Peltier Blocks

[zer0signal667]

Originally Posted by Jabo

Go like this : rad->pump->CPU->gp->rad
Radiator does not benefit from high pressure nearly as much as water blocks and unless your pump dumps massive amounts of heat into water there's no point in placing rad between pump and CPU block - we want block to enjoy the highest pressure in the system

The order is not going to affect the pressure drop that each device creates in the loop. Similarly, the water temperature does not vary that much around the whole loop and you won't see a significant cooling advantage in any certain order of devices.

[Butcher]

You should go whichever order gives you the shortest and straightest tubing runs. That gives the best flow and thus the highest pressure. As pressure in a given compenent is a factor of flow and the design of that component, and flow is uniform throughout the loop (unless it's leaking), different order doesn't inherently alter either.
I'd expect someone who lectures people about thermal dynamics in such an arrogant tone as you do Jabo to know this.

[Jabo]

Originally Posted by Les

Evidence please.
It is a new concept to me.
Think is crap but I am open to elucidation

Pump adds kinetic energy to water. This energy is being lost bit by bit at each piece of kit (pressure loss).
Play with Bernoulli Law equations (using simplified set of condtions as described in your favorite physics compendium ) -> as pressure drops velocity drops *hint, hint* - just play with supplied calculator to check it if ya want Les.
I am not able to produce any other evidence and I deem this to be as conclusive as possible.

[Jabo]

Originally Posted by Butcher

You should go whichever order gives you the shortest and straightest tubing runs. That gives the best flow and thus the highest pressure. As pressure in a given compenent is a factor of flow and the design of that component, and flow is uniform throughout the loop (unless it's leaking), different order doesn't inherently alter either.
I'd expect someone who lectures people about thermal dynamics in such an arrogant tone as you do Jabo to know this.

See above and learn basics first and stop calling me arrogant when I simply state facts of life (physics) discovered tens of years before your birth. I do not make any personal remarks nor I criticise anyone - I merely add my 2 pennies to discussion on a given subject
Peace Butcher!

[Jabo]

Originally Posted by Incoherent

It's a compromise. Agree to a point but with the added surface area comes a lower water velocity and hence lower turbulence for a given pump. Add the thermal gradient of copper and it can easily end up way inferior.

Not necessarily, if you decrease 'headroom' (pins height) you will create very much more velocity due to decreased cross-sectinal area (see Bernoulli Law again) even though surface area increased - again quite nice trick employed here - mre surface area but smaller stream cross sectional area. Pins in hit&miss pattern would create very much yummy turbulence I suspect (to calculate this .... forget it) that the pressure drop of such a block would be even more than the smallest nozzle size Cathars produce but then again with taller pins one could have an universal block like latest swiffy offering but providing much better performance.
Thermal gradient SPREAD would be superior with regards to max attainable overclock and flux spikes dissipation capacity.

The basic rule for all water block designs is the bigger thermal exchange area the better and the higher energy density dissipation per blocks area unit the better. Inertia of a block is an added bonus.


Anyone knows how much does a 60mm diameter copper ball cost?
and how much would it cost to solder all this pins to its surface (BillA? curious how would it cost to produce in USA)?

[Butcher]

Um, bernoulli's law states the the energy in the fluid is conserved before and after the constriction, thus there is no loss of energy according to this and so the pressure is not affected by the position of the component in the loop.

[Jabo]

Originally Posted by Butcher

Um, bernoulli's law states the the energy in the fluid is conserved before and after the constriction, thus there is no loss of energy according to this and so the pressure is not affected by the position of the component in the loop.

Totally wrong I am affraid Butcher.
Not energy but mass is conserved.
Well, if you consider mass to be a form of energy (E=mc^2) then yes, you are right

If what you say was true we wouldn't need pumps in our loops. A little hand crank would be nuff to increase energy state of our WC loops and voila, perpetuum mobile (sp?) in its finest

It is a tad confusing and it took me some time to get through the lot of it. What really helps me is the fact that all systems thrive to assume the lowest energy state possible and to get ANY work done energy MUST be inserted into a system and if this work is to be sustained energy needs to be added continually I am affraid.

[Butcher]

Bernoulli's Law states that energy is conserved, if energy is not conserved, then Bernoulli's law does not apply.
Also, if mass is conserved, then energy is conserved according to E=mc² - c is contant after all. Try again.
You still haven't offered a reasonable explaination of where the energy in the loop is actually going (hint: think friction).

The fact of the matter is, flow is uniform throughout the system. For a given flow rate and density, water will flow through an aperture of a fixed size at a fixed speed. Changing where it is in the loop won't change the flow, nor will it change the size of the aperture. Water doesn't compress very well, nor does it expand or contract much with heat, so the density can be assumed to be constant throughout the system (especially given the temperature difference is on the order of 0.1-0.2C). I'm really not seeing where this huge difference in pressure is coming from.

[Butcher]

Additionally, Bernoulli only works for non-vicious fluids, which water isn't.

[Jabo]

Originally Posted by Butcher

Additionally, Bernoulli only works for non-vicious fluids, which water isn't.

Bernoulli Law is a special case but demonstrates a principle and it can bve used for viscous fluids if it is assumed that distance between non-restriction and restrion parts of a loops is so small taht viscous losses do not have any effect.

If theoretical approach does not work try an experiment at home. hav an open loop with thrre blocks and a rad. One setup has block just after the pump and the second has same block as the last thing on the line before discharge. Which one is better for this block? Flow is constant through both systems.

Pressure loss in a loop is basic fact and you are right that it is contributed to viscous loses. This is basically where using ernoulli velocity decreases despite aperture size decrease and to make it all tick pressure must drop.

Oh, and this one should be able to answer some of your questions I hope *hint* mass conservation*end hint*

[Jabo]

Here is something to keep you occupied with quite precise head loss calcs.
This index page.
Again, applies to simple pipes only. Forget about calculating head loos for a block or rad (search this forums for BillAs answer and reasoning).
The best way is to measure it

[Butcher]

That link just proves what I already stated - that speed is dependent on flow and the size of the aperture. Flow is uniform throughout the system, thus speed is not affected by the position in the loop.

[Jabo]

Originally Posted by Butcher

That link just proves what I already stated - that speed is dependent on flow and the size of the aperture. Flow is uniform throughout the system, thus speed is not affected by the position in the loop.

I rest my case here. I I failed to explain it uo till now I give up.
Try the experiment I described.

Flow is uniform, speed is not uniform, presure is not uniform, vesels aperture is not uniform.

It 1.00 am here and I must get back to work to get anything done t'nite still

[Incoherent]

Originally Posted by Jabo

Anyone knows how much does a 60mm diameter copper ball cost?

LOL.
That is one heavy block.

Originally Posted by Jabo

Not necessarily, if you decrease 'headroom' (pins height) you will create very much more velocity due to decreased cross-sectinal area (see Bernoulli Law again) even though surface area increased - ...

The basic rule for all water block designs is the bigger thermal exchange area the better and the higher energy density dissipation per blocks area unit the better. Inertia of a block is an added bonus.?

Jabo, I'm inclined to agree with you. I was not envisaging a block quite that large however...
Reducing the "headroom" can be done with low profile blocks too. But. The surface area is vastly higher, offsetting it perhaps.
I've always lÃ*ked the idea of a spherical block. Never followed it through though.

[Butcher]

Originally Posted by Jabo

I rest my case here. I I failed to explain it uo till now I give up.
Try the experiment I described.

Flow is uniform, speed is not uniform, presure is not uniform, vesels aperture is not uniform.

It 1.00 am here and I must get back to work to get anything done t'nite still

How is the aperture not uniform? If I put my waterblock at the start or end of the loop, it's not going to change the shape of the block in any way.

What I'm saying is - in a given component (e.g. my waterblock), the speed is dependent on the aperture and the flow rate (as given by the link you provided). The flow rate is uniform throughout the system. The aperture of the waterblock is constant - I'm not changing the block itself, just it's position in the loop, thus the aperture does not change. Therefore, whether placed right after the pump, or right before it, the speed is the same.

[Groth]

No Jabo, you've got it wrong. Since flow is the same through all components in a series system, order will not affect their performance. It is true that the first bit will have higher pressure (at both inlet and outlet), but the pressure drop across it is what is related to its flow rate. I think you are looking at the pressure relative to atmospheric, and thinking that higher pressure there implies higher water velocities, but what matters in the delta-P from inlet to outlet.

And Bernoulli's equation is about energy conservation - the conversion of pressure energy or kinetic energy or gravitational potential energy into an equal amount of one of the others. It also relies on conservation of mass indirectly, in that it is built on conservation of volume of an incompressible fluid.

[kizzap]

okay while we are on the topic of coldplates will the video card need a coldplate as well seeing that the pelt is around the same size as the core?

[8-Ball]

YES!

And the pelt is not the same size as the core. The pelt is roughly the same size as the chip package, but the core is much smaller.

You NEED to spread the heat from the core to the whole of the pelt.

On top of that, you need to clamp a pelt MUCH more tightly than you need to clamp the assembly to the core. This can only be done using a separate coldplate.

8-ball

[Jabo]

Originally Posted by Groth

No Jabo, you've got it wrong. Since flow is the same through all components in a series system, order will not affect their performance. It is true that the first bit will have higher pressure (at both inlet and outlet), but the pressure drop across it is what is related to its flow rate. I think you are looking at the pressure relative to atmospheric, and thinking that higher pressure there implies higher water velocities, but what matters in the delta-P from inlet to outlet.

And Bernoulli's equation is about energy conservation - the conversion of pressure energy or kinetic energy or gravitational potential energy into an equal amount of one of the others. It also relies on conservation of mass indirectly, in that it is built on conservation of volume of an incompressible fluid.

I always pathetic at explaining my ideas and even worse at what I am trying to do here.

From what you are saying groth it transpires that doesn't matter the pressure and the only thing which counts is delta-P, right?
My use of basic Bernoulli was to say the least unfortunate and to imprecise for the level of knowledge of members here,

Butcher ->yes you are right, blocks aperture doesn't change I was taling about whole loop where vessles diameter changes from one block/rad/reservoir etc to the other. Velocity of coolant decreases after each unit in a loop due to kinetic energy of coolants paticles being used to combat viscous forces+gravitational forces.

[Jabo]

Right, my final contribution to this discussion

The link below is pro-level water-cooled heatsink thermal calculations for simulation purposes,

Enjoy

[8-Ball]

The way I see it is as follows.

Flow rate through a restriction is a function of the pressure drop across the restriction.

Given a number of restrictions between the outlet and inlet of a loop, the total pressure drop is additive.

As an example, at a given flow rate, the pressure drop across the cpu block may be x, y for a gpu block, and z for all of the tubing.

Thus, to maintain this flow rate, you need a pump capable of providing head equal to x + y + z at the given flow rate.

Beyond that, the order of the components does not matter, as the pressure drop across a given component will always be the same for a given flow rate.

Is that correct or not.

In other words, if you have pressure x at the block inlet and pressure x-dp at the outlet, would the resulting flow through the block be any different if the pressure was x+dp at the inlet and x at the outlet. The pressure drop is still dp, so it shouldn't matter.

If the flow rate through a block is the same in these two scenarios, then given that the geometry doesn't change, the velocity of coolant shouldn't change either.

8-ball

[Jabo]

Originally Posted by 8-Ball

The way I see it is as follows.

Flow rate through a restriction is a function of the pressure drop across the restriction.

Given a number of restrictions between the outlet and inlet of a loop, the total pressure drop is additive.

As an example, at a given flow rate, the pressure drop across the cpu block may be x, y for a gpu block, and z for all of the tubing.

Thus, to maintain this flow rate, you need a pump capable of providing head equal to x + y + z at the given flow rate.

Beyond that, the order of the components does not matter, as the pressure drop across a given component will always be the same for a given flow rate.

Is that correct or not.

That is 90% correct. You forgot that the lower the pressure the lower the dP (head loss/pressure drop) and conversly the stronger the pump the higher the dP valu is going to be! That's why it does make a differnece where you place your components

In other words, if you have pressure x at the block inlet and pressure x-dp at the outlet, would the resulting flow through the block be any different if the pressure was x+dp at the inlet and x at the outlet. The pressure drop is still dp, so it shouldn't matter.


If the flow rate through a block is the same in these two scenarios, then given that the geometry doesn't change, the velocity of coolant shouldn't change either.

8-ball

Yes, with higher pressure value dP s going to be higher and to maintain flow rate velocity has to go up. Head loss curves are non-linear and rise steeply after some point

[8-Ball]

Originally Posted by Jabo

That is 90% correct. You forgot that the lower the pressure the lower the dP (head loss/pressure drop) and conversly the stronger the pump the higher the dP valu is going to be! That's why it does make a differnece where you place your components

Could you possibly clarify a little.

If you use a stronger pump, then yes the pressure drop would be higher, for obvious reasons.

OK, just to clear this up.

Two scenarios.

A pump,
one long length of tubing,
one short length of tubing,
a waterblock.

In the first scenario, the short length of tubing is before the block and the long section after. Lets make it really long, so there is a real difference in pressure)

In the second scenario, the long section is first, then the block, then the short section.

Would this have any affect on flow rates through the block? I would say not, since the OVERALL pressure drop for the loop should be the same, and so for a given pump, the flow rate would be the same in both scenarios, thus the velocity in the block should also be the same.

Granted the absolute pressure values will be different, but it is the actual pressure drop from one side to the other which drives flow, yes?

8-ball

[BillA]

8-Ball
I'm concluding that Jabo is a (techno) troll
ALL posts by others are somewhat wrong
ALL posts by Jabo are incomplete
ALL clarifications require more yet of same
no one understands except Jabo

I certainally do not understand

[8-Ball]

Oh well.

Back to thesis.