"Radius" by BigBen2k

[Alchemy]

Originally posted by myv65
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.

I think you misread. That's its viscosity at 25°C.

Rather big difference five degrees makes. Viscosity is incredibly temperature-sensitive.

Alchemy

[morphling1]

Wow being optimistic bb2k, for 40m/s to be pushed through 4*5mm2 openings you need 2880 L/h pump, and if you're thinking of pushing that kind of flow rate through 20mm2 opening , you'll need "slightly" different type of pump

[myv65]

Originally posted by Alchemy
I think you misread. That's its viscosity at 25°C.

Rather big difference five degrees makes. Viscosity is incredibly temperature-sensitive.

Alchemy

Could well be as my texts are at work and I was trying to recall from the top of my head. Details would appear to be like hair in that there is less up there with each passing day. LOL.

[bigben2k]

Ah yes, that was in feet per second, oops, my bad!


Anyways, I figured that the nozzle is either going to be 1/2 inch diameter, or 3/16:

3/16 nozzle: the opening of such an area is 17.8 mm^2, and that leaves an exit through the fins that can best be described as 8 channels (1 by 5) so 40mm^2, for a ratio of entrance to exit of 0.45 .

Flow speed is 14.2 m/s (46.5 fps). Pressure drop over one inch is 11.7 feet of head.


1/2 nozzle: the opening is 126.7 mm^2, with an exit of 16 channels, 80mm^2, for a ratio of 1.6 .

Flow speed is 8 m/s (26.1 fps). Pressure drop over one inch is 0.1 foot head (1.2 inch).

The above ratios are the smallest and largest, respectively.


Here's the bug: what am I trying to achieve with this inpingement? Since the flow rate is the same everywhere, the exit speed will be higher than the entrance, with a 1/2 nozzle. Using a 3/16 nozzle, the entrance speed is higher, but the exit speed is lower.

So if I go with a 3/16 nozzle, I'll have high-speed (and turbulent) flow within a very small area, which has a diameter of about 5 mm. The nozzle will have to be kept short, given the very high pressure drop.

Note that the turbulence that occurs due to the 90 deg flow bend hasn't been accounted for, yet.


Other progress:

Round saw blades:
The 20mm saw blade, if used to cut 5mm deep, will extend 13.2mm beyond the point of cut, in either direction.
25mm: 15 mm
32mm: 17.2 mm

Which makes the blades pretty useless except for the central cuts, which leaves a center square post.


Turbulator thoughts:

Turbulators at the bottom of the channel: using a 1.0mm drill bit, drill a hole every other 1.0mm, to a depth of 0.5 mm.

Turbulator at the fin: don't leave the fin tips at an angle, cut them off at 90 degrees. Necessary?

[edit]: added depth-of-cut to bottom-of-channel turbulator.

[myv65]

Still seems awfully optimistic. Make it and measure it. Reality is the best and most objective means of determining true flow rates. Once you've tested both options to determine true flow and performance, the better one will be clear.

[Dix Dogfight]

@BB2K
Why not use a straight inlet?????
To eliminate a bend in the piping where the water velocity is high.

Nice to see you are making progress with your design.
I'm looking forward too see it in action.


cheers

EDIT: It might not matter but you are aiming for a top performer right?

[bigben2k]

Originally posted by Dix Dogfight
@BB2K
Why not use a straight inlet?????
To eliminate a bend in the piping where the water velocity is high.

EDIT: It might not matter but you are aiming for a top performer right?

The problem is that this 1 1/2 (1.5) tee is about 3.5 inch wide, at the branch. Since there's only 8 inches in width in the case (which I assume is typical), that would only leave a couple of inches for a tube to bend, with a "straight inlet". That bend would make it all much worse for flow (I'll be using thick braided 3/4" tubing).

The inner tube would be bent in the largest radius possible. I'm even considering using a drain trap tee, which has a nice curve from the branch to one of the runs (ends).

The only thing I have to add about it is that the water return should exit downward, to make draining this thing easier.

I am aiming for a top performer, yes, but I have A LOT of calculations to go through! I was hoping to do that Saturday, but I got caught up in all of the other calculations:

Fin set #2 starts (the pointy tip) at radius 2.5 mm
Fin set#2 taper ends at 3.3mm.

Fin set #3 starts at radius 4.6mm, and the taper ends at 6.4 mm.

fin set #4 does not appear in this design, because it starts at radius 9.0 mm, and the whole fin pattern has been reduced (temporarily) to 7.5, pending my thermal calculations.

I made a number of drawings (which I'll try to post this week), so that I can visualize the inpingement. I have more work to do on that too, using some info found here:
http://www.electronics-cooling.com/h...01_may_a2.html

and
http://widget.ecn.purdue.edu/~eclweb/jet_benchmark/

To be explored: using 4 nozzles/holes, instead of one.

Myv65: I'm not done with the numbers yet, to even say that I'm redy to experiment. I'd rather spend more time going through the numbers, until I get this thing to a point where I have a fair idea of how well it will cool. I know that ya'll just can't wait to see it in action, but unless someone is going to run the thermals for me, ya'll are going to have to wait!


I still need help on the Reynolds calculation.

[bigben2k]

Reynolds calculation

v=14.2 m/s
a=0.001 m
b=0.005 m
A=0.000005 m^2
P(or C)=0.031 m

d'=0.00167 by 8 channels= .0134

So Reynolds = 14.2 * .0134 / 8.9*10-4

= 107 ?

(didn't I say I needed help )

[bigben2k]

Originally posted by bigben2k
2" outer hose ugly? I don't think so... I'll pick up a foot of the stuff this weekend. We can apply our artistic interpretations then!

Originally posted by morphling1

Ok, I'll be waiting

Well, here it is, in all it's glory...

[bigben2k]

And the pic (DOH!)

(If you look REAL close, you can see me!)

[Dix Dogfight]

/ Wierd mode ON/
If you hadn't said it was a piece of vinyl tube. I would have guessed it to be a screenshot from the motionpicture Ghostbusters. It looks like a ghost trapped inside some kind of forcefield(with a little imagination). If you say you are in the picture. I can only deduct that you (BB2K) then must be the ghost and therefore you are inside the tube IN MY COMPUTER!!!!!!!
Are't firewalls supposed protect against all evil?????
/ Wierd mode OFF/

Errrrrrrrrrrrrrrrrrrrr

[myv65]

Originally posted by bigben2k
Reynolds calculation

v=14.2 m/s
a=0.001 m
b=0.005 m
A=0.000005 m^2
P(or C)=0.031 m

d'=0.00167 by 8 channels= .0134

So Reynolds = 14.2 * .0134 / 8.9*10-4

= 107 ?

(didn't I say I needed help )

Sorry Ben, but it's time to get yourself a decent fluids text. You are mixing "hydraulic diameter" and "characteristic dimension". A hydraulic diameter applies to calculating an equivalent diameter for a non-circular flow pathway. It's really useful when working with channels, which are typically rectangular or v-shaped (or something else comprised of straight sides). Before you try calculating and applying a Reynolds number, ya really ought to know what it represents (and why one would bother to calculate it).

Once you do that, you'll find that pressure drop across an orifice is not something that's analytically defined. There are general relationships that use coefficients dependent upon the shape of the orifice. The relationships are based on classical physics while the coefficients are the "as measured fudge factors" determined through careful lab experimentation over the years. You're really, really putting the horse ahead of the cart right now. You can run all the "calcs" you wish, but until you get the understanding behind them they are of little value.

Honestly you will be hard pressed to find existing data to determine how your intended entrance will perform with your pump (and the rest of your system). You can take pot shots at it, but the only way to be confident that you're even within +/- 20% is to build and measure. Even then, you'll only be just starting to find out what is really "best".

Find out more about orifice calcs, pressure drops, and Reynolds numbers. Rough out where you think you need to be (range of +/- 50%). Start prototyping some stuff to see where you really are.

[bigben2k]

All right. It turns out that Kryotherm has this calculation already, and that the dynamic viscosity (which is labelled kinematic viscosity) of water at 20 deg C = 1.006 E-6. At 30, it's .805 E-6 .

The thermal conductivity W/(m*K):
Water@20: 0.599
Water@30: 0.618

That's for a flow rate of 1 m^3/hr (264.2 U.S. gallons/hour).

Reynolds comes out at 15279, for a heatsink (cooled by water at 20 deg C), where the "forcer mounting" (whatever that is) is set to "on the top of the heat exchanger". The other option is too fuzzy, I'll have to read the help file. Heatsink is roughly the area of the first set of fins, before fin set#2, so fin width = 1.5 mm, channel width (I averaged between 1.0 and 2.0) to 1.5, length = 5 (from radius = 2.5 mm). Copper Thermal conductivity adjusted to 391 W/(m*K).

The resulting "Heat Exchanger thermal resistance" (K/W) = 0.420 .


Those are preliminary numbers. I'm just playing with Kryotesc today.

[myv65]

Might want to verify the units on viscosity. Using units of N-s/m^2, my tables show 1080*10E-6 at 290K and 959*10E-6 at 295K. As Alchemy noted, my first value should have read 8.9*10E-4 at 25°C. If you're using the same units, it sounds like you are off by a factor of 10. If you're wondering about the "10E-6" vs "10E-4" or the "K" vs "°C", it's the difference between a table in one book (my heat transfer text) and a graph in another (fluids text).

[Les]

Originally posted by myv65
If you're using the same units, it sounds like you are off by a factor of 10.

My only checks on the Kryotherm numbers suggest they are correct. eg
http://www.coolingzone.com/Content/D...iz/default.htm
However .......

[bigben2k]

Thanks Les.

The above link suggests that the "Dynamic Viscosity" of water at 21.11 deg C is 980 E-6 kg/ms .

KryoTesc shows this:
(the unit is fuzzy)

[bigben2k]

Now if I can only get Windows 2000 Pro to open the Russian help file...

[edit]
Got it. I added "Cyrillic" to "Regional settings".

[bigben2k]

[edit]
rambling removed
[/edit]

[Alchemy]

Originally posted by bigben2k
Thanks Les.

The above link suggests that the "Dynamic Viscosity" of water at 21.11 deg C is 980 E-6 kg/ms .

KryoTesc shows this:
(the unit is fuzzy)

Well, that dynamic viscosity is alright.

What do you mean "the unit is fuzzy"?

I can tell that those kinematic viscosity measurements don't coincide with either kinematic or dynamic viscosity units, if you didn't know that already.

I'm with myv65. We've given you our best guesses as to how this device might pan out, but aside from extensive computational modeling, you're not going to learn much more through theory.

I still think the basic idea of micro-channels is inferior to high-flow, low pressure drop design, but I would be happily surprised if you proved me wrong.

Alchemy

[bigben2k]

Thanks, but I'm not done yet.

I want to try using Kryotherm, and seperate the different rings, where the number of fins changes i.e. within radius #1, 4 fins, within radius #2, 8 fins, within radius #3, 16 fins, then equate each one with a square heatsink out of kryotherm.

Besides, I haven't figured out the entrance area yet...

The Kryotherm unit appears to be blocked out (bad programming?). One really can't tell what "10-6 *m/s" really means.

Note: this isn't relevant. Kryotherm displays this value, it's not a variable that can be changed.

[gone_fishin]

Originally posted by bigben2k
Thanks Les.

The above link suggests that the "Dynamic Viscosity" of water at 21.11 deg C is 980 E-6 kg/ms .

KryoTesc shows this:
(the unit is fuzzy)

Waterloo shows this, 9.5131E-4 kg/m.s for the dynamic viscosity of water at 21.11C.

[bigben2k]

9.5131E-4 kg/m.s is 951.31 E-6 kg/m.s .

Thanks G_F.

[myv65]

Originally posted by gone_fishin
Waterloo shows this, 9.5131E-4 kg/m.s for the dynamic viscosity of water at 21.11C.

This is why I said you might want to check the units (and why I listed the units I use). Since F = m * a, Newtons = grams (or kg) * m/s^2. If you replace N in my units above with kg-m/s^2, the resulting units will be kg/m-s. Just realize that when you plug all the units into an equation, such as for Reynolds number, you must be certain that all unit conversions are done.

[bigben2k]

Thanks for the pointer Dave!

(that's what I figured, but I removed my ramblings!).

On to the numbers...

[bigben2k]

Here's a hand drawing of the bottom turbulator:

It's a 1 mm drill bit hole, every 2 mm, 0.5 mm deep.

[bigben2k]

Here's another drawing of the cross-section:

[Ion]

Don't we see the main problem with cpu cooling here AGAIN?
We have limited space around our cpu, so we can't put the radiator directly on it, instead we use water as a transfering medium to transfer the heat from the cpu to the big radiator (simplified and not counting on all factors).
But now we're even having troubles transfering the heat from the center of the waterblock to the rest of it in order to use contact area in a good way.
Do you get my point? bigben2k had a rather nice design, but not good enough right where it counts. As I see it, the best design to use in the middle of the chip would be fins in a row, like the heatsinks om grapic cards memory. And that would spoil the whole radius idea. (And would probably be very hard to manufacture with small enough fins)
The main goal must be this: To as fast as possible transfer all heat from the center of the block, to a bigger surface. The rest of the block, or later, the radiator.
And this obviously can't be done much better with copper than we are already doing.

We must use heatpipes!

(don't you think?)

[bigben2k]

Well, unless you're trying to pitch a new product, I disagree. In fact I disagree anyways! If you're going to go with heatpipes, you might as well go with phase change cooling, and that's a whole different ballgame.

In order to "as fast as possible transfer all heat from the center of the block, to a bigger surface" is accomplished with fins. Copper is a limitation, but the only affordable solution.

The fin pattern may be optimal in a straight-through design, but that assumes an even heat source. This fin pattern has the same characteristics as a straight one, with one fundamental difference: the coolant speed is much higher in the center, where most of the heat originates. Optimal flow, for an optimal fin pattern.


Update: 3/16 is the nozzle of choice, for the moment. Still running the thermal calculations (spent a weekend with a cold).

Utabintarbo and I are going over minor details for the block, mount and seperating res. I've spec'ed out most parts, and will start purchasing soon. Polycarbonate top.

In the mean time, here's an example of Utabintarbo's excellent work: an AMD Athlon, rendered with all the pins!

[Albigger]

whoah! awesome rendering job there. i HAVE to ask though - why would you need the backside of a cpu modeled in 3D like that?


anyway - can't wait to see your design as a final product. I know you've put a lot of work into it.

So you are going to use a 3/16" nozzle to increase water velocity right over the center?

[bigben2k]

3/16 is in the works, but that'll be about the size of it, yeah.

Flow rate will still be in excess of 250 gph.

As for the render, no the pins aren't necessary: that's a tribute to Utabintarbo (too much time on your hands, buddy?).