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Engineering › Fluid Mechanics › Pipes ›

Forces on Pipe Bends

The forces created in Pipe bends by a combination of static, dynamic,and frictional forces.

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Contents

Overview
Theory
Page Comments

Overview

Fluids flowing round bends in pipes suffer from an increase in turbulence and a consequential increase the head lost in the pipe. They also produce forces on the bend which are examined in this section.

Theory

The Total or Resultant Thrust on a Pipe Elbow will be made up of :

The Static Pressure Force;
A Dynamic Force due to the change in direction of the fluid flow. This is the product of the mass of fluid passing through the bend per second times the change in velocity;
The frictional forces on the Pipe;
The weight of the elbow and the fluid contained in it.

In general these last two are small and can be neglected.

23287/force_on_pipe_bends_1.png

Considering the $90^0$ elbow shown in the diagram.

Force in Direction X:

$R_X=p_1a_1+\frac{wa_1{v_{1}}^{2}}{g}$

Force in Direction Y:

$R_Y=p_2a_2+\frac{wa_2{v_{2}}^{2}}{g}$

Resultant Force R:

$R=\sqrt{{R_{X}}^{2}+{R_{Y}}^{2}}$

R has a direction measured from the $X$ axis of :

$\theta=\tan^{-1}\frac{R_Y}{R_X}$

Example:

[imperial]

Example - Example 1

Problem

A pipe $AB$ of 1 ft. diameter carries water at 12 ft./sec. down to a Power-house. At $A$ and $B$ there are frictionless expansion joints and the pipe which weighs 40 lb./ft.run, is mounted on frictionless rollers, except where it is tied in the concrete block $C$ .

23287/force_on_pipe_bends_2.png

At $A$ the head is 120 ft. and $f$ for the pipe is 0.0075.

Determine the horizontal and vertical components of the force on the block $C$ due to the pipe line.

Workings

The length of the short vertical section is ignored and so:

$Z_A-Z_B=200\sin30^0=100 ft.$

Applying Bernoulli at $A$ and $B$ :

$\frac{p_a}{w}+\frac{{v_{A}}^{2}}{2g}+Z_A=\frac{p_B}{w}+\frac{{v_{B}}^{2}}{2g}+Z_B+h_f$

$v_A=v_B$

And:

$h_f=\frac{4\times 0.0075\times 200\times 12^2}{2\times 1\times 32.2}=13.4\;ft.$

$\therefore \;\;\;\;\;\frac{p_B}{w}=120+100-13.4=206.6\;ft.$

The static pressure in the $X$ direction is:

$p_A\times a\times \cos30^0=120\times 62.4\times \frac{\pi\times 1^2}{4}\times 0.866=5093\;lbs$

(where $a$ is the area)
The static pressure in the $Y$ direction is:

$- p_B\times a\times \sin30+p_B\times a$

$=- 120\times 62.4\times \frac{\pi\times 1^2}{4}\times 0.5+206.6\times \frac{\pi+1^2}{4}\times 62.4=7182\;lbs$

The weight of water per sec. is:

$W=\frac{\pi\times 1^2}{4}\times 12\times 62.4=588\;lb./sec.$

The Dynamic Force in the $X$ direction is:

$\frac{W}{g}\times v_A\cos30^0$

$\frac{588}{32.2\times 12\times 0.866}=190\;lb.$

The Dynamic Force in the $Y$ direction is:

$\frac{W}{g} \left ( v_A\sin30^0+v_B \right )$

$=\frac{588}{32.2}\left ( -6+12 \right )=110\;lb.$

The weight of pipe plus contents:

$=200\times 40 +200\times \frac{\pi\times1^2}{4}\times 62.4$

$=17,800\;lb.$

The reaction on the rollers is:

$17,800\sin 30=F=8900\;lb.$

Component in the $X$ direction is:

$F\cos 30 =7705\;lb.$

Component in the $Y$ direction is:

$- F\sin 30 =- 4450\;lb.$

Thus the Total force in the $X$ direction is :

$5093+190+7750=12,988\;lb.$

And the Total force in the $Y$ direction is:

$7182+110-4450=2842\;lb.$

Solution

The Total force in the $X$ direction is $12,988\;lb.$
The Total force in the $Y$ direction is $2842\;lb.$