Laminar Flow

I'm not smart enough to do the calculations, but I think you'll need to supply the velocity of the pipe for the big brains.

OOoops!

2 knots of speed or 2.3 mph ( I think).

Too much fried shrimp and crawfish etoufee...........cholesterol over load is slowing brain function.

nm

2 knots or 2,3 Mph?

Why the question mark, the conversion is correct. 1 Knot = 1.1507794 mph - approximately.

Who's talking about knots is on the ocean, or at least moving on water. The Mile there is the nautical mile. And one Knot equals 1 mile per hour or re: to the land 1,151 Mph. The question here is laminar flow in water. Not in the air. Just to clarify.

I don't think you've clarified anything. They are different units. A (statute) mile is 5280 ft.

A nautical mile is 6076 feet.

who is referring to knots (abrev. kn and or/kts) uses the Mile and that is the nautical mile. And since the question has knots in it... The nautical mile is still used for navigation on water and in the air and affiliation to the knot unit is evident. I have put the question mark there because the Mph is misleading since the knot reader, only reads miles and that are Nautical miles. And the NMph or nMph or Nmph is not commonly used. Check e.g. on a airplane flight monitor. We have no more info but a guess could be that the pipe is a part of a construction under water on a boat, or an obstacle in a river. This is a guess post forum? And indeed both are different units, that use the same abbreviation but mean different distances or speeds, depending on where you use it.

I hope this is a better approach?

Sorry, I don't know who, "Who" is. And the rest of your post is unintelligible to me.

I don't believe we are adding anything to the discussion. Bye.

When is a knot not a knot "on land or sea or in the air".

I do not think that you will get laminar flow anywhere in the pipe at this high flow rate with water as media. Reynolds number will be much higher for laminar flow. You may try calculations of reynolds number. It should come less that some 4000.

In fact with liquids with viscosity near to water, it is very difficlt to find laminar flow in actual field. It can exist only with high viscocity liquids.

(It is what I understand)

gsuhas

For flow in a pipe, changes from laminar to turbulent over Reynolds number 2300 - 4000, but for bodies moving through fluid, as in this problem or eg particles settling in water, the Reynolds numbers involved are quite different.

Netmaker - I'll see if I can find something that adds to what others have posted, but most of the stuff I remember is concerned with finding the forces involved, rather than the distances you're looking for.

Cheers............Codey

codey,

have to agree, its kinda is backwards. I remember trying something that I didn't finish, but can't remember why I quit, maybe thats why?

p911

You could try checking your lecture notes

Or bribing sue with that NO food...

Lecture notes?

I haven't been inside a classroom since 1972.

This is actually something I need to use in my business.

Sorry Sue, but you'll never make it on NCIS.

........seafood is all gone too .......

,

nm

netmaker:

At the speed at least you may not have to worry about natural frequency of the pipe to create vortex shedding issues.

I look to see what I have. Its usually stuff such as laminar/turbulent (Reynolds number) flow for inside the tube for thermals when I wrote shell & tube spec. programs. but I believe I attempted to wrote for the outer years ago but deemed it unnecessary see what I can come up with.

I believe Stokes Law would have something one can use.

p911

phoenix,

Given the 2" dia. and the velocity of only 2 knots, would it be conceiveable (using Stokes Law) that the flow would resume laminar at less than 5" from the rear of the pipe?

Is it my understanding from the video you suggested, that the FASTER the water moves around the cylinder, the SHORTER the turbulent area behind the cylinder is?

Thank you.

nm

Look into the book, particularly chapter 1

http://books.google.ca/books?id=-lbnuyzAXOkC&pg=PA510&lpg=PA510&dq=wakes+of+cylindrical+bluff+bodies+at+low+reynolds+numbers&source=bl&ots=YcQlU2E38C&sig=jlk9PTY8CW59wRq4TJUHv7z_67M&hl=en&ei=wpavS_C3PIXeNdnX2JkF&sa=X&oi=book_result&ct=result&resnum=10&ved=0CDEQ6AEwCTgK#v=onepage&q=wakes%20of%20cylindrical%20bluff%20bodies%20at%20low%20reynolds%20numbers&f=false

Just a preliminary observation at this stage. Something everyone has overlooked so far is that there is no laminar flow to break up ahead of the pipe or to return to behind the pipe as the surrounding water is stationary - well that's what I assumed when I read it.

If that's the case then you are asking at what distance does the stationary water begin to be disturbed and after what distance does it return to a natural state?

Frank

Well, I guess you do have a point. Either way, I'd still need to know these distances if possible?

Thank you.

nm

Something everyone has overlooked so far is that there is no laminar flow to break up ahead of the pipe or to return to behind the pipe as the surrounding water is stationary - well that's what I assumed when I read it.

not quite true,

Do not know how critical, there are what they call micro boundry layers across the object.

since it could be minor, it can be ignored?

p911

Are u knots?

There is a basic question: is your pipe full in the fluid or only part of is submerged? In the second case literature is not very rich since usually this kind of flow was investigated especially for heat exchangers where the pipes are fully in the fluid.

However if you try on google "flow transversal to cylinder" the links can bring you at least a good feeling about what happens when a cylinder has a relative movement with respect to surrounding fluid. In general the theories and the experiments were done with static cylinders and moving fluids. It can be assumed - but with some reserves - that the moving cylinder will generate same patterns in a static fluid.

In front if there is no fluid movement we can not use the notion of "laminar" since there is no flow. What can be but mentioned is that at the front "edge" of the cylinder" a boundary laminar layer will form and depending on the relative speed the layer will go turbulent and leave the cylinder till full cavitation.

If the cylinder moves you can imagine it as kind of round "piston" pushing the fluid aside. This means that in front the disturbance will be rather short and that the behind one will progress with speed growth.

Download the free, Open Source CFD software package, OpenFOAM- I'm not sure if this is available for a Windows. This problem is their first example problem in the manual. When you reset the example to your constraints, run the calculations, look at the pressure distribution, not the flow lines. You will have a high pressure area on the leading side and a low pressure area on the trailing side. The extent of the pressure regions should give you a feel for the distances you are looking for. The program can also give you a feel for whether you stay in the laminar region or not. The finer the mesh you use, the more accurate your results will be.

Cwarner,

Nope, it doesn't seem to work with Windows.

Thank you anyway.

The whole thread comes down to this;

At 2 knots, how far back will the turbulence continue behind a 2 Meter long x 2" hollow sched. 80 aluminum beam, fully submerged.

Thank you.

Give me a few days & I will run a quick study for you- I'm playing catchup on a couple of projects right now...

Warner,

Most appreciated.

Thank you for the interest in my question.

IN A NUTSHELL..........................................

I am looking to replace the square tubing on this design of beam net with 2" round sched. 80 marine grade aluminum.

I just need to get a good estimate as to what new (TURBULENT) forces might be introduced or eliminated.

* This design ( Misago beam) has the netting set back from the beam by as much as 30".

** The new design will have the netting as close as 8-10".

Does this make it any easier?

Thank you all.

Thank you for the explanation. If you trawl that net, You could consider a oval shape pipe or if you want to have it performing better at higher speeds, your best result will be with a spool (I hope this describes the form) model pipe. Look at how rudders and wings are made. At the sharper side, your net would be able to enjoy the laminar flow, because of the smoother transients. You can form the pipe into a better shape, with rollers or a press to flatten it.

DMV,

We are experimenting with an steel angle welded to the fornt of a round pipe to give it a "bow" or "arrow head" configuration.

I am NOT an engineer by any means and most of the 30 years i have in this business was trial and error until it worked Good -'nuff.

Times are changing now and some of the sampling work being done is extremely deep.

I am looking for better designs of the frame work on these types of gear. The nets are easy, its the delivery systems ( dihedral, polyvalent , baffled and slotted type doors and such) as well as these deep water beams that makes or breaks a successful research trip.

When you drop a gear off the stern of an R/V , in 5-7K of water, there's a whole lot that can happen just on the way to the sea bed.

Thank you for the input .

Before heading to the test tank down at Texas A&M near Houston, one could learn a lot just by comparing the different profiles in a CFD program. Ultimately, tank tests (or sea tests) are going to be required to confirm the output of any theoretical calculations, but you can seriously narrow down the design choices (and understand why) by having a look at the problem with computer simulation. And, when budgets are tight, computer simulations are a whole lot cheaper than tank tests!

There is another program out there, Flow3D, that works in Windows that you may be able to find for something like $20 on the Internet- or maybe cheaper. This software is a whole lot easier to use (and to misuse, if you don't know what you are doing) than what we were using 20-30 years ago...

Warner,

I did not know A&M had a test tank.

I'll try to find the program. There are some other apps I might use something like for also as we do some dihedral trawl doors for midwater sampling as well. Those things are more like airplane wings than anything else.

But back to the point of all of this, we have made beams from round to square to arrow head shaped. Not being able to get down to the 2K mark on the bottom, I just guess at what is happening by the rubbing and wear on the skids.

Its just that I can see more of the deep water work coming up and some of these agencies do not have the deck gear for handling the Misago or Iruka type benthic beams. I need to find a more efficienct and lighter way.

I do know that the turbulence does affect the way the gear stays on bottom. Too much turbulence and the frame start rocking back and forth. Net placement is an issue and when we start putting the netting closer to the beams, then we might see another problem develop.

Software can help, but I guessing that trial and error just might win over here.

Thank you for the input.

nm

One other thing you must bear in mind- water at depth has different properties than at the surface, which are going to affect the results. One of the issues with tank tests is getting the right fluid properties to actually match the real world conditions.

It has been many years since I have used the tank at Texas A & M- I assume it still exists, but I might be wrong about that...

you are right about that. I had the use of a tank for free. If your project is interesting enough that is how universities in Europe welcome you. My status is more like a high reward for what I have been doing and no public domain info. Call me. I wish I had my simulator programs with me, but they were company property. I try to help people within my limits and do not often charge for it. I have used a program before, but that was mainly based on Bernouilly and Stokes' law and this is not working better for netmakers' problem, because his object has to much drag.

D,

Understood.

thank you for the quick lesson in hydrodynamics.

I'll think on this for awhile before heading to the machine shop.

nm

I would think you could attach the beam via a load cell or a couple of (much cheaper) spring balances.
You could then actually measure the forces.
Call me a cynical old cat if you like, but I'll guess that all the calculations in the world will prove far less than a bit of measurement.
Del...
Now where's my tool kit?

Greetings ol' Cat,

Seems the thread always strays away from the intended question.

I only needed to know if the turbulence of the 2" beam at 2 knots would be greater than or less than 5" behind the beam.

I believe the answer is yes.

I also think we are going to weld a bow plane on the round beam ( pic shown) and that should help kill the turbulence.

That tool kit would not last around my grand kids!

nm

Do not be surprised every engineer thinks that he thinks better than the OP and his ideas are the best in the world. It is an unfortunate aspect of this profession but some times productive. I am no exception!

With respect to your intention to weld a "bow" in front it is a good decision but the the welding of either a bed plate or of 2 plates at the other side will even more reduce the turbulence after since the flow will have the trend to follow the wall and will be less turbulent. A single plate will not affect the point of flow leaving the tube. The presence of a wall will reduce the free volume behind.

You have the pictures you were provided with the tube/beam alone, above sketch shows you why the back angle is a help to reduce the turbulent flow. Dimensions are in mm (1"=25.4mm).

It would be good if you try for comparison the three possibilities one after the other.

I would appreciate if you make those tests to have your results.

Good luck

I only needed to know if the turbulence of the 2" beam at 2 knots would be greater than or less than 5" behind the beam.

I believe the answer is yes.

Is this intended to be tongue-in-cheek? Yes, it is very likely to be greater than or less than, although there is a small possibility that it could be equal.

The turbulence behind a 2" pipe at 2 knots will extend well beyond 5".

No tongue and cheek intended. I am no where near smart enough to run that on this crowd.

Getting to the gist of this:

If I place a Bow on the nose of the beam and a smaller angle on the aft end, the turbulence (However much it would be) will be reduced significantly?

Is this a correct assumption?

The tail should be longer than the head. So in Nick Name's very good drawing, the flow would be from left to right, and as he said, the tail has the bigger effect on reducing turbulence.

nick name:

hats off to you. you graphical address the issue quite directly, No scale on lt but still a GA

p911

The profile is the one I proposed so that the scale was not necessary if you think about geometry, if you copy the pictures you can increase the size and read the values on the left scale. I increased to a maximum the steps so that the zones will not be so coarse.

Thanks for the appreciation

Nick

proportional, of course

Actually, the scale is very easy to determine, if you think in terms of pipe diameters. The leading effect is so many diameters ahead of the leading edge, the wake is significant so many diameters behind the trailing edge.

One question I have- the OP is going to be towing this at depth, in which case there will be a significant difference in water density than what one encounters on the surface. It is not clear that the analysis took this in to account.

The answer to your questions:

- between the centre line of the 2" profile and the front limit a distance of 200mm≈4*d/ was considered and behind 400mm≈8*d. I took less in the front because I knew that the front wave will be shorter and much more behind to cover the whole possible wave evolution.

- If the bulk modulus of water is considered in a first approximation as constant versus pressure (it is in fact a function of p and p^2) the differential equation for the pressure is [p(z)-p(0)]/E= e^(z/L) where p(0)= pressure at depth z=0; E= bulk modulus of sea water =2.34 E9 N/m² and L= E/[ρ(0)*g] a characteristic depth ρ(0)=1030 kg/m^3; g= 9.81m/s². "e" is the basis of natural logarithm. L=2.34E9/(1030*9.81)=2.316E5m.

The specific mass at a pressure p(z) will be ρ(z)=ρ(0)*[1+(p(z)-p(0))/E]. For a depth of 10.000m (32808ft) the ratio z/L= 0.0432 and if we approximate e^(z/L)=1+(z/L)+(1/2)*(z/L)^2+(1/6)*(z/L)^3 the ρ(10000)=1.045*(ρ(0)). The difference is -I think- small enough to be neglected! For the computations the surface density was considered.

In order to better recognize the apparition of cavitation via negative pressures it was assumed that the static pressure is also the pressure at water surface (1E5 Pa).

If other questions appear I shall be glad to answer.

Good job done

GA

All in all I think I have a partial answer to the problem.

My situation is that these gears will always be changing from 2Mrt. to 10Mtr. and from depths of 1M to 1000M.

There is probably no real way to get one true answer.

However, after looking over some of these replies and maybe spending some time on the internet, I'll find a happy median somewhere in the middle.

Right now what we have is working. I just wanted to see if we could make something work better. The turbulence issue has always bothered me.

Thank you all for the input.

nm

hi Netmaker. Think about rudder shapes. They work in these depths' and speed range. Submarines' dept is classified but some public secrets talk of pelagic zone reach. Check also the shape of the sub itself and see how fish you catch are build. And your net is a long tail.

Thank you. I will look into this too. there might be some public information posted.

nm

I have no background in this, but I've been thinking about it.

I think the fish simile is good. The suction on the sides and back of the pipe are caused by the shape in the front that gives the water a sideways component as it passes along. This is similar to the Bernoulli jet, as the water speeds up around the volume of the obstacle, it gains kinetic energy and consequently losses pressure, even to becoming negative. Depending on the speed, that suction can pull water in from both sides with fairly high velocity behind the object, causing a wake.

It would seem that a shape that got gradually fatter toward the center and then gradually thinner again would do the job. Something that would not add a high lateral velocity to the water, and then not leave a void as the water returned inward.

I have no idea how to calculate these effects though.

What you describe is a symmetrical hydrofoil. Such profiles are used especially for their low loss among other applications for fans, wind turbines and aeroplanes. For those applications they are not symmetrical but present a slight curvature in order to generate a pressure difference between under and upper sides via a speed difference.

For the application it could be cheaper to use a tube with a corner in front (standard profile at 90°) and a bend metal sheet or two welded sheets to form the back side. The angle should be about 20..30°. It is not ideal but will be enough and does not ask for a too complex manufacturing.

Thank you.

We tried the tubing. It seemed to work well enough with the heavier gears.

My concern was with the lighter stuff we are going to be using now.

Thanks again.

nm

What is the difference between the Bernouli Effect and the Venturi ( ?) effect?

Both were doing almost the same, Both sound Italian, but Bernouilli was a Hollander and Venturi a Italian. They were both examining the effect of restrictions in pipes.

Venturi was a lot younger and studied the findings of Bernouilli.

Probably he had a more appreciative name or communication started to become better I guess. But the Italian Exhausts sound better.

I assume you are concerned with drag.

In water, the coefficient of drag of a cylinder is very close to 1.2 at Reynolds numbers from 1 x 10² to 1x 10⁵. At about 2 x 10⁵, (where separation goes from laminar to turbulent) the Cd drops to about .4. In your case, the RN will be about .5 x 10⁵, so you are operating in the high drag regime. In this regime distance A is much greater than distance B (by a factor of 5 or more).

You could put turbulence stimulators on the forward surface of the pipe to reduce drag (by transitioning to turbulent separation at a lower speed) but this is probably getting a bit exotic. If you could rig up a towing test, with the cylinders stabilized (e.g., by a foil on a long arm trailing behind) then you could try something like a length of 1/8" welding rod attached at about 60 degrees above and below the leading edge of the cylinder.

The drag of various shapes in water is discussed at length in "Aerohydrodynamics of Sailing" by Marchaj.

Like this?

nm

Here's part of a page copied from Marchaj.

In the middle pic, you can see the wake size from laminar separation. In the right pic, you can see the wake size from turbulent separation (at a higher Reynolds number). In the middle pic, there are two little circles at the start of separation. If you placed something a little forward of those points, (like a piece of welding rod welded along the length of the tube) you could promote turbulence, and reduce drag, counter-intuitively.

When I mentioned a tail, it is just to allow testing without having the tube flail around. The tube could be rigidly mounted to a slide which would be mounted to the underside of a small flat-bottomed boat (like a Jon boat) , which you could tow. The slide would be instrumented to measure drag force. You'd want the test piece length about the width of the jon boat, and you'd want to have a foot or so separation from the boat's bottom. The towing distance would need to be pretty great to get undisturbed water behind the tow boat. On the other hand, if your propelling boat is pretty good sized, the jon boat could be off to the side, (towed from a boom) so it is in clean water, or possibly even being pushed (towed from a boom forward from the bow).

For this testing to make sense you'd have to do it mainly for fun and curiosity, I'd think -- you could get a lot of hours into such stuff.

If you enclosed the tube in a foil-shaped fairing, the drag would be reduced to about a tenth of a standard round tube (i.e down to a Cd of about .1) If you used such a tube in your rig, you'd need to take steps to reduce the probablity of fluttering... you'd want to make it hard for the tube to change its angle of attack.

→ Perry, "The Chemical Engineer's Handbook", any edition.