Pressure Drop Across a Change in Diameter

[nlis12]

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!

 

[dRic2]

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! :(

 

[nlis12]

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! :(

Thank you very much for your reply.
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]

ΔH is the pressure drop written as a lenght. Are you familiar with Bernoulli equation? https://en.wikipedia.org/wiki/Bernoulli's_principle scroll down until you find the paragraph about "total head"

 

[dRic2]

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 length thus they divided it but $\rho$ and $g$(that are constant).

This is the simplest explanation I can think of.