Trying to visualize amount of airflow turbulence and drag in pipe

[sciguyjim]

I was told recently that ribbed tubing can have the effect of decreasing the effective diameter of a tube by as much as 50% due to drag. If I have a ribbed tube 4" in diam with ribs .25" wide and .25" deep, then is it accurate (even a little) to estimate that the eddies causing the drag extend at least 1" from the walls (about 50% of tube has major eddies)?
Could you estimate, how much would the eddies (and drag) be reduced if I reduced the wall ribs to 1/20 the original depth? Assume I have a tube about 4" I.D. and an airflow of about 500 CFM.
I'm trying to form a mental image of the amount of turbulence and laminar flow in the tube, and how much the drag could be reduced by smoothing the ID of the tubing. It's just an interest I've been obsessed with lately. I've done a lot of searching on the web and at my library and can't find what I'm looking for. Thanks.

 

[poetix99]

Friction factors account for varying amounts of "rouhgness" in a pipe and give a relative indication of the effect that you are talking about: for a given pressure drop through a pipe, the flow will decrease as the wall friction increases. Check an undergraduate fluid mechanics text or look at Crane Technical Paper No.410. Published "friction factors" are typically for uniform, RANDOM roughness, not for regular features such as ribs. You might be able to get an equivalent roughness for this kind of pipe from another source.

I think that in the extreme, some of the reduction of flow rate for a pipe with internal ribs is due to a reduction of the flow area; distinct from the roughening effect, but maybe I am not accurately picturing the geometry that you're describing.

For what it's worth, it is not accurate to picture "eddies" slowing down the flow only near the walls of a rough pipe, like a wake behind a body in a free stream flow. Internal flows are a little bit different. The eddies do exist, but they have the effect of tending to make the velocity more uniform across the entire pipe diameter (except for very near to the wall where it will always be zero velocity). The average velocity is lower, of course, because of the increased drag.

 

[Rollerblades]

You can detect turbulence by using a die in the flow.

 

[sciguyjim]

I wish I could use a dye, but I'm dealing with air not water, and it's totally enclosed in opaque plastic tubing.

Let me try to draw a picture:

--------------^^^^^^^^^^^^---------------------
| valve
Airflow>>> ))))) obstruction
| valve
--------------^^^^^^^^^^^^---------------------
Smooth walls Corrugated
region

The "ribs" in the corrugated region resemble the top line in my diagram. They begin at the inside wall and go outwards about .25" (they don't reduce the internal area for airflow, they actually increase it a little bit.) I guess you'd actually call them elongated pits rather than raised ribs. Ok?
I found a book on basic fluid mechanics at my library. It was the most basic book on the subject they had. Although it's full of equations, and not as much calculus as the more advanced books, it seems my example is too simple, or unusual. I can't find anything even remotely similar in the book, or online.
I understand the basics of how the eddies and laminar flows in the tube behave, I'm just trying to get a feel for the change in "size" or amount of turbulence in the tube if I can smooth out the corrugations. If I could measure the airflow or a pressure change I would but I don't have that capability. All I have is the few numbers I mentioned in my original post. I have no experience doing anything like this so I'm having a hard time visualizing the change in airflow caused by a given change in tube structure. I hope I made myself clear enough. Thanks.

 

[poetix99]

I take it that the two threads you have posted in this forum (1-28015 and 1-29182) are interrelated. I'll also try not to read non-existent details into your "sketch" (which I think was pretty good, given what you had to work with), but the sketch does raise more questions for me than it has answered.

This might appear to be somewhat "stream-of-consciousness" thinking aloud on some possible approaches to a calculable solution:

Could you model the obstruction(s) and the terminal valves as effectively being a single valve? guesstimate a C(sub)v for the cowled and uncowled versions.

The obstruction could be modeled as an (annular) orifice.

How much of a restriction does the obstruction represent when compared to the terminal valves? Is the C(sub)d of the obstruction really going to make a difference in the back pressure?

I don't know, in any case, how one relates a drag coeff. for a free stream flow to a discharge coeff. for an orifice in an internal flow field. My mind migrates toward relating the obstruction to a "K" factor, ala pipe fittings (see Crane TP410). You might browse Crane 410, and look at chack valves and butterfly valve just to get a sense of the order of magnitudes, and the relative change resulting from a (somewhat) more aerodynamically smooth shape (butterfly vs. check).

What is the total length of the tube? (Think of this in terms of equivalent diameters (3.75 in.) for the rectangular channel.)

What length of the tube is ribbed? How many equiv. diameters?

Part of the difficulty for you in wanting to obtain a CALCULABLE solution is that some of the flow phenomena that are (or might be) occurring in the tube do not have enough length to fully develop (the ribbed tube velocity profile vs. the smooth tube upstream), or are not adequately separated from the next feature (such as the obstruction and the valves) to be easily characterized in a calculable form. This is why for, geometrically complicated fixtures (such as valves) the essential data is empirical.

The pipe friction, as indicated by a friction factor (see for example Crane410 again) could be anywhere from 2X to 10X greater for the ribbed tube vs. the smooth tube, depending on the Reynolds number (I don't have air viscosity handy to me right now.)

All of this leaves you with no answer, but, I hope, some food for thought.

 

[sciguyjim]

Hi poetix99,
"Could you model the obstruction(s) and the terminal valves as effectively being a single valve? guesstimate a C(sub)v for the cowled and uncowled versions."

Yes, the terminal valves do work together as if 1.
Please explain what is meant by "Cv for cowled and uncowled versions". The system is never run with the ends open. It only operates when fully connected at both ends.

"The obstruction could be modeled as an (annular) orifice.

How much of a restriction does the obstruction represent when compared to the terminal valves? Is the C(sub)d of the obstruction really going to make a difference in the back pressure?"

The obstruction can be modeled as an annular orifice. It vents a little air at low flow conditions but basically acts as a high drag, sort of concave surface facing the airflow at high flow rates.
The obstruction has maybe the same area facing the flow as both valves combined. So it essentially takes up half the tube diameter (it sits between the 2 butterfly valves.)
I believe the Cd of the obstruction is a very large source of backpressure, especially at maximum flow rates. At this time, it is essentially the only thing sitting in the flow since the valves are wide open and no longer interfere with airflow.

"I don't know, in any case, how one relates a drag coeff. for a free stream flow to a discharge coeff. for an orifice in an internal flow field. My mind migrates toward relating the obstruction to a "K" factor, ala pipe fittings (see Crane TP410). You might browse Crane 410, and look at chack valves and butterfly valve just to get a sense of the order of magnitudes, and the relative change resulting from a (somewhat) more aerodynamically smooth shape (butterfly vs. check)."

My lack of experience in this field inhibits my understanding of this paragraph.
The "valves" are 2 butterfly valves. I don't consider them aerodynamically smooth, especially when partly open so air is essentially hitting angled flat plates and finding its way around them. When closed, the little airflow is mostly removed by the obstructing concavity. When open, most air goes through the valves and the obstruction is really an obstruction.

"What is the total length of the tube? (Think of this in terms of equivalent diameters (3.75 in.) for the rectangular channel.)
What length of the tube is ribbed? How many equiv. diameters?"

The total length of the tube is about 3' and most is smooth inside. It is 3" in diam except at the outlet where it stretches to a 3"x5" oval. So the total length of the tube is about 10 diameters.
The length of the ribbed section is about 4", and is in the large oval 3"x5" end of the tube, immediately in front of the valves and central obstruction.

"Part of the difficulty for you in wanting to obtain a CALCULABLE solution is that some of the flow phenomena that are (or might be) occurring in the tube do not have enough length to fully develop (the ribbed tube velocity profile vs. the smooth tube upstream), or are not adequately separated from the next feature (such as the obstruction and the valves) to be easily characterized in a calculable form. This is why for, geometrically complicated fixtures (such as valves) the essential data is empirical."

Again, I didn't know this because of my lack of experience.

What is "Crane 410"? If it's a book, my local library can borrow it from anywhere in the US. I will however need the full title and author's full name.

Can you give me formulas for calculating the Reynold's numbers for the smooth and ribbed sections? I have seen diagrams of the amount of turbulence in flows with different Reynold's numbers for air passing a spherical object. This could give me a basis for comparison. I've seen formulas for Reynold's numbers for simple flow arrangements, but I don't know how to adjust for smooth vs ribbed walls (I suppose I could assume the sphere is smooth in one case and ribbed in the other, but I don't know how to adjust the formula for this either.)

Thank you for helping me with all this detail, I really appreciate it!
Jim.