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Old   March 6, 2014, 14:19
Default pressure drop [Urgent]
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Hi all,

I have a probleme in the comprehension of pressure drop,

I have two relationship but i haven't the explication and origin :

1) deltaP= lamda*(L/D)*(V^2/2)* Rho(density)


2) deltaP= - lamda*(L/D)*(V^2/2)

Please someone can explain
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Old   March 7, 2014, 17:48
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You mean pressure drop in pipes?
If so, then your first equation is the well known Darcy-Weisbach equation:–Weisbach_equation

But in the form of head loss and not pressure drop (as you stated), so it should be written as:

H=amda*(L/D)*(V^2/2)* Rho(density)

and it refers to head losses in straight circular pipes.
Note the following relationship also:

The second equation doesn't look right to me.
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Old   March 10, 2014, 09:51
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Thanks for your response,

I choose to calculate the head loss in a pipe but in presence of a turbine!

So i know that to calculate the head loss we use the relationship:
deltaP= lamda*(L/D)*(V^2/2)* Rho(density)

But in the resolution of an exercice in the solution I have this one : deltaP= - lamda*(L/D)*(V^2/2)

So I don't undestund the difference between this two relations !!
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