Optimal rad setup: pressure

[gmat]

Originally posted by Cova

soon gmat will likely post up the formula that shows how to calculate them all based on one-another.

Hahaha well i wont since i totally agree with you. I only come to dirty formulas when ppl make wrong statements...

About pressure there's a lot to be said. Of course we dont care about it in our pumps.
Be we should care about it in our waterblocks and rads (and seals, for obvious reasons). It has a lot of influence on heatt transfer coefficient, together with fluid velocity. Very few ppl actually tested this out apart from this guy at overclockers.com.

[jtroutma]

Gmat:
Just to be specific; our water system are hydo systems where we move water. What I was refering to when I stated "hydrolic" was oil based systems. YES water can be used as well but on the systems that I have delt with, a hydrolic oil was used (probably due to its pressure properties)


Also I should have been more clear when I made the accusation:

Head Pressure = NO resistance; how much water can you force up that outlet with nothing attached.

What I should have said was that with nothing attached, the pump will put out a large amount of volume of water with a fininte amount of pressure due to the large volume of water trying to make it through that small opening.

Someones going to dispute me again.... i just know it

[MeltMan]

It sounds about right to me. Pressure only develops when there is resistance. Just like in an electrical circuit. What happens to the Voltage if you have direct short? (no load). Oh it drops to none.

[Quickmcj]

Man this is no good for me

I understand some of this you write....but gmat man those links you gave, and the math..........that's not easy for a stupid Dane

I will not try to agu against any of you, cause I cant explain myselft good enough in english and therefore wont start any debate I cant anticipate in

But when you are "done" wont you please make an "easy reading" for people like me, that aint that good at technical english ?

By dooing this, you and all others in here, will then know what to buy or do, based on all this theory and eventually experiments that you will share

You all are very wise...there are really many thoughs in this tread

[Dix Dogfight]

I'm still amazed that people actually think that a lower flow through the WB will increase the temp by a few degrees.

Some simple math:

big heatload: 200W=200J/s
specific heat of water: 4180J/(kg*K)
low flow:2L/s
rho of water: 1kg/L

increase in temp of the water in WB will be:

dT=200/(4180*2*1)=0.0239K

So the temp will increase 0.02 degrees when it passes through the WB.

Can we now please stop thinking that the water can stay to long in the WB because it will heat up too much.

The same argment can be made for the rad.
Consider a system in equilibrium = rad dissipates the same amount of heatload as the cpu puts in.
Then the temperature difference of the water, from entrance to exit in the rad, will be the same as calculated above for the WB.

Therefore we can conclude that low or high flow doesn't produce big/small temp differences of the water in the system.
Therefore any BS about faster flow in rad makes smaller tempdiff and thus produce much better cooling is a load of crap.

any takers

[gone_fishin]

Quote by Dix Dogfight

"Therefore we can conclude that low or high flow doesn't produce big/small temp differences of the water in the system.
Therefore any BS about faster flow in rad makes smaller tempdiff and thus produce much better cooling is a load of crap." end quote

Clarify something on your stance. When you state temp differences of the water, what are you comparing the water temp to? The ambient temp or the cpu temp or the inlet/outlet temp of the rad or block? If you mean high flow doesn't produce small temp differences between the inlet/outlet of either the rad or block then you are sadly mistaken.

[Dix Dogfight]

gone_fishin:
I'm talking about the tempdiff between inlet/outlet.
And how can i be "sadly misstaken" when simple physics say so?

Have you actually measuerd the tempdifference between inlet and outlet? Or can you show me where the simple formula is wrong please.

EDIT:
Hmm I just realized that a flow of 2L/s is actually ALOT.
EHEIM 1250 produces 0.3 L/s with no resistance.
so lets take 0.1L/s instead which gives a deltaT of 0.5 which is still to small to make a difference and remember that this is with a 200W heatload.

[gone_fishin]

Originally posted by Dix Dogfight
gone_fishin:
I'm talking about the tempdiff between inlet/outlet.
And how can i be "sadly misstaken" when simple physics say so?

Have you actually measuerd the tempdifference between inlet and outlet? Or can you show me where the simple formula is wrong please.

EDIT:
Hmm I just realized that a flow of 2L/s is actually ALOT.
EHEIM 1250 produces 0.3 L/s with no resistance.
so lets take 0.1L/s instead which gives a deltaT of 0.5 which is still to small to make a difference and remember that this is with a 200W heatload.

In my testing a rise by .5C in the water changes the cpu temp. I'm getting a through system flow rate of 360gph (1gal = 3.9liter) so in liters that's 24lpm or .4lps. An average system usually gets 70 - 80gph or .08lps
You're trying to solve something by equation, but I have a real high flow system which keeps the water outlet temps only 2.5C higher than the room ambient at full load, overclocked.
Higher flow is better for watercooling whether it's in a nuclear powerplant, a car, or a computer. Read the third link that gmat posted a few posts back, very clear and informative.

[gmat]

Originally posted by Dix Dogfight

low flow:2L/s

2 notes
1 - 2l/s is an amazingly high flow. Reaching 0.5l/s is already uncommon. Usually we got about 100l/h in our systems that is 0.2l/s, that with a *very* good pump such as the E1250.

2 - If you plan to include inlet vs outlet temperature differences in your equations, do so but properly. You must take the logarithmic (ln) sum of those temperatures.
I'll save you a lot of work and headaches by telling you higher flow leads to better results (higher transfer rates).

(edit) oh and thanx to gone_fishing for the experimental figures

[Cova]

Ahh..., another day at the office. Another day to bitch about differences in flow-rate that 90% of us don't have the tools to measure (myself included) to get our CPU 1/2 a degree cooler.

Originally posted by jtroutma:
What I should have said was that with nothing attached, the pump will put out a large amount of volume of water with a fininte amount of pressure due to the large volume of water trying to make it through that small opening.

Someones going to dispute me again.... i just know it

A pump will always produce a finite/constant amount of pressure (unless you start altering it's voltage or something, but thats beyond this discussion) - regardless of the volume of water it is pushing, or how many blocks, rads, and other things are attached to the pump that is trying to stop the flow.

Once you know how much pressure the pump is creating, and how much resistance to flow the system will exert, you can calculate the volume of water that will move through the system. Obviously when there is no resistance to flow (pump sitting in a bucket with no tubing) the volume of water moved will be large. But the pump doesn't magically change the way it reacts to backpressure the moment you put it into an in-line system with some restrictions - it still tries just as hard as it ever has to move water (same pressure exerted on the outlet), though again obviously less water moves.

I'm not disputing what you are saying, just trying to clarify it. Our pumps react the same whether they are in a bucket with no resistance, or pushing water through a cooling loop that involves 10 computers with 10 blocks and 10 rads. Just that the ~5 PSI of pressure that our pumps provide doesn't move much water when you put that kind of restriction on it.

Originally posted by MeltMan:
It sounds about right to me. Pressure only develops when there is resistance. Just like in an electrical circuit. What happens to the Voltage if you have direct short? (no load). Oh it drops to none.

If you have a short, voltage does not drop to zero. The battery continues to supply a voltage to the circuit. Without voltage no current would flow, without current flowing your wires would not melt and fuses would not blow. And if that was true, we wouldn't need circuit-breakers and such to protect against shorts. Though - with a short in the circuit it's likely that the battery or one of the connections to it will quickly fry. Once that happens there is no longer a complete circuit - now voltage is zero.

[MeltMan]

If you have a short, voltage does not drop to zero. The battery continues to supply a voltage to the circuit. Without voltage no current would flow, without current flowing your wires would not melt and fuses would not blow.

Ah, but you forget, there is resistance in the wire which allows the current to flow. If there were a way to have a PERFECT no ohm load, then there would be nothing on the voltage.

[Cova]

Originally posted by MeltMan


Ah, but you forget, there is resistance in the wire which allows the current to flow. If there were a way to have a PERFECT no ohm load, then there would be nothing on the voltage.

There is a chemical reaction occuring in the battery that produces a voltage. Simply connecting the 2 terminals of the battery (even with a perfect 0-ohm load) does not magically stop that chemical reaction and dispell the voltage. A 0-ohm load would fry things, not get rid of voltage. Of course, it would be virtually impossible to measure the fact that there is no voltage, because a 0-ohm load would short your volt-meter as well

[MeltMan]

Yes, but there is still no such thing as a 0 ohm load. My fluke meter will measure resistance within its own test leads.

[bigben2k]

Maybe I can shed some light...

0 ohm = full load = closed circuit = short = superconductor

infinite ohm = no load = open circuit

Batteries have the potential to provide 1.5V (AA). The potential is there, wether it is used or not. If it is used, it can be measured. If it is not used, then it is just potential, but is still measured in volts.

A battery can only supply so much power. If it is shorted, then the voltage would drop some, as the short tries to draw everything out of it. The voltage would drop to where the amperage is high, but to where the maximum power is outputed. (outputed?)

Batteries also have an internal resistance, but I guess that doesn't mean anything here.

[NoSoupForYou]

Very interesting discussion - check out the overclockers.com board for some more if your a glutton for punishment.

Increasing the water flow rate through the rad does one thing, and one thing only. It increases the heat transfer coefficient between the coolant and the walls. This in turn decreases the delta-T between the coolant temp and the radiator wall temp required to 'push' a given amount of heat into the rad/fins.

Generally speaking however, the delta-T between the coolant and the walls is pretty small to start with. If you think about it, the radiator has a much larger internal surface area than the WB, and will thus have a much smaller delta-T as well. I would be surprised if for a 100W load the coolant to wall delta-T exceeded 2C in even a small radiator. I would bet the delta-T is lower in most setups.

So, the best that can be hoped for by increasing flow is to reduce this delta. Even with a 2C delta and doubling the heat transfer coefficient, the net change is 1C - not a lot. Note that if the CPU load is not changed, there will be no change in the heat actually dissipated by the radiator. But the heat will be dissipated with a lower coolant temperature. Since the CPU temperature is directly related to coolant temperature, this will lower CPU temps (in addition to any CPU temp reductions directly related to the increase in flow in the WB).

The whole concept of temp increase/decrease of the coolant as it passes through the block/rad is a red herring - its pretty much irrelevant. Yes, an increase in flow will result in a smaller coolant temperature change. But as described above, the change is minimal to start with, and is certainly much less than the coolant temperature increase above ambient at equilibrium conditions (in most setups anyway).

gmat - LMTD is a little over the top here - the temp change in the fluid is small enough that approximating it as constant won't introduce significant error.

[Dix Dogfight]

gmat:

2 - If you plan to include inlet vs outlet temperature differences in your equations, do so but properly. You must take the logarithmic (ln) sum of those temperatures.

Heating up a specific amount of water with a constant heatload produces a LINEAR rise in temp. If you think you somewhat trick that into beeing logarithmic I suggest you restudy your thermodynamics.

I'll save you a lot of work and headaches by telling you higher flow leads to better results (higher transfer rates).

I didn't make any contradictary statement that I can recall.

And please stop asuming all others in theese foums to be total morons. There are actually alot of intelligent people here. That doesn't get a headache over a few differential equations like you do.
Some of us actually work with them all day long and are happy and smiling anyway.

gone_fishin:

You're trying to solve something by equation.

Not trying to solve a problem. I'm trying to redirect our joint efforts on what matters and kill a few general "thinkerrors" that that a few persons do.

Higher flow is better for watercooling whether it's in a nuclear powerplant, a car, or a computer

Again I newer made a contradicory statment.
I stated that the tempdiff as a function of flow has a small impact on the cooling.

Higher flow is better but because of other factors.
NOT because it makes a lesser tempdiff in the WB.

[NoSoupForYou]

Dix Dogfight,

I think if you are going to have an attitude, learn to clarify a bit better.

Higher flow IS better because it makes a lesser tempdiff in the WB. Not tempdiff as you are calling it (difference between inlet and outlet temperature) but tempdiff between the coolant and the WB itself.

Tempdiff from inlet to outlet is always a function of heat input and flow rate only. As I said earlier, using it in this discussion is a red herring.

[gmat]

Thanx noSoup for replying

Originally posted by NoSoupForYou
Tempdiff from inlet to outlet is always a function of heat input and flow rate only.

And thats the one that needs to be logarithmic in order to be included in the general heat equation (see formulaes above). That's not a trick from me, i'd like all things in life to be as simple as 2+2, but they are not. I wont repost any more formula, please refer to the links i provided before.

As I said earlier, using it in this discussion is a red herring.

Yes !
It's a purely local and dynamic effect which is irrelevant here. (would be relevant from a "rad design" point of view, where one wants to optimize rad design... OT here).

Originally posted by Dix Dogfight
...total morons. [...] That doesn't get a headache over a few differential equations like you do.

1 - if you're here to flame ppl and dont understand humour go to [H] you dont belong here.

2 - actually i hate diff. equations and i hate those pesky thermodynamic laws. One needs to live with them though... Thats just a matter of taste, just like some ppl hate coding in assembly...

[jtroutma]

Cova:

Yes I agree with your clarification. I just couldnt put it into words that well yesterday

gmat:

I am not sure about this but I have a feeling that NONE of us like Differental Equations (Lord knows I hate them!!! ) However your right, we do have to deal with them unless we all want to stay in "theory" world.

As for the faster flow throught a radiator = more heat disappation; I am still not convinced. Could be true but still has some holes in it.

One other thing. I have been reading around the threads and mentioning that we need to heat the water before it enters the radiator in order to make the radiator more efficent.....
Here is my take on that: YES a radiator WILL be more efficient the higher the Delta-T between ambient air and the water temp. However, making the radiator more efficent by heating the water more than necessary only adds more heat to the system and thus heats the whole system up. This is obviously bad; heat = bad Our radiators in our system WILL only be running at say 10% efficency because we want our water temps AS CLOSE TO AMBIENT AS POSSIBLE!!! A radiator running at say 50% efficency could remove say 2 degrees from a 15 degree delta-T water but that still leaves 13 degrees delta-T which goes right back to the block and gets more heat. OUr Radiators running at say 10% efficency will only be able to remove say 1 degree from a 8 degree delta-T which leaves only 7 degrees delta-T water.
NOTE: DO NOT HOLD ME TO THESE NUMBERS!! This is for theory example ONLY! Just to give us an IDEA of what is going on, not the actual figures.

Alright another wave of friendly flames come forth!

[gone_fishin]

Qote by Dix Dogfight,
"Again I newer made a contradicory statment.
I stated that the tempdiff as a function of flow has a small impact on the cooling.

Higher flow is better but because of other factors.
NOT because it makes a lesser tempdiff in the WB."end quote

I still don't understand what you are trying to convey.
You state that higher flow makes a lesser temp diff in the wb but that's not why higher flow is better.
In my observance higher flow keeps the water temp lower so as it enters the block there is a greater potential for the water to absorb heat. If you run ice water through the waterblock then the temps will go down in the cpu so if higher flow causes lower water temps to enter the block is this not the same thing?

[gmat]

Originally posted by jtroutma

I am not sure about this but I have a feeling that NONE of us like Differental Equations (Lord knows I hate them!!! ) However your right, we do have to deal with them unless we all want to stay in "theory" world.

Hehe thats why CFD and fast computers exist

As for the faster flow throught a radiator = more heat disappation; I am still not convinced. Could be true but still has some holes in it.

Well look again:
http://www.stewartcomponents.com/adv...tem_basics.htm

They explain it all pretty clearly. Obviously watercooling has been thought about in cars long before PC's.. And i'm pretty sure these ppl know what they're talking about. Dont they ?

[Cova]

One other thing. I have been reading around the threads and mentioning that we need to heat the water before it enters the radiator in order to make the radiator more efficent.....

You're probably getting this from mostly my posts scattered around the board - but you're misinterpreting what I've been trying to say (and it really should be the topic for another thread)

Our rads run really inefficiently because we have a low delta temp between the water and ambient air. To get the rad to dissipate more heat (be more efficient) we need to increase that delta, but by doing so we also increase the temp of water passing through the CPU block, which is a bad thing. This is the basic water-cooling dillema - we need cold water in the CPU, and hot water in the rad, and in a typical computer cooling system, the water is within a couple degrees of the same temp everywhere in the system.

But I never said I want to heat the water - why would one start putting excess heat into the system that is not required. I want to move heat around in the system, I just haven't figured out how to do it yet.

[jtroutma]

My only problem with that argument Gmat with that link is that.... those are cars and engines running at temperatures that we never want to get even close to.

Our systems run at much lower temperatures and much closer to ambient than a car radiator system runs at. For a car cooling system, I agree. But I think that with our system we need to also take into account that we have much lower heat sources, lower efficency radiators, AND water temps much closer to ambient.

That is where I am running into a problem with that argument.

[jtroutma]

Cova:

Clarification recieved and acnowledged!

[gmat]

Originally posted by jtroutma
My only problem with that argument Gmat with that link is that.... those are cars and engines running at temperatures that we never want to get even close to.

Our systems run at much lower temperatures and much closer to ambient than a car radiator system runs at. For a car cooling system, I agree. But I think that with our system we need to also take into account that we have much lower heat sources, lower efficency radiators, AND water temps much closer to ambient.

That is where I am running into a problem with that argument.

And whats the difference with cars ?
1 - our rads are the same
2 - pumps have about the same rate
3 - car cooling gets less efficient due to boiling water problems. But one wants to avoid this in most engines...
4 - from an engineering standview those systems are *exactly* the same, same formulas, same setup, same problems, etc...
Please explain clearly the difference.

Dont forget the Q=UAdT ...

(edit) hint: our dT is lower. That means our Q (heat tranfer) is lower. And so what, work at a given dT and see what reducing U (direct factor of flow) or A (area) does to Q...