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Railgun Project | Current Question

  1. Jun 7, 2009 #1
    Hello, as stated I'm working on a railgun. At this point, I'm in the design phase. I'm currently trying optimize the performance of the gun within the constraint of my resources.

    As you may or may not know, a railgun consists of two rails short-circuited by a conductive projectile which is propelled down the rails by an electromagnetic force, in this case the Lorentz Force.

    My rails, both of them, will be pure copper, 1 cm in width, 0.5 cm in height and 40 cm in length. They will be 0.5 cm apart.

    The projectile, armature, will be 0.5 cm high, 0.5 cm wide and 1 cm long with a hole 0.25 cm in diameter drilled from the rear end 0.75 cm deep. This hole shifts the center of pressure behind the center of gravity, giving it aerodynamic stability. It will be made of aluminum.

    To power this system, I will use a bank of capacitors. Each of them will most likely be as follows: 3300 uF (.0033 farads) at 400 volts. This yields 264 joules.

    Could someone help me figure out the current, in amperes, said capacitor would generate within this circuit? The capacitors will be connected directly to the rails. Your assistance would be enormously appreciated. I can then establish how many capacitors I need in series and/or parallel to achieve the highest projectile speed possible by plugging the value (in amperes) in a few equations.

    Thanks in advance for your time.

    axi0m
     
  2. jcsd
  3. Jun 7, 2009 #2

    vk6kro

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    You would need to know the Equivalent Series Resistance of the capacitors.

    This is not usually shown on the capacitor, but you can measure it with the right equipment.

    Set an Audio signal generator to 1000 Hz sinewave and maximum output.

    Put a 1000 ohm resistor and a 10 ohm resistor in series (10 ohm at ground end) across the signal generator output.

    View the voltage across the 10 ohm resistor on an oscilloscope. Get a full scale deflection if possible. The voltage will be small so you need an oscilloscope.

    Now place the capacitor across the 10 ohm resistor.The reactance of the capacitors should be negligible at 1000 Hz and a perfect capacitor would reduce the oscilloscope deflection to zero.
    However it will probably not read zero due to internal resistance and you can use the parallel resistor formula to work out what the resistance is, relative to 10 ohms.
    If the voltage drops to half, the resistance would be 10 ohms.

    This is a good way to sort through old capacitors to see which ones should be thrown out.
    Equipment often fails because this resistance becomes too high.
     
  4. Jun 8, 2009 #3
    Thank you very much for your time and thorough response. In fact, however, I do have access to the datasheet for all potential capacitors. It specifies the Max ESR as 33.0 milliohms at 125 Hz. My application will only involve DC, so I'm not sure if this ESR, being specified with a frequency, will be compatible. If so, can I simply use this ESR value of 0.0033 ohms as the R value in any standard formula for current?

    Edit:
    Additionally, if the above is true, would combining identical capacitors in parallel simply increase capacitance without changing the ESR?
     
    Last edited: Jun 8, 2009
  5. Jun 8, 2009 #4

    vk6kro

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    The other resistance in the circuit will affect the final current but, yes, that resistance would be a good starting point. Those ESRs seem very low, though.

    Unfortunately, you have to measure this resistance at a frequency because it is in series with a capacitor. The idea is to measure it at a high enough frequency so that the reactance of the capacitor doesn't matter. It is mainly the resistance of the connections and of the plates of the capacitor, so it probably doesn't vary much in these frequency ranges from the DC value.

    The voltage will drop rapidly, of course, so the current will as well.

    Unless the capacitors are new, those figures would need checking. The setup I described works well and is easy to set up if you have access to the gear. If you include the external wiring across the 10 ohm resistor, you might get a more realistic figure.

    Putting the capacitors in parallel still leaves the capacitors in series with their individual ESRs.
    If your ESR figure is still correct, external resistance is probably going to be more important.

    Let's see: 400 volts across 0.033 ohms is 12000 amps. Hmmmm!
    Wouldn't last long, but it would make quite a spark and possibly weld all the parts together.
    Absolutely lethal of course.
     
  6. Jun 8, 2009 #5
    Again, thanks for your help. I'm still curious though, when adding capacitors in series or parallel, would I multiply the ESR by the total number of the capacitors?

    For example, adding two caps in series at 450 volts each with 28.8 mOhms each, would it yield the following?

    I = (900)/(0.0576)
     
  7. Jun 8, 2009 #6

    vk6kro

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    Yes, I think that is right. The voltages and the resistances both add up when the capacitors are in series.
    Current = 900 / 0.0576 = 15625 Amps.
    So, you don't get extra current by doing this.


    If they are in parallel, the resistances also appear to be in parallel

    Current = 450 / .0288 = 15625 amps for each capacitor.
    So twice this for two capacitors is 31250 Amps.
    So the combined resistance is 450 / 31250 = 0.0144 ohms
    Which is like two 0.0288 ohm resistors in parallel.
     
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