Battery Effect In waterblocks Need A Suuggestion

[Sapientiea]

SINCE87 you are right MG is the metal to use!

greetz sapientiea

[Althornin]

Originally posted by Since87


But, if the top and bottom of your waterblock are aluminum and copper, and they are bolted together using threads in the aluminum, then the electrical resistance between them is likely to be on the order of 0.001 Ohms or less.

Off the top of my head, I'll say the desired voltage differential between the copper and aluminum is 2V. Maintaining 2V across a 0.001 Ohm requires 2000 Amps. So doing this would require dissipating 4 KW in the joint between the copper and the aluminum. Kind of makes corrosion a moot point.


BTW, aluminum is more electronegative than zinc. You'd need to go to magnesium as your sacrificial anode.

Ok, but you are missing a few things:
The limit is the salt bridge, not the direct connection between the two metals....ergo, the galvanic cell will not be producing a high current, and you will not need to send a high current in reverse.
You can easily add resistance to limit your current throughput to reasonable levels - just add a resistor in series, and you are done.

As for the Zinc/magnesium, you are correct - i looked at the potentials in seawater by accident - the potentials are very different

[Since87]

Originally posted by Althornin
Ok, but you are missing a few things:
The limit is the salt bridge, not the direct connection between the two metals....ergo, the galvanic cell will not be producing a high current, and you will not need to send a high current in reverse.

No, it is you who is missing things.

It seems to me I've explained things such that anyone who is truly trying to understand can. However, I'll try briefly one more time.

Assume you have a D-cell battery.

If there is no conductor between the positive and negative terminals of the battery, then very little reaction will occur in the battery. There is no need to apply a voltage to the battery to stop the reaction from occuring.

If there is a copper wire connecting the two terminals of the battery, then a very energetic reaction will occur in the battery. In theory you could use an external power source to maintain the 'natural' voltage of the battery between the terminals. However, considering the amount of power that would need to be dissipated in the copper wire to maintain that 'natural' battery voltage, it's pointless to try.

Only in the unusual case (WRT WC systems) where there is a 'high' resistance between the terminals of the battery, can an external power source be of any value.

Yes, the electrolyte in a watercooling system is much less conductive than the electrolyte in a battery. However this is beside the point in discussing the usefulness of an externally applied power source. It is the resistance of the conductive path outside the coolant that matters when discussing external power sources.

A somewhat relevant link.

[Althornin]

Originally posted by Since87
[b]No, it is you who is missing things.

It seems to me I've explained things such that anyone who is truly trying to understand can. However, I'll try briefly one more time.

the problem is you are wrong, AFAIK. You ignore one limit (ion concentration and salt bridge) to focus on the other ("leakage path").

Assume you have a D-cell battery.

If there is no conductor between the positive and negative terminals of the battery, then very little reaction will occur in the battery. There is no need to apply a voltage to the battery to stop the reaction from occuring.

thanks for the patronizing "lesson". I think i said the same thing about 15 posts ago.

If there is a copper wire connecting the two terminals of the battery, then a very energetic reaction will occur in the battery. In theory you could use an external power source to maintain the 'natural' voltage of the battery between the terminals. However, considering the amount of power that would need to be dissipated in the copper wire to maintain that 'natural' battery voltage, it's pointless to try.

Im sorry, but you are not correct here.
The output of the galvanic cell is limited by the ion concentration in the salt bridge, not only by the resistance of the connection between the two metals.
Ergo, you are incorrect wrt to power required.
I believe that i have been quite clear in saying this, yet you continue to ignore my statements. If you could direct your content to showing me the flaw in my thinking rather than repeating yourself, maybe we could move on.

You seem to be neglecting the electrolyte in your considerations (ion concentration...)
from your source:

Electrolyte factors that have a major influence on bimetallic
corrosion are composition, pH and, in particular, electrical
conductivity, which affects both the intensity and distribution
of corrosion.
The severity of corrosion often increases with increasing
electrical conductivity of the electrolyte because, in practice,
high conductivity is often caused by the presence of aggressive
ions such as chloride, or by acid or alkali.

I dont think that the water in most cooling systems is a very effective electrolyte, and ergo, the current generated by said galvanic cell will be minimal. All that is needed to prevent said coprrosion is an equal and oposite voltage and current.

[Since87]

Originally posted by Althornin
If you could direct your content to showing me the flaw in my thinking rather than repeating yourself, maybe we could move on.

You seem to be neglecting the electrolyte in your considerations...

I don't know how to show you the flaw in your thinking without repeating myself.

I am neglecting the electrolyte in my considerations, because it is irrelevant to the power dissipation involved in trying to prevent galvanic corrosion using an external power source.

The goal, in adding an external power source, is to stop any flow of current through the electrolyte/coolant. If there is no current flow through the electrolyte, then there is no electrical power dissipation in the coolant. Ergo, the electrical conductivity of the electrolyte is irrelevant for the purposes of this discussion.

I'm going to go through this all once again. Please tell me what you dont understand, or disagree with.



Figure 1 represents aluminum and copper in an electrolyte, with no electrical path between the metals, aside from the electrolyte. (Equivalent to a battery sitting unused.) After initial creation, this system will reach a state of equilibrium at which point the voltage between the two metals will be that defined by the difference in the metal's electrode potentials. The resistance of the electrical path through the electrolyte is represented by Rbat. No current flows through the electrolyte in this system because there is no conductor available to complete the loop, and Galvanic corrosion is nonexistant. (Though there may be other types of corrosion going on.)

Figure 2 represents the same system as Figure 1 with the addition of an electrical conduction path other than through the electrolyte. This is the typical situation with a copper and aluminum waterblock where the two metals are bolted together. (Equivalent to a battery with its terminals shorted.) The existance of the electrical path outside the electrolyte (Rs) allows current to flow through the coolant.

Now suppose the resistance through the coolant (Rbat) is 1,000,000 ohms and the resistance of the metal to metal connection is 0.001 ohms. (Probably reasonable numbers to within an order of magnitude or two.)

The current flow in the system will be:

2 Volts / (1,000,000 ohms + 0.001 ohms) = 2 microamps

The power dissipated in the system will be:

2 Volts * 2 microamps = 4 microwatts

Fairly small current and power dissipation, but enough to eventually allow the aluminum to get corroded through.

Are you with me so far?



Figure 3 represents a system that attempts to prevent the Galvanic corrosion using an external power source. By connecting an external power source so that the voltage differential across Rbat is zero, the Galvanic corrosion can be stopped.

BUT, the voltage applied by the external power source is applied across Rs. Therefore the current through Rs is:

2 Volts / 0.001 Ohms = 2000 Amps

And, the power dissipated in Rs is:

2 Volts * 2000 Amps = 4000 Watts

(The resistance Rbat is totally irrelevant to this calculation, because the voltage of the external power source has been selected specifically to provide for zero voltage differential across Rbat, and therefore zero current flow through Rbat.)

[Althornin]

Thank you for actually responding to my point this time, it was helpful in pointing out a flaw in my thinking.

However, nothing prevents the addition of a current limiting resistor, as i said before, like so:


Now, If we make Radd some high ohm value, like 1000, then we end up with
2 volts / 1000.001 ohms = .002amps
2 volts * .002amps = .004 watts.

Unless i am missing something.

[Since87]

You haven't calculated all the currents and voltages correctly.

I've redrawn the picture with the added resistor, because it makes the math a bit easier without changing anything of significance.



The voltage drop across Rs is 2 microvolts.

So the only difference between this setup and my Figure 2 above is that the voltage differential across Rbat is:

2V - 0.000002V = 1.999998 volts (instead of 2.0 volts)

The current through Rbat will be 1.999998 microamps instead of 2.0 microamps. The rate of Galvanic corrosion will only be slowed by 0.0002%.

You simply can't have a significant effect on the current through Rbat without either; dissipating a prohibitive amount of power in Rs, or increasing the value of Rs substantially. If you can increase the value of Rs substantially, then you can as easily make it effectively infinite.

[Althornin]

ah thats right.
Damn, i took circuit analysis 4 years ago, i've already forgotten alot.
thanks.

[DigitalChaos]

a much easier and cheaper sollution would be to just use redline water wetter.

i have had a cup with a chunk of copper and aluminum sitting next to eachother in a mixture of water wetter for several years now. The aluminum gained an anodized like coating in the first few days, and there has been NO corrosion.

but the same cups containing only distilled or tap water broke down pretty fast.