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Why is there a pressure loss with friction?

by Anony-mouse
Tags: friction, loss, pressure
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Anony-mouse
#1
Apr17-08, 06:07 AM
P: 61
If a fluid has reached steady flow, ie constant velocity, yet there is friction created by the shear stress of the wall, i dont understand how the pressure can drop. I tried using bernoulli to see how this would happen, but i cant see it. If the velocity doesnt decrease, i dont understand how the pressure can?
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Astronuc
#2
Apr17-08, 06:33 AM
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Bernoulli's equation is simply an energy balance equation, and in Bernoulli's pressure drop is considered. The pressure drop can be shown with the momentum balance equation.

Pressure is simply force per unit area. Fluids 'move' from high pressure to low pressure. The shear forces, imposed by the boundary or pipe wall, oppose fluid motion so the pressure must drop consequently. Force represents a change in momentum.

In fluid mechanics, one has the Navier-Stokes equations: continuity, momentum and energy, although some use Navier-Stokes equation in reference to the momentum equation.
Anony-mouse
#3
Apr17-08, 06:50 AM
P: 61
since pressure if force/area, and the friction is considered as the force, then if a larger force (friction) is applied to the fluid, would the pressure increase?

also could you please help explain how if the velocity gradient (du/dx) is less than 0, the pressure gradient (dp/dx) will be greater than 0 and vice versa.

Astronuc
#4
Apr17-08, 07:09 AM
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Why is there a pressure loss with friction?

The greater the friction (and viscosity) the greater the resistance (opposition) to motion. As friction increases, the pressure drop must increase, and more work will have to be done on a fluid for a given flow rate.

The sign on the differentials depends on the coordinate system (reference) or orientation used, e.g. is the coordinate system (in a pipe flow) measured from the center of the flow outward, or from the pipe wall inward.

What does du/dx < 0 mean? Simply as the distance/position increases, then u is decreasing. If du/dx increases, then u is increasing, i.e. the fluid is accelerating with x.

dP/dx < 0 means a pressure drop as x increases, or dP/dx > 0 means pressure increases with x. Then one has to look at what causes the Pressure to increase or decrease.
Anony-mouse
#5
Apr17-08, 09:01 AM
P: 61
thanks for the help, makes sense now :D
silk101
#6
Apr19-08, 12:38 PM
P: 2
Pressure will decrease slightly with friction. Take a centrifugal compressor assembly for example. After the compressor has compressed the air the air will slightly lose some of its pressure because it has to do work to bend round the diffuser. This pressure drop is not extremely high but still exists.
ank_gl
#7
Apr19-08, 01:22 PM
P: 735
pressure loss at bends is a minor loss and so are the losses at fittings etc. Friction loss is one of the major losses
makethings
#8
Apr19-08, 01:55 PM
P: 94
If the problem is laminar pipe flow, you can look at Poiseuille's law to find the pressure drop.
stewartcs
#9
Apr20-08, 10:47 PM
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Quote Quote by ank_gl View Post
pressure loss at bends is a minor loss and so are the losses at fittings etc. Friction loss is one of the major losses
Although they do classify pressure losses at bends and fittings as minor, they can contribute significantly to the total loss.

CS


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