Frictional Losses for Branched Pipes

[SKahle89]

I am trying to calculate the total differential head across a pump (to size it), and I am having conceptual issues calculating the frictional losses on the discharge side. I'll ask a much simplified question that should help.

Assume I have a 2" pipe carrying a liquid. The 2" pipe then branches into three 1" pipes. Those three pipes are each identical and discharge into three separate reactors that are at the same pressure and height. Essentially, each of the 1" lines experiences an equal flow and thus the same frictional losses.

If I want to calculate my total frictional losses, how to I take into account the friction of the three 1" pipes. My instinct was to add them up:

F,total = F,2" + F,1"a + F,1"b + F,1"c

I understand that the friction in the 2" section will not be the same as the sum of the 1" sections; I want to know what the frictional losses are for the TOTAL discharge side of the pump. Any advise is greatly appreciated.

 

[BigInch]

Add only one of the 1" pipes.

You need to get the highest pressure drop along any one possible route from the pump to any reactor. Since each route is the same, pick one.

If you were plowing a field, which would you rather use? Two strong oxen or 1024 chickens?" - Seymour Cray (1925-1996), father of supercomputing
***************

http://virtualpipeline.spaces.live.com/

 

[Bribyk]

Add them like electric resistances. F,total = F,2" + (1/F,1"a + 1/F,1"b + 1/F,1"c)^-1, F2 in series with the equivalent resistance of the 1" sections which are in parallel. Your pressure drop along all three 1" pipes will be the same, regardless of the reactor pressure, elevation changes, etc..., the flow rates will change to keep it constant. So a simpler equation is: F,total = F,2" + F1"(a=b=c)/3.

 

[Unotec]

Just calculate the frictional loss on one 1", that will be your total pressure loss (you already divided your 2" flow/3)

<<A good friend will bail you out of jail, but a true friend
will be sitting beside you saying ” Damn that was fun!” - Unknown>>

 

[Latexman]

Does the liquid go to one reactor at a time (this usually means one flow meter in the 2" line), or does the liquid go to all three reactors at the same time (this usually means three flow meters, one flow meter in each 1" line)?

Good luck,
Latexman

 

[BigInch]

Latexman

Given: "Essentially, each of the 1" lines experiences an equal flow and thus the same frictional losses."

If you were plowing a field, which would you rather use? Two strong oxen or 1024 chickens?" - Seymour Cray (1925-1996), father of supercomputing
***************

http://virtualpipeline.spaces.live.com/

 

[Latexman]

BigInch,

That is true for both scenarios. Thanks, but I'm waiting for the OP to answer.

Good luck,
Latexman

 

[SKahle89]

Latexman,

The liquid goes into all three reactors at the same time; this system is diluent delivery system for a run-away reactor emergency. There are flow meters and control valves on each of three 1" lines.

Everyone else,

You all have provided me good answers, but can you help me out conceptually? Right now I'm looking at the system from the pumps point of view. Why is it that the pump only feels the resistance (due to friction) of only one of the 1" pipes? My gut tells me that the pump has to muscle the fluid through each of the 1" pipes... so it's the sum.

I can also argue with myself (in you all's favor) that if the pressure at the point of divergence is the same in each of the three 1" pipes... then the pressure drop from that point to the reactor discharge is the same for 1" each pipe. So for pump sizing the total frictional losses is just the F2" + F1" pipe, and it doesn't matter what happens in is the other pipes so long as the pressure drop is not greater (from a pump sizing prospective). Obviously more fluid will take the path of least resistance, but that is why I have control valves and flow meters to iron that out. Someone slap me in the face with puzzle piece that makes it all make sense.

 

[Unotec]

If the system has two lines closed and only one 1" open, things change quite a bit. Now you are trying to put the full flow through a 1", not three. That is the most pressure drop your system will see

<<A good friend will bail you out of jail, but a true friend
will be sitting beside you saying ” Damn that was fun!” - Unknown>>

 

[BigInch]

Think about it like this.

The pump doesn't "feel" anything other than the head needed immediately at its discharge point into the pipe. It provides that head, if it can, and really dosn't give a flip about anything else.

If you were plowing a field, which would you rather use? Two strong oxen or 1024 chickens?" - Seymour Cray (1925-1996), father of supercomputing
***************

http://virtualpipeline.spaces.live.com/

 

[SKahle89]

BigInch,

10-4, Thanks. For the record I would take the chickens.

 

[katmar]

Do not confuse pressure with work. The way you have described the pressure drop is correct, but the work done by the pump is the sum of the flows through each of the 1" lines.

People are often more comfortable with the electrical analogy. If you connect several light bulbs across a 12 volt battery the voltage remains 12 volts (to a first approximation) irrespective of the number of bulbs, but the current and therefore the power is dependent on the number of bulbs.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[vzeos]

SKahle89,
The way I see it, a fully flowing 2” pipe has sufficient capacity to feed 6 – 1” pipes and one 0.5” pipe under steady state conditions, so the 2” pipe is flowing below capacity. The pipe resistance in each pipe is proportional to flowrate in the pipe, and the flowrate in the 2” pipe should be the sum of the flowrates in the 3 - 1” pipes. Now, the three 1” pipes are in parallel with each other and they are in series with the 2” pipe. Resistances of pipes in series are added directly. Parallel resistances are added as follows:

1/[&radic;]k [sub]total[/sub] = 1/[&radic;]k [sub]1[/sub] + 1/[&radic;]k [sub]2[/sub] + ...

The analysis of k does not lend very much insight to the pipe designer, but if you compare ?P = kQ[sup]2[/sup] to Darcy's liquid flow equation you'll see that k=cfL/D[sup]5[/sup], then,

1/?{ cfL/D[sup]5[/sup] }[sub]total[/sub]= 1/?{ cfL/D[sup]5[/sup]} [sub]1[/sub]+ 1/?{ cfL/D[sup]5[/sup]} [sub]2[/sub]+ ...

and

1/?{ 1/D[sup]5[/sup]}[sub]total[/sub] = 1/?{ 1/D[sup]5[/sup]}[sub]1[/sub] + 1/?{ 1/D[sup]5[/sup]}[sub]2[/sub]+ ...

Where:
C =constant
f = friction factor
Q = flow rate
L = length
D = diameter
?P = pressure drop