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Registered Member #5323
Joined: Fri Jun 15 2012, 02:14PM
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Posts: 104
A high voltage capacitor rated at 3.75 farads is discharged through a 1,500v spark gap. Can the amount of energy discharged through the spark gap be expressed in Joules or coulomb or some other familiar unit? If so, how?
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
Before and after the discharge the energy stored in the capacitor can be calculated using E = 0.5 x C x V^2 The difference went to heat, light and sound in the gap, and a little energy went to electromagnetic losses in the circuit. So it is measured in Joules,
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
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GR, you can lead yourself astray by thinking in terms of "the energy discharged through the spark gap" instead of "the electrical energy in the charged capacitor". 4.2 MJ of stored energy isn't necessarily hard on the switch -- it all depends on the current and duration of the discharge, and resistance of the switch.
Let's suppose your capacitor is a bank of supercaps, as Signification suggested. That's the only kind that can store 4.2 MJ (1172 Wh) in about 200 kilograms instead of tons.
The fastest possible discharge, limited by the capacitor's own ESR, is on the order of 1 second. When your spark gap fires, its voltage immediately drops to a tiny fraction of that 1500 volts -- almost like closing a mechanical switch. Current could well exceed 5000 amps, but more than 95% of the voltage drop and power dissipation is inside the capacitor due to ESR. Capacitor will end up slightly warmer than before the short-circuit discharge.
I ran some numbers based on this Maxwell datasheet: Sig's (3000 F, 2.5 V) x 450 has the right energy & approx. the right mass, but wrong total C and V: 6.7 F, 1125 V. A much closer fit is BCAP2000: 2.7 V, 2400 F initially, let's say they've been worn down to 2085 farads, for 7.6 kJ per unit. We need a bank of 556 to meet the total energy goal, and that works out to 3.75 farads and 1501 volts. From specific energy figure in datasheet, total mass is 202 kg. ESR spec is 0.35 milliohms per unit, 0.195 ohms for whole string. So the RC product, with no external resistance, is 1.05 seconds.
At full voltage, initial short-circuit current would be 7.7 kA. That matches the value given in datasheet, which warns that it's abusive to the capacitor and more of a safety thing (use wires thick enough not to melt).
Datasheet gives both "impedance matched" and "usable" specific power values, that work out to about 5.1 and 2.5 kW per capacitor. The former is consistent with initial current of 4 kA, initial load voltage = 1/2 of initial capacitor voltage, and RC = 2.1 seconds. There's no load that can receive more power from that capacitor. (Perfect short circuit would have zero voltage drop, so zero power).
At the "usable" power level, 14% of capacitor voltage is lost in max. ESR and 86% gets to the load. Load resistance for whole string is 2.15 ohms, initial current 1.1 kA, RC 7.5 seconds. I think similar capacitors, in modules stacked to maybe 500 V, are used in some city buses with hybrid propulsion systems.
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Spark discharge expressed in Joules?
