Relationship between Flow Rate and Pressure


by Katalyze
Tags: flow, pressure, rate, relationship
Katalyze
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#1
Jun20-12, 05:51 PM
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Hi,

I can't go out and measure this system up atm. But I have 36m^3/s flow rate at 400kpa. How would I go about predicting the flow rate at 500, 600, 700 and 800 kpa through the same pipe? I think it's a linear relationship but want to be sure. Thanks.
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Q_Goest
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Jun20-12, 06:16 PM
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Flow through a pipe can be determined using the Darcy-Weisbach equation or one of the other equations for irreversible pressure drop. These equations are generally restricted to small changes in density, so if you have a gas, the density change needs to be less than about 20% for the equations to be accurate. For density changes in excess of that, we generally break up the pipe into sections and recalculate properties as the pressure changes.

The DW equation is:

where
hf is the head loss due to friction (SI units: m);
L is the length of the pipe (m);
D is the hydraulic diameter of the pipe (for a pipe of circular section, this equals the internal diameter of the pipe) (m);
V is the average velocity of the fluid flow, equal to the volumetric flow rate per unit cross-sectional wetted area (m/s);
g is the local acceleration due to gravity (m/s2);
fD is a dimensionless coefficient called the Darcy friction factor.[citation needed] It can be found from a Moody diagram or more precisely by solving the Colebrook equation. Do not confuse this with the Fanning Friction factor, f.
Ref: http://en.wikipedia.org/wiki/Darcy%E...sbach_equation

Note that pressure drop (ie: head loss) is a function of the velocity squared, so doubling the flow rate doubles velocity which quadruples pressure drop. However, the friction factor f, also changes depending on Reynolds number, which may or may not change significantly as flow rate changes. But as a general rule of thumb, pressure drop changes as a function of the square of flow rate, assuming the change in density is relatively small. So this holds well for water, but less well for compressible gasses.
Katalyze
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#3
Jun20-12, 06:24 PM
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Thanks for the reply. I forgot to mention the application is water. Furthermore the pipe is the same so some factors should be cancelled out.
Furthermore this equation does not seem to relate Pressure to Flowrate. Only Flow rate to head loss. Am I to assume that 400kpa is effectively 40m head loss?
However is there a way to equate scenario one (400kpa) to scenario 2 (500kpa)? Because the DW equation does not seem to be a mass balance. Furthermore if I'm trying to evaluate the velocity using the pressure, I can't really use Re number because that requires flow rate to be known.

boneh3ad
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Jun20-12, 06:49 PM
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Relationship between Flow Rate and Pressure


The better approach would be to use the Hagen-Poiseuille relation, which describes flow rate in a pipe based on the pipe size, the fluid properties and the pressure drop. I will skip the derivation, but it can be easily derived from the Navier-Stokes equations, meaning it is a momentum balance. The relationship is:

[tex]\Delta P = \frac{128\mu L Q}{\pi d^4}[/tex]

where:
[itex]\Delta P[/itex] is the pressure drop [[itex]\mathrm{Pa}[/itex]]
[itex]\mu[/itex] is the fluid viscosity (easy to look up for water) [[itex]\mathrm{Pa}\cdot\mathrm{s}[/itex]]
[itex]L[/itex] is the length of the pipe over which the pressure drop takes place [[itex]\mathrm{m}[/itex]]
[itex]Q[/itex] is the volumetric flow rate [[itex]\mathrm{m}^3/s[/itex]]
[itex]d[/itex] is the pipe diameter [[itex]m[/itex]]

This also assumes the following:
Steady state
Irrotational
Axisymmetry

These assumptions are valid assuming you are sufficiently far from the pipe entrance or the pipe is sufficiently lengthy that the error induced by the entrance is minimal and as long as the pipe is straight. If your pipe has a bend in it, then you will not have accounted for that.

If you have a more complicated system of pipes with bends and elbows in it, then you are probably better off using the Bernoulli equation modified with the Darcy-Weisbach equation and other correction factors for the head loss in elbows and bends.

Quote Quote by Q_Goest
Note that pressure drop (ie: head loss) is a function of the velocity squared, so doubling the flow rate doubles velocity which quadruples pressure drop. However, the friction factor f, also changes depending on Reynolds number, which may or may not change significantly as flow rate changes. But as a general rule of thumb, pressure drop changes as a function of the square of flow rate, assuming the change in density is relatively small. So this holds well for water, but less well for compressible gasses.
The head loss is a function of velocity squared and the friction factor. For laminar flow, the friction factor varies linearly with 1/Re, so the head loss (or pressure drop) varies linearly with velocity, just as the OP surmised. For turbulent flow, the relationship is more complicated. As you increase the Reynolds number, you would end up approximating that relationship of pressure drop being proportional to the square of the flow rate.

It would be fairly easy to Taylor expand something like the Haaland equation (approximation of the Colebrook equation) about Re and determine the actual relationship, though I would imagine this is beyond the OP's concern. You would just need to expand it about the Reynolds number you expect to see in your situation.

EDIT: I remember for certain that for very high Reynolds number you approach the velocity squared relationship.
Katalyze
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#5
Jun20-12, 06:59 PM
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The issue is that I don't know any pipe details atm so I can't put in L, d etc. I need to have a mass balance so I can use ratios to define a proportional relationship.
boneh3ad
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Jun20-12, 07:05 PM
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Quote Quote by Katalyze View Post
The issue is that I don't know any pipe details atm so I can't put in L, d etc. I need to have a mass balance so I can use ratios to define a proportional relationship.
Without knowing those you are kind of dead in the water since you can't even determine if the relationship is linear quadratic or somewhere in between without that.

If you just assume the flow satisfies the requirements for Hagen-Poiseuille flow (most importantly that it is laminar) then you can easily cancel out the [itex]L[/itex] and [itex]d[/itex] terms and get your answer without knowing them. Keep in mind that Hagen-Poiseuille flow is equivalent to the Darcy-Weisbach equation assuming laminar flow where [itex]f_D=64/\mathrm{Re}[/itex].

Your flow may or may not be laminar though, and if it isn't, then your case gets hairier.
Katalyze
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Jun20-12, 07:20 PM
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So assuming laminar flow. fD=64/Re does not seem to relate pressure to flow rate in any case. How would I incorporate pressure and flow rate into that equation?
Q_Goest
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Jun20-12, 08:55 PM
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Quote Quote by Katalyze View Post
Furthermore this equation does not seem to relate Pressure to Flowrate. Only Flow rate to head loss.
Head loss is the same as pressure drop. Head loss is simply the pressure equal to a column of the fluid h high. Relate head loss to pressure drop using
P = rho*g*h
Because the DW equation does not seem to be a mass balance.
DW equation is not a mass balance equation and I don't understand why you want a mass balance equation. DW is a steady state equation where fluid is not 'stored' in any section of pipe, it just flows through. The DW equation therefore determines pressure loss (ie: head loss) from a constant flow rate. Note that flow rate is not explicitly written into the equation, velocity, friction factor and Reynolds number is. I've attempted to provide you an interpretation to help you with the calculations. If you'd really like to understand the specifics of how to determine irreversible pressure loss in piping systems, I'd suggest starting with the manual I posted online here:
http://www.physicsforums.com/showthread.php?t=179830
Furthermore if I'm trying to evaluate the velocity using the pressure, I can't really use Re number because that requires flow rate to be known.
Just understand that to double flow rate, you need to double velocity which, when squared, will quadruple the head loss (pressure loss). As I'd mentioned above, this isn't a perfect relationship since friction factor is a function of Re, but it's a reasonable rule of thumb as long as you understand the limitations.
Quote Quote by boneh3ad View Post
The better approach would be to use the Hagen-Poiseuille relation
The HP equation assumes a "long" pipe such that flow is laminar which is not generally the case for the vast majority of industrial applications. It IS however, applicable to very long pipelines such as cross country lines for natural gas where it's used quite a bit. It isn't used as often in industry since the DW equation is much more general and can be applied equally well to both long and short pipelines.

See also: http://en.wikipedia.org/wiki/Hagen%E...uille_equation
Quote Quote by boneh3ad View Post
The head loss is a function of velocity squared and the friction factor. For laminar flow, the friction factor varies linearly with 1/Re, so the head loss (or pressure drop) varies linearly with velocity, just as the OP surmised. For turbulent flow, the relationship is more complicated. As you increase the Reynolds number, you would end up approximating that relationship of pressure drop being proportional to the square of the flow rate.

EDIT: I remember for certain that for very high Reynolds number you approach the velocity squared relationship.
Correct. I tried to emphasize that in my previous post. The relationship Q = C * P2 is not particularly accurate but is reasonable for turbulent flow which covers most of the typical piping systems.
Katalyze
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#9
Jun20-12, 09:09 PM
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Quote Quote by Q_Goest View Post
Correct. I tried to emphasize that in my previous post. The relationship Q = C * P2 is not particularly accurate but is reasonable for turbulent flow which covers most of the typical piping systems.
Thank you. I assumed it was proportional to the flow rate. However I wasn't sure if it was linear or quadratic. It seems I'll have to use a more mathematical approach using proportion instead of recalculating the friction factors etc.

This equation seems to be backwards. Wouldn't it be P = C Q^2?
Q_Goest
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Jun21-12, 06:21 AM
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Quote Quote by Katalyze View Post
This equation seems to be backwards. Wouldn't it be P = C Q^2?
Sorry, my bad. You're right. I was in a hurry.


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