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We are pulsing 20A, 12V current for electrolysis through a 6% KOH solution between two steel plates. Because of steel shortages, we have switched to a steel plate that is very slightly thicker. We are suddenly seeing 40C more heat in the system. The thicker steel plate should mean less resistance and therefore less heat. So we're thinking that the heat comes from the KOH solution. Our engineer says that if we increase the KOH to 12%, it will allow more current per pulse, so more heat. But I'm thinking that greater conductivity in the water would mean less resistance and less heat. He says it would be better to drop the KOH to 3% to reduce the current flowing per pulse and therefore reduce the heat. What would be the correct approach and, more importantly, why?

foolishmuse
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    Is the new plate the same alloy as the last one? For example, 1040 is 1.6E-7 Ohm-m, while 4340 is 2.5E-7. 316 stainless is 7.4E-7. – vir Jun 16 '21 at 23:11

1 Answers1

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If the voltage is fixed, the lower the resistance, the more current and the more heat.

$$P = \frac{V^2}{R}$$

Where P is the power dissipated by a circuit element, V is the voltage across its terminals, and R is the element's effective resistance at the given voltage.

Math Keeps Me Busy
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  • "*The thicker steel plate should mean less resistance and therefore less heat*" if your current is stays the same. Which it is not as per this answer, What might make it stay the same as you fiddle with R? If the R was actually a wire and the load was dominating and determining the major current draw. But in your example the plate is the load, is dominating and its resistance therefore determines the current. \$P=I^2R\$. So less resistance = more current = more heat due to the square – DKNguyen Jun 17 '21 at 00:52
  • Thanks. I always thought that it was resistance that caused heat, but you're saying that current itself causes heat. Is this correct? – foolishmuse Jun 17 '21 at 15:22
  • It is the product of current and voltage that creates heat. Current in a superconductor, which does not require a continuous voltage, creates no heat, so current by itself is only part of the story. – Math Keeps Me Busy Jun 17 '21 at 15:28