Answer to Question #147860 in Mechanical Engineering for Janet

Question

A 400 mm diameter pipe carries water under a head of 30 m with a velocity of
3.5 m/s. If the axis of the pipe turns through 46⁰, calculate the magnitude and
the direction of the resultant force at the bend

Answer 1

Diameter of pipe is 400 mm, velocity of fluid=3.5 m/s, Head of flowing water=30 m

$\theta=46 ^0$ ,

Cross sectional area of pipe=$\pi\times d^2/4= \frac{3.14\times .4\times .4}{4}=0.1256 m^2$

Pressure=$P=\rho \times g\times h= 1000\times 9.8\times 30=2.94 \times 10 ^5 Pa$

At the site of bend there two force act one horizontal and other vertical force

Force acting in horizantal direction

Rate of flow = AV= 0.1256$\times 3.5=0.4396$ =Q

$F_x=\rho Q(v_1-v_2cos 46^0)+ P_1A_1-P_2A_2cos 46^0$

$F_x=1000\times 0.4396 (3.5-2.43)+ 2.94 \times 10^5\times 0.1256- 2.94 \times 10^5 \times 0.1256 \times 0.694)$

$F_x=470.372+3.693 \times 10 ^4-2.56 \times 10^4=11800.372$ N

$F_y=\rho Q(v_1y-v_2sin 46^0)+ P_1A_1y-P_2A_2ysin 46^0$

$F_y=1000\times 0.4396 (0-2.43)+ 0- 2.94 \times 10^5 \times 0.1256 \times 0.719)$

$F_y=-316.22-26550.08=-26866.3$ N

$F=\sqrt{F_x^2+F_y^2}=\sqrt{(11800.372)^2+(-26866.3)^2}=29343.599$ N

and for the direction we have

tan$(\phi)= \frac{F_y}{F_x}=-2.726$

$\phi= 69.85^0$