Question

$A$ ${45}^{\circ }$ bend of uniform diameter of $80cm$ in horizontal plane. The magnitude of the force (in kN,) rounded off to two decimal places) exerted by the flowing water on the bend for following data is____.
(i) Flow rate in pipe $=0.8m^3/sec$.

(ii) Pressure at inlet $=20kPa$

(iii) Friction across the bend in negligible.

• A. 8.5

Answer 1

The correct option is A 8.5
Let, force exerted by the water on bend be $R_x$ and $R_y$
$\sum F_x=p_1{A}_1-p_2{A}_{2}cos{45}^{\circ }+R_x=pQ{V}_{2}cos{45}^{\circ }-\rho Q{V}_1$ ...(1)
$\sum F_y=0-p_2{A}_{2}sin{45}^{\circ }+R_y=\rho Q{V}_{2}sin{45}^{\circ }-0$ ...(2)
$p_1=20kPa,A_1=A_2,V_1=V_2,Q=0.8m^3/sec.,d=0.8m$
Applying Bernoulli`s equation 1 and 2
$\frac{p_1}{{\gamma }_w}+\frac{V_1^2}{2g}+Z_1=\frac{p_2}{{\gamma }_w}+\frac{V_2^2}{2g}+Z_2$
$\because V_1=V_2$ & $z_1=z_2$
$\therefore p_1=p_2=20kPa$
$V_1=V_2=\frac{Q}{\frac{\pi }{4}{d}^2}=\frac{0.8}{\frac{\pi }{4}\times (0.8{)}^2}=1.6m/sec$
From equation (1)
$20\times {10}^3\times (\frac{\pi }{4}\times {0.8}^2)(1-cos{45}^{\circ })+R_x=1000\times 0.8\times 1.6(cos{45}^{\circ }-1)$
$R_x=-3319.4N=-3.32kN$
From equation (2)
$-20\times {10}^3\times \frac{\pi }{4}\times {0.8}^{2}sin{45}^{\circ }+R_y=1000\times 0.8\times 1.6sin{45}^{\circ }$
$R_y=8013.7kN=8kN$
$\therefore$ Resultant force $=\sqrt{R_x^2+R_y^2}=8.66kN$