A couple questions regarding pressure drop

  • Thread starter kosig
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  • #1
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Main Question or Discussion Point

First, I am very confused as to the difference, if any, between head loss and pressure drop/loss. Can someone help me?

Also, I know that the Darcy-Weisbach equation for head loss due to friction is hf=fL/D*V2/2g

and you can plug that in to get an equation for delta p that is
dp=fL/D*pV2/2

And this is dimensionally correct. My company says that for our applications all flow will be turbulent Re>4000 and the equation for the friction factor can be approximated to
f=(.0337*v.25)/(V.25*d.25) Which Im assuming is from the Colebrook equation.

So my problem is, when I plug this into the equation for delta p I get an equation which is not dimensionally correct. Why? And what does it mean?
 

Answers and Replies

  • #2
minger
Science Advisor
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They are often times used somewhat interchangably. Head typically has units of length, while pressure has the typical force/area units. The terminology probably comes from the Bernoulli Equation, we can be written such that each term has units of pressure
[tex]
p + \rho\frac{V^2}{2} + \rho g z = C
[/tex]
or of units of length:
[tex]
\frac{p}{\rho g} + \frac{V^2}{2g} + z = C
[/tex]
You'll find that often times in pipe flow or industries such as that, you'll find pumps and suchs in terms of head.

As far as what they physically mean, they are essentially the same thing, one divided by specific weight.
 
  • #3
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Alright that makes sense. What effect does a change in height have on pressure drop? For instance, I have a tube that is carrying a fluid and the tube starts at a pump and connects at a filter below. The tube has a large radius bend.
 
  • #4
what is the diffrance between pressure drop and head loss,exactly?thanhs
 

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