Waterblocks effectiveness in terms of power dissipation

[Since87]

myv65 mentioned something a while back that eventually sunk in to the point I could make use of it.

Flow rate (L^3/T) multiplied by pressure drop (F/L^2) = units of F*L/T (power). It's as simple as that. Measure this stuff in metric and the unit conversions are a little easier, but it's not so bad in imperial units.

In other words:

If you take a flowrate (say liters per minute) through a waterblock and convert it into cubic meters per second...

m^3 / s

...and you take the pressure drop (say meters of H2O) across the waterblock at that flowrate, and convert it into Pascals (or equivalently Newtons per square meter)...

N / m^2

...and you multiply that flowrate by that pressure drop...

( m^3 / s ) * ( N / m^2 ) = N * m / s

...you get Watts.

1 Watt = 1 N * m / s

For lack of a better term, I'm calling this power dissipated in water blocks 'Hydro Power'.

I was curious as to what the thermal resistance of waterblocks would look like graphed vs 'Hydro Power'. BillA was kind enough to let me have the dataset which three of the graphs in his article Waterblock Bench Testing Results are based on. From this data I produced the following graph:



About the graph:

It's important to note that the horizontal axis of the graph is on a log scale. Each major vertical line represents a factor of 10 increase in 'Hydro Power' over the vertical line to the left. (The chief benefit of using a log scale is that it makes the differences in the curves of each WB standout more clearly.)

Also important to note, is that the Watts referred to in "C/W" is completely separate from Watts of 'Hydro Power'. The former refers to heat transferred from the CPU to the water. The latter refers to power which is supplied by a pump, and is dissipated in the WB.

The 'Hydro Power' really is dissipated in the WB and results in additional heat in the coolant which the radiator must remove. Just as a certain amount of electrical power dissipation can be the result of, a 'high' voltage and 'low' current, or a 'low' voltage and 'high' current; a certain amount of 'Hydro Power' dissipation can be associated with different ratios of flowrate and pressure drop. The relevant flowrate to pressure drop ratio for a particular block cannot be determined from this graph. (Refer to BillA's graphs for that information.)

So what?

That's what I'm hoping to find out from you guys. Everything I know about fluid dynamics I learned from studying sailing, model rocket design, or watercooling on the forums. No formal training whatsoever. I'm curious about what explains the differences in these curves, but all I can do is speculate.

Particular things I'm curious about:

The WB75 block has a vastly steeper slope than the other WB's suggesting that at about 10 Watts of 'Hydro Power' its cooling performance will exceed that of the TC-4. 10 Watts of 'Hydro Power' would take (by overclockers standards) a monster pump to produce, and the total heat put into the system by such a pump would most likely outweigh the benefit of the WB thermal resistance achieved. However, maybe understanding this big difference in slope would enable someone to design a WB with the same slope but shifted substantially to the left.

Why do the Swiftech blocks show so much curvature on this graph compared to the others? My speculation is that: At low 'Hydro Power', the flow of water through the block has little effect on the water near the baseplate - it just loops from inlet to outlet. At medium 'Hydro Power', the inlet flow is impinging on the base and the cooling is therefore substantially improved. At high 'Hydro Power', a lot of the applied 'Hydro Power' is being used in maintaining eddies as shown below in red.



The outgoing water has to fight it's way past the eddies and the jet and a lot of the applied 'Hydro Power' is wasted in generating turbulence that is of little benefit for cooling. Also, warmed water from the eddies gets mixed into the jet on the way down.

My speculations on the Swiftech curves are very much uneducated guesses, so please feel free to point out their faults or suggest alternative explanations.

I've got some other thoughts on this stuff, but it's after midnight and this post is long enough already.

Let me know what you think.

[Les]

Differences in Spreading Resistance. is the explanation I use for much of the behavior shown in the plot.. In its simplest form, bp thickness and area.
My many plots of C/W v "h" (related* to Velocity) and C/W v PD illustrate my interpretation of Spreading Resistance influence
E.g..



I,also, have no formal training in Fluid Dynamics.

* Usually expressed as a power relationship (e.g Sieder Tate) . In the impingement case I still like to heretically imagine it "proportional to Thrust(massflow rate x vel)

[bigben2k]

What you're referring to, I believe, is the hydraulic properties. Yes, it can be measured in Watts, but has nothing to do with the heat dissipation: it's just a measure of the power used to produce that kind of flowrate at that kind of pressure drop.

What makes the curves different is this: each part of a waterblock will be made up of different size openings, which can be equated into their hydraulic equivalents, in round tubes.

Knowing that, you should be able to imagine that the curve would be dramatically different if, for example, you compared a Swiftech block (open area, center inlet) versus Radius (center inlet, with varying number of channels, and channel width).

In other words, any block can be represented as a bunch of tubes, of varying diameters. If you have a bunch of small tubes, then the pressure drop will increase dramatically, with the flow rate. IF the block is equivalent to one large tube (and it never is THAT simple), then the pressure drop would follow standard tubing specs.

Use Darcy to calculate those pressure drops.

[murray13]

Someone please tell me if I'm wrong but isn't this thermodynamic efficiency vs. hydrodynamic efficiency. Or put another way cooling vs. flow.

I know it's over-simplified but it fits.

[myv65]

Now, now, Ben. This is very minor, but definitely does have to do with heat transfer. It's also something that a lot of people don't get or refuse to believe is true.

The gist of this is asking how much of the total energy contained in the fluid stream gets converted from "fluid" energy (pressure * flowrate) to thermal energy while in the block.

Personally, I don't think it matters much for a few reasons. One, at "typical" pump power levels this is relatively small vs the power of the chip(s). Two, decreasing this number generally means a less effective waterblock (though as these graphs show there is some variation in how efficiently different blocks use fluid energy). Three, there's only so much you can do and the block isn't the only thing that turns fluid energy into thermal energy.

I've said it before and I'll say it again, virtually all power you put into the pump shaft eventually turns into thermal energy that the radiator removes.

Whether you've got an engineering background or not it should not take a big leap of faith to realize that energy doesn't magically disappear. If you put energy into a pump, that energy goes into the water. If you put energy into water, you must get it back out again. At steady-state, the rate of energy entering and leaving must balance.

[bigben2k]

Ah yes, indeed!

I should have been more clear: the hydraulic power is not related to how much heat the coolant will absorb, i.e. for a given hydraulic power applied, different blocks will perform differently.

But indeed, this hydraulic power is induced in the coolant, in the form of heat. Relatively small, but should be understood while designing a waterblock.

Big thanks to Since87, for posting an applied example!

[myv65]

FWIW, if you look at a hypothetical example of 2 gpm and a pressure drop through the block of 25 inches of water (7.6 lpm, 6221 Pa), the related power is only ~0.78 watts. That may be representative of a fairly high-flow, low-restriction block. Doubling the delta-P doubles the energy to ~1.6 watts. In this general range, all the blocks on the graph fall within 0.02 °C/W (or about 1.2°C for a 60 watt load).

Better to merely forget such minor details and simply buy the only block that really stands out from the rest, and it isn't on the above graphs.

[Since87]

Originally posted by myv65

Better to merely forget such minor details and simply buy the only block that really stands out from the rest, and it isn't on the above graphs.

Didn't you see that tiny speck in the lower left corner of the graph?

Actually, if I remember correctly, I don't think WW even shows up on the chart with this scaling.

I'm not so much interested in which of these blocks is better, as I am interested in whether anything applicable to new block designs can be learned.

[bigben2k]

Many things can be learned.

If you look at Radius, you should be able to see that as the flow rate increases, the pressure drop should increase relatively quickly. In a Swiftech type block, the increase wouldn't be near as dramatic.

Of course I haven't calculated it yet, but that's what I expect.

What I did do though, is allow the channels to split up, at different radius: this minimizes the overall pressure drop. If I had stuck with say, 8 channels, straight out (a la morphling1), within the same fin diameter, the pressure drop would have been greater. This was also done in order to maintain a fin-to-channel ratio between what Cathar has calculated as being ideal, which is 0.75 to 1.5. So I maximized fin efficiency, and minimized the pressure drop.

That's the lesson, for me. Of course there are many other factors in block design, so it comes down to balancing it all. My inlet is still not functional.

[BillA]

for those interested in the wb's pressure drop, there is a lengthy thread here
kind of beaten all to death

nooo, NOT 'hydro power' - just a pressure drop

Ben -
you been sniffing epistemological swamp gas again ?
(or sharing your bottle with Since87 ?)
joking of course (?)

[BillA]

Originally posted by bigben2k
. . . . .
My inlet is still not functional.

oooohh
just noticed, sorry 'bout that Ben
a bigger pump perhaps ??

[bigben2k]

I must be...(sniffing epistemological gas), but it must have happened after that whole "galvanic corrosion" fiasco!

I won't switch pumps: that's what I got, and by golly, I'll make it work!

(sorry, I should have put the "my inlet is still not functional" in another paragraph).

Be Kewl

[Since87]

Originally posted by myv65
Doubling the delta-P doubles the energy to ~1.6 watts.

Is this right? If so, I'm missing something.

Doubling the pressure increases the flowrate. (Generally as the square root of the pressure change.)

So if the power dissipation (X) in the first case is:

X1 = dP1 * Q1

then the power dissipation in the double pressure case is:

X2 = ( 2 * dP1) * ( Q1 * 2^.5 )

X2 = ( 2 * 2^.5) * ( dP1 * Q1 )

X2 = ( 2.828 ) * ( X1 )

???

[bigben2k]

myv65 meant that for the same flow rate, if you double the pressure, the power is doubled. (which works, verified). He didn't mean for the same block.

(Your calculations are otherwise correct)

[myv65]

Originally posted by bigben2k
myv65 meant that for the same flow rate, if you double the pressure, the power is doubled. (which works, verified). He didn't mean for the same block.

Ding, ding!

[Les]

myv65

I post many graphs.

Am I posting stupidity?

[myv65]

Originally posted by Les
myv65

I post many graphs.

Am I posting stupidity?

LOL, stupid as it sounds, I have no idea. The only graph I pulled up was the first one of Since87's. I generally surf with graphics disabled and yes, I have broadband. Call it a quirk. In any case, I'll give your graphs a gander later tonight and if I have anything to say rest assured that I will say it.

[Since87]

Originally posted by unregistered

nooo, NOT 'hydro power' - just a pressure drop

Ben -
you been sniffing epistemological swamp gas again ?
(or sharing your bottle with Since87 ?)
joking of course (?)

Bill,

You know very well that I have my own bottle of epistemological swamp gas.

Ok, WHAT is "NOT 'hydro power' - just a pressure drop"?

[Bill and I have been discussing this for some time, and I don't know if we're arguing about something or just babbling at each other in different languages. myv65, if you can step in and translate I'd appreciate it greatly.]

[myv65]

One of the greatest impediments to sharing knowledge is a common set of definitions. Even among engineers like myself, different folks call the same thing by different names. In this particular instance, I don't really side with either you or Bill.

In your favor, "power" is the correct word given to things that have units of length*force/time. This is volumetric flow rate * deltaP. Bill uses deltaP vs flowrate to characterize blocks and is not particularly concerned with what this means in terms of power (me putting words in his mouth, mind you). This is likely because of the reasons I outlined above for ignoring this aspect of blocks.

To me it really comes down to the basics of doing an energy balance on the "box", which is the water block in this discussion. So long as you take into account all the energies involved and clearly state what they are and how they're determined, it's all good. Your thread here discusses one of the energies involved that most people haven't got a clue about and I'm glad to see it debated a little.

[Since87]

myv65,

I wasn't wanting you to take sides, I've just observed that when Bill says something repeatedly and vehemently, there is something there worth getting, and I'm not getting it. It's frustrating.

Anyone got a guess at why the WB75 has such an odd slope. Total dog at low power, relatively outstanding performance at high power. (Assuming a short linear extrapolation is reasonably accurate)

[BillA]

ok, one last time (I promise myself)
the dP in the wb is but one of a number through the WCing system
but does it in particular offer any unique insight into:
1) a wb's performance, or
2) a (unique) means of differentiating between wbs

wrt #1:
as a rule, I find a measurement always preferable to a calculated derivative value, for all the obvious reasons
- a pressure drop can be the difference between 2 measurements, or a single measurement with a differential pressure transducer (which is how I do it)

Since87, I know you do not wish to accept that this topic was extensively discussed in the thread I linked to above - but if you review it you will see that Tecumseh plotted the sqrt of the data and non-linearity was apparent at both 'ends' of the individual data sets
- this is indicative of an experimental error, which was also discussed at length and the error sources fairly well understood
-> the upshot is that the very low flow rates (specifically) have a significantly higher 'uncertainty' due to both the magnetic flow meter's performance at low stream velocities and the rounding errors because I was indicating only to 0.01psi (!)

so YOU need to be very prudent in any conclusions based on the 'tail end' data points
(this is all in that thread, nothing new being stated here)

so does the product of the pressure drop and the flow rate indicate something ?
sure, etc. etc.

wrt #2:
is that something unique ?
in an absolute sense; certainly
is it significant ? (significantly better/more informative than only the dP ?); to me no
the additional complexity is quite beyond all but the engineers

Since87, your degree is showing - which is quite ok,
but not all can follow even my simplistic presentations (having not Dave's skill)
- and I would ask you this ? (which to me is quite obvious, but then I'm rather involved in this)

is not the difference in flow 'configurations' kinda obvious ?
hence also their individual responses to changes in the flow rate ?

well, perhaps not to others (but it sure is clear to me !)

[Since87]

Ok Bill,

I reread that thread much more thoroughly and I realize now why you've been frustrated with me. (At least to a much greater extent than I did before.) A lot more of the stuff I've been hashing over was discussed there than I had realized. I'll go stand in the corner with a dunce cap on as soon as I'm done writing this.

My main interest in this has been 'useful turbulence' vs 'useless turbulence'. The heat generated by flow through the waterblock was a secondary or tertiary concern for me. Because that thread started out focused on the heat generation, my eyes glazed over and I didn't pay attention.

Quoting Tecumseh:

Look Specifically at the curves for the Cooltech WB75 and
the LiquidCC Surge. Notice how the curves cross at 4 lpm.
Now look at the Power Dissipation vs Flow curves above.
These two blocks have almost identical head-loss curves
and so dissipate internally the same amount of power over
the entire flow range.

What we are seeing is the effect of turbulence kicking in at
about 4 lpm. At flows below 4 lpm the LiquidCC Surge has a
lower C/W, but at flows greater than 4 lpm the Cooltech WB75
becomes a lot better than the Surge.

What this shows is that with identical head-loss (power dissipation),
the power is "partitioned" more for useful
turbulence in the WB75 at greater than 4 lpm. The WB75
is a spiral design. The Surge is a 3-fold labyrinth.

There you go. Blocks with identical head-loss and flow demonstrating
the onset of "useful" turbulence.

This was exactly what I was interested in, and I missed it.

Bill, I don't think I have anywhere near the 'intuitive feel' for what's going on with the flow in blocks that you do. I'm starting to see that in a practical sense:

useful turbulence vs wasteful turbulence =
useful power consumption vs wasteful power consumption =
useful pressure drop vs wasteful pressure drop

However, a lot of things that are clearly obvious to you, definitely aren't to me. Thanks for bearing with me.

Now where did I leave my dunce cap...

[SysCrusher]

Originally posted by Since87

Why do the Swiftech blocks show so much curvature on this graph compared to the others? My speculation is that: At low 'Hydro Power', the flow of water through the block has little effect on the water near the baseplate - it just loops from inlet to outlet. At medium 'Hydro Power', the inlet flow is impinging on the base and the cooling is therefore substantially improved. At high 'Hydro Power', a lot of the applied 'Hydro Power' is being used in maintaining eddies as shown below in red.



The outgoing water has to fight it's way past the eddies and the jet and a lot of the applied 'Hydro Power' is wasted in generating turbulence that is of little benefit for cooling. Also, warmed water from the eddies gets mixed into the jet on the way down.

There is a way around this that I have found. Keep the tip of the jet closer to the base and the depth of the pool should be roughly the same height of the opening of the jet. That keep those "eddies" in check. Then you will start to see an increased benefit for cooling. Then again swiftech can't predict what pump a customer will use with the broad spectrum of pumps available.

I like to see BillA test nozzle diameters from 1/4 to 1/8 and their height in a block plus find a way to take up space inside the block to keep the water from pooling.

[bigben2k]

Actually, inpingement (in theory) works best with the distance betweed 2d and 12d (d being the diameter). [edit] Wrong. Wrong, wrong, wrong... [/edit]

I had a similar problem with Radius (actually, I still have that problem...). It's not because of the jet, it comes from the opening's geometry.

[Les]

Originally posted by bigben2k
Actually, inpingement (in theory) works best with the distance betweed 2d and 12d (d being the diameter).

.

News to me .

EDIT:: Flomerics calculator maximizes at ~ 2.0 -2.5D but this is not for submerged jet impingement and in no way considers cases outside the calculators limits.