Answer to Question #147860 in Mechanical Engineering for Janet

Question #147860
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
Expert's answer

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"

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