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Change in Pipe Section

Pressure head loss due to change in pipe size or shape


The sudden changes or the introduction of bends or valves disturbs the flow pattern along the pipe. The result is an increase in turbulence and this increases the frictional losses.

Sudden Enlargement

h_l\:=  \frac{(v_1\:-\:v_2)^2}{2g}

Note: The pressure just inside the larger pipe is the same as that in the small pipe!

Momentum at section (1) = Mass \inline \times Velocity = \inline \frac{w}{g}\;a_1\,v_1\times \:v_1

Momentum at section (2) = \inline \frac{w}{g}\:a_2\:v_2\times \;v_2

Drop in momentum

= \frac{w}{g}\:a_1\:v_1\,(v_1\:-\:v_2)

for this to happen there must be a force opposing motion

The magnitude of the force = \inline p_2\,a_2\:-\:p_1\,a_1\:-p_1(a_2\:-\:a_1) = a_2\:(p_2\:-\:p_1)

\therefore\;\;\;\frac{p_2\:-\:p_1}{w} = \frac{v_2(v_1\:-\:v_2)}{g}

Applying Bernoulli:

\frac{p_1}{w}\:+\:\frac{v_1^2}{2g}\:= \frac{p_2}{w}\:+\:\frac{v_2^2}{2g}\:+\:h_l
\therefore\;\;\;h_l = \frac{v_1^2\:-\:v_2^2}{2g}\:-\:\frac{p_2\,-\,p_1}{w}
\Rightarrow h_l= \frac{v_1^2\:-\:v_2^2}{2g}\:-\:\frac{v_2(v_1\:-\,v_2)}{g}
\therefore\;\;\;h_l\:=  \frac{(v_1\:-\:v_2)^2}{2g}

Sudden Contraction


The coefficient of Contraction is defined as :

where aC is the area of the choke and a is the area of the pipe
The head lost is given by:

k is a constant and in this case is approximately 0.5. The value of \inline k varies with the application and a selection are shown in the table at the end of this section.

Loss of head, hL is given by:




\therefore\;\;\;h_l = \frac{1}{2g}\;\left( \frac{v}{c_0}\:-\:v \right)^2

=\:\left( \frac{1}{c_c}\:-1 \right)^2\;\frac{v^2}{2g}=k\times \frac{v^2}{2g}

Head Loss At An Obstruction (e.g. A Valve)

The general equation for the head loss due to an obstruction, including bends, elbows, gate values, etc, is

  • k is the constant associated with the obstruction
  • v is velocity

Some typical values for k are:

Fitting or Obstruction Value for K
Standard 45 degree Elbow 0.5
Standard 90 degree Elbow 1.0
Globe Value - fully open 10.0
Gate Valve - fully open 0.2

\therefore \;\;\;\;\;h_L=\frac{v^2}{2g}\left ( \frac{A}{A-a}-1 \right )^2=\frac{v^2}{2g}\left ( \frac{1}{C_C} -1\right )
=k\times \frac{v^2}{2g}