- #1

nlis12

- 32

- 3

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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- Thread starter nlis12
- Start date

- #1

nlis12

- 32

- 3

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!

- #2

dRic2

Gold Member

- 875

- 233

##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! :(

- #3

nlis12

- 32

- 3

##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!

- #4

dRic2

Gold Member

- 875

- 233

- #5

dRic2

Gold Member

- 875

- 233

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.

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