Compressible Flow Analysis

[RJB32482]

I just want to check my engineering analysis on a nitrogen system.
Nitrogen is stored in an accumulator tank at 100 psig. It exits the accumulator then splits off in 3 branches. Each branch is designed for a different flow rate (60 CFM, 25 CFM, and 120 CFM). Here is how I am going about the analysis:
Calculate pressure drop through all 3 lines at design flow. Take the highest value of the 3. Then calculate the pressure drop from the tank to the branch point at the flow of 205 CFM. Add the 2 pressure drops. If its below 100 psi, we should be able to get the necessary flow. If not, then we need more pressure at the tank.

Thanks.

 

[trashcanman]

It turns out that your assumptions are not quite correct. All 3 lines will have the same pressure drop, just different flow rates. For instance, if line 1 has a pressure drop (pd) of 10 psi and line 2 had a pd of just 5 psi, why would the fluid go into line 1 when it has an easier time in line 2? Answer: The flow will split up proportionally to balance out the pressure drop exactly equal. That is the whole basis for calculation of fire sprinkler flow rates. All heads flowing have the same pressure drop, just different flow rates.

 

[israelkk]

What is connected on the exit of each branch? Is it opened to the atmosphre?

 

[BigInch]

RJB, you are correct, provided that you are aware that the 120 CFM branch, the 60 CFM branch and the 25 CFM lines will not have the same end pressures. A backpressure control valve will have to hold the pressure drops in each branch to your calculated end pressure values.

If you do not hold the end pressures as calculated using your method, ie, all end pressures are equal, then the systems will flow as stated by trashcanman.

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[sailoday28]

Israelkk-has made a good point. If open to the atmosphere, is it possible for choked flow in any one of these pipes?

Regards

 

[BigInch]

Possible but highly unlikely.

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[israelkk]

If it is opened to the atmosphere it will definitly be choked (100/14.7 = 6.8 >>> ~2).

 

[RJB32482]

OK, so how would i go about finding the pressure needed in the tank to get the exact flow rates I need for all of the branches? If all of the lines will have the exact same pressure drop (100 psig) then would I just use the resistance factors (line, valves, etc) to find how much flow I will get from the branches at a certain length?

Thanks.

 

[RJB32482]

Will they all have the pressure drop that is the highest of the 3 at the certain flow rates?

Thanks.

 

[BigInch]

Israel, I meant open to atmos.

RJB,

In order to determine flowrate in a pipe or branching pipes in a network, you must know two of the following 3 items,

1.) upstream pressure,
2.) downstream pressure,
3.) flowrate

If you think you know all three, you don't (its overspecified) and you must relax one variable.

The branch end pressures really depend on what is downstream of the branches. If you have control valves, you can set them to control the backpressure according to the pressure drops you calculate for each branch using each branch's flowrate, just as you have proposed orginally. If the calculated end pressures are not being held by equipment, each branch will have the end pressure of the downstream equipment. If open to atmosphere, the system will flow as trashcanman has said and the pressure drops in each branch will not be as you have proposed in your original statement.

If you set the tank pressure and the flowrates of each branch, as you have proposed to do in the original post, then you must control the downstream pressure at each branch ending to guarantee you will get your proposed flowrates.

If you know tank pressure and all three branch ending pressures, you must calculate the appropriate flowrate in each branch that results in the known pressure drops over each branch in the system, using the facts that at the branch node, the sumation of flows in and out must be = 0 and all branch end pressures at that node are equal.

Tell us what's downstream and maybe we can figure out what you will get.

Going the Big Inch!

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[israelkk]

If it a tank then the pressure will decrease as the flow continues. Is the upstream pressure constant (somtimg continuously supply gas to the tank) or the tank has a definitive amount of gas and is discharged to the atmosphere?

 

[RJB32482]

Thanks.

Downstream of all the three piping outlets is basically atmospheric pressure. Let's just say for the three branches:
Branch 1: 35 feet of 2" ID pipe (60 CFM)
Branch 2: 40 feet of 3" ID Pipe (25 CFM)
Branch 3: 20 feet if 1"ID pipe (120 CFM)

Pipe from tank to branch point: 60 feet of 3"ID pipe

(I am really designing this, but this is just looking at the system).

So how would I find what pressure I need in the tank to satisfy all of these flow rates (at minimum). Higher flows won't hurt the process.

Can't I backcalculate all 3 branches and assume the highest upstream pressure is what is needed at the branch point? Then backcalculate from the branch to the tank to get tank pressure?

Thanks.

 

[BigInch]

What you've listed is not possible.
Your pressure at a common branch point for the 3 branches is equal. With atmospheric pressure at the outlets, the pressure drops of each pipe are equal.
With equal pressure drops in each branching segment, the smaller diameter pipes will tend to have less flow than the 3", and you have the 3" with the least flow, the 1" with the highest flow.

You will have the same pressure at the branch and the outlets of the branching segments. Therefore the pressure drop of each branch line is the same, but the diameters are different, therefore the flows will be different.

|======1"=======Po
tank =+Pb =====2"=====Po
|======3"=======Po

From tank to branch pressure drop = Qt * Kt * (Pt^2-Pb^2) where,
Qt = tank flow
Pt = tank pressure
Pb = pressure at the branch point
Kt = Press loss coefficient in tank line as a function dP

At branching point,
Q1" + Q2" + Q3" = Qt

where Q1"= flow in 1" diameter branch, etc.

In each branch i,
Qi = Ki * (Pb^2-Po^2), for i = 1,2,3

Po is known = 14.696 psia
You have 6 remaining unknowns, Qt, Q1, Q2, Q3, Pb, Pt

You will have to write the equations for gas pressure drops for the tank line and each branch, including compressibility effects if you wish. They are nonlinear in terms of P. To solve the nonlinear system will requires a nonlinear technique, such as Newton Raphson method.








Going the Big Inch!

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[Latexman]

I'll be glad to plug the above information into SiNet at work tomorrow. It should take just a few minutes.

Good luck,
Latexman

 

[dfish67]

If the flow rate is important why not choke the flow through orifices of known diameter? For choked flow the flow rate is only a function of area. It seems that you have plenty of pressure to play with so line sizing up to the orifices would not be critical.

 

[RJB32482]

OK Thanks to all for the help. But I am trying to find out at the minimum what pressure I need at the tank to get at least the flow I want in each line. So I'm figuring I can:
-calculate the pressure drop in the 3 branch lines at the design flows. The highest of the delta P's will be the minimum pressure I need at the branch. I can then calculate the other two branches using that maximum pressure drop to get the flows from those 2 other branches. So for example
Branch 1 delta P=10 psi
Branch 2 delta P=15 psi
Branch 3 delta P=20 psi
So I need at least 20 psi at the tiepoint. Calculate branch 1 and 2 at 20 psi drop to get new design flow rates.
-add these 3 flows up and get what flow I need from the tank to the branch point to satisfy all branches. Calculate deltaP from the tank to the branh at that flow. Say thats 40 psi.

So I need at least 40+20=60 psi from the tank.

Does this sound reasonable in engineering design?

Thanks

 

[Zapster]

Perhaps I made the wrong assumption in that the flows given by RJB32482 were CFM values at std conditions. I just assumed 14.7 psig and 65 degF for the CFM.

No it is not reasonable for 60 psig at the tank, it seem like you would need something like 20 psig in the tank. Your bottleneck is the one inch line. If you enlarged the 1” line you could do it with less than 10 psig in the tank. With 60 psig in the tank, you will have choked flow; and in addition, much greater flow than you want.

 

[sailoday28]

I suggest that the general precedure outlined by BigInch be followed with the following modifications---After the pressure Pb is established:

1.Use adiabatic flow with friction--FANNO line in each branch. BigInch method seems to be for isothermal flow,
2. Don't necessarily assume exit from each downstream pipe is atmospheric pressure. With FANNO or isothermal flow choked exit flow is possible FANN0, M=1 and isothermal M=1/sqrt gamma)

Regards

 

[BigInch]

I wouldn't even say its isothermal. As it seemed RJB was having a hard time getting past the network flows, I was just outlining the general procedure for network analysis. A specific method for determining pressure drop in each branch must be used for the user's specific flow conditions, laminar, turbulent, thermal considerations, liquid, gas or 2 phase flows, etc.

Going the Big Inch!

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[Latexman]

Using 60 F and 14.7 psia as standard conditions and:

Main - 60 feet of 3"ID pipe
Pipe 1 - 35 feet of 2" ID pipe + 1 branch tee + 1 exit
Pipe 2 - 40 feet of 3" ID Pipe + 1 run tee + 1 exit
Pipe 3 - 20 feet of 1"ID pipe + 1 branch tee + 1 exit
Pipes 1-3 exit to atmosphere

With no controls and achieving the minimum of:

Pipe 1 - (60 CFM)
Pipe 2 - (25 CFM)
Pipe 3 - (120 CFM)

You need 17.3 psig at the start of the Main. The junction of Pipes 1-3 is at 7.62 psig. The calculated flows are:

Main - 2082 scfm
Pipe 1 - 535 scfm
Pipe 2 - 1427 scfm
Pipe 3 - 120 scfm

Pipe 3 sets the flows and pressures.

Good luck,
Latexman