angled bend in a horizontal plane, find the resultant force extended on the bend if the pressure at inlet and outlet of the bend are 24.525 N $cm^2$ and 23.544 $N/cm^2$

Given :

1] Dia of bend, D = 300 mm = 0.3 m

2] Area, $A = A_1 = A_2 = \frac{\pi}{4} (0.3)^2$ = $0.07068m^2$

3] Discharge, Q = 0.30 $m^3/s$

4] Velocity, $V = v_1 = v_2 = \frac{Q}{A}$

$= \frac{0.30}{0.07068}$

= 4.244m/s

Angle of bend, $\theta = 90˚$

$p_1 = 24.525 N/ cm^2$

$= 24.525 \times 10^4 N/m^2$

$= 245250 N/m^2 $

$p_2 = 23.544 N/cm^2$

$= 23.544 \times 10^4 N/m^2$

$= 235440 N/m^2$

Force on bend along x axis,

$\therefore$ $F_x = 1000 \times 0.30 [4.244 - 0] + 24.5250 \times 0.07068 + 0$

= 1273.2 + 17334.3

$F_x = 18607.5 N$

Force on bend along, y-axis,

$F_y = pq [v_1y - v_2y] + (p_1 A_1)_y + (p_2 A_2)_y$

$\therefore$ $v_1y = 0, v_2y = v_2 = 4.244 m/s$

$\therefore$ $(p_1 A_1)_y = 0$

$(p_2 A_2 )_y = - p_2 A_2$

= -235440 x 0.07068

= -16640.9

$\therefore$ $Fy = 1000 \times 0.30 [0 - 4.244] + 0 - 16640.9$

= -1273.2 - 16640.9

Fy = -17914.1 N

Resultant force, $F_R = \sqrt{(F_x)^2 + (F_y)^2}$

= $\sqrt{(18607.5)^2+(17914.1)^2}$

= - 25829.3N

$\therefore$ $tan \theta = \frac{F_y}{F_x} = \frac{17914.1}{18607.5} -0.9627$

$\therefore$ $\theta = tan^-1 (0.9627)$

$\theta = 43˚ 54'$