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Forces Due To Pipe Bends

The momentum change and the unbalanced internal pressure of the water leads to forces on the pipes
The force diagram in figure is a convenient method for finding the resultant force on a bend. The forces can be resolved into X and Y components to find the magnitude and direction of the resultant force on the pipe.

hydrualics 9

V1= velocity before change in size of pipe, ft /s (m/s)

V2= velocity after change in size of pipe, ft /s (m/s)

p1= pressure before bend or size change in pipe, lb/ft2
(kPa)

p2= pressure after bend or size change in pipe, lb/ft2
(kPa)
A1= area before size change in pipe, ft2 (m2)

A2= area after size change in pipe, ft2 (m2)

F2m= force due to momentum of water in section 2 V2Qw/g

F1m= force due to momentum of water in section 1 V1Qw/g

P2= pressure of water in section 2 times area of section 2 p1 A1

P1= pressure of water in section 1 times area of section 1 p1 A1

w= unit weight of liquid, lb/ft3 (kg/m3)

Q= discharge, ft3/s (m3/s)

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If the pressure loss in the bend is neglected and there is no change in magnitude of velocity around the bend,then

R=2A[(wV2/g)+p]cosine of angle between pipes

where R resultant force on bend, lb (N)

p= pressure, lb/ft2 (kPa)

w= unit weight of water, 62.4 lb/ft3 (998.4 kg/m3)

V= velocity of flow, ft/s (m/s)

g= acceleration due to gravity, 32.2 ft/s2 (9.81 m/s2)

A= area of pipe, ft2 (m2)

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Kanwarjot Singh

Kanwarjot Singh is the founder of Civil Engineering Portal, a leading civil engineering website which has been awarded as the best online publication by CIDC. He did his BE civil from Thapar University, Patiala and has been working on this website with his team of Civil Engineers.

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3 comments on "Forces Due To Pipe Bends"

cathy says:

how would u find the change in velocty and why?

rajesh kumar chauhan says:

pls send us calculation of weight & force of water

Chaz says:

Um, why is vQw/g being treated as Force? vQw would be units of force, same as pA=P, not vQw/g. I don’t understand why you’d divide by g ….

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