Building a capacitor bank capable of pulsing 16000 A DC

[trini]

well, considering that i only need 150 A, isn't there some way i could get that kind of current without needing the capacitors at all?

 

[vk6kro]

A car battery provides currents like that to a starter motor in a car.

Alternatively, there are apparently DC welders available. There was a similar thread a couple of months ago that dealt with this.

The car battery would be the easy one.

 

[trini]

ok i have updated the design.

i will use 14 car batteries in series to get to 168 V DC. The coil will be 4 cm long, 5 layers high, making the N/l ratio 5000 using 1 mm wire[(40 x 5)/0.04]. This equates to a 9.69 Ohm resistance, making the current through my coil 17.337 A.

17.337 x 168 = 2913 J/s = 174,752 J/min

the volumetric heat capacity of copper is 3.45 J/cm^3/K
Using l = 367 m, r = 0.0005 m
c = 12,661 J/K

By holding the current while i heat the powder, i will get an even stronger final B field in my magnet. I need to heat the powder to 200 C throughout but not over that, then let it cool. I can quickly heat up the metal using an IR lamp
so if i run my Current for 1 minute, this will equate to a 13.8 K rise in my coils.

I will use jumper cables to connect the coil to the batteries, and heat the powder using a magnetron. I will need to perform tests to determine heating rates of the magnetron, what would be the best way to measure the core temperature of my powder? i was thinking a digital thermometer with a probe may work, but i need something that will give me fast readouts, as i can't let the powder get even 1 degree above 200( i may just heat to about 190 to be safe, i only need to get it to 175, to work, but 200 is optimum)

 

[gary350]

ok i have updated the design.

i will use 14 car batteries in series to get to 168 V DC. The coil will be 4 cm long, 5 layers high, making the N/l ratio 5000 using 1 mm wire[(40 x 5)/0.04]. This equates to a 9.69 Ohm resistance, making the current through my coil 17.337 A.

17.337 x 168 = 2913 J/s = 174,752 J/min

the volumetric heat capacity of copper is 3.45 J/cm^3/K
Using l = 367 m, r = 0.0005 m
c = 12,661 J/K

By holding the current while i heat the powder, i will get an even stronger final B field in my magnet. I need to heat the powder to 200 C throughout but not over that, then let it cool. I can quickly heat up the metal using an IR lamp
so if i run my Current for 1 minute, this will equate to a 13.8 K rise in my coils.

I will use jumper cables to connect the coil to the batteries, and heat the powder using a magnetron. I will need to perform tests to determine heating rates of the magnetron, what would be the best way to measure the core temperature of my powder? i was thinking a digital thermometer with a probe may work, but i need something that will give me fast readouts, as i can't let the powder get even 1 degree above 200( i may just heat to about 190 to be safe, i only need to get it to 175, to work, but 200 is optimum)

Car batteries are over kill most batteries are rated 600 to 1000 cold cranking amps. I just bought a 750 amp battery 2 weeks ago $78.00 plus tax. So your going to have $1201.00 worth of batteries with 14 in series. A D size flashlight battery produced 8 amps. Many of the new D size rechargable batteries will produce 30+ amps. I know of some D batteries that produce 50 amps. The battery on my RC model airplane produces about 40 amps.

 

[vk6kro]

You can get non-contact thermometers that would give a reading in about a second.

But I'd agree that 14 car batteries are not a good idea.
All for one great splat? You could buy a DC welder for that and have something useful afterwards.

Why not redesign it so you use only one car battery? You probably have easy access to one car battery and a set of jumper leads.

This thread has been going for nearly a month. Why not just do something?

 

[gary350]

Why not design it to use capacitors. You can get more amps out of a cap bank than batteries.

 

[trini]

thanks for the continued replies peeps, correct me if I'm wrong, but the reason i thought i'd need 14 batteries is because i need a minimum of 16.57 A in a 9.69 ohm resistance = 160 V minimum.

also i have to wait for some orders to arrive before i can proceed anyway (mould, mould release agent) and i still have to go get my magnetrons, so I'm using this time to work out all the kinks possible.

I am not an electrical engineer so this is very unchartered territory for me. also, by following the thread you see why it is a good thing i waited, i went from needing a $1500 capacitor and charger setup, which could have potentially killed me, to needing a car battery or perhaps even just a few D batteries.

while it is true that i can get more amps out of a cap bank, my material achieves 95% saturation above 1600000 Am. anything above that will not make my residual B field material much stronger anyway.

let me include a bit of permanent magnet theory for those of you who are unfamiliar:

when making a compression molded magnet, one first compresses the powder(epoxy coated) to a certain density (in my case 7.6 g/cm^3), then saturates the magnet powder in the desired direction. The powder is then heated so as to activate the epoxy which then spreads and holds the powder particles in place.

when i use AC, after i pulse, the magnet's internal b field will balance itself out and end up at its Br value of about 0.872 T, so no matter how many amps i pump into it, once that current stops it's still going to end up at around 0.872 T(this also explains why your magnet will only be 'supercharged' for a while before going back to it's original Br value), so when i begin heating, I'm 'gluing' this 0.872 T into place.

if i use a dc current, i can hold the B field above 0.872 T(what this means is that the arrangement of magnet particles are aligned at this point such as to achieve a field over 0.872. remember, even when you 'supercharge' your magnet, all you're doing is temporarily realigning the particles inside each domain), and 'glue' it into place while it is in this 'supercharged' state. so DC clearly has it's advantages.

 

[gary350]

You get a continious stream of amps out of a battery, you don't need that to charge a magnet. A pulse will do the job. It is like compairing a fire hose to a shotgun blast. You can charge a lot of amps into a capacitor bank over several seconds, when it discharges all those amps come out in a micro second like a shotgun blast. The huge pulse of amps produces a giant magnet pulse in the coil. 1 pulse is all it takes to charge a magnet. It is not much different than a flash on a camera the flash goes off in a micro second and produces plenty of light to take a picture. If your building your own magnet you can pulse the cap bank over and over if you want. If you dump a bunch of iron filings on a sheet of cardboard then pulse the coil with the cap bank all the iron filings instantly line up. It takes only 1 pulse to line them up. I'm not sure your idea of continious magnet field will do a better job or not. If the magnetic particles are mixed in with thick glue then the glue will slow down their movement they will not line up very quick.

 

[trini]

the glue is dry until i begin heating, so it won't affect alignment. what you're saying is true with regards to a pulse doing the same job, however there is a limit to the B field of a magnet. This limit is reached when all the particles in all the domains line up in the same direction. According to the information i have been provided with, a field of 16,000 Am will line up 95% of particles in my magnet. even if i pulsed 1000 A through my coil continuously, my final B field would not be very much stronger than if i used 17 A.

What i will concede however is that a pulsed current may 'shake' any particles which don't align themselves initially into place, but i can achieve the same effect by flicking the power on and off a few times before holding it.

EDIT:
quick question, if i have a power supply which delivers 10kV with an internal resistance of 4Mohm, would i still get 10000/9.69 = 1000 A when i connect it to a 9.69 ohm resistor, or does the internal resistance of the supply somehow limit this?

 

[vk6kro]

quick question, if i have a power supply which delivers 10kV with an internal resistance of 4Mohm, would i still get 10000/9.69 = 1000 A when i connect it to a 9.69 ohm resistor, or does the internal resistance of the supply somehow limit this?

The internal resistance of a power supply appears in series with the voltage of the supply.

So, it appears in series with the load as well.

Current = 10000 volts / 4000009.69 = 2.5 mA

 

[trini]

Ok so due to school starting back recently and a huge delay in getting everything ready, i had to put this off for a while. Now that i have some better wire to use, i have new requirements to work with.

I got AWG#34 wire, which did wonders in reducing my required current to just 2.4 A, however the resistance of this wire is 235 ohm. Now this implies(since I'm using DC) that i need a voltage or 564 V.

Any ideas on a DC power supply to drive this thing?

564V 2.4 A(minimum)

 

[mheslep]

... Now this implies(since I'm using DC) that i need a voltage or 564 V.

Any ideas on a DC power supply to drive this thing?

564V 2.4 A(minimum)

This is simply to charge the capacitors right? Any kind of bench ~2kw high voltage supply is going to cost you probably $3/Watt. However, since you SHOULD be using an isolation transformer anyway, better yet just get a variac (1.5kw, $100 tops) and then follow it with a http://www.tpub.com/neets/book7/27c.htm" (~800V breakdown , 4A diodes, minimum) . The resulting pulse train will charge the caps nicely. You'll have to start the variac at very low voltage to not trip the breaker on the grid side of the variac, or put in some other kind of current limiter.
I don't recall seeing as part of your plan a way to safely dump the charge on these capacitors without firing the entire mechanism, say into a resistive load. Hopefully you've thought of that. Those caps should no more be left alone and idle with that kind of energy than you would leave cups of gasoline lying around. Think about a fire starting for instance, an electrical fire energized by the capacitors. Pulling the plug out of the wall or flipping the breaker won't stop that.

Enclose the entire output of the variac / diode bridge properly. If you are not absolutely sure about the risks and safety measures required for HV forget the variac/bridge and buy the certified DC supply. If that's not affordable drop this project and move on.

 

[trini]

Oh no this isn't for the capacitors, I'm using dc to magnetise the material, so you gave me my answer in your first paragraph lol, thanks!

 

[mheslep]

Oh no this isn't for the capacitors, I'm using dc to magnetise the material, so you gave me my answer in your first paragraph lol, thanks!

Please also see my edits above.

 

[Bob S]

The drawing in post #62 is good, because it minimizes unnecessary air gaps, but I do have some comments:

1) the two coils are very hard to wind, because the breaks between the two halves of the core are in the middle of the coils.
2) The breaks in the core are where the magnetic field is highest, so will have a maximal effect on the field.

The breaks should be moved 90 degrees to the center on the right and left of the sample, because that is where the magnetic flux is least. This would permit winding the coils on single pieces of the core.

I do not understand your drawing. Are you magnetizing across the diameter?

Are you planning on inserting the sample into the opening in the side of the complete assembled core including windings? It might be more efficient magnetically to place the sample in a wound coil, and inserting this assembly into the magnetic core without coils.

Bob S