# Forces Due To Pipe Bends

Posted in Hydraulics | Email This Post |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.

V_{1}= velocity before change in size of pipe, ft /s (m/s)

V_{2}= velocity after change in size of pipe, ft /s (m/s)

p_{1}= pressure before bend or size change in pipe, lb/ft^{2}

(kPa)

p_{2}= pressure after bend or size change in pipe, lb/ft^{2}

(kPa)

A_{1}= area before size change in pipe, ft^{2} (m^{2})

A_{2}= area after size change in pipe, ft^{2} (m^{2})

F_{2m}= force due to momentum of water in section 2 V_{2}Qw/g

F_{1m}= force due to momentum of water in section 1 V_{1}Qw/g

P_{2}= pressure of water in section 2 times area of section 2 p_{1} A_{1}

P_{1}= pressure of water in section 1 times area of section 1 p_{1} A_{1}

w= unit weight of liquid, lb/ft^{3} (kg/m^{3})

Q= discharge, ft^{3}/s (m^{3}/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[(wV^{2}/g)+p]cosine of angle between pipes

where R resultant force on bend, lb (N)

p= pressure, lb/ft^{2} (kPa)

w= unit weight of water, 62.4 lb/ft^{3} (998.4 kg/m^{3})

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

g= acceleration due to gravity, 32.2 ft/s^{2} (9.81 m/s^{2})

A= area of pipe, ft^{2} (m^{2})

how would u find the change in velocty and why?

pls send us calculation of weight & force of water

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 ….