
#1
Jun2012, 05:51 PM

P: 5

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. 



#2
Jun2012, 06:16 PM

Sci Advisor
HW Helper
PF Gold
P: 2,878

Flow through a pipe can be determined using the DarcyWeisbach 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 crosssectional 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. 



#3
Jun2012, 06:24 PM

P: 5

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. 



#4
Jun2012, 06:49 PM

P: 1,438

Relationship between Flow Rate and Pressure
The better approach would be to use the HagenPoiseuille 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 NavierStokes 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 DarcyWeisbach equation and other correction factors for the head loss in elbows and bends. 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. 



#5
Jun2012, 06:59 PM

P: 5

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.




#6
Jun2012, 07:05 PM

P: 1,438

If you just assume the flow satisfies the requirements for HagenPoiseuille 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 HagenPoiseuille flow is equivalent to the DarcyWeisbach 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. 



#7
Jun2012, 07:20 PM

P: 5

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?




#8
Jun2012, 08:55 PM

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PF Gold
P: 2,878

P = rho*g*h http://www.physicsforums.com/showthread.php?t=179830 See also: http://en.wikipedia.org/wiki/Hagen%E...uille_equation 



#9
Jun2012, 09:09 PM

P: 5

This equation seems to be backwards. Wouldn't it be P = C Q^2? 


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