# Pressure Drop Across a Change in Diameter

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## Main Question or Discussion Point

Greetings,

I have very little experience with fluid dynamics and I was wondering how I can calculate a pressure drop across an abrupt change in diameter of the piping used for water.

Any help is appreciated!

Thanks!

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dRic2
Gold Member
I looked my notes on fluid dynamics by my professor and I found this formula

$ΔH = \frac {(V_1-V_2)^2} {2g}$

Then $ΔP = \rho g ΔH$ where $\rho$ is the density of the fluid.

I'm sorry that I can't give you any reference, but my notes are in italian... and I don't know the name of this formula in english! :(

I looked my notes on fluid dynamics by my professor and I found this formula

$ΔH = \frac {(V_1-V_2)^2} {2g}$

Then $ΔP = \rho g ΔH$ where $\rho$ is the density of the fluid.

I'm sorry that I can't give you any reference, but my notes are in italian... and I don't know the name of this formula in english! :(
But I wonder what the H stands for?
I think your equation determines the pressure drop due to a change in potential energy.
Unfortunately, my system is perfectly horizontal, or I can assume Delta H is zero, so I have no losses due to potential energy changes. (I think)

Regards!

dRic2
Gold Member
dRic2
Gold Member
I think your equation determines the pressure drop due to a change in potential energy.
Unfortunately, my system is perfectly horizontal
No, H is a misleading letter but It doesn't refer to the heights necessarily. It's just that egineers like to work with the pressure drop as it was a lenght thus they divided it but $\rho$ and $g$(that are constant).

This is the simplest explanation I can think of.