Search

Forces Due To Pipe Bends

Posted in Hydraulics | Email This Post Email This Post |
Print Friendly, PDF & Email

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)

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)

More Entries :
Comments
  • cathy September 27, 2010 at 2:57 am

    how would u find the change in velocty and why?

  • rajesh kumar chauhan July 8, 2011 at 8:02 am

    pls send us calculation of weight & force of water

  • Chaz November 6, 2012 at 2:25 am

    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 ….

  • Post a comment