# 13.7.1.2: Acceleration Pressure Loss

The acceleration pressure loss can be estimated by
$-\left. \dfrac{dP}{dx} \right|_a = \dot{m} \, \dfrac{dU_m} {dx} \label{phase:eq:accPL} \tag{35}$
The acceleration pressure loss (can be positive or negative) results from change of density and the change of cross section. Equation (35) can be written as
$-\left. \dfrac{dP}{dx} \right|_a = \dot{m}\, \dfrac{d} {dx} \left( \cfrac{\dot{m}} {A\,\rho_m} \right) \label{phase:eq:accPLa} \tag{36}$
Or in an explicit way equation (36) becomes
$-\left. \dfrac{dP}{dx} \right|_a = {\dot{m}}^2 \left[ \overbrace{\dfrac{1}{A} \, \dfrac{d} {dx} \left( \dfrac{1} {\rho_m} \right)} ^{\text{pressure loss due to density change}} + \overbrace{\dfrac{1}{\rho_m\,A^2} \dfrac{dA} {dx}} ^{\text{pressure loss due to area change}} \right] \label{phase:eq:accPLae} \tag{37}$
There are several special cases. The first case where the cross section is constant, $$\left. dA \right/ dx = 0$$. In second case is where the mass flow rates of gas  and liquid is constant in which the derivative of $$X$$ is zero, $$\left. dX \right/ dx = 0$$. The third special case is for constant density of one phase only, $$\left. d\rho_L \right/ dx = 0$$. For the last point, the private case is where densities are constant for both phases.

### Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.