A couple questions regarding pressure drop

Click For Summary
SUMMARY

The discussion clarifies the distinction between head loss and pressure drop in fluid dynamics. Head loss, represented in units of length, is derived from the Darcy-Weisbach equation, while pressure drop is expressed in force per area units. The equation for pressure drop, dp = fL/D * pV^2/2, is dimensionally correct when using the appropriate friction factor, which in turbulent flow can be approximated by f = (0.0337 * V^0.25) / (V^0.25 * d^0.25), likely derived from the Colebrook equation. The discussion emphasizes that both terms are often used interchangeably, though they represent different physical concepts.

PREREQUISITES
  • Understanding of the Darcy-Weisbach equation for head loss
  • Familiarity with fluid dynamics concepts, particularly turbulent flow
  • Knowledge of the Bernoulli Equation and its implications
  • Basic grasp of dimensional analysis in physics
NEXT STEPS
  • Research the Colebrook equation and its application in calculating friction factors
  • Explore the implications of height changes on pressure drop in fluid systems
  • Study the relationship between head loss and pressure drop in various fluid flow scenarios
  • Learn about the practical applications of head loss and pressure drop in pump and piping system design
USEFUL FOR

Engineers, fluid dynamics researchers, and professionals involved in the design and analysis of piping systems and fluid transport applications.

kosig
Messages
9
Reaction score
0
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 into 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 I am 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?
 
Engineering news on Phys.org
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
<br /> p + \rho\frac{V^2}{2} + \rho g z = C<br />
or of units of length:
<br /> \frac{p}{\rho g} + \frac{V^2}{2g} + z = C<br />
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.
 
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.
 
what is the diffrance between pressure drop and head loss,exactly?thanhs
 

Similar threads

Replies
1
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 29 ·
Replies
29
Views
9K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 20 ·
Replies
20
Views
11K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
11
Views
11K