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Pressure Loss In A Horizontal Pipe.
#1
Posted 23 April 2016 - 07:40 AM
I would like to know when there is pressure loss when a fluid flows through horizontal circular pipe after displacing the length L. Will the velocity drop along with loss of pressure or remains constant
due to friction?
Case -1
If velocity doesn't drop along with pressure loss due to friction, why do friction cause only loss in pressure but not in velocity.
What is the basis for it?
Why is it happening in that way. Why can't pressure remains constant and velocity gets lost due to friction. Please explain the logic behind it.
Case -2
If the velocity drops along with pressure loss, how I compute drop in velocity for a particular loss of pressure.
Please shed some light and help me out. Thank you.
#2
Posted 23 April 2016 - 11:38 AM
I would like to know when there is pressure loss when a fluid flows through horizontal circular pipe after displacing the length L. Will the velocity drop along with loss of pressure or remains constant
due to friction?
Chalapathi,
At first, you should specify the state of the fluid...gas or liquid?
#3
Posted 23 April 2016 - 11:52 AM
#4
Posted 23 April 2016 - 12:17 PM
Chalapathi,
Then the case-1 would be happened; the velocity remains simply constant due to liquid incompressibility and no change in density. The pressure cannot remain constant because the liquid flouid is flowing...!
#5
Posted 23 April 2016 - 12:42 PM
Hi chalapathi,
1. As Mr. Fallah stated, for liquid velocity remains almost constant for a constant cross section of pipe. In simple, mass coming in the pipe is coming out at the same rate without any accumulation (i.e. mass flow rate is constant). For liquids, density is almost constant for constant temperature neglecting change in density due to pressure . Now you can calculate velocity v = mass flow rate (kg/sec) / density (kg/m3) / cross sectional area of pipe (m2). Hence for a constant mass flow rate & constant area, velocity remains almost constant for liquid.
2. You assumption for velocity and pressure drop together will never happen. Pressure and velocity are inversely proportional. If velocity drops, pressure will rise and wise versa in liquid flow.
#6
Posted 23 April 2016 - 12:57 PM
Thank you for your responses. The reason why I thought both will be lost is because, both pressure and velocity are the forms of mechanical energy delivered by the pump. In reality due to friction if energy is lost, the energy in the form of pressure and velocity should be affected.
That's only reason I felt so.
#7
Posted 23 April 2016 - 01:27 PM
Pressure is a driving force which pushes the liquid for the given pressure difference.
Velocity V = sqrt( 2 x (pin-pout-pdrop)/ density))
As Pin-Pout increases, velocity will increase and velocity will decrease if Pdrop increases. Hence for constant cross section, for fixed inlet and outlet pressure, flow will start increasing and there by increasing the drop due to friction etc. At a point, velocity and pressure drop will come to an equilibrium and remains constant until the resistance was unchanged.
#8
Posted 23 April 2016 - 06:54 PM
Hi ,
I found this document very informative.
Hope this helps
Breizh
#9
Posted 25 April 2016 - 04:14 AM
Hi Chalapathi,
Conceptually speaking,
1. Practically, Velocity and Pressure both drop as the fluid keeps flowing. In real cases, there are viscous forces of fluid acting on every layer of the fluid. So this causes one layer to pull the adjacent layer of fluid back, by virtue of attractive molecular forces between the two layers and also with the pipe internal boundary. This causes the pressure exerted by fluid and the velocity of the fluid also to decrease.
2. But the drop in velocity is continually compensated due to existing pressure difference in the fluid.
This means, the fluid's pressure energy, by virtue of its velocity gets lost, into compensating for any fluid velocity lost and any potential energy changes. This is what we call the Bernoulli's theorem which is nothing but a modified form of the Conservation of Energy equation.
3. This equation is written as
P1 + rho1 * (V1^2) / 2 + g*H1 = P2 + rho2 * (V2^2) / 2 + g*H2
Where P=pressure, V=velocity of fluid in that cross section, rho = density of fluid in that cross section, H=height of fluid cross section from reference line.
4. So, as explained, in the above equation, P reduces in order to compensate for any changes in v or H in general and vice versa.
5. But in a real, non-ideal case, when we have the frictional and viscous forces come into play, the velocity also has a possibility of reducing, due to fluid layers pulling adjacent fluid layers back by intermolecular forces.
6. So, the pressure reduces, because the pressure is the driving force by which the fluid is converting all pressure energy into velocity of the fluid which is what makes the fluid move.
And, The velocity also reduces due to presence of friction, but that reduction in velocity is usually not very high and almost negligible value depending on the pipe, material, flow parameters etc.
Regards,
Shantanu
Edited by shantanuk100, 25 April 2016 - 04:15 AM.
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