Pressure drop in pipes and Velocity

AI Thread Summary
Head loss in pipes directly causes pressure drops, affecting fluid velocity. When an object is placed inside a pipe, it reduces the cross-sectional area, requiring the fluid to move faster to maintain continuity, which results in a pressure drop. The average velocity in a pipe can remain constant if the pressure gradient balances fluid friction, but introducing an obstruction increases friction and decreases velocity. Bernoulli's equation does not account for unrecoverable pressure losses due to friction, so the Darcy-Weisbach equation is recommended for calculating pressure drops in incompressible fluids. Understanding these principles is crucial for analyzing fluid dynamics in piping systems.
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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
 
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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.
 
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.
 
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.
 
gmax137 said:
Another note for clarity: The continuity applies to the mass flow rate (not the volumetric flow rate)...

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
 
RagingSezz said:
For pipes head loss causes pressure drop. If there is lower pressure at the end of the pipe, is the velocity lower?

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

What I'm trying to figure out is if the average velocity in the pipe is constant
pressure gradient is balanced by fluid friction, therefore velocity is constant(nothing is pushing fluid, so no acceleration)

And how does it change if you place an object inside the pipe?
by placing an object, you have increased friction(equivalent to saying that head is lost at a restriction), so lesser velocity.

Does velocity only change with a change in area? Assume turbulent flow
yes, if the flow is incompressible & there is no pressure gradient.

turbulent model has nothing to do with continuity
 
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?
 
I was thinking about friction accounting or pressure drop but Bernoulli law has nothing to do with friction doesn't it?
 
nhpn said:
I was thinking about friction accounting or pressure drop but Bernoulli law has nothing to do with friction doesn't it?
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 https://www.physicsforums.com/attachment.php?attachmentid=10471&d=1184948043" helps.
 
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