Peltier Blocks

[BillA]

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


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

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

[Jabo]

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

[BillA]

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 ?

[Jabo]

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'.

surely there is a forum of your peers somewhere ?

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

[BillA]

Originally Posted by unregistered

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 spirit so mean (or incapable ?), that YOU require PRE-payment

surely there is a forum of your peers somewhere ?

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

[zer0signal667]

Originally Posted by Jabo

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'.

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.

[Butcher]

Originally Posted by unregistered

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

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

[Jabo]

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

[WAJ_UK]

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

[Butcher]

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.

[BillA]

Originally Posted by Jabo

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

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