I have forgotten
my Password

Or login with:

  • Facebookhttp://facebook.com/
  • Googlehttps://www.google.com/accounts/o8/id
  • Yahoohttps://me.yahoo.com

Pipe Head Loss

The pressure head lost due to flow through pipes and other losses.
+ View other versions (3)

Key Facts

Gyroscopic Couple: The rate of change of angular momentum (\tau) = I\omega\Omega (In the limit).
  • I = Moment of Inertia.
  • \omega = Angular velocity
  • \Omega = Angular velocity of precession.

Overview

In general the flow of liquid along a pipe can be determined by the use of The Bernoulli Equation and the Continuity Equation. The former represents the conservation of energy, which in Newtonian fluids is either potential or kinetic energy, and the latter ensures that what goes into one end of a pipe must comes out at the other end. However as the flow moves down the pipe, losses due to friction between the moving liquid and the walls of pipe cause the pressure within the pipe to reduce with distance - this is known as head loss.

Note: Only Incompressible liquids are being considered.

Head Lost Due To Friction In The Pipe

Two equations can be used when the flow is either Laminar or Turbulent. These are:
Darcy's Equation For Round Pipes
For Round Pipes:

where
  • h_f is the head loss due to friction [m]
  • l is the length of the pipe [m],
  • d is the hydraulic diameter of the pipe. For circular sections this equals the internal diameter of the pipe [m].
  • v is the velocity within the pipe [m/s^2]
  • g is the acceleration due to gravity
  • f is the coefficient of friction.

This equation can be expressed in terms of the quantity flowing per second.
h_f=\frac{4flv^2}{2g\;d}=\frac{4fl}{2g\;d}\times \left ( \frac{4Q}{\pi d^2} \right )^2
Darcy's Equation For Non Circular Pipes
where \mu=Wetted area / Wetted perimeter
The Chezy Equation
where
  • i is \frac{h_f}{l}
  • m is \frac{A}{P} or wetted area or wetted perimeter

and
C=\sqrt{\frac{2g}{f}}
Reynolds Number
R_N=\frac{Density \times\,Velocity\times\,Diam. \,of\,pipe}{Viscosity}

Which equals 2,300 at the point where the flow changes from Laminar to Turbulent. This is known as the Critical Velocity

If Ln hf is plotted against Ln v

22109/img_diag_17.png
+

The critical velocity is the velocity at which the change over from laminar to turbulent flow takes place.

Consider a circular pipe running full

V = C\:\sqrt[]{m\:i}
\therefore\;\;\;V^2\;=\:\frac{2g}{f}\:\:\frac{A}{\rho}\:\;\frac{h}{l}
\therefore\;\;\;h\:= f\;\frac{P\:L}{A}\:.\:\frac{V^2}{2g}
\therefore\;\;\;h = f\;\frac{\pi\,d\,l}{\pi\,\displaystyle\frac{d^2}{4}}\;.\;\frac{V^2}{2g}

Which can be written as The DARCY EQUATION

i.e. Head lost
= 4\:f\frac{l\:V^2}{2\:d\,g}

Note: f in the Darcy formula is not an empirical coefficient

For Laminar flow
f\:=\:\frac{16}{R}
The change over from laminar flow to turbulent flow occurs when R = 2300 and is independent of whether the pipe walls are smooth or rough.

Laminar Flow

When the flow is laminar it is possible to use the following equation to find the head lost.

The Poiseuille equation states that for a round pipe the head lost due to friction is given by:

h_F=\frac{32\times \mu\;l\;v}{\rho\;g\;d^2}

22109/img_0001_16.png
+
Flow occurs because the force across AB is more than that on AB and hence it is possible to write down the following equation.

(P_1-P_2)\times \pi\; r^2=-\mu\;\frac{dv}{dr}\times 2\pi r l

Integrating
-\left [ v \right ]_{v_r}^{v_0}=\left ( \frac{P_1-P_2}{l} \right )\left [ \frac{r^2}{4\mu} \right ]_{r_0}^{r_1}
\therefore \;\;\;\;\;v_r=v_0-\left ( \frac{P_1-P_2}{l} \right )\frac{r^2}{4\mu}

But when
$r$= \frac{d}{2}\;\;\v=\0

\therefore \;\;\;\;\;v_0=\left ( \frac{P_1-P_2}{l\succ } \right )\frac{d^2}{16\mu}
\therefore \;\;\;\;\;v_r=\left ( \frac{P_1-P_2}{l\succ } \right )\frac{d^2}{16\mu}-\left ( \frac{p_1-p_2}{l} \right )\frac{r^2}{4\mu}
Thus:
v_r=\frac{p_1-p_2}{l}\times \left ( \frac{d^2}{16\mu}-\frac{r^2}{4\mu} \right )

But:
Q=\int_{0}^{\frac{d}{2}}v_r\times 2\pi\; r\;dr

\therefore \;\;\;\;\;\;Q=\int_{0}^{\frac{d}{2}}\frac{P_1-P_2}{l}\times \left ( \frac{d^2}{16\mu}-\frac{r^2}{4\mu} \right )\times 2\pi\; r\; dr

=\frac{p_1-p_2}{l}\times \frac{2\pi}{4\mu}\int_{0}^{\frac{d}{2}}\left ( \frac{d^2}{4}-r^2 \right )r\;dr
=\frac{p_1-p_2}{l}\times \frac{2\pi}{4\mu}\left [ \frac{d^2}{4}.\frac{r^2}{2}-\frac{r^4}{4} \right ] _{0}^{\frac{d}{2}}
=\frac{p_1-p_2}{l}\times \frac{2\pi}{4\mu}\times \frac{1}{4}\left [ \frac{d^4}{8}-\frac{d^4}{16} \right ]
=\frac{p_1-p_2}{l}\times \frac{2\pi}{4\mu}\times \frac{1}{4}\times \frac{d^4}{16}

Now the mean velocity is given by:-
v_m=\frac{Q}{\pi\;d^2/4}
=\frac{p_1-p_2}{l}\times \frac{\pi\;d^4}{128\mu}\times \frac{4}{\pi\;d^2}
From equation (5)
The head lost due to friction is given by:

h_f=\left ( \frac{p_1-p_2}{\omega} \right )=\frac{p_1-p_2}{\rho\;g}

Substituting from equation (6) for P_1-P_2

h_f=\frac{32\;\mu\;l\;v}{\rho\;g\;d^2}

Example:

[imperial]
Example - Example 1
Problem
Water is siphoned out of a tank by means of a bent pipe ABC, 80 ft. long and 1 in. in diameter. A is below the water surface and 6 in . above the base sof the tank. AB is vertical and 30 ft long; BS is 50 ft. long with the discharge end C 5ft. below the base of the tank.



If the barometer is 34 ft. of water and the siphon action at B ceases when the absolute pressure is 6 ft. of water, find the depth of water in the tank when the siphon action ceases. F is 0.008 and the loss of head at entry to the pipe is:

=0.5\frac{v^2}{2g}

Where v is the velocity of water in the pipe.

22109/img_0007_9.png
+
Workings
Bernoulli's Equation is:

\frac{p_1}{w}+\frac{v_1^2}{2g}+Z_1=\frac{p_2}{w}+\frac{v_2^2}{2g}+Z_2+L
where L=Losses
Applying Bernoulli for the pipe length AB. Note that the pressures quoted in the question have been expressed in ft. of water and therefore:

Applying Bernoulli for the whole pipe length:

From equations (1) and (2)

41\displaystyle\frac{1}{2}-34=\frac{4\times 0.008\times 50\times v^2}{2g\times 1/12}

From which:
v=5.02\;ft./sec.
Substituting in equation (1)

39\displaystyle\frac{1}{2}+h=41\displaystyle\frac{1}{2}+\frac{5.02^2}{2\times 32.2}\left ( 1.5+\frac{120\times 0.008}{1/12} \right )
\therefore \;\;\;\;\;\;h=7.6\;ft.
Solution
The depth of water is h=7.6\;ft.

Please note that further worked examples on this topic will be found in the paper on "Hydraulic Gradients"