Q.198. Fluid Mechanics

Question 198.
A horizontal 600mm diameter pipe carries water with a velocity of 3m/s. The water is at a pressure of 294.3 kN/m^2. The pipe is fitted with a bend that deflects the water through an anglel of 75 degrees. Calculate the resultant force on the bend and its direction.

Answer 198.
Resulting force in horizontal direction, Rx
= p A (1 - cosθ)
= 294.3 x 10^3 * π (0.6/2)^2 * (1 - cos75°)
= 61675 N

Resulting force in vertical direction, Ry
= p A sinθ
= 294.3 x 10^3 * π (0.6/2)^2 * sin75°
= 80376 N

Resulting force, R
= √[Rx^2 + Ry^2)
= √[(61675)^2 + (80376)^2]
= 101312 N.

Let the resultant make an angle β with the horizontal.
=> tanβ = Ry/Rx = (80376) / (61675)
=> β = 52.5° with the horizontal.

http://www.engineeringtoolbox.com/forces-pipe-bends-d_968.html

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