Relates pressure drop $\begin{array}{l}\Delta p = p_{in} - p_{out}\end{array}$ on the choke with the flowrate through the choke $\begin{array}{l}q\end{array}$ arising from fluid friction with choke elements (ISO5167):

(1)	$\begin{array}{l}\displaystyle \Delta p = p_{in} - p_{out} = \frac{ \rho \cdot (1- \beta^4)}{0.125 \, \pi^2 \, d^4 \, C_d^2 \, \epsilon^2} \cdot q^2\end{array}$

where

$\begin{array}{l}\rho\end{array}$	fluid density
$\begin{array}{l}d\end{array}$	orifice diameter
$\begin{array}{l}D\end{array}$	pipe diameter
$\begin{array}{l}\beta = \frac{d}{D}\end{array}$	orifice narrowing ratio
$\begin{array}{l}C_d\end{array}$	discharge coefficient
$\begin{array}{l}\epsilon\end{array}$	expansion factor

Derivation

Assume steady-state, incompressible, inviscid, laminar flow in a horizontal pipe (no change in elevation) with negligible frictional losses with incharge fluid velocity $\begin{array}{l}v_{in}\end{array}$ and discharge velocity $\begin{array}{l}v_{out}\end{array}$ at the orifice exit.

The mass conservation (equivalent to continuity equation):

(2)	$\begin{array}{l}\displaystyle \rho \, q = \rho \, v_{in} \, A_{in} = \rho \, v_{out} \, A_{out} \Leftrightarrow v_{in} = \frac{q}{A_{in}}, \, v_{in} = \frac{q}{A_{out}}\end{array}$

where

(3)	$\begin{array}{l}\displaystyle A_{in} = \frac{\pi \, D^2}{4} \, A_{out} = \frac{\pi \, d^2}{4}\end{array}$

Bernoulli's equation reduces to an equation relating the conservation of energy between two points on the same streamline:

(4)	$\begin{array}{l}\displaystyle p_{in} + \frac{1}{2} \rho v^2_{in} = p_{out} + \frac{1}{2} \rho v^2_{out}\end{array}$

(5)	$\begin{array}{l}\displaystyle \Delta p = p_{in} - p_{out} = \frac{1}{2} \rho v^2_{out} - \frac{1}{2} \rho \, v^2_{in} = \frac{1}{2} \rho \, v^2_{out} \cdot \left[ 1 - \frac{v^2_{in}}{v^2_{out}} \right]\end{array}$

(6)

$\begin{array}{l}\displaystyle \Delta p = \frac{1}{2} \rho \, \left( \frac{q}{A_{in}} \right)^2 \cdot \left[ 1 - \frac{A^2_{out}}{A^2_{in}} \right] = \frac{\rho \, q^2}{2 \, A^2_{in}} \cdot \left[ 1 - \frac{d^4}{D^4} \right] = \frac{\rho \, q^2}{2 \, A^2_{in}} \cdot \left[ 1 - \beta^4 \right] = \frac{\rho \, q^2}{2 \, (\frac{\pi \, d^2}{4})^2} \cdot \left[ 1 - \beta^4 \right] = \frac{8 \, \rho \, q^2}{\pi^2 \, d^4 } \cdot \left[ 1 - \beta^4 \right]\end{array}$

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Pipeline Choke @model

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