I was bored this Sunday morning and decided to try to calculate the power output of a proton pack. Lets see if this exercise is going to work. If there are any engineers feel free to correct this most likely faulty calculation. I'm an architect, not an electrical engineer or a nuclear physicist.
The canon lists the electric output of Ghostbusting equipment as respectively 1 kV for the Trap (movies) and 3 kV for the Proton Pack (GBTVG). Let's look at the wattage and amps.
Curium apparently yields about 4 watt per gram when used as fuel by heat generated through decay in radioisotope thermoelectric generators. Basically it glows red hot because of radiation and constantly generates heat. If we assume the packs have about a 1kg sample, this means 4000 watts of constant power. So a 3 kV proton pack powered by a kilogram of Curium fuelled RTG, so consuming 4000 watts would mean it's only drawing about 1.33 amps.
3kV and 4000 watts is akin to a large electric generator. For example a fridge uses about 1500W, a jacuzzi about 2800W. This kind of power could power smaller cyclotrons such as those used in medical devices, X-Ray machines, radiotherapy machines, etc. But the power -output- of these are around 50-100 watts maximum, like a light bulb. Cyclotrons are very inefficient in converting the power they consume into the power they produce.
Obviously that is too little power for what we see the beams in Ghostbusters do.
Judging from the proton beam destruction of the bar in the Sedgewick Hotel scene I'd guess the power output is around 50 kW. Similar to the destruction a 300lbs man would cause if crashing into that bar running at about 22 mph. A strong firehose is about 30 kW so that sounds about right.
But cyclotrons are very inefficient.
http://accelconf.web.cern.ch/AccelConf/ ... 1_talk.pdf
Depending on the size the efficiency varies from (small cyclotron) less than 1% to (large cyclotron) around 14%. Lets take 3%.
That means we would need to consume 30x the power needed to be able to output 50kW. So we would need 1.5 megawatts of power. 1500000 watts at 3000 volts which would draw 500 amps.
1.5 MW of power can power around 200 homes simultaneously. As a comparison, the worlds largest particle accelerator, the Cern large hadron collider uses about 120 MW of power. That is a serious electricity bill.
Anyhow so we could say the proton pack consumes about 1.5 million watts. That means the 4000 watts generated by using the 1kg of Curium as fuel in a radioisotope thermoelectric generator would be nowhere near enough. That leads me to think that the Curium maybe isn't used as a simple heatsource through decay, but is used as actual nuclear reactor fuel through fission. The fission of 1 g of nuclear fuel (like Curium) per 24h liberates about 1 MW of power. That would mean there is a miniature nuclear reactor in the proton pack.
Edit: lol, this is what Wikipedia says
"All isotopes between 242Cm and 248Cm, as well as 250Cm, undergo a self-sustaining nuclear chain reaction and thus in principle can act as a nuclear fuel in a reactor.
245Cm and 247Cm have a very small critical mass and therefore could be used in portable nuclear weapons, but none have been reported thus far."
Then again parts of the proton pack could be nuclear powered and parts could be RTG powered. Here's a list I found of parts.
Ok my mind is fried, I should never have done this on a Sunday morning.