Velocity change in pipe due to abrupt diameter change 1

[jball1]

I would appreciate help with what seems like it is probably a pretty easy fluid mechanics problem. I need to calculate the velocity change in a pipe due to an abrupt diameter change (see attached figure). I have the velocity in the larger pipe, and need to calculate the velocity in the smaller pipe.

I need the velocity in the smaller pipe in order to calculate the Reynolds #, which I need in order to calculate the friction factor, which I need in order to calculate the damping coefficient, which I need to plug into my FEM... this is the first time I have dipped my toe in the world of fluid mechanics since I graduated 10 years ago, so I am a little rusty. Any help would be much appreciated. I am doing some googling, and am coming up blank...

 

[pmover]

so, if Q1 = Q2 with 1 being the larger diameter and 2 being the small diameter and Q=VA, an engineer ought to be able to solve this problem.

time to scrap those rusty brain cells . . .

 

[LittleInch]

It's more basic than you think.

Ratio of areas.

But friction and pressure losses on a sharp edged nozzle are quite different.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[jball1]

I know I know... I'm in panic mode a bit. Having a really hard time showing that this design is acceptable, and so we're grasping at straws... I'm trying 3 diff things simultaneously and they will prob all be deadends. Anyways, I appreciate the help!

 

[jball1]

Ok so I'm a little stuck again. I've calculated my Reynolds # (20,000) as well as my relative roughness (5e-5), and now I'm using the Moody chart to calculate a friction factor. I can use the chart itself with the calculated Reynolds # and Relative Roughness to determine the friction factor, but it also appears that I can calculate the friction factor from the Darcy Weisbach equation that is listed on the lefthand side of the chart below. However, the Darcy-Weisbach equation has another an unknown (h = head loss) in it.

So I guess what I am asking is, are the friction factor curves on the below chart calculated from the Darcy-Weisbach equation? Or are they calculated using another equation, and then you can use the Darcy-Weisbach equation to calculate the head loss (not a number I need, just trying to figure out how the equations are intended to be used)


Moody Chart in Imperial Units:

https://www.pumpfundamentals.com/php_pages/friction/moody.pdf

 

[jball1]

I can get the friction factor (the number I need) from the Moody Chart using the numbers I now have, I just want to make sure I am not misunderstanding how the chart is intended to be used.

 

[danschwind]

Well, theoretically you can obtain your friction factor from the D-W equation if you are dealing with a known head loss.

But, as we usually want to find the head loss, we calculate the friction factor with other equations. I suggest you take a look at the Colebrook-White equation (

https://en.wikipedia.org/wiki/Darcy_friction_factor_formulae#Colebrook–White_equation).

Daniel
Rio de Janeiro - Brazil

 

[Latexman]

Why not do both (use chart and equation) and see if the friction factor agrees? If not, let us know.

Good Luck,
Latexman

 

[katmar]

You would get the friction factor from the Moody Chart, or if you want to calculate it from a formula, the original formula is the Colebrook-White equation mentioned by danschwind. Note that the C-W equation applies to turbulent flow only (RE > 4000). A disadvantage of the C-W equation is that it requires an iterative solution, but it converges very rapidly. At the top of this page you can find the FAQ section and in the Flow in Pipe section you will find faq378-1236 by member quark. This discusses some of the newer explicit equations which do not require iterative solutions.

For laminar flow (Re < 2000) the friction factor is given by the Hagen-Poiseuille equation. In the critical zone between 2000 and 4000 there are some equations (such as Churchill) that give answers, but flow in this regime is not stable and you should not give too much credence to the numbers generated by these equations.

Back-calculating the friction factor from the Darcy-Weisbach equation would typically be done when you are not sure of the condition (i.e. roughness) of the pipe, but you are able to measure the pressure drop.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[IFRs]

Do you care what the pressure drop is?
If not, just use pmover's advice: if Q1 = Q2 with 1 being the larger diameter and 2 being the small diameter and Q=VA.
If you care what the pressure drop is, the abrupt change may be small compared to the length of pipe before and after...

 

[katmar]

It is good that IFRs has brought us back to the original problem. You have said that you are a bit rusty with this stuff so this is just a reminder to keep both the friction and Bernoulli effects in mind. Because the downstream pipe diameter is smaller, the velocity will be higher and the energy to accelerate the fluid can only come from the pressure energy. In your case the downstream pressure will be lower than the upstream pressure because of the friction effects in the sudden contraction and also because of the conversion of pressure energy to kinetic energy (i.e. Bernoulli). See the discussion section in this example from my web page.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[shussain786]

Hi,

In this problem I'd use crane 'flow of fluids'. Go to page A-26, use sudden contraction with angle 90deg. Calculate resistance coefficient (K) in formula 2.

This will allow you to find head loss (pressure loss, KE energy loss), hL=(Kv^2)/2g.

Thereafter apply Bernoulli equation where hL is known, v1 is known, change in pressure should be already known (otherwise you won't know its velocity) and you are left with finding v2 (what you want to know).

NOTE: remember bernoulli equation conditions must be met (don't use incompressible bernoulli equation with compressible fluid)


Kind regards,
Sadik