Advise about distribution of jets

[dGawD]

Im about to make a jet block but still have a few questions regarding the distribution of jets. Which design do you think will allow better thermal transfer and thus better results? Also do you think that having 2 outputs will make it a lot less restrictive (i know this way ill be losing the "cascade-alike" effect but im planning on having others blocks in my system so i want as much water flow and pressure as i can).

Pics. (Made in AutoCad, holes are 2.5mm, barbs are 15.25mm. Let me know if you want some 3dmax renders and ill get to work )

[FireCrack]

I'm no expert on the subject at all, hell i just registered here to see the pictures! But i can still give my input i suppose.

I'd say that wich one's better would depend on the shape of the base, but generaly any diference would be neglible

[Long Haired Git]

Second design has 33 holes each 2.5mm holes, right?
Have you done the maths?
I can't see how so much surface area for flow can have holes acheiving anywhere near enough velocity to run jet impingement.
My maths shows a jet velocity of just 0.5m/s @ 5 LPM, and a Reynolds number of just 1500 @ 5 LPM. My pressure drop maths are flawed so I can't use them, but according to my Excel spreadsheet you need to flow like 21 LPM to get the jet velocity to what you need.

[BillA]

go look at Hoot's new wb on OCF

[dGawD]

Amazing the work hoot has done, very nice indeed.
But still, i cant do any circle shapes cause all of this work ll be done with a X-Y only mill and curves, well, they re forbbiden as i was told
So, as the desing has to be square like, and could or not have 2 outlets, what do u think is the best way to distribute the jets, like design 1 or 2.

p.d: Yea yea i know, read dont make us waste our time The thing is i have read about the 2 designs having excellent results but have to stick with 1.

[Long Haired Git]

dGawD, did you read my post?
If you run this block with any typical pump used in WC, then it will perform like a dog.
You need smaller jets or far fewer jets to be an impingement block.

[MadHacker]

Originally Posted by Long Haired Git

dGawD, did you read my post?
If you run this block with any typical pump used in WC, then it will perform like a dog.
You need smaller jets or far fewer jets to be an impingement block.

perhaps he read it but didn't understand it...
I myself don't understand what it means but i understand the conclusion in the above quoted post...

[dGawD]

I did read it and started looking for what u said before i could answer you,
So your saying that if i use 2.5mm pipes i should make fewer jets (cause i already bought the pipes and really cant go and buy another size, too much trouble here where im living). Lets say 19 or 21 jets, do you think it d make any difference, and it would be too restrictive?




edit/ Dont take into consideration the number of holes in the pics. They are just for u to have an idea

[flatline]

may i ask what the water jets will be hiting?

"cups" like cascade?
"slits" like ww?
"cut pins ++++" like alfacool block
"cut pins XXXX" like roscals paper looks at
"other" ??

i think it may give a beter idea of where jets shud be

tnx

[JFettig]

I would suggest you use the top one as your baseplate, if you group jets that close they wont have water flow around them to escape.
The main thing is that you need water to be able to escape the cups faster than it goes in for good impingment, otherwize theres no real purpose.

Jon

[dGawD]

Ok, i understand, gonna redesign it and see what comes out.
Flatline? Which one do you think perform better? I was thinking of doing cups like the cascade but now im not that sure

[flatline]

im no expert but after seeing the numbers on the alfacool block and as that block has had comments about its offset inlet and restricted out let in another therd and that alfa cool use the +++ pins not XXX sed to be better
(nice flow chart of terbulence on a linked artical somewhere)

so id say XXX pins or if u going jet/cup what about makeing pipes cone shaped bouth internaly (tiny file?) and externaly (no idea how)

just some ideas

[Long Haired Git]

dGawD - you can do the maths. The volume of liquid flowing through the rig is constant.
Say you're running 1/2" system and you want 3 litres per minute.
That equates to 52,000 mm3 per second.
So, you work out the SA of the hole inside each jet using Pi*(R^2), and times that by the number of holes. So, a 2.5mm diameter jet hole (inner diameter) has a surface area of Pi*1.25*1.25 mm2 = 4.9mm2, and all 20 jets thus sum up to a total of 98.2 mm2.
To put that number in perspective, a 12mm ID tube (ie, 1/2") has an internal SA of 126.7mm2.
Back to our 52,000 mm3 per second, you divide that by the number of mm2 that can pass through at one, ie the total SA of the jets, and you get the velocity of liquid required as mm per second. That number is 530 mm per second, which is 0.5m per second.
So, the speed of the coolant "rushing" out of the jets and "slamming" into the cups is basically the same speed (0.41m/s) as it would come out of a straight 1/2" tube.

"Reynolds" number is a number to describe liquid flow. For jets, I've heard (no reference) you really want to see R > 7000. You too can search and find the formulae for Reynolds number online, but my calcs show R for your rig of 941 @ 3 LPM.
In other words, woof woof.

Oh, the above calcs are for 20 holes at 2.5mm.

Now the next calc you need to be able to do is to calc the back pressure required to get the jet velocity up high enough to form good jets. I cannot get my numbers to publish numbers referenced by others. I get pressure heads for guesses at Cascade and G4 jets of 55m of H20, and that's just not right.
Speaking of which, I'd love someone to work a simple example here for the 20 x 2.5mm holes show how to get "right" numbers.

[Cathar]

LHG, get a hold of the pressure drop calculator from www.pressure-drop.com

[dGawD]

Damn nice explained LHG.
I searched the formula but there are a few things i dont quite understand yet. What do they mean by viscosity (could it be that its always 10 -6kg/ms) and density, or how can i find the mass velocity. (last year of school, didnt even heard the word reynolds -_- , but im a quick learner)
Also, how did u come up with the 52.000mm3 number if i may ask?

Cathar, that program is great i think i found my answer there, but still i wish to know the theory of reynolds.

[Long Haired Git]

Google and ye shall find.
Oh, I'm so nice:
dynamic viscosity = 0.0009 @ Around 25C (Table at http://www.engineeringtoolbox.com/24_596.html)

52,0000 mm3 is a mistake, BTW, its 50,000 for 3LPM and 52k for 3.12 which was a number I just happened to be working with in my Excel spreadsheet. I plan on posting the spreadsheet for all to use once I get it right. You can use "Goal Seek" in Excel to work out the flow in LPM.

Anyhoo, math is simple. f LPM = f/60 L/sec. A liter is 1,000,000 mm3 (from Google), so now we have 16,666.666666 mm3 per second per LPM. 3 LPM thus equals 50,000 mm3 per second.

In fact, working the entire equation through to get velocity in meters per second, I get:
velocity in meters per second = (1000*flow_in_LPM)/(15*Pi*internal_diameter_of_tube_in_mm * internal_diameter_of_tube_in_mm * count_of_jets )

So, in your example of 20 jets each 2.5mm @ my 3LPM:
v = (1000*3)/(15*3.14159*2.5*2.5*20)
v = 3000 / 5890
v = 0.51 meters per second.

Cathar - haven't done yet, but worked out my problem. I converted from Newton per meter square to meters of H20, but forgot "per meter squared". Still not right, but I'm closing in...

[dGawD]

Saw that site, but missed that, must be the hours :P
I finally understood it, thanks a lot.

[Long Haired Git]

Uggg, still can't get it.
Yes, I could use pressure drop, but I want my spreadsheet to do the calcs as then I don't need the app, and I can also include my own stuff.
Cross referencing the application (not reverse engineering) I have all the same numbers for dyn viscosity, density and I have triple-checked my conversion from Pascal to mH20.
Still not right.
The app takes friction into consideration, whereas my spreadsheet doesn't, but the spreadsheet is returning numbers 1/10th mine.


I mean, the equation is:
delta Pressure = ( Density/2 ) * ( v2^2 - v1^2 )
Me and the app agree density = 998.2 kg/m3
Me and the app agree dyn viscosity = 0.001002
Me and the app agree v2 = say, 9m/s

I thus get 40427.1 pascals.
Conversion factor is 0.000102 mH20 per Pascal
Hence 4.12 m of head.

The application (with zero roughness) gets 0.05 "bar", which converts to 0.5m of head using the apps own calculator. Obviously I'm wrong as none of these blocks could work if they were as restrictive as my calcs, but I am blind to my mistake....

Uggg, where am I going wrong?

[lolito_fr]

9m/s => 4.12m H20
Agree. This is simply pressure energy converted directly into kinetic energy (not taking into account frictional losses)
The 0.5m would account for the additional frictional losses in the nozzles (not the only ones, either!!)
Does this make more sense?

[Long Haired Git]

No, it doesn't, because 4.12m of head is a lot bigger than 0.5m.
If I got 0.412m of head and they got 0.5m, I could assume friction.
But I am a factor of ten **larger**.

[lolito_fr]

Both values are correct, but each figure is only giving you part of the total losses. You have to add the two values.

The equation you posted delta Pressure = ( Density/2 ) * ( v2^2 - v1^2 ) is based on Bernoullis equation , which is true for a non viscous liquid, ie no friction losses: perfect world.

The app (I suspect) is calculating only the friction losses, and for this you need another equation such as Hazen Williams.

To illustrate, if you have a 4.12m high column of water in a tube, tank or whatever and you have a hole in the bottom, then in an ideal frictionless world the water will be forced out of this nozzle at 9m/s. (Bernoulli)
If you connect a pipe to this hole, then in the real world friction will slow the water down. To get back up to 9m/s, you will have to make the water column higher by adding another 0.5m (in this case).
So you now need 4.62m for 9m/s.

[Hoot]

Woof Woof!

My current block with 61) 1.65mm ID nozzles (3/32" tubing) does not cool as well as my WhiteWater clone. Now I see why. Thanks guys. To avoid major rework, I'm going to try sliding smaller telescoping nozzles with an ID of .87mm (1/16" tubing) inside each one and retest. This will be a compromised effort from the start since the smaller tubing has an OD of 1.575mm and slide freely inside the larger tubing. They will drop down and rest against the bottom of the cups, but I'm guessing that the water pressure may lift them slightly to allow it to escape. "Hydraulic Lifters"

My results will be over on the original thread

I'm a hack, but I love playing in the mistique of not doing the math. Just gut feeling.

Hoot

[Long Haired Git]

Hoot, I'd strongly suggest blanking off some of those jets.
61 jets @ 0.87mm results in
2.3 m/s jet velocity
0.27mH20 head
Reynolds = 1991
@ 5 LPM.
Really need the velocity to be bigger than that in my ever humble opinion.

[Hoot]

Originally Posted by Long Haired Git

Hoot, I'd strongly suggest blanking off some of those jets.
61 jets @ 0.87mm results in
2.3 m/s jet velocity
0.27mH20 head
Reynolds = 1991
@ 5 LPM.
Really need the velocity to be bigger than that in my ever humble opinion.

Thanks for the input. I kind of went half way in this first step to assess the impact before going further. In 38 of the jets furthest from the core, I inserted short lengths of the smaller tubing and decided to retest. The AS5 has only had 5 hours to settle so far, but already the core temp as read with an external MAX6657 has dropped 1C. At 124 estimated watts using this on-line calculator in a 90nm core, the cooling challenges eclipse those of my former Tbred running only 70W. Add to that, the fact that I'm running a triple cascaded set of blocks (CPU to NB to GPU), eheim 1250 and Camaro heater core with a 17.6cm fan, head luckily is only .13m using 12mm ID tubing and all 12mm ID barbs. Here's a diagram of the change.



In a day or two, I will put the smaller tubing in all the jets and retest. I love experimenting, but I try to have a life besides. I sure appreciate the input and did not mean to hijack this thread. I don't always get the quality input at OCforums that is available here.

Hoot

[Les]

The influence of "Total Orifice area" for typical centrifugal pumps was discussed here.
The optimisation achieved by the "G series" was discussed here .