Flow calculation

[777gne]

The above diagram represents 2 pipe circuits, the header in bold and a utility circuit (thin blue line).

The pressure at points P1 and P2 is known, and the header flow rate is known.

The pressure at P3 is assumed to be equal to P1 (close enough to call equal).


Can the flow rate through the utility circuit be calculated? The piping configuration between P3 and P4 is defined and a pressure drop can be determined easily enough for any given flow rate, however the flow rate depends on the differential between P1 and P2, and the difference of P4 and P2 (correct?).

 

[IRstuff]

Student posting is not allowed.

TTFN

FAQ731-376

 

[BigInch]

What happened to the image?

http://virtualpipeline.spaces.msn.com

 

[Artisi]

yes it can, although a complex calculation if you want accuracy.

 

[777gne]

This isn't a student post. This is an existing chiller loop where a small tubular hxx is tapped off the header without a booster pump. I'm trying to get an idea of how much glycol water is moving through the line due to the differential pressure. I didn't design the thing but I have to deal with it now.

Artisi, any input that you can provide would be great. If I can get within 10% of the actual flow that would be excellent.


p.s.

I still see the image above, I'm just using a free hosting service so I don't if it works for everyone or not.

 

[BigInch]

Since you know the pressure at P1 and P2 and the flow through the header between P3 and P4, its not too complex as there are a sufficient number of known variables to simply solve for the flow in the small line.

You just need an equation of flow written in the form of
Q = (P2-P1)^n * K
or
where K is a function of ID and Pipe roughness and Pipe length, all known values (I presume).

http://virtualpipeline.spaces.msn.com

 

[katmar]

If you know the flowrate through the header, and you can measure the amount of piping between P4 and P2 then it is simple to calculate the pressure drop from P4 to P2. You know the pressure at P2, therefore you know the pressure at P4.

You now have the pressure differential from P3 to P4, and you have the geometry of the utility line. Calculating the flowrate is simply an iterative process of assuming a flowrate and calculating the pressure drop until the calculated pressure drop matches the actual.

If the flow through the utility line is a significant fraction of the main flow you may want to go back to step one and re-calculate the pressure drop from P4 to P2 with the combined flow and repeat the above process.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[BigInch]

If P2 and P4 can be assumed to be equal, as are P1 and P3, just calculate flow as I said before.

If you can't make P4 = P2, then you must somehow proportion the unknown flows using an iteration as has already been suggested.

You say you know the flowrate in the header, but don't say at what point that flow is known, so its not easy to tell you exactly how to proceed. Is it known between P1 and P3, or between P3 and P4?

If you know the flow between P1 and P3, then that flow must be proportioned between utility pipe P3-P4 and header pipe P3-P4.
-----------------------------------------------------
If you know the flow in header pipe P3-P4, and apparently you cannot make P4 equal to P2, then you must find the flow in utility pipe P3-P4 by iterating two flow equations.

Equation 1 will be the flow equation for Q utility pipe given the pressure drop of P3 to P4.

Equation 2 will be the flow equation for the header pipe P4-P2, with a flow equal to the sum of the known header flow (Q)between P3-P4 + the unknown flow in the utility pipe.

Solve Equation 1 for P4 using a guess for that flow, then
Solve Equation 2 for P4 using the known flow - the guess.
If both P4 values are not equal, then change the guess a little and see if the difference in the P4 values gets smaller or larger and adjust your guess accordingly.

http://virtualpipeline.spaces.msn.com

 

[vzeos]

777gne

This is a network problem and ordinarily would be handled by the simultaneous solution of the two pipe loop equations as has been pointed out. However, you say that you know piping configuration between P3 and P4 and the pressure drop between P3 and P4 can be determined easily enough. I see no reason why you can’t calculate the flow between P3 and P4 by using the formula below. But be careful about making any other assumptions or drawing conclusions about the header flow. The system still must obey Kirchhoff’s laws: 1) the pressure at any node is unique, and 2) the flow going into any node is equal to the flow leaving the node (steady state).


Q= C x [√]((dP / (Ro x L)) x (D ^2.5 ) x 1/[√]f

C = 68.04138 for Darcy
C = 34.02069 for Fanning
f = friction factor, Darcy or Fanning
Q = gpm
L = feet
D = inches
dP = psid
Ro = lb/cu ft

 

[trashcanman]

Why not cut to the chase and strap a magnetic flow meter on the line?

 

[Artisi]

" trashcanman (Mechanical) 5 Oct 07 4:28
Why not cut to the chase and strap a magnetic flow meter on the line? "

I just came on the offer the same suggestion.

As this is an operating system - I can only agree this is the only true and accurate method to establish the flow rate/s

 

[Matador]

Yes it can be calculated, just visualize an elecrical circuit in parallel. The voltage drop across the branches is the same, in this case the pressure drop acroos each branch will be the same.