Question

A 30 cm diameter pipe carries water under a head of 20 meters with a velocity of 3.5
m/s. If the axis of the pipe turns through 450
, find the magnitude and direction force
on the bend.

Answer 1

Given, diameter of pipe=30 cm,Head =20 m and velocity of flowing is 3.5 m/s

Axis of pipe turn by 45^o now

Area of pipe flow =$\frac{\pi}{4}d^2=\frac{3.14}{4}\times (0.3)^2=0.07065 \frac{m^3}{s}$

Now pressure with which fluid flowing is

$P_1= \rho gH=1000\times 9.8\times 20=196000 N$

Now,

According to question force along x-axis will be

$F_x=\rho Q(v_{1x}-v_{2x})+P_1A_{1x}+P_2A_{2x}$

$F_x=1000\times (0.07065\times 3.5)(3.5--3.5 cos 45)+196000\times 0.07065+196000\times 0.07065 \times 0.707$

F_x=4310 N

Similarly we will get force in y-direction as

$F_y=\rho Q(v_{1y}-v_{2y})+P_1A_{1y}+P_2A_{2y}$

$F_y=1000\times (0.07065\times 3.5)(3.5--3.5 cos 45)+196000\times 0.07065+196000\times 0.07065 \times 0.707$

F_y= 23890 N

Net resultant force

$F=\sqrt{ (F_x^2)+(F_y^2)}$ = 24275 N

and the direction of force

$\theta= tan^{-1} \frac{F_y}{F_x}=79.77 ^o$