Question
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.
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
Resulting force in horizontal direction,
Resulting force in horizontal direction,
= p A (1 - cosθ) Rx
= 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.
= 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.
No comments:
Post a Comment