"Radius" by BigBen2k

[bigben2k]

Right on Bob.

I got a PM from Dave, and he pointed out that because the fins don't run across the core (like Cathar's), the fins do not have the same heat dissipation effect. The baseplate thickness is also becoming an issue, where I was shooting for 2 mm, the heat dispersion from the fin pattern doesn't match the heat spread within the baseplate. That's also a reference to the "heat gradient" factor.

As Alchemy so importantly pointed out, turbulent flow may not be achieved. I was under the impression that it could be achieved in one of two ways: turbulators or plain high speed. Can you tell me more? What about Utabintarbo's suggestion?

The 5 mm gap between the top and the fins defeats the structural integrity. I agree that it would give much more flow, but I can also see that most of the flow would occur above the fins, and as such, would decrease the cooling effect.

I got an e-mail from Utabintarbo this morning. The saw blade diameter seriously limits the amount of cutting that it can be used for, because it will either cut into other fins, or into the block's outside wall, with its 20, 25 or 32mm diameter.

With some of the info above, I'm going to temporarily reduce the fin pattern radius to 7.5 mm (15mm diameter), pending my calculations. I'll look into some more ways of improving the flow, with turbulence in mind.

Someone started a thread in the liquid forum, about socket mounts. Check it out

It seems to me that a socket mount is possible, using the Maze3 scheme. I will spec this out later, but that mount is pretty close to what I was aiming for.

Many thanks to BillA, for putting me back on track, as usual. I have much work to do on this block, and I really need to go through the calculations before getting more into it, like I have been here.

[Alchemy]

Originally posted by bigben2k
As Alchemy so importantly pointed out, turbulent flow may not be achieved. I was under the impression that it could be achieved in one of two ways: turbulators or plain high speed. Can you tell me more? What about Utabintarbo's suggestion?

Well, turbulence is determined by a dimensionless value called a "Reynolds number." The exact range between turbulent and laminar flow is disagreed upon in some circles.

According to McCabe, Smith, and Harriott, turbulent flow will occur at Reynolds numbers above Re~24,000. Turbulence can be forced by obstructions in the flow as long as the Reynolds number is well above Re=2,100. Below that, there's no way to avoid laminar flow.

Sieder and Tate define turbulent flow for significant heat transfer to be abover Re=100,000.

In the inner channels, you're going to have Re=2600. In the outer channels, Re=1300.

This makes for an extremely inefficient design - the increase in surface area these channels create won't make up for this.

My *rough estimate* of the heat transfer coefficient between the copper channels and the the fluid is h~4 kW/m^2 C.

Under the very best of conditions, you could get perhaps a 3 degree C difference between the copper walls and the fluid. This might still be acceptable, but I don't think this is going to outperform most WB's on the market.

Alchemy

[nicozeg]

Ben, look at the good side, what alchemy said could be interpreted as: Your block sure is not going to start a revolution, but it could perform just de same that the best currently available.

But I suspect from those numbers, 3ºc between water and copper would be extremely well for 100w heatload. Let`s guesstimate some numbers:

1.5ºc delta in TIM
2.5ºc " inside copper
3ºc " in water
4ºc " in rad.
---------------
10ºc delta temp between cpu and cooling air.

That equals 0.1 C/W

[bigben2k]

Thanks for the guestimate! I need to be able to calculate that, which shouldn't be complicated at all. Knowing thickness and thermal properties, I should be able to build such a gradient.

About Alchemy's post: I don't know if the numbers are different with the flow reversed, I'm really having a hard time putting together a mental picture (lack of sleep?!?).

I am working on some changes. The flow in the center is still a problem, so right now I'm looking at solutions where the flow is reversed, where the inner tube is actually an outlet.

What I gather from Alchemy is that the high speed route isn't practical, and that turbulator(s) would be more practical, given a common range of flow rates. I may yet pull an ace out of my sleeve: what's clear to me is that as soon as one gets into turbulators, there is a potential for a sweet spot, where the turbulence becomes in tune within the channel, within a very narrow range of flow rates.

More later.

[Alchemy]

Originally posted by bigben2k
Thanks for the guestimate! I need to be able to calculate that, which shouldn't be complicated at all. Knowing thickness and thermal properties, I should be able to build such a gradient.

About Alchemy's post: I don't know if the numbers are different with the flow reversed, I'm really having a hard time putting together a mental picture (lack of sleep?!?).

I tried to ignore the effect of the water jet striking the copper just above the core, so I'd imagine my numbers would be more accurate if you made a design that avoided that effect.

If you do design the flow striking the center of the block from above, you'd get better performance than my k=4 estimate.

The basic problem with the design is that you're forcing the flow into channels so narrow that the fluid can't become turbulent.

By the by, Reynolds number is equal to linear velocity multiplied by channel diameter multiplied by density, all divided by viscosity. So to get turbulent flow by increasing flow only, you'd have to increase flow rate a hundred times.

Re = V * D * rho / mu

I am working on some changes. The flow in the center is still a problem, so right now I'm looking at solutions where the flow is reversed, where the inner tube is actually an outlet.

Well, that would be much easier for me to model mathematically, but will probably worsen performance.

What I gather from Alchemy is that the high speed route isn't practical, and that turbulator(s) would be more practical, given a common range of flow rates. I may yet pull an ace out of my sleeve: what's clear to me is that as soon as one gets into turbulators, there is a potential for a sweet spot, where the turbulence becomes in tune within the channel, within a very narrow range of flow rates.

That's my suggestion.

Alchemy

[MadDogMe]

The thing that most worried me about your fin design was that they 'branched out', you'd loose pressure like crazy with each branching, I was going to post before but was unsure of the size/area the barb would cover, if it was small it would of resulted in a single channel branching into four!.

Why not try the calculations on the 'circle cross cut many times style'?, so easy to manufacture!, if the outerwall is in the way, get rid of it!, incorporate it into the plate above(and the O ring cjhannel) 2mm will still be enough to countersink some flush bolts into won't it?. if not: don't get rid of it, just lower it enough for the blades '9 o'clock' point to miss...

I think you're worrying to much about the fins when the real revolution is in the outlet and full balanced use of radial flow...
********************************************

PS. what is the definition of 'Impingment'?...

[morphling1]

Originally posted by Alchemy
I
By the by, Reynolds number is equal to linear velocity multiplied by channel diameter multiplied by density, all divided by viscosity. So to get turbulent flow by increasing flow only, you'd have to increase flow rate a hundred times.

Re = V * D * rho / mu

As I know Reynolds number is v*d/kinematic viscosity

v.... velocity [m/s]
d... diameter [m]
kinematic viscosity [m2/s]

if the chanell isn't round you have to take the equivalent hidraulic diameter d' = 4*A/C
A... cross area [m2]
P... circumference [m]

so for bb2k rectangular chanell d'=2ab/(a+b)
So you can see that narrow rectangular chanell isn't realy too good for introducing turbolence round chanell is much better.
Also to manny channells (to big cross area) reduce water velocity
and again lower Reynolds. So in your case the flow would be lamilar IF the fins would be long and that kind of design. But with central nozzle and short paths I realy don't think that there would be lamilar flow in that kind of block

[myv65]

Originally posted by morphling1
As I know Reynolds number is v*d/kinematic viscosity

v.... velocity [m/s]
d... diameter [m]
kinematic viscosity [m2/s]

if the chanell isn't round you have to take the equivalent hidraulic diameter d' = 4*A/C
A... cross area [m2]
P... circumference [m]

so for bb2k rectangular chanell d'=2ab/(a+b)
So you can see that narrow rectangular chanell isn't realy too good for introducing turbolence round chanell is much better.
Also to manny channells (to big cross area) reduce water velocity
and again lower Reynolds. So in your case the flow would be lamilar IF the fins would be long and that kind of design. But with central nozzle and short paths I realy don't think that there would be lamilar flow in that kind of block

I'm pretty good with fluids, but will not go so far as to guarantee all that follows is 100% correct. Consider another "educated viewpoint".

First of all, your equation is the same as Alchemy's. The two viscosity values differ only in that one already includes the density value.

Second, the hydraulic diameter equation applies to converting odd shaped channels to an equivalent sized round one when calculating pressure drop vs flow rate, but not for calculating the Reynolds number. The Reynolds number is based on a "characteristic" dimension, which is open to some interpretation. In a round passage, it is diameter. In a square passage it is leg length. In rectangular (or any other odd) passage, it is not so easily estimated.

When the aspect ratio gets large (one rectangle leg vs the other), the characteristic dimension tends to become dictated by the short leg. eg, if you had two channels, one 1 cm by 10 cm and the second 1 cm by 100 cm, and defined flow in terms of liters/cm of width, each channel would have pretty much the same Reynolds number despite having dramatically different hydraulic diameters.

Alchemy has been talking about laminar flow. This is an area that I believe is greatly misunderstood and rarely have I seen what I feel are correct statements made about it in various forums. The Reynolds will tell you the flow regime (with some room for intepretation), but only for a uniform, undisturbed run of plumbing. Laminar flow will be upset by any change in cross sectional area, shape, or sometimes even deviation from a straight path. Laminar flow takes time (rather distance) to develop following a change in the flow passage. This time/distance may be exceedingly short if the Reynolds number is very low, but still exists. The very thought of having completely laminar flow in most blocks that I've seen is absurd.

All you need to do is take a look at what defines laminar flow. In my understanding, it is "flow along streamlines" (insert Ghostbusters joke here). In layman's terms, this means that a given "particle" of fluid remains in the same precise 2-d location relative to the the "side walls" that define its pathway. Turbulent flow means that the true path of any given particle may not be analytically determined (ie: it's random). These are broad definitions and there is a possibility for both conditions to co-exist in a given setup (at different locations or different times). Specifically, because fluid at a surface essentially has zero velocity, you'll always have a laminar region within the boundary layer of a turbulent flow regime. The thickness of this laminar portion will vary with Reynolds number. Perhaps this is what Alchemy has been driving at.

I'll admit, it's been a long time since I took a hard look at the distinctions between laminar and turbulent flow, but that's how I remember it. I'd appreciate any comments/corrections.

[morphling1]

Ups I missed that Alchemy didn't use kinematic viscosity, but rather dynamic viscosity; kinematic visc. = dyn. visc. / denstity.
So that's the same equation.

But I agree that lamilar flow cant' occur in waterblock. As I know lamilar flow is actualy very hard to achieve, but there is rather more turbolent or less turbolent flow. Unideal surface quality has a lot to do with this too.
Definition of Laminar flow is following: this is the flow where particles are moving in infinetly thin layers which slides between themself without mixing .

With turbolent flow particles are moving iregulary in all directions

I hope I wrote correctly so it is understandable (english isn't my native language)

[bigben2k]

Many good comments, thanks to all.

What's obvious here is that turbulence from speed is directly proportional to flow rate. If I had a target Reynolds number, which I don't right now, I'd have a better idea of where I'm headed. It'll come, in time.

MadDogMe: the flow speed is reduced in the outer channels on purpose. The speed of the coolant is highest where it is important for it to be. Having more channels in the outer area reduces the overall pressure drop, or flow restriction.

There is a certain logic behing round channels: since a circle is the optimal shape for the greatest area with the smallest perimeter, the boundary layer is reduced to a minimum.

As for Radius' Reynolds number, estimated to range between 2600 and 1300 throughout the design, as Alchemy says, there's some error. In the design, I was hoping that the 90 degree bend from the center inlet would have some effect.

What I'm shooting for (and forgot/put-aside in this design) is the addition of turbulators specifically to create vortices over the baseplate and fins. I've been doing a lot of reading about vortex flow meters, and I'm picking up gobs of good information.

Alchemy: I'm going to refer you to Cathar's block Click me! . I think you'll see some similarities, and some differences. Maybe you could give us some comments on it?

Some notes:
-1mm baseplate
-channels/fin width = 1 mm
-Center inlet, dual outlet
-7 fins, 8 channels.
-channel depth = 5 mm

Myv65: You wrote "Laminar flow will be upset by any change in cross sectional area, shape, or sometimes even deviation from a straight path". That reminded me that the channel width in this design varies from 1.0mm to 2.0mm. I still plan on adding more turbulence, but I'd like to know your opinion on this varying channel width, as it relates to turbulence.

To throw the discussion into turbulence, I'll post a link to this most interesting article by Mike Larsen.

In the mean time, I'll post a link to Nicozeg's thread, and his excellent waterblock design.

Click me!

Some notes:
-the base of the block has ridges, to create turbulence.
-The top is conical, not flat.
-there are no fins



Here's a preliminary idea of a revision to Radius.

[bigben2k]

Googl'ing through the web, it seems that turbulent flow is achieved somewhere between Reynolds 2000 to 4000, as far as I've searched (opinions DO vary!).

Here are some of the links that I found.

http://www.mas.ncl.ac.uk/~sbrooks/bo...p07/node9.html
("Turbulent Reynolds stress" for Dave!)

http://www.wikipedia.org/wiki/Reynolds_number
(basic definition)

http://www.ichmt.org/abstracts/MECT-...tracts/2-2.pdf
(You'll like this one Dave!)

http://www.nag.co.uk/simulation/Fast...html/node8.htm
("Theory of laminar and turbulent flow")

http://psdam.mit.edu/2.000/Administr...ulent-Flow.pdf
(simplistic, but there)

http://www.efm.leeds.ac.uk/CIVE/CIVE..._turbulent.htm
(dye experiment)

http://www.sigmaxi.org/amsci/article...demenos-5.html
(a medical perspective!)

http://wuche.wustl.edu/~sato/flowtrans/flowtrans2a.html
(Reynold's experiment)

http://www.icase.edu/Dienst/Reposito...e/TR-99-33/pdf
("Streamwise Vorticity Generation in laminar and turbulent jets")

http://home.olemiss.edu/~cmprice/lectures/turb.html
(some basics about turbulent flow)

http://www.uts.com/products/tkintro.html
(A software)

http://www.me.mtu.edu/courses/me328/...328formula.pdf
(A formulae roundup)

[myv65]

Originally posted by bigben2k
Myv65: You wrote "Laminar flow will be upset by any change in cross sectional area, shape, or sometimes even deviation from a straight path". That reminded me that the channel width in this design varies from 1.0mm to 2.0mm. I still plan on adding more turbulence, but I'd like to know your opinion on this varying channel width, as it relates to turbulence.

To throw the discussion into turbulence, I'll post a link to this most interesting article by Mike Larsen.

I have never spoken (at least that I know of) to Mike Larsen. I'll say only a couple of things about that article, which I did read quite some time ago.

One, those diagrams look suspiciously similar to those that exist in the fluids text I used in college. [edit start] Glad to see credit was given to the source. [edit end]

Two, all those diagrams have the same basic premise. They all pertain to an "infinite flow field". This means the conditions upstream of the "x=0" or the "entrance point" are uniform. Once you toss out this premise (as will happen anytime you go through a bend or real life fitting) things get a whole heckuva lot more interesting and complicated.

To put it bluntly, I thought it was a decent discussion but rather misleading. Call it the difference between understanding and applying concepts. Pretty much all students that make it through introductory fluids will understand the concepts of laminar/turbulent, entrance length, etc. Not too many really get an appreciation for the practical aspects of how these things exist in real life.

A lot can happen when trying to write technical stuff for a non-technical audience. Believe me I know. Perhaps Mike has a solid grasp on all of this and didn't wish to bog down the readers. I can't say for certain.

[utabintarbo]

Originally posted by myv65
I have never spoken (at least that I know of) to Mike Larsen. I'll say only a couple of things about that article, which I did read quite some time ago.

One, those diagrams look suspiciously similar to those that exist in the fluids text I used in college. Makes me wonder why there's no bibliography covering what are obviously copywrited diagrams.

From the end of page 2:

All images courtesy of "Fundamentals of Heat and Mass Transfer" by Incropera and De Witt.

Just so nobody gets the wrong idea!

Bob

[myv65]

I heartily retract point one from above. My eyes and brain may not be good enough to recall that part of the text, but at least I remembered the diagrams.

[BillA]

and the above is why, despite the passage of 50 years, that
"Flow of Fluids through Valves, Fittings, and Pipe"
- Crane, Technical Paper No. 410
is still a reference today, in print, and as interactive software
(~$450 as I recall, I use the book @ $40)

as an experimentalist, it is most informative

EDIT:
I do know Mike, he posts on OC and OCAU as Aesik
one needs to note the intro
"the basic physics behind flow in a water cooling system"
it was written for those with no technical education - how far can a single article go

for some background the OC threads on 'turbulence' are illustrative
here are a few words by Mike (Aesik)
(and the usual misbehaving types)

[myv65]

Originally posted by unregistered
I do know Mike, he posts on OC and OCAU as Aesik

So the "has a solid grasp on all of this and didn't wish to bog down the readers" is the correct answer.

Thanks for chiming in, Bill.

[bigben2k]

For what it's worth...

What I got from Mike's article is that the boundary layer restricts the flow. In a circular cross-section, the flow is higher in the middle. I also deduced that this would be similar in non-circular channels. Turbulence is wanted for improved heat dissipation. Putting it all together, what we want is turbulence at, or near the boundary layer.

Upon closer examination, it seems that the turbulence provided by high speed flow alone, doesn't fit with what's needed to improve cooling.

[utabintarbo]

Originally posted by bigben2k
Putting it all together, what we want is turbulence at, or near the boundary layer.

Perhaps I've misinterpreted what I've read (or skimmed ), or perhaps I am merely saying what you said in a different way, but is turbulence not sought to reduce the boundary layer? I think one causes the other, no?

Bob

[bigben2k]

Originally posted by utabintarbo
Perhaps I've misinterpreted what I've read (or skimmed ), or perhaps I am merely saying what you said in a different way, but is turbulence not sought to reduce the boundary layer? I think one causes the other, no?

Bob

No, that's the whole issue.

If you look at the first graph in Mike's article, there's what is labelled as "Buffered Layer" and more importantly "Laminar sub layer".


This is why a rough surface (sandblasted?) is better, because it helps break up this laminar sub layer.

[bigben2k]

Incidentally...

The surface roughness adds to the flow restriction, as per this formula:

f=0.2083 (100 / C) ^ 1.812 * (Q^1.852 / d^4.8655)

Where
f = friction loss per 100 feet in feet of water
C=roughness coefficient (Hazen-Williams factor)
Q=flow rate (gpm)
d=inside diameter of pipe (inches)

Source: http://www.ppfahome.org/pdf/pvcpipewaterspec.pdf

For PVC piping, a value of 150 is used for C. A steel pipe would have a C value of about 100, and a rusty one about 80.

While on the topic, the optimal tubing size should allow a flow speed of no more than 8 fps (feet per second), otherwise the tubing becomes a significant restriction. A flow speed of 5 fps is the guideline for 1 inch ID and higher.

[myv65]

Originally posted by bigben2k
While on the topic, the optimal tubing size should allow a flow speed of no more than 8 fps (feet per second), otherwise the tubing becomes a significant restriction. A flow speed of 5 fps is the guideline for 1 inch ID and higher.

This guideline works fine for industrial pumps, but would cause most aquarium style pumps to struggle. Low speed is the weak pump's friend. Save the high speed for where it is needed, and that isn't in the tubing.

[bigben2k]

Agreed.

I ran the numbers last night, and 1/2 inch tubing will have a flow velocity of about 5 fps, with a 3 gpm flow rate.

If I shoot for 5 gpm, I will have to go to 3/4 inch tubing, because 5 gpm in 1/2 inch is 8 fps!!!

[bigben2k]

Well, it's official.

It's going to 3/4 inch tubing.

Now all I have to figure out is why my pump has 1/2 fittings... Otherwise, the pump is at its max P/Q curve with a flow of about 250 gph (4 gpm), at which point the head is 11 feet. That converts to 0.42 atmospheres, or 4.4 meters of water, or 323 mm of mercury, or 0.43 bars, or 43 kPa, or 6.25 psi.

I'm going to try to use PVC (CPVC) parts to construct that "cube-res", starting with this:

[bigben2k]

Here's a diagram:

As I explained it, the tee (1 1/2, threaded at all ends) is capped with 1 1/2 to 3/4 threaded adaptors at the in/outlet. The inside pipe could either be a 3/4 piece of PVC piping (which I'd bend with heat) or the tube (it's possible to mount a barb on both sides of the adapter). The branch, which goes to the block, would have a 1 1/2 barb screwed into it.

[bigben2k]

Here's an overview of it all (not to scale):

[edit]
(updated pic, with more details)
[edit]
(closer to scale)

[BillA]

so you would couple "fitting" and "high flow" ?
truly Ben, you do not listen (or are incapable of comprehending ??)
just post, and post, and post some more

have fun, too much for me (I'll leave you in peace)

[utabintarbo]

Originally posted by unregistered
so you would couple "fitting" and "high flow" ?
truly Ben, you do not listen (or are incapable of comprehending ??)
just post, and post, and post some more

have fun, too much for me (I'll leave you in peace)

I think what "Mr. Sunshine" is referring to is the barbs (allegedly) necessary at the intake side (to block) of the separator tee. If there is a way to perform the transition without "fittings", this would be preferable, no? I am referring only to the intake side - the "exhaust" side is less relevant due to your use of an air trap/reservoir.

Perhaps you could look into allowing the intake hose to go thru the separator tee, and simply sealing around it at its entrance point. Not the most reliable solution, I admit. Hmmm...


Bob

[bigben2k]

He he.

What "M. Sunshine" might have missed is that the tee is 1 and 1/2 (1.5) inch in size, which has an ID of about 1 7/8 (1.875) inch.

Last I checked, a tee of that size wasn't significantly restrictive, even theoretically, given a 4 gpm flow!

edit: calculated, the pressure drop across the run of a tee, 1 1/2 inch at 4 gpm is approximately the equivalent of a quarter inch of head.

Given that the expected head is about 11 feet, I'm not going to loose any sleep over it...

[bigben2k]

Originally posted by morphling1
As I know Reynolds number is v*d/kinematic viscosity

v.... velocity [m/s]
d... diameter [m]
kinematic viscosity [m2/s]

if the chanell isn't round you have to take the equivalent hidraulic diameter d' = 4*A/C
A... cross area [m2]
P... circumference [m]

so for bb2k rectangular chanell d'=2ab/(a+b)
So you can see that narrow rectangular chanell isn't realy too good for introducing turbolence round chanell is much better.... but with central nozzle and short paths I realy don't think that there would be lamilar flow in that kind of block

It's Saturday... I've tested my airtrap (with some success: the thin wall vynil tubing just collapses, otherwise all air is cleared within 10 seconds).

I'm recalculating the different radius' involved: boy was I off!

Doing a quick recalc of Reynolds# at center, assuming the last design posted (center square post):
-opening size (for trial): 5mm diameter, aka 3/16 inch.
(At 4 gpm, I loose 2 feet of head!)
-Velocity: 40 m/s
-4 openings measuring 1 by 5 mm (still not accurate)

if the chanell isn't round you have to take the equivalent hidraulic diameter d' = 4*A/C
A... cross area [m2]
P... circumference [m]

so for bb2k rectangular chanell d'=2ab/(a+b)

(I assume that the "C" in the equation is "P"?)

where
a=.001
b=0.005
A=0.000005
P(or C?)=0.031

so
d'=.00167 (using: 2ab/(a+b) )
multiply by 4 (4 channels)
d' becomes .0067

so Reynolds = ... What is "dynamic viscosity"?

[myv65]

Originally posted by bigben2k
-Velocity: 40 m/s

And just what pump do you think will manage this? How pray tell did you arrive at this number? You'd need at least 80 meters of static head to generate a velocity of 40 m/s. Maybe you missed this musing of mine a while back.

Water's dynamic viscosity varies with temperature (as do all liquids). At a reference temperature of 20°C, the dynamic viscosity is about 8.9*10-4 N-s/m^2.