Watercooling: Does Waterflow speed matter?

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kleinwl

Senior member
May 3, 2005
260
0
0
cyclonWizard,

I agree that the Nusselt number will increase as the flow velocity increases in normal flows... which will effect the heat transfer coefficent. My point is that in the reynold calculation that you have shown is overly simplified and does not simulate the actual effects of the jet turbulence.

The flow is laminar exiting the nipple of the waterblock... I think we can both agree with. However, the nipple acts as a trip, increasing the turbulence of the water flow. The boundary layer seperates and the jet of water impacts directly upon the waterblock.

In an impinging jet flow, most turbulation models do not accuratly predict the mean velocity and reynold stress preditions in the near-wall region of the flow. At distances well removed from the point of impingement the Nusselt number predictions are accurate... however due to failure to show acceleration of the flow along the impingement plate, the heat transfer predictions are poor. This is due to the assumptions in 2 equation turbulence models. The jet flow, after exiting the nipple, undergoes streamline curvature and is subjected to strong deceleration as the fluid nears the impingment plate. After the jet impinges on the plate, the fluid accelerates radially, forming a 3 dimensional flow field. Turbulent vortices generated in the jet shear layer will be streatched parallel to the plate after impingement, altering the turbulent length scale of the flow. 2 equation turbulence models assume a linear stress-strain relation and a length scale based on thin shear layer approximations. Therefore, these equations can be expected to perform poorly near the point of impingment where these assumptions are not valid. At locations well removed from the point of impingement, the flow approximates that of a 2D boundary layer and the 2 equation models of turbulence perform reasonable well. [copied without permission from the CFD society of Canada] http://www.cfdsc.ca/english/benchmarks/cfd95/node1.html

I believe, based on the above the although velocity will or can have some effect on the turbulation of the flow and thus the heat transfer... the equations that you are showing are incorrect... and in fact that the effects of a velocity increase are negligible when compared to the rapid decelleration and turbulation caused by the jet hitting the waterblock.

And yes... I am an engineer... even if my spelling sucks... they need a spell checker in this thing!!
 

CycloWizard

Lifer
Sep 10, 2001
12,348
1
81
Originally posted by: kleinwl
And yes... I am an engineer... even if my spelling sucks... they need a spell checker in this thing!!
First, turbulation is not a word: turbulence. Second, dynameter isn't a word. It doesn't even sound like diameter. You may very well be an engineer, but your contributions don't indicate that your expertise is in the area of discussion. I can readily solve 2-D boundary layer problems analytically. However, that doesn't mean that they're relevant to the discussion at hand. The long and short of this discussion was well summed up by Mday:
No, there would not be a point where the speed of the water would cause any more heat than cooling. However, there is a point where the speed of the water would make the cooling process less efficient. Furthermore, there is also a speed of the water which would cause leaks due to the forces placed at the joints.
This is all that needs to be said. For modeling efforts, you can make them as overcomplicated as you'd like and achieve an extra 0.01% accuracy - feel free to do so. Maybe you can even get a publication out of it or something. For myself, I prefer models that give accurate, analytical solutions so I can see exactly which parameters are useful to vary to optimize the system. Feel free to use CFD, but there is no guarantee your results are any more accurate than my own. Based on the assumptions that you put forth here, I very much doubt that your results would be as accurate as mine.
 

Dough1397

Senior member
Nov 3, 2004
343
0
0
yawns... you guys are really smart.... i read so much and yet i understand so litttle....
 

kleinwl

Senior member
May 3, 2005
260
0
0
turbulence:
noun: unstable flow of a liquid or gas
noun: instability in the atmosphere
noun: a state of violent disturbance and disorder (as in politics or social conditions generally)

dynameter = diameter... give me a break!

I think this discussion is done. I explained the practical way of optimizing a system... but you were more interested in showing off with textbook formula. I said, if you really want to technical then you need to treat the system as the 3D system it actually is. But, no... that's too much. so do whatever you want.
 

CycloWizard

Lifer
Sep 10, 2001
12,348
1
81
Originally posted by: kleinwl
turbulence:
noun: unstable flow of a liquid or gas
noun: instability in the atmosphere
noun: a state of violent disturbance and disorder (as in politics or social conditions generally)

dynameter = diameter... give me a break!

I think this discussion is done. I explained the practical way of optimizing a system... but you were more interested in showing off with textbook formula. I said, if you really want to technical then you need to treat the system as the 3D system it actually is. But, no... that's too much. so do whatever you want.
Turbulence != turbulation. I never said turbulence is not a word.

The method you described is NOT practical for anyone in this forum except probably you and myself. The method I described, while not exact (neither is yours), can be accomplished without the use of software that costs thousands of dollars. I'd also be willing to bet that the results obtained using my methodology would be within 1% of the experimental results. But, even without all my handwaving arguments, I can say with absolute certainty that the trends I stated will be 100% correct: increasing the fluid flowrate WILL increase heat transfer. At some point, it's not worth increasing it any further. The point at which the tradeoff is no longer beneficial is in the eye of the beholder, not in a CFD solution.
 

nailzer

Junior Member
Aug 11, 2005
17
0
0
Originally posted by: CycloWizard
Originally posted by: check
if we are assuming the water is going to be a X temp coming out of the radiator no matter what, then yeah I agree the faster the better.

But the faster you pump the water the faster is is going through the radiator, therefore the warmer the water will be when it gets to your waterblock on your cpu...

it's begining to sound like a calculus max/min problem to me >_<

Maybe my thinking is ass backwards though...
Actually, the faster you move the water through the radiator, the more it will cool down. Generally speaking, the faster a fluid flows, the more quickly it will equilibrate with its surroundings. This is because as the particles/molecules move past each other at higher and higher rates (corresponding to higher and higher velocities), they have greater numbers of interactions. These interactions cause the energy to be transferred from one particle to another, so the more interactions you have, the greater uniformity of temperature you will have.


Uhh no. The faster you move water through a radiator the less heat transfer will occur. If sufficent heat transfer does not happen at the radiator the whole system heats up. There has to be a speed that accomidates the size of the cooling system with the amount of fluid present, and the physical size of the radiator.
 

CycloWizard

Lifer
Sep 10, 2001
12,348
1
81
Originally posted by: nailzer
Uhh no. The faster you move water through a radiator the less heat transfer will occur. If sufficent heat transfer does not happen at the radiator the whole system heats up. There has to be a speed that accomidates the size of the cooling system with the amount of fluid present, and the physical size of the radiator.
Uhh, no. While what you're saying might make sense in a batch system where there is not continuous looping flow, this is not the situation at hand in a watercooling system. The thermal boundary layer within the fluid will decrease as the velocity increases, thereby increasing overall heat transfer. This is day one of transport phenomena - basic film theory. If you disagree, I can't say anything but that you're wrong.
 

nailzer

Junior Member
Aug 11, 2005
17
0
0
WOW! Great! So I can go out to my car, take the radiator fan off, as long as I keep the RPMs up, and the water and antifreeze is moving really really fast, I'll never have to worry about it over heating. Great way to get a little more horsepower too! But if it overheats and blows up I'll send you the bill, OK?
 

CycloWizard

Lifer
Sep 10, 2001
12,348
1
81
Originally posted by: nailzer
WOW! Great! So I can go out to my car, take the radiator fan off, as long as I keep the RPMs up, and the water and antifreeze is moving really really fast, I'll never have to worry about it over heating. Great way to get a little more horsepower too! But if it overheats and blows up I'll send you the bill, OK?
In almost no auto design does the radiator fan lessen the produced horsepower by any significant margin, even in the instance of a belt-driven fan. In my car, it's electric, so nothing at all would be gained. If you knew much about many cars, you'd realize that the radiator fans on non-belt-driven models doesn't run most of the time - it's thermally controlled to only turn on when the natural convection caused by the car's own velocity is insufficient to remove all of the heat generated by the engine.

You are confounding several different heat transfer resistances and trying to keep them all under the same umbrella. Unfortunately, the fan does not aid the heat transfer from the coolant mixture to the metal. It simply lessens the resistance to heat transfer at the surface of the radiator by, again, decreasing the thermal boundary layer thickness. The fact that you recognize the importance of a fan in creating high velocities for thermal transfer but can't see the mirror image for the hot side of the heat exchanger is clearly demonstrative of your lack of understanding of the situation.

However, since I'm sick and tired of being talked down to by high schoolers, I am going overboard to demonstrate just how right I am. Yes, my modesty is lacking, but ignorant impudence can only be overcome with education.

http://it.che.wustl.edu/~reillyma/watercooling/velocity-heatflux.htm
 

nailzer

Junior Member
Aug 11, 2005
17
0
0
Umm yes there is horsepower loss, from the fan belt and fan, clutch or not, and if you take off the mechinical water pump and replace it with an electric water pump you would see and even greater gain.
I get tired of being talked down to by someone that thinks their technobabble is in any way impressive, and actually has no clue what they are talking about.
 

CycloWizard

Lifer
Sep 10, 2001
12,348
1
81
Originally posted by: nailzer
Umm yes there is horsepower loss, from the fan belt and fan, clutch or not, and if you take off the mechinical water pump and replace it with an electric water pump you would see and even greater gain.
I get tired of being talked down to by someone that thinks their technobabble is in any way impressive, and actually has no clue what they are talking about.
Um, I answered your jackassery in my first paragraph of my previous post:
In almost no auto design does the radiator fan lessen the produced horsepower by any significant margin, even in the instance of a belt-driven fan. In my car, it's electric, so nothing at all would be gained. If you knew much about many cars, you'd realize that the radiator fans on non-belt-driven models doesn't run most of the time - it's thermally controlled to only turn on when the natural convection caused by the car's own velocity is insufficient to remove all of the heat generated by the engine.
Simply because you havea belt-driven fan does not mean all cars have them. Simply because it's belt-driven does not mean it significantly impacts horsepower. If you knew much about fan performance, you would know that at most it's putting out 1-2 hp of work, so it might draw up to 3-4 hp. Even on my four-banger, that's negligible.
 

Jmmsbnd007

Diamond Member
May 29, 2002
3,286
0
0
Originally posted by: nailzer
Umm yes there is horsepower loss, from the fan belt and fan, clutch or not, and if you take off the mechinical water pump and replace it with an electric water pump you would see and even greater gain.
I get tired of being talked down to by someone that thinks their technobabble is in any way impressive, and actually has no clue what they are talking about.
Because you don't understand it, that somehow means that he's wrong, right?
 

nailzer

Junior Member
Aug 11, 2005
17
0
0
q]Originally posted by: CycloWizardIn almost no auto design does the radiator fan lessen the produced horsepower by any significant margin, even in the instance of a belt-driven fan. [/L][/quote]

Then you should look up "Flex Fan"

"Flex Fans pull additional air through the radiator at low RPMs when the speed of the vehicle is not enough to cool the engine's coolant system. At higher RPMs, when the vehicle's speed forces ample air through the radiator, the fan blades flatten out, resulting in less power being used to spin the fan. Engine performance and gas mileage are thus improved."

http://www.perma-cool.com/Catalog/Cat_page20.html
 

CycloWizard

Lifer
Sep 10, 2001
12,348
1
81
Originally posted by: nailzer
Then you should look up "Flex Fan"

"Flex Fans pull additional air through the radiator at low RPMs when the speed of the vehicle is not enough to cool the engine's coolant system. At higher RPMs, when the vehicle's speed forces ample air through the radiator, the fan blades flatten out, resulting in less power being used to spin the fan. Engine performance and gas mileage are thus improved."

http://www.perma-cool.com/Catalog/Cat_page20.html
And you actually believe that this will save you more money in gas than the price of the fan? If so, I have a bridge to sell you...
 

Lazien

Member
Jul 18, 2005
128
0
0
Originally posted by: nailzer
q]Originally posted by: CycloWizardIn almost no auto design does the radiator fan lessen the produced horsepower by any significant margin, even in the instance of a belt-driven fan. [/L]

Then you should look up "Flex Fan"

"Flex Fans pull additional air through the radiator at low RPMs when the speed of the vehicle is not enough to cool the engine's coolant system. At higher RPMs, when the vehicle's speed forces ample air through the radiator, the fan blades flatten out, resulting in less power being used to spin the fan. Engine performance and gas mileage are thus improved."

http://www.perma-cool.com/Catalog/Cat_page20.html
[/quote]



http://www.edmunds.com/reviews/list/top10/103164/article.html
http://www.edmunds.com/ownership/maintenance/articles/105528/article.html
http://www.consumerreports.org/main/det...d=113261&ASSORTMENT%3C%3East_id=333137

I also read a study done by the EPA which stated that there is currently no gasmilage devices that actually work. (can't find it right now though)

With this mentality wouldn't You think that for say $5-10 any car manufacturer would put that fan in just to show a higher gas milage rating for their car? Damn bro with gas prices as high as they are that would be a SURE selling point for the insignificant cost.

If You read these You'll notice that the increased drag on Your car from having the windows rolled down will actually give You worse gasmilage than using the AC as stated in the above article, so I wouldn't go pulling off My AC belt to save that milage either lol

My 2 cents
 

Spencer278

Diamond Member
Oct 11, 2002
3,637
0
0
Originally posted by: nailzer
Umm yes there is horsepower loss, from the fan belt and fan, clutch or not, and if you take off the mechinical water pump and replace it with an electric water pump you would see and even greater gain.
I get tired of being talked down to by someone that thinks their technobabble is in any way impressive, and actually has no clue what they are talking about.

Yes you do lose power from the belts that power the water pump. Of course you will lose power from the belts that drive the alternator.
 

Intelia

Banned
May 12, 2005
832
0
0
I do alot off racing my son drives I build the engines . They produce right around 600Hp. So you can imagine the heat . We use a High volum pump because a stock pump won't cool it. Than we have to restrict it so as to get the right flow otherwise the water goes threw the rad. to fast and won't cool. Get the point. Zinn2b
 

CycloWizard

Lifer
Sep 10, 2001
12,348
1
81
Originally posted by: Intelia
I do alot off racing my son drives I build the engines . They produce right around 600Hp. So you can imagine the heat . We use a High volum pump because a stock pump won't cool it. Than we have to restrict it so as to get the right flow otherwise the water goes threw the rad. to fast and won't cool. Get the point. Zinn2b
:roll:

If the point is that anecdotal 'evidence' trumps actual science in this thread, then I got the point a while back.
 
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