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Calculating required pressure to maintain flowrate

by bugatti79
Tags: flowrate, maintain, pressure, required
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bugatti79
#1
Nov7-12, 01:22 PM
P: 660
Folks,

I seek just a very rudimentary idea of what pressure is required to pump oil through a very small hole of 1mm diameter to maintain a flow rate of 0.5L/min. See attached.

I would imagine that under gravity that the flow rate would be very low because of the viscosity of the oil thus to maintain a this flow rate significant pressure would be required to push it through. Lets assume the oil container bore is 10mm diameter.

Any thoughts?

Regards
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darkside00
#2
Nov7-12, 03:20 PM
P: 83
Can treat it like a restriction orifice. It will consider the discharge coefficients depending on thickness, viscosity, temperatures, diameters, reynolds #, etc. to give more accurate #'s

The head pressure required might make your container quite high?
bugatti79
#3
Nov9-12, 12:36 PM
P: 660
Quote Quote by darkside00 View Post
Can treat it like a restriction orifice. It will consider the discharge coefficients depending on thickness, viscosity, temperatures, diameters, reynolds #, etc. to give more accurate #'s

The head pressure required might make your container quite high?
Well I have come across this link

http://en.wikipedia.org/wiki/Orifice...ugh_an_orifice

but it is only for inviscid flow, I am interested in viscous flow like oil. I guess a simple hand calculation is not possible...?
Basically I would like to argue the point that a very large pressure pump would be required to pump oil through a 1mm diameter at a rateof 500ml/min....

Vadar2012
#4
Nov11-12, 11:36 PM
P: 208
Calculating required pressure to maintain flowrate

I think you're confusing what is involved in the viscous and invicsous flow assumptions. Any value of viscocity is going to make a contribution to the shear forces at the wall due to viscosity of the fluid. Making the assumption of inviscid flow depends on what area of the flow you are interested in. If you're after a value inside the boundary layer, close to the wall, then you might need equations to describe viscid flow. Otherwise their contributions to the flow somewhere else is usually negligible in problems like this one.

It really depends on the Reynolds number of the flow. For example, Air at high speeds is going to have more of an effect on the flow than oil at low speeds, even though oil has a much higher value than air.
bugatti79
#5
Nov12-12, 01:18 PM
P: 660
Quote Quote by bugatti79 View Post
Folks,

I seek just a very rudimentary idea of what pressure is required to pump oil through a very small hole of 1mm diameter to maintain a flow rate of 0.5L/min. See attached.

I would imagine that under gravity that the flow rate would be very low because of the viscosity of the oil thus to maintain a this flow rate significant pressure would be required to push it through. Lets assume the oil container bore is 10mm diameter.

Any thoughts?

Regards
Quote Quote by Vadar2012 View Post
I think you're confusing what is involved in the viscous and invicsous flow assumptions. Any value of viscocity is going to make a contribution to the shear forces at the wall due to viscosity of the fluid. Making the assumption of inviscid flow depends on what area of the flow you are interested in. If you're after a value inside the boundary layer, close to the wall, then you might need equations to describe viscid flow. Otherwise their contributions to the flow somewhere else is usually negligible in problems like this one.

It really depends on the Reynolds number of the flow. For example, Air at high speeds is going to have more of an effect on the flow than oil at low speeds, even though oil has a much higher value than air.
Well I would be interested in low speeds,so it looks like a simple hand calculation is not possible?
darkside00
#6
Nov12-12, 10:30 PM
P: 83
For a simple/rough and incompressible calculation using bernoullis principle:

http://en.wikipedia.org/wiki/Bernoulli%27s_principle

E.g. (v1^2)/2 + g*z1 + P1/p = (v2^2)/2 + g*z2 + P2/p

This obviously ignores the losses but gives an idea. Also, assuming your pouring to atmosphere and using conservation of energy to get the initial velocity (pVA)in=(pVA)out


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