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:)

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:)

LOL. :D
That is one heavy block.


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.

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.

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.

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?

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

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, :rolleyes:

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.

Right, my final contribution to this discussion :)

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

Enjoy

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

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

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

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

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

Oh well.

Back to thesis.

can the bullshit Jabo

90%
ok, give us the numbers illustrating 90%

got that, 90%
show it
the f**king numbers, yours, not a link to someone else's work

BillA, pay me for the time to do all the calcs and I'll be happy to provide you with the most comprehensive analysis on the subject.
No need to get so exicted :(
I just do not have time to do it, sadly and that is why I use other people's results (pubilshed as open source on the net) to back up my writing here.
Take it or leave it, choice is yours.
But before you hail me The Bullshiter of The Century try to prove me wrong please :)

not going to grant what has not (yet) been earned

to date you have provided critical commentary, only
allusions, aspersions, inferences, mis-attributions over and over

based on what you have posted, your qualifications are extremely well masked

others here have contributed of their efforts, but your efforts are so valuable, and your sprit so mean (or incapable ?), that YOU require PRE-payment

surely there is a forum of your peers somewhere ?

No, I have family to provide for and time spent doing this extensive calculations would mean no money for my family, as simple as that.
It so easy to criticise isn't it BillA?

At least I am not judging anyone here what you seem to enjoy to the extremes.
Do not pass judgement where you have no reason nor right to do so, especially if it considers somebody personally, not ones 'theories'.

are you sugestting anything here or is it again the language barrier playing up here?

we are all evaluated based on our actions
review your posts, they are as described
my conclusions are above

you may not be familiar with the procedures of A.S.T.M. (and why should you be, look it up if you are);
any criticism may be made, so long as a possible solution is also proposed

now to hear some rail bird (an outside fence sitter) express constant criticisms, without ever a solution, is to listen to the baying of a jackass in the morning

Jabo, focus on adding to the store of knowledge
or support your family and leave us fools in peace
-> you are not contributing anything here Jabo

What kind of calculations do you plan on doing with no data or numbers of any sort to work with? All you need to do it provide some sound theory or basis for your claims.

Consider this: an electrical circuit with a battery supplying voltage V, and two resistors of values X and Y ohms. If you wire up the resistors to the battery in series, the circuit as a whole will have resistance (X+Y) Ohms (assuming the wires have negligible resistance). So by Ohm's Law, current through the circuit is I=V/R which is the same as V/(X+Y). If you then switch the order of the resistors in the loop, you could say that total resistance is now (Y+X), and we know that's the same as (X+Y). Now, despite changing the order of devices in the loop, the resistance is still the same and the current is still I=V/(X+Y).
Now I realize that there are some differences (non-linearity in flow-pressure relations, but at constant flowrate it shouldn't even matter), but does the same principle not apply to a watercooling loop? Battery voltage V is equivalent to pump pressure, current is equivalent to volumetric flowrate, resistance R is equivalent to flow resistance. Each resistor has a set voltage drop in the circuit just like each watercooling device has a set pressure drop, regardless of its position in the loop.

That does seem to be the case bill. I guess since his time is so precious, we better not waste it by answering him. ;)

All I can say is that I did not want it to come out this way...:( Well, $hit happens and obviously I am not a PR guru :)
I just wanted to contribute here as much as I can, it didn't work, tough luck.
See ya around.

Jabo

Jabo I've read what you have written a few times now and still don't understand what you are getting at.

You seem to be suggesting that a water blocks performance is not just a function of the flow rate but also related to the absolute pressure in the block. This suggests to me that you are talking about the density of the water. i.e. the higher the absolute pressure, the greater the density of the water, thus the water needs to travel at a higher velocity to maintain the same flow rate. Is that right?

In my limited experience I have always dealt with water as incompressible and I wouldn't have thought that there would be much change in the specific volume of the water at the kind of pressures we are dealing with. However, I may have completely misunderstood. If so, any clarification would be appreciated

Water is very hard to compress in it's liquid form, at the sort of pressures we deal I doubt you can even measure to density change.

no desire to run you off Jabo,
just to shift your commentary to something more useful and productive

many valid engn considerations are glossed over, some through ignorance, others because of cost constraints, others because they have been found to be not significant; but when no data is provided, the references are scrambled eggs, and the terminology is obtuse, then few will be swayed by any amount of words