Watercooling: Does Waterflow speed matter?

[daines1]

Maybe not too technical, but here it goes:

Say you have a water cooling system for a heat source (CPU, VPU, Car Engine). Would there be a point where the speed of water running through the system cause more heat than cooling?

Would the friction of the water cause more heat than cooling?

daines1

 

[EightySix Four]

At some point I'm sure it would, but it would have to be going ridiculously fast.

 

[CycloWizard]

The speed of the water through the loop will definitely affect your temperatures. The faster the water is traveling, the more heat it can remove and the lower your temperatures will be. There are a few ways to look at why this is the case, but the simplest explanation is that increased velocity will result in an increase in the convective heat transfer (and therefore, the related heat transfer coefficient).

The friction increase due to increased velocity will not add any significant amount of energy (heat) to the water. This phenomenon (viscous dissipation) can be neglected unless you're using a very viscous substance like a liquid polymer. Since water is very 'thin'/not very viscous, this effect is trivial.

In short, pump the water as fast as you can. 😛 I also recommend avoiding the Cases & Cooling forum, as there is an abundance of misinformation to be had there.

 

[scsi drv1]

I would also take into factor the medium used for tranportation. Pending on how the water was transported to the object in mind it would be effected by conduction of heat through the medium.

 

[Calin]

Originally posted by: daines1
Maybe not too technical, but here it goes:

Say you have a water cooling system for a heat source (CPU, VPU, Car Engine). Would there be a point where the speed of water running through the system cause more heat than cooling?

Would the friction of the water cause more heat than cooling?

daines1

There would be a point where the water flow so fast it will heat because of that.
Let's try to estimate it.

Let's not talk about speeds, but heights. I drop 1kg of water from 1 km high. Its energy (potential energy) is 10N * 1000m = 10 000 joules.
Now water has 4.something Joules per gram caloric capacity, let's say 5 000 Joules per kilogram. As a result, the water will heat itself 2* Celsius, or some 3* Fahrenheit.
The speed is v=sqrt(2*g*h), so v=sqrt(20000), so 141 m/s, 500 km/h or 310 miles per hour.

Now, if you throw in the system water at 300 miles per hour and it will come to a complete stop on the cooler, its temperature increase is 3 degrees Fahrenheit.

Use the fastest water speed available, as long as the noise factor is under control

Calin

 

[Parkre]

You also want the water to be turbulant, not laminar. The hoses and elbows already do this, but using a wire mesh inside the tubing will create more turbulance and increase the coeffecient of convection....But I think everyone already knows that.

 

[dguy6789]

Waterflow is important, but you don't need anymore waterflow than what would come from an Eheim 1250 to be effective. What is more temperature changing is the waterblock the water is going through.

 

[Calin]

As long as increased waterflow brings you no temperature advantage on the processor die, increasing waterflow in meaningless. If you want to have your water increased by a single Celsius degree by flowing over the processor die, then (assuming a 170W overclocked processor), it needs to release 170J of energy every second. Let's call it 40 calories every second, and you need 40 grams of water every second to be heated 1 degree Celsius. These 40g (or 40 ml) a second adds up to 2.4 liters per minute or some 0.65 gallons a minute. However, if your water can increase its temperature not 1 but 10 Celsius degrees (18 Fahrenheit) the water flow can be reduced ten times to less than a tenth of gallon per minute, or six gallons an hour.

In the light of the great heat absorbtion properties of water, indeed much more important is the actual piece that transfer heat from processor to water.
EDIT: that piece is usually called heat exchanger. However, computer industry have its own nomenclature (waterblock I think)

 

[daines1]

Thanks for the reply. I was just sitting at work and the question popped into my mind.

daines1

 

[tinyabs]

I think it is similar to blowing a spoon of soup where the harder you blow, the faster it become cool.

 

[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...

 

[Check]

Originally posted by: Calin

There would be a point where the water flow so fast it will heat because of that.
Let's try to estimate it.

Let's not talk about speeds, but heights. I drop 1kg of water from 1 km high. Its energy (potential energy) is 10N * 1000m = 10 000 joules.
Now water has 4.something Joules per gram caloric capacity, let's say 5 000 Joules per kilogram. As a result, the water will heat itself 2* Celsius, or some 3* Fahrenheit.
The speed is v=sqrt(2*g*h), so v=sqrt(20000), so 141 m/s, 500 km/h or 310 miles per hour.

Now, if you throw in the system water at 300 miles per hour and it will come to a complete stop on the cooler, its temperature increase is 3 degrees Fahrenheit.

Use the fastest water speed available, as long as the noise factor is under control

Calin

How does the translational kinetic energy get turned into heat energy though in the situation?

 

[rgwalt]

Increasing the water flow rate won't necessarily reduce your temperatures. There are two sides to the heat transfer equation, and increasing the water flow rate quickly creates a diminishing returns situation where all you accomplish is reducing the temperature rise of the water instead of reducing the processor temperature.

R

 

[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.

 

[CycloWizard]

Originally posted by: rgwalt
Increasing the water flow rate won't necessarily reduce your temperatures. There are two sides to the heat transfer equation, and increasing the water flow rate quickly creates a diminishing returns situation where all you accomplish is reducing the temperature rise of the water instead of reducing the processor temperature.

R

The water temperature and processor temperature are inherently related. As you increase the water flowrate, the CPU surface will asymptotically approach room temperature (as will the water in the loop). When its outer surface is at room temperature, then you will also have the minimum temperature of the processor itself.

 

[Calin]

Originally posted by: check

How does the translational kinetic energy get turned into heat energy though in the situation?

Friction - you transform kinetic energy in heat like on the brakes of your car (if you have disk brakes, try to take their temperature even after some easy use. After long and hard braking, the water can boil on them - even after washing your car on automatic washers is recomended (if you have disk brakes) to brake a couple of times to eliminate the water from the disk brakes.

So, it could be friction, it could be by viscosity (which is again some kind of friction).

Some big trucks have "oil brakes" - the power shaft will have blades that rotate against some blades on the chassis. When oil is pumped, the blades will stir the oil, and the kinetic energy of the truck is turned in heat. Big advantage - nothing wears (but the oil get HOT)

 

[Calin]

Originally posted by: CycloWizard

Originally posted by: rgwalt
Increasing the water flow rate won't necessarily reduce your temperatures. There are two sides to the heat transfer equation, and increasing the water flow rate quickly creates a diminishing returns situation where all you accomplish is reducing the temperature rise of the water instead of reducing the processor temperature.

R

The water temperature and processor temperature are inherently related. As you increase the water flowrate, the CPU surface will asymptotically approach room temperature (as will the water in the loop). When its outer surface is at room temperature, then you will also have the minimum temperature of the processor itself.

However, there is a "heat transfer resistance" between the processor core (where heat is generated) to the inside of the waterblock (where water take that temperature). This is the core-to-heatspreader area, the heatspreader-to-waterblock area and waterblock-to-water. By reducing the temperature of the water in the waterblock to ambient, the actual temperature of the waterblock might be ambinent + 1F, the heat spreader could be ambient + 4F, and the core ambient+5F. Remember, when the temperatures of two materials in contact are the same, no heat exchange take place! You need different temperatures to have heat transfer.
The cooler the cooling fluid is, the cooler the core - but the processor will be hotter than the cooling liquid

 

[CycloWizard]

Originally posted by: Calin
However, there is a "heat transfer resistance" between the processor core (where heat is generated) to the inside of the waterblock (where water take that temperature). This is the core-to-heatspreader area, the heatspreader-to-waterblock area and waterblock-to-water. By reducing the temperature of the water in the waterblock to ambient, the actual temperature of the waterblock might be ambinent + 1F, the heat spreader could be ambient + 4F, and the core ambient+5F. Remember, when the temperatures of two materials in contact are the same, no heat exchange take place! You need different temperatures to have heat transfer.
The cooler the cooling fluid is, the cooler the core - but the processor will be hotter than the cooling liquid

You're right that this heat transfer resistance exists. However, this needs some clarification. Within the heat transfer surfaces (heat spreader, water block), the thermal conductivity that occurs is governed by Fourier's law of heat conduction, which states that the heat flux is proportional to the temperature gradient across the part under consideration. The proportionality constant is the thermal conductivity of the metal (very high for copper, silver), which will be approximately constant over the small temperature ranges considered in these systems. So, then, if the thermal gradient across the heat spreader/waterblock (read: temperature difference between the water and the core) is very high (water is much cooler than the core), the heat flux will be very high. The larger this gradient is, the greater the flux will be. Thus, this heat transfer resistance is not fixed and may be readily decreased by using a colder and colder heat transfer fluid. This is exactly why liquid nitrogen and similar cooling techniques are so effective at generating uber-low temperatures.

Now, when considering the liquid-waterblock interface, the heat transfer is no longer governed by Fourier's law. It's governed by a simple convective heat flux: q=h(Tw-Tf), where q is the heat flux, h is the heat transfer coefficient, Tw is the wall temperature, and Tf is the fluid temperature. The heat transfer coefficient increases with velocity according to the Chilton-Colburn analogy (Prandtl Number varies with Reynolds Number). Thus, if you ramp up the velocity, the Reynolds Number (ratio of inertial to viscous forces) increases proportionally, as does Prandtl number (ratio of resistance to energy transfer to momentum transfer). The relationship between velocity and heat transfer coefficient is, therefore, somewhat complex, but they're strongly related.

So, by increasing the velocity in the loop, you increase the heat transfer coefficient. When this is increased, the heat flux from the waterblock approaches that of the fluid. When this happens, the thermal gradient from the core to the fluid decreases.

 

[kleinwl]

Pardon me... but this seems like a silly discussion of a simple practical design problem.

The turbulence of the water is important to eliminate boundary layer effects (ie. non-moving water acting as an insulator inside a wetsuit). However due to the crazy in/out ports of a typical water block there is no way that the water will be laminar... thus velocity is imaterial (including wire-mesh ideas) to turbulation (above ridiculously low levels like .00001 mm/hr). (I would actually recomend a redesign of the average waterblock due to the impingment/poor flow distribution.... but what can I do).

The flow velocity is important to keep temperatures in the waterblock as low as possible... however, like posted above, this is a point of dimensing returns. Keeping water flow high enough that the water does not increase in temperature (significantly) inside of the waterblock is sufficent. Heat transfer is directly related to the Temperature differential. One or two degrees will not make an appreachable difference.

Coolant heating due to the waterpump work should be insignificant. Since I expect that no one is running a 200W waterpump... the only points of significant heat transfer should be the cpu/gpu. I also assume that the radiator is correctly sized to cool the water sufficently before recirculating.

Ok, so what does this mean?
If you want the minium processor temps:
1) Make sure that the waterblock is well connect to your cpu (silver 5 / maximize contact area).
2) Measure water temp in/out of your waterblock... if it is higher than a few degrees... increase water flow velocity
3) Measure water temp out of your radiator... if it is singificantly higher than ambient then consider increasing the air flow and/or radiator size.

My experience:
MSME - Thermal / Fluid Dynamics
PE
8 years automotive engineering (RAD/EGC/CAC/Intake Manifolds/HVAC systems/etc).

 

[CycloWizard]

Originally posted by: kleinwl
Pardon me... but this seems like a silly discussion of a simple practical design problem.

The turbulence of the water is important to eliminate boundary layer effects (ie. non-moving water acting as an insulator inside a wetsuit). However due to the crazy in/out ports of a typical water block there is no way that the water will be laminar... thus velocity is imaterial (including wire-mesh ideas) to turbulation (above ridiculously low levels like .00001 mm/hr).

Let me stop you here. First, there is a quantity called turbulent intensity which you're ignoring in this analysis. Unfortunately, it is the controlling factor in the temperature of your processor when all is said and done. It depends on the velocity. In other words, the fluid isn't simply turbulent or laminar: it can be in the transition region between the two, slightly turbulent, extremely turbulent, or a number of other things.

All of these things are known after proper calculation of the dimensionless Reynolds Number (Re). Re = L*u*d/v, where L is the characteristic length scale of the flow (twice the hydraulic radius in this case), u is the fluid velocity, d is the density of the fluid, and v is the fluid viscosity. Since all of the terms except velocity will be constant, velocity will be the deciding factor in all of this. As a rule of thumb, flows are turbulent when Re>2300. Since the hydraulic radius in the systems we're considering here are extremely small (as this increases velocity), it is indeed possible to have laminar flow if the volumetric flowrate is sufficiently low, though you're correct that it's undesirable.

Ok, so what does this mean?
If you want the minium processor temps:
1) Make sure that the waterblock is well connect to your cpu (silver 5 / maximize contact area).
2) Measure water temp in/out of your waterblock... if it is higher than a few degrees... increase water flow velocity
3) Measure water temp out of your radiator... if it is singificantly higher than ambient then consider increasing the air flow and/or radiator size.

This is all well and good unless you're trying to design a flow loop. These things can all be calculated a priori without too much effort. Obviously anyone can just keep cranking up the volumetric flowrate until he achieves the best temps he can get, but that isn't really the point of this discussion as far as I can tell.

 

[kleinwl]

CycloWizard,

I understand completely that the Reynolds number is the definition of turbulence. However, the point that I was trying to make is that the waterblock itself creates significant amount of turbulation to the flow. The entrance effects of the waterblock will serve to turbulate the flow significantly.

Trying to calculate the Reynolds number using the hydraulic dynameter will be incorrect (unless you are calculating the dynameter experiementally). Once the flow leaves the nipple and hammers into the waterblock face, the water will turbulate quickly. So even though you can calculate the reynold number easily in the tubes and nipple and predict if the flow is laminar or turbulent or even in the transition state that will all be insignificant to your cooling performance (although not to your system pressure drop) once the fluid is actually in the waterblock. Yes, I am assuming a reasonable water flow velocity... since at significantly low velocities the flow could be laminar (though with significant recirculation issues). However, I expect in any practical application this will not be the case.

What I am trying to say is that due to the entrance and exit effects of a waterblock, I would not be able to predict the effective Reynold number (at a given flow) without expermintation. Because of that, I suggested that people with water cooling systems could quickly check to see if their systems are optimized using temperature probes at various locations, rather than trying to hypothetically caculate the system performance.

To answer the OPs question: Waterflow maters... but there is a optimial speed... beyond which it is counterproductive.

 

[CycloWizard]

Originally posted by: kleinwl
CycloWizard,

I understand completely that the Reynolds number is the definition of turbulence. However, the point that I was trying to make is that the waterblock itself creates significant amount of turbulation to the flow. The entrance effects of the waterblock will serve to turbulate the flow significantly.

Re isn't the definition of turbulence - it's simply the easiest way to determine the heat transfer coefficient a priori. Simply put - you CAN calculate the hydraulic radius a priori. Indeed, if you could not, it would be meaningless to have any sort of correlation for heat transfer coefficients based on Re or Prandtl Numbers. Since there are entire books devoted to this subject, I will reiterate that you CAN do this with a high degree of accuracy.

What I am trying to say is that due to the entrance and exit effects of a waterblock, I would not be able to predict the effective Reynold number (at a given flow) without expermintation. Because of that, I suggested that people with water cooling systems could quickly check to see if their systems are optimized using temperature probes at various locations, rather than trying to hypothetically caculate the system performance.

Simply because you're unable to predict it does not mean that it cannot be predicted. I have done just such calculations for the design of a watercooling system posted in a recent watercooling thread in the HT forum.

It's all well and good if you don't know the answer, but please don't imply that since you don't know it, no answer exists. Worse - don't give misinformation. This is simply irresponsible, and is the common action of the poster in the Cases & Cooling forum, which is why I now avoid that forum. In other words, it's ok to say that you don't know the answer to a question.

 

[kleinwl]

CyclonWizard,

It is not to say that the turbulence in a waterblock could not be predicted... however using the Reynold number as you stated would be incorrect. Unless you compensate for the enterance and exit effects in the waterblock, using a hydraulic dynameter that is only the dynameter of the waterblock would be incorrect.

What I would do if I wanted to calculate percisely the turbulent effects of the enterance and exit of the waterblock is to run CFD on the waterblock (Fluent/StarCD/etc).... otherwise, using the geometerical shape compensations to modify the Reynold number would be somewhat in error. Certainly, anything a CFD program could do could be replicated using books and manuals... along with experiemental results to correlate and correct for computational errors. All of this is quite possible... the downside is that it is beyond the capabilities of most of the people posting in this forum.

So yes... you can do such a calculation for a waterblock and caculate the reynolds/pradtl number for such a system. However... I would be very interested in knowing what you would use to cacluate the effects of the jet expanding outwards from the nozzle... splashing on the waterblock face (disrupting the jet from the nozzle slightly)... turbulating in the corners of the waterblock... and finally flowing into the exit nozzle....

Visualizing the flow alone inside the waterblock inherently tells me that the water will be highly turbulated and little futher calculation needs to be done. If you want to make a project of it and cacluate percisely what the turbulation will be, that is fine.... but I would be more interested in using that time and effort into redesigning the waterblock to reduce the pressure drop while maintaining the heat transfer effects of the turbulations (perhaps by adding fin turbulators).

However... for my sake and pleasure... could you please demonstrate how you would calculate the turbulence inside of a waterblock?

 

[CycloWizard]

You're right - most of it is beyond the capabilities of most posting in this forum. Does that mean it can't be done? Hardly. Will the results achieved using my methodology be exactly correct? Certainly not. They're based on correlations, which have errors inherent to them. Will it give a reasonable approximation? Yes - that's the entire point. Will entrance effects be really as important as you make them out to be? No, because the system is turbulent. If the flow were laminar (which we've both agreed that it should not be in such systems), then the entrance effects would be significant.

Simply put, if you want an exact temperature profile everywhere within the waterblock, you can use CFD. However, since I doubt the guy asking the question has access to such software (nor do 99.9% of the other people in this forum), this is irrelevant. What I have stated is correct, whether you would care to acknowledge it or not: increasing the flow velocity will increase the heat transfer coefficient and decrease core temperatures to an asymptotic limit. If you would care to dispute this simple argument, please feel free to do so, but this time using real words, not 'dynameter', 'turbulating', and other thomfoolery. I find it extremely hard to accept criticism from someone claiming to be a fluid mechanics engineer who can't even come close to the proper spelling of the word 'diameter'. As I said, I've already performed said calculations and supplied a link in the previous thread on this subject in this forum. The provided solution is not given for a perpendicular-impinging waterblock, but for a parallel flow configuration as that is the simpler case. Since no one expressed interest, I didn't waste my time solving the more complex case for perpendicular flow.

 

[Mday]

Originally posted by: daines1
Maybe not too technical, but here it goes:

Say you have a water cooling system for a heat source (CPU, VPU, Car Engine). Would there be a point where the speed of water running through the system cause more heat than cooling?

Would the friction of the water cause more heat than cooling?

daines1

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.

The only thing you have to worry about is that the cooling mechanism (usually a radiator) is removing heat from the water as fast as the water is heated. This is the only real dependency on the speed of the water flow you have to worry about. All other effects are negligible.

Frankly, by the time the friction of the water moving reaches an extent to which it is really generating heat relative to the heat source, your tubing would come lose from the forces involved.