Calculating temp rise caused by electric arc in an enclosed space

[pixelsnpings]

Summary:: Given a known closed space/apparatus ( e.g. constant volume, pressure, density, current, temperature, voltage, spark gap distance - let me know if I missed something) how would I compute the change in gas temperature.

Hello,

Given a known closed space/apparatus ( e.g. constant volume, pressure, density, current temperature, voltage, current, spark gap distance - let me know if I missed something) how would I compute the change in gas temperature. Obviously, temperature is not the only perameter that would change, I'm just focused on that parameter for now.

Thanks!

 

[berkeman]

Welcome to PF.

I would think that a first approximation would be to calculate the energy delivered to the gas (the electrical power to generate the arc multiplied by the arc duration in seconds). That energy would go into the light from the arc and the heating of the gas. I'm not sure in what proportions, though.

[Note -- Thread moved from the EE forum to the Thermo forum for better replies]

 

[tech99]

Summary:: Given a known closed space/apparatus ( e.g. constant volume, pressure, density, current, temperature, voltage, spark gap distance - let me know if I missed something) how would I compute the change in gas temperature.

Hello,

Given a known closed space/apparatus ( e.g. constant volume, pressure, density, current temperature, voltage, current, spark gap distance - let me know if I missed something) how would I compute the change in gas temperature. Obviously, temperature is not the only perameter that would change, I'm just focused on that parameter for now.

Thanks!

I presume you measure the power supplied to the arc, and assume this is the power supplied to the gas. Then you apply the ordinary heating formula using the Specific Heat Capacity of the gas. For air at constant volume (sealed box) the figure I have found is 0.718 kJ/kg K, and at constant pressure (vented box) 1.01.
So supposing we have 1 kg of air in a sealed box and we apply 1kW of energy for 10 seconds, energy supplied is 10 kJ. Then the temp rise will be 10/0.718 = 13.9 Kelvin.
https://www.engineeringtoolbox.com/specific-heat-capacity-gases-d_159.html

 

[pixelsnpings]

@tech99, That is exactly what I need. Thank you so much!