Question

The angle of a reducing bend is 60° ( that is deviation from initial direction to final direction ). Its initial diameter is 300 mm and final diameter is 150 m and is fitted in a pipeline , carrying a discharge of 360 liter/s . The pressure at the inlet is 2.943 bar . The Friction Loss in the pipe bend may be assumed as 10 percent of kinetic energy at exit of the bend . The fluid Flowing through the pipeline is water . Determine the force exerted by the fluid (water) on the reducing bend

Answer 1

Answer:

=−6322.2N

Explanation:

Find the force exerted by water on the bend if the intensity of pressure at inlet to bend is 8,829 N/cm2 and rate of flow of water is 600 lit/s.

Given:

Angle of bend, θ=45˚

Dia of inlet, D1=600mm

=0.6m

Area A1=π4D21=π4×(0.6)2

= 0.2827m2

Dia. of outlet, D2=300mm

=0.3mArea A2=π4D22=π4(0.3)2

=0.07068m2

Pressure at inlet, P1=8.829N/cm2

=8.829×104N/m2

Q=600lit/S

=0.6m3/s

v1=QA=0.60.2827=2.122m/s

v1=QA2=0.60.07068=8.488m/s

Applying Bernoulli's equation at section (1) and (2),

p1pg+v212g+z1=p2pg+v222g+z2

(z1=z2)

∴ p1pg+v212g=p2pg+v222g

8.829×1041000×9.81+2.12222×9.81=p2pg+8.48822×9.81

9+0.2295=p2pg+3.672

p2pg =9.2295−3.672

p2pg =5.5575m of water.

∴ p2=5.5575×1000×9.81

p2=5.45×104N/m2

Forces on the bend in x and y direction are given by equation.

Fx=PQ[v1−v2cosθ]+p1A1−p2A2cosθ

1000×0.6[2.122−8.488cos45˚]+8.829×104×0.2827−5.45×104×0.07068cos45˚

=[−2327.9+24959.6]−[2720.3]

Fx=19911.4N

Now,

Fy=pq[−v2sinθ]−p2A2sinθ

=1000×0.6[−8.488sin45˚]−5.45×104×0.07068×sin45˚

Fy=−3601.1−2721.1

=−6322.2N