# Pressure drop in pipes and Velocity

1. Jul 30, 2008

### RagingSezz

For pipes head loss causes pressure drop. If there is lower pressure at the end of the pipe, is the velocity lower? What I'm trying to figure out is if the average velocity in the pipe is constant. And how does it change if you place an object inside the pipe? Does velocity only change with a change in area? Assume turbulent flow

2. Jul 30, 2008

### jaap de vries

head loss does not cause pressure drop, it IS pressure drop. By laws of continuity, the velocity must be equal if cross sectional area is constant. In other words, if the flow into a pipe is the same as the flow out the velocity must be equal (or at least integrated over the area). The losses typically come from he wall roughness of the pipe and this will cause a pressure drop. Since changes in area of the pipe can change the velocity and thus the static pressure, it typically makes more sense to keep track of the total pressure.

If you place an object in the pipe you effectively reduce the cross area and thus the fluid has to move faster through the remaining area to maintain continuity. Therefore the pressure will go down.

3. Jul 30, 2008

### Staff: Mentor

Note for clarity: if you place an object into a pipe, acting like an orifice, you will decrease the flow through the pipe vs if there was no orifice, unless you speed up the pump. This is a result of the larger pressure drop through the pipe.

4. Jul 31, 2008

### gmax137

Another note for clarity: The continuity applies to the mass flow rate (not the volumetric flow rate) so if you have heating or cooling taking place (like in a heat exchanger tube) the density changes and so the velocity can change even for a constant area.

Edit - and if you are looking at the flow of gas vs. liquid (ie, compressible vs incompressible) the density and hence velocity may be quite variable even without heating or cooling.

5. Jul 31, 2008

### stewartcs

Strictly speaking this is true, but since liquids are generally treated as incompressible substances, the density is assumed to be constant. Hence, the volumetric flow rate holds.

CS

6. Aug 2, 2008

### ank_gl

No, lower the pressure at exit, more the velocity.

pressure gradient is balanced by fluid friction, therefore velocity is constant(nothing is pushing fluid, so no acceleration)

by placing an object, you have increased friction(equivalent to saying that head is lost at a restriction), so lesser velocity.

yes, if the flow is incompressible & there is no pressure gradient.

turbulent model has nothing to do with continuity

7. Dec 17, 2008

### nhpn

I have a similar question too.

I have a horizontal pipe, with P1 on one end , and P2 on the other end ( imagine a fan blowing air through a pipe to atmosphere). The pipe has constant crossectional area. I find the Bernoulli equation is not applicable for this case?! I'll show why shortly
The Bernoulli equation is: Ro*V1^2 + g*h1+ P1= Ro*V2^2 + g*h2+P2
By continuity equation, A1=A2 -> V1=V2.
h1 =h2 by assumption that the pipe is place horizontally
Plug in the equation, the only thing left is pressure: P1=P2, which I think is weird, since if P1=P2 how does the fan blow air?
Some of you suggested here that's Ro may change, but I could not see the reason behind since incompressibility assumption in this case is pretty common.
Any idea?

8. Dec 17, 2008

### nhpn

I was thinking about friction accounting or pressure drop but Bernoulli law has nothing to do with friction does'nt it?

9. Dec 17, 2008

### Q_Goest

Right. Bernoulli's equation does not address unrecoverable pressure loss. Use the Darcy Weisbach equation for calculating pressure drop when the fluid is incompressible or when changes in density are less than about 20%. See if this reference helps.