Calculate Friction factor K for "Pipe connect to a Pipe body" or fitting

[Anonymous]

Dear everybody and Mr Jon,

If we have to connect a pipe to another pipe by drilling a hole on pipe body then we seal the connect position like picture below:


So, the connect position look like a fitting with K factor.

My question are:

1. How to calculate the factor K with diameter a,b and height to connect h.

2. The connect position like a fitting now, so that mean our pipe a,b will shorter a little bit. How long is it?

3. Have you ever calculate it by Autodesk CFD? And how to do that? 

Thanks and Best Regards.
HS

[Anonymous]

Does the following article answers your doubt?

https://knowledge.autodesk.com/support/cfd/learn-explore/caas/CloudHelp/cloudhelp/2014/ENU/SimCFD/fi...

Regards,
Rajdeep

[Anonymous]

Thanks for your help Rajdeep,

But your link is a friction factor K along the pipe, and my question is fiction factor K for local ( special position) in the systems and it is not a standard T fitting.

THanks you

[Anonymous]

I am not getting what exactly you want to achieve. Do you know some theoretical formula for such calculations? Because, in handbooks such parameters are given directly for different types of joints using some correlations or like Moody diagram.

One way of what you want to achieve is : As you mentioned word "special position or local", I think then you can find the reynolds number in that position using CFD results. Then, use the formula given above or some another formula which relates Re and f. That is the only way I can think of.

Regards,

Rajdeep

[AstroJohnPE]

Hi SulNT,

Are you trying to calculate the K in this equation for the pressure drop in a fitting?

where Dp is the pressure drop, r is the density, and V is the velocity.

If so, the CFD analysis will give you all of the values except for K. You can calculate  the K factor from the equation. Of course, you need to use the proper equation (there are several forms) and decide whether to use the upstream or downstream velocity.

[Anonymous]

Dear Rajcoep.88 and Mr. AstroJohnPE,

Thank you both so much. Of course, I knew about this formula before, but in this case I'm not for sure when we apply this formula for fitting. Please give me some advice ( suppose I call velocity: in b side is V1, out a side is V2, out b side is V3):

1. If we have a standard fitting, so we will have a constant friction factor K regardless velocity V2 and V3 ?? is it right?

2. And because the fitting will have one side in ( V1) and two side out ( V2 and V3), how we apply this formula to find factor K ??

3. I'm wonder how we can apply our fitting in real-life. Because we don't know exact value of V2, V3. That mean the value V2, V3 we set (random) above ( point 2) to find K, I'm wondering that value V2, V3 correct or not?

I'm afraid if we set a random V2, V3 then we have a wrong value of K. Then when we apply our fitting with K in real-life , we will have a difference value V2,V3.

Thanks and Best Regards.
HS

[Anonymous · Accepted solution]

Lets break your case into few parts.

1) If V1 is your inlet velocity, then in CFD you give velocity at inlet or flow rate if you know. At other two points you will give Pressure Gage Static =0. Run the simulation. You will get pressure drop which will be pressure at Inlet. Density you know for fluid. Then you know Inlet velocity (if you do not know inlet velocity but have flow rate then use formula Q=AV, where A is area of pipe at inlet). As you know all parameters, you can calculate K mentioned in above post. This will be friction factor for whole system not a single component.

2) Now, run second simulation without T joint only straight pipe. Calculate K again by using above procedure. As you will get different pressure drops, the value of K will change. The difference in Pressure drop between two simulations is indication of how much T joint contributes in friction factor.

I hope this helps.

Regards,

Rajdeep 

[AstroJohnPE · Accepted solution]

Hi HS,

You have a tee. So it may be helpful to look at the pressure loss coefficients for tees to get an idea of what you need to do. Since there are two branches, there are two values of K depending on whether you want to calculate the pressure drop from 1 to 2 or from 1 to 3.

Based on the tables published by ASHRAE (American Society of Heating, Refrigerating and Air-Conditioning Engineers) for duct design, the coefficient for the path 1 to 2 depends on the ratio of the areas (A2/A1) and the ratio of the flow rates (Q2/Q1).

• So you need to create a 2D table and supporting calculations to cover all ranges of areas (A2/A1) versus flows (V2/V1), and then create multiple tables that cover the range of h (probably h/D).

According to ASHRAE, the coefficient for the path 1 to 3 only depends on the ratio of the flow (V3/V1).

• In your case, it may also be affected by h (or probably h/D), so you either need a 1D table based on flows (V3/V1), or a 2D table table for different values of h (or h/D) versus flow (V3/V1).

Because you are calculating the k for various combinations, whoever is using these results will pick the combination that fits their case. You don't have to worry about whether your "random" choice of V2 and V3 is wrong because you will be covering the entire range.

All you need to worry about is whether your CFD analysis is accurate.

[Anonymous]

Dear Rajdeep ,

I used to run CFD like you advice before but It doesn't work, The pressure drop maybe not the same value ( just a little difference) and the V2 allways equal zero. The reason is the fluid never go to top branch with no reason, it will go strainght by gravitation. Unless at V3 we have block the way or we have some pressure push back the water run to the top branch.

Thanks

[Anonymous]

Dear Mr. John Holtz,

It is interesting if we can divide the T-fitting part like that. Step by step I think we should do:

1. We will block side 2, that mean V2 = 0, then We find the pressure drop from 1 to 3 by CFD, then we get K2.

2. We will block side 3, that mean V3 = 0, then We find the pressure drop from 1 to 2 by CFD, then we get K1.

And my question is value of PRESSURE DROP ( deltaP) through the tee fitting if both V2 and V3 not equal zero.

Is it equal deltaP1 + deltaP2 (with deltaP1 and deltaP2 calculate by K1,K2 above and by velocity through side 2,3 above)

I think we will have some % error rate. Because when we block side 3 , the fluid will reverse from side 3 and fluid go back to side 2. It will take more pressure drop than usual normal ( when V2 and V3 not equal zero). And the % error rate base on size of pipe and velocity input at side 1. Should we accept this % error rate?

And my another question is K-factor is a constant with size of T-fitting, notwithstanding with flow ,is it right ?

Thanks and Best Regards.

HS

[Anonymous]

I ran 2D simulation. One Straight Pipe and other with T pipe. Pressure drop difference is observed. Straight pipe 215Pa and T pipe 3806Pa.

But, one thing is strange rather. If you observe flow rate balance when simulation is running, then initially it will be balanced. I gave 12LPM as inlet (P1 end as per your diagram) Boundary Condition. So sum of flow rate from P2 and P3 end was equal to inlet P1. But, then suddenly flow rate at P2 started reducing towards 0 and P3 started increasing, then again flow rate at P2 started increasing and P3 is was increasing. Now flow rate at P3 became Inlet (P1) 12LPM plus flow rate calculated at P2 by simulation. Then I observed velocity vectors so they show that flow comes in from T branch instead of going out. It means if you give P=0 BC at two surfaces CFD will consider only one as outlet. That is the reason flow is not divided.

So for dividing the flow you have to give flow rate or velocity as BC instead of P=0 at P2 or P3.

[Anonymous]

Or may be you can run with P2 and P3 with P=0 till it reaches convergence and then see the flow distribution. Does the distribution is in accordance with the area or not that you can analyse or you could have some idea what is happening? Today I will run T joint case till convergence and see what happens or may be you can try with 3D model.

[AstroJohnPE · Accepted solution]

I think I would make a 4 part assembly, where one of the parts (yellow in my diagram) is just the tee, and the other three parts are long inlets and outlets so that the boundary conditions are far enough from the tee to allow the flow to be fully developed at the tee.

Using the boundary conditions shown, you can vary V3 from 0% to 100% of V1. Of course, V2 is the balance. (Actually, I should be talking flow rates instead of velocities.) The software will calculate the velocities and the pressures at 1, 2, and 3 at the tee. Those calculated pressures and velocities will let you calculate K2 and K3 for the given set of conditions.

And to answer one of HS's question, there is no "pressure drop for the tee". There are "two different pressure drops for the tee" depending on whether you are working with branch 2 or branch 3.

[Anonymous]

Dear Rajcoep.88,

I used to simulation CFD and wait until it convergence that model in 3D before I post this thread. The brand side 2 is allways Q2 =0. If we don't have some pressure at side 3, we won't ever have Q2 > 0. And even I have Q2, I will consider P3 I set will right in real-life or not, or it just a random value. And After that I will don't know how to find K...:'( it isn't simple like build it in 3D model then run CFD..

Thank you BTW, I think Mr AstroJohnPE is advice the best solution.
Have a good day.

HS

[Anonymous]

Dear Mr John Holtz,

Now, I totally understand your advice. But I need to confirm for using your advice with Autodesk Simulation CFD. I will follow below:

1. First I make a 4 part assembly, where one of the parts (yellow in my diagram) is just the tee, and the other three parts are long inlets and outlets so that the boundary conditions are far enough from the tee to allow the flow to be fully developed at the tee

2. Then I will set flow rate Q1 at (1)

3. Then I set flow rate Q3 at ( 3) ( equal 0 to 100% of Q1)

4. And Pressure P2 = 0 at (2) 

Now we will simulate with Autodesk Simulation CFD, when it reaches convergence. We will have P1, Q2, P2, P3. But the folmula to find K is:

where Dp is the pressure drop, r is the density, and V is the velocity.

That means:

Then

My trouble is know exact how to find Delta P2. Is it equal P1 - P2 ( if Q3 isn't equal zero ).

The same trouble with find Delta P3 to calculate K3.

Finally I will set step by step Q3 = 80%Q1; 60%Q1; 40%Q1; 20%Q1, then I compare K2 in 4 case, K3 in 4 case to decide which value K2 and K3 I will apply for my T-fitting.

I'm looking forward to hearing from you.

Thanks and Best Regards.

Hoang Su

[AstroJohnPE · Accepted solution]

Hi Hoang Su,

Yes, your description is correct.

The pressure difference delta P2 will always equal P2 - P1 regardless of what the flow Q3 is. You will use a plane at the exit 2 and inlet 1 of the tee to get the average pressure at those faces.

Then for the straight through flow, from 1 to 3, you will calculate K3 using delta P3 = P3 - P1, where the average pressures are calculated at the outlet face 3 and inlet face 1 of the tee. This is true regardless of what the flow is in the side branch 2.

Let us know how it turns out.

For your reference, I have attached the table from ASHRAE that I mentioned in an earlier post. Design 6-8, item d (with the 90 degree angle between the main and the branch) applies to your case with h=0. You should be able to get similar results.

[Anonymous]

Dear Mr John Holtz,

With your idea and your instruction, I took almost a month to calculate my part ( my tee fitting ). And I have done it with some conclution below:

- The Friction Factor K for Tee Fitting isn't linear, it can go up then down from left to right of the ASHRAE table. That why you have to carefully to "diagnose" the value of Factor K even you have two same value on left and right of the table, the value between maybe higher or lower than the same value left and right.

- Each point ( ratio point) have a limit flow, if flow higher that limit, that point will change alot of value Friction Factor K. If under limit flow, and with the same ratio flow, you can have the same Friction Factor K, but with higher limit flow, and the same ratio flow, you have difference K. So carefull and you need to know your tee very well

- Some point ( ratio point) with a "slow" flow ( with me that flow is under 90 litre an hour), that mean the pressure drop will very small, the computer maybe have some error, you can have two difference value of Friction Factor K with the same condition when you simulation

Thank you and Best Regards.

Hoang Su

[Anonymous]

For Pressure calculation you have to use Bernoulli.
Secondly K is Coefficient of discharge and not the friction factor.