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Topic: Making a needle glow (1000°C) by Battery (Read 4351 times) previous topic - next topic

jazzar

#15
Jun 17, 2010, 04:05 pm Last Edit: Jun 17, 2010, 05:06 pm by jazzar Reason: 1
Ok lets put same data together:

Power supply I'd suggest 2 9V-Blocks, parrallel, each 280mAh.
Total: 0.56Ah => 5.04Wh

My capacitor has an impedance of 0.08Ohm,
capacitance of 1000uF = 0.001F
and will be charged with 9V (10Vmaximum rating)

So Ohm's law:
U = R * I suggests I can draw: 9V/0.08Ohm = 112.5A

with: C * U = Q
0.009 A*s or Coulomb

I will get a discharge with the duration of:
0.009A*s/112.5A = 0.00000072 seconds

*found an error here, now correct:

Energy of a capacitor:
E = 0.5 * U² * C
0.0405J

So in theory I just need to connect 24692 capacitors in parallel :-)
(and thats without loss in cables etc.)

Would be great if someone could look over these calculatons and tell me if I'm on the right track!

Cheers!

P.s. the batterys would have the energy of 5.04Wh * 3600 = 18144 Ws = 18144J
=> ~18 charges

CowJam

As someone who burnt a hole in a pocket using nothing but two AA cells and a set of keys, you might want to try shorting different batteries.

It's probably quite dangerous, but you could try putting a pin across the terminals of a 9v battery and seeing what happens.

jazzar

The pin will get hot... but not red glowing (note here that I didn't try this but consider unlikely due to the nature of batterys - also as described above) and then again I want to precisely control when and where the needle begins glowing, speaking of which, my pocket would definately not be on the list :-)

artjumble

Is 1000°C really a requirement, or are you really just looking for 'glowing red hot'?

If red hot is the goal, then wouldn't a thinner material be a possibility? Seems to me that a small filament should take much less power to get glowing.

Udo Klein

Depends on for how long it is required to glow. If you use an old foto flash circuit to charge a sufficiently big capacitor you can even evaporate it.

I did something like this. 400uF at 300V will store 24J. This is enough to evaporate some part of your needle.

The interesting question though is: what exactly is your usecase?

Udo
Check out my experiments http://blog.blinkenlight.net

weirdo557

why do you need the needle anyway? perhaps there is a better way of doing whatever you hope to accomplish

doublet

Another cool effect: charge a big cap, then short-circuit it by putting some steel wool between the two legs... Awesome AND reuseable!
Sorry God members, I'm an atheist.

jazzar

#22
Jun 17, 2010, 09:28 pm Last Edit: Jun 17, 2010, 09:33 pm by jazzar Reason: 1
Hm I'm basically trying to expand my horizon :-)
finding out what is possible... getting to know the limits of different applications, at the moment I'm playing around with different extremes.

In  the end I might build a mini-exhibit featuring a miniature jacobs ladder (high V low A) glowing needle (high A low V) a laser... stuff..

So yes it has to be a needle , if you want people to be interested in your project, using recognisable elements is important.

Can anyone enlighten me on how to convert Joules to glowing needle degrees? :-)

And another question: From several high current appliations I've seen people always use some kind of charging resistor. To prevent overheating I guesss, but is there any formula I can calculate such a thing with?

P.s. 1000°C isn't necessary but a nice red-orange glow would do. Also this is not inteded for continual use just one "flare" (I will call it: JANIO: Jazzars Awesome Needle Instant Oxidizer^^)

AWOL

Quote
Can anyone enlighten me on how to convert Joules to glowing needle degrees?


I think you need to know the specific heat capacity of the substance you're trying to heat.
"Pete, it's a fool looks for logic in the chambers of the human heart." Ulysses Everett McGill.
Do not send technical questions via personal messaging - they will be ignored.

P_Wood

Quote
Can anyone enlighten me on how to convert Joules to glowing needle degrees?


Here is the equation you need:
Q = m*C*(T2-T1)
note: (T2-T1) is generally written as deltaT
Q = Heat energy(Joules)
m = mass(grams or Kg depending on which val used for 'C')
C = specific heat capacity(Joules/(gram*degreeCelsius) or Kg...

So you want to find 'Q'. You need to know the 'C' of your needle. I would just look for specific heat of steel. And you need to know its mass. As for the change in temp... I don't know what temperature makes things glow so you will have to guess. I think its completely possible, but figuring out how to discharge the capacitor(s) in an effective way might be challenging. Be sure to post some pictures of this when all said and done.

halley

Well, I think it's possible to get a filament to glow given a small single-cell source.  I've seen these:



And apparently this isn't a joke:


jazzar

#26
Jun 17, 2010, 10:11 pm Last Edit: Jun 17, 2010, 10:14 pm by jazzar Reason: 1
Hmm specific heat capacity of steel is 0.46 J/g*K that means I need 0.46 Joules of energy to heat one gram steel by one Kalvin

So from 25°C to 1000° that makes (average needle about 7.5g):
(0.46 J/g*K) * (7.5g) * (975K) = 3363.75J

seems a bit high...

jazzar

@ halley ...wow just.. wow...


@ P_Wood Thanks :-)

AWOL

Quote
(average needle about 7.5g)


What sort of needle is that heavy? A sailmaker's?
"Pete, it's a fool looks for logic in the chambers of the human heart." Ulysses Everett McGill.
Do not send technical questions via personal messaging - they will be ignored.

P_Wood

Quote
seems a bit high...

Seems about right. Though, depending on the amount of time it takes, heat may radiate from the needle into the air before it ever gets to glow.

Question: What is the resistance of this needle? :P

Quote
What sort of needle is that heavy? A sailmaker's?

lol ;D

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