pressure drop [Urgent]

stage81 · March 6, 2014, 13:19

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)

and;

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

Please someone can explain

lovecraft22 · March 7, 2014, 16:48

You mean pressure drop in pipes?
If so, then your first equation is the well known Darcy-Weisbach equation:
http://en.wikipedia.org/wiki/Darcy–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:
deltaP=H*rho*g

The second equation doesn't look right to me.

stage81 · March 10, 2014, 08:51

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 !!

March 6, 2014, 13:19	pressure drop [Urgent]	#1
stage81 Member Join Date: Sep 2010 Posts: 60 Rep Power: 15	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) and; 2) deltaP= - lamda(L/D)(V^2/2) Please someone can explain

March 7, 2014, 16:48		#2
lovecraft22 Senior Member lore Join Date: Mar 2010 Location: Italy Posts: 460 Rep Power: 18	You mean pressure drop in pipes? If so, then your first equation is the well known Darcy-Weisbach equation: http://en.wikipedia.org/wiki/Darcy–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: deltaP=Hrhog The second equation doesn't look right to me. stage81 likes this.

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March 10, 2014, 08:51		#3
stage81 Member Join Date: Sep 2010 Posts: 60 Rep Power: 15	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 !!