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Marx Power Measurement with CuSO4 resistive divider

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Proud Mary

Sun Apr 26 2009, 08:36AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

The output of a Marx generator is to be applied to a resistive load consisting of aqueous CuSO4.

Given knowledge of the volume and concentration of the solution, and allowing for thermal losses, would it possible to calculate the power dissipated in the load by the rise in temperature of the solution - measured perhaps with an in situ platinum resistance wire.

And: could the amount of power dissipated be quickly determined by some kind of automated (or quick) measurement of the amount of electrolysis products?

Would the normal 'laws of electrolysis' apply in a linear fashion to large magnitude, fast rise-time pulses, where local pyrolytic decomposition of water might occur as an artefact of plasma electrolysis?

And any other thoughts on the electrochemistry of this set up would be welcome.

Simple answers are best for simple minds like mine!

Bored Chemist

Sun Apr 26 2009, 08:53AM

Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022

"Given knowledge of the volume and concentration of the solution, and allowing for thermal losses, would it possible to calculate the power dissipated in the load by the rise in temperature of the solution - measured perhaps with an in situ platinum resistance wire."
Yes. If you put a small heating element into the liquid you can use that to calibrate the system.
I suspect that you will end up measuring rather small changes in temperature and so you will need a rather sensitive thermometer. Unfortunately, sensitive electronics and large pulsed currents don't go well together. You might have some serious shielding problems.
With a CuSO4 solution the usual choice is for Cu electrodes because this avoids polarisation effects. The change that the electric current would produce is that one electrode would gain weight at the espense of the other. Also, all the weight change would tell you is the number of coulombs of charge passed through the system (give or take errors that would arise because other current carriers would also have an effect under that brutal conditions of a short high- current discharge). You can calculate that from the voltages and caps in the Marx gen much more easilly than you can measure the masss of copper trasnsfered.

Proud Mary

Sun Apr 26 2009, 10:10AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

Thanks for your interest, Your Boringness!

I could see that the rise in temperature would probably be very small, but I suppose one could have an aggregate over a number of shots, and the apparatus could be placed in a calorimeter jacket, so the temperature of the solution need not be measured directly.

I hadn't thought weighing the amount of copper deposited as a very practical solution either - a lot of fiddling with milligram quantities long after the fact.

What about changes in the optical properties of the solution - refractive index, optical rotation, spectrometer, or even some kind of colorimetry - perhaps using an indicator of some kind, where the emergence of a colour, or its density occurred in proportion to the underlying electrolytic reaction?

aonomus

Sun Apr 26 2009, 10:39AM

Registered Member #1497 Joined: Thu May 22 2008, 05:24AM
Location: Toronto, Ontario, Canada
Posts: 801

Part of the problem with using optical methods is that usually cells for measuring processes are fairly small so that the entire cell is consistent (no area effects due to concentration of reactants being uneven). Refractive index, rotation, and absorbance spectroscopy require the solution to be homogenous for accurate measurements. If you want to make a good measurement you would fire the marx for a single, allow diffusion to occur (or mix via agitation, at no time removing solution by a stir rod, etc), measure, and repeat.

Another problem is that with such a high voltage output from the marx, a small cuvette would result in arc-over through air possibly, or corrona would waste some of the energy, thus to get all the energy into the cuvette, you would need a long narrow cuvette, further hampering diffusion.

In voltammetry there is something called diffusion limited current, essentially once you exceed the potential required for redox reactions to occur, the local analyte immediately adjacent to the electrode gets plated onto the electrode, creating an area of effectively zero concentration. Current is limited to how fast the analyte can diffuse from bulk solution towards the electrodes. Since the marx will have a high current pulse, you would basically want to maximize both electrode surface area, and electrolyte concentration.

Having said that, measuring the weights of the electrodes is probably the best so long as you clean them thoroughly to prevent salts from increasing the weight, and if you have a scale that is sensitive enough to detect changes on the order of fractions of a mg. I have my doubts about the actual size of mass change being large enough to fall into the 10's of mg range, although that is dependent on the marx generator. Hopefully staying up till 6:30AM hasn't hampered my ability to science.

Wait what? :D

Proud Mary

Sun Apr 26 2009, 10:55AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

Thanks for your opinion, gentlemen. I was trying (as ever!) to look for an alternative approach to an old problem, but perhaps chemistry is not it.!

Bored Chemist

Sun Apr 26 2009, 03:25PM

Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022

Let's just be clear about this, To a fairly good approximation, apart from warming up slightly, the solution doesn't change. Some copper leaves the anode and goes into solution. Exactly the same amount leaves the solution and turns up at the cathode.

The mass of copper transfered depends on the details of the generator, but lets guess at 100 Joules and 10KV. That gives 0.01 Coulombs. Divide that by Faraday's constant which is about 100000 Coulombs per mole and you have something like a ten millionth of a mole. Halve that because copper ions are doubly charged then multiply by the molar mass of copper. You are talking about 3 micrograms of copper. Good luck weighing that. Even with a thousand shots you would need quite a good balance to measure the change in mass.

The heating effect should be a better bet. The same 100 Joules that I guessed earlier would produce a (barely) measurable temperature change in a cup full of water ( or copper sulphate solution).

Proud Mary

Sun Apr 26 2009, 03:31PM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

Thank you for your excellent explanation. Not a very practical idea then, but worth having thought about all the same.

aonomus

Sun Apr 26 2009, 04:14PM

Registered Member #1497 Joined: Thu May 22 2008, 05:24AM
Location: Toronto, Ontario, Canada
Posts: 801

What about emitted light? Perhaps a CdS cell can be calibrated and some math be done to compensate for the surface area at distance?

Bored Chemist

Tue Apr 28 2009, 05:14PM

Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022

"What about emitted light?"
Good questtion. What do you think will emit light?

aonomus

Tue Apr 28 2009, 07:52PM

Registered Member #1497 Joined: Thu May 22 2008, 05:24AM
Location: Toronto, Ontario, Canada
Posts: 801

I meant fire the marx into a sparkgap and measure emitted light. Not off a CuSO4 solution... it was late.

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