Tee Flowrates - Not a normal direction.

[sheeldz]

Been reading on pressure losses and flowrates within fittings, and have came accross a rather interesting problem. Our system has an existing Tee that the flow does not come in from the run, but infact from the branch. The system was designed so that the right hand branch was shut off, for expansion later. The overall question is that when we switch on the right hand branch, which way will the flow be more inclined to flow to.

Intuitvely, I belive the right hand new route will be less likely for flow due to the configuration (large height gain and head losses) but it should, according to bernoulli, be the same flowrate (different pressure).

Crane 410 has basic K values for through a run, and through a branch - can I safely assume the K value being used for the left hand turn as it is now, will be the same for the right hand turn when the right hand turn is opened and the left hand turn is closed, and does anyone have any input into the flow prediction I have suggested above?

Flowrates are around 30'000 BOpD at this point.

Thanks,
Mark

 

[BigInch]

Which downstream side has the lowest total energy head? You can't use Bernoulli unless you know the flow in each branch and that's what you're trying to determine.

That's two unknowns, so you need two other equations to solve the system, i.e. the total drops from the inlet of the tee to the outlets of each of the branch pipelines.

http://virtualpipeline.spaces.msn.com

 

[bduane]

The flow will divide in the tee so the pressure drop from the tee to the end of the pipe is equal in both lines including any elevation difference. There will be one unique answer for any inlet flowrate. You need to be careful though if the pipes discharge at different elevations - you will get zero flow out of the pipe that discharges at the higher elevation until the flow in the pipe that discharges at the lower elevation creates an equivalent headloss equal to the elevation difference. This flowrate can easily be calculated.

I routinely use the excel solver routine to solve this problem since I take into account both frictional losses due to the straight piping and minor losses due to fittings. I use the Hazen & Williams equation for the straight pipe which gives me headloss as a function of Q^1.85 and the minor losses are a function of Q^2, therefore, it is not directly solveable and requires iteration.

Equation 1: Qin = Qout1 + Qout2 (you know Qin and you want to solve for either Qout1 or Qout2)

Equation 2: Hloss1 + Static Change1 = Hloss2 + Static Change2

Hloss1 is a function of Qout1 and Hloss2 is a function of Qout2

Substitute Qin - Qout2 for Qout1 in equation 2 and iterate for Qout2 for a known value of Qin

If the 2 pipes discharge at the same elevation and you can ignore the minor losses, then the flow will divide approximately:

Q1/Q2=((L2/L1)^0.5)*((D1/D2)^2.5)*(F1/F2)

where F1 and F2 are the friction factors (or C factors) in pipes 1 and 2 and L1 and L2 are the pipe lengths. Units doesn't matter as long as they are the same for both pipes.

bduane

 

[sheeldz]

The new connectionn is basically a new line that goes from the existing tie in, to an exact replica of the replaced pumps, so they is no elevation change, and the overall number of fittings between the two points is similar (if not identical). The inital analysis seems to show that the shutting of the valves will be no easier either way, and both will have the same resitance.

The above formula is similar to what I had worked out from first principles. Good to see some similarity between the what I had in my mid and what you have provided.

The problem I have is that most of the litrature only deals with flows going from the run to the run or branch.

Interestingly (maybe) is that from the original commision report and detailed design when coming up witha fitting pressure loss, the Tee was taken to be similar to a 90 degree bend as the back up the right hand turn was taken to be negliable. Not sure if that is best practise, but it was in 197x ... Thanks.

 

[BigInch]

bduane,
I don't think your statement is correct.

The flow will divide in the tee so the pressure drop from the tee to the end of the pipe is equal in both lines including any elevation difference.

There are only two conditions for n pipes forming a junction, the energy head at the center of the tee is common for all joining pipes and the flow into the joint must equal the flow out of the joint. Each pipe forming the junction will have THE energy drop along its length and the corresponding flowrate to make those two conditions true. The energy drop of each joining pipe corresponds to the pipes elevation difference, fluid density and the head lost to the flowrate in each one. Energy drops of the outlet pipes do not have to be equal. Have a look at a typical "three reservoir" problem.

http://virtualpipeline.spaces.msn.com

 

[bduane]

BigInch:

I agree with your assessment - probably should have said: "the flow will divide so that the energy gradelines are the same at the junction node of the tee" as you indicated.

Bduane