Melting gold using resistive heating.

[Amol Gupta]

I agree. The main issue with doing it this way is that, assuming the crucible is made of pure graphite, there's no way it could possibly produce enough resistance. Graphite is a pretty good conductor.
I have no idea how the guy in the video managed to melt aluminium, perhaps he faked it. I did some rough calculations: based on a crucible with a 2cmx2cmx2cm cube-shaped interior giving a volume of 8cm^3 (which would fit 100g of gold with about 7mm height left at the top) the resistance from one side to the other, or from top to bottom, would be in the order of less than a milliohm, when empty. To achieve even 0.5ohm you'd need walls thinner than paper. And it gets even worse when you fill it with gold, a highly efficient conductor of current!

Perhaps if it were some kind of highly resistive graphite clay...
Or, as you say, simply use a long thin heating coil wrapped many times around a non-conductive crucible, with lots of insulation. Trying to reinvent the wheel is fun but alas there's always a reason why it hasn't been done before.

@Amol Gupta :
Here are some of the calculations I started on but I abandoned them when I started calculating the resistance of the crucible and realized it wouldn't work. Maybe you will find something useful here for your calculations since I spent time on it:

Using the UK supply for example purposes- we have 240V supply and for the sake of argument we can assume it's fused at 13A as per normal (it's possible to get up to 16A if you wire in directly but that's illegal unless you're an electrician).

Example:

Supply is 240V and can give max 13A

Assume a starting temperature of 20degrees Celcius
Gold melts at 1064 degrees Celcius
So our gold will need to be heated by 1044 degrees to reach melting point ("deltaT")

Gold has a specific heat capacity of 129 J/kg ("c")
Gold has a latent heat of fusion of 64500J/mol ("L")

We want to heat and melt 100g
so the mass of our gold =0.1kg ("m")

Gold has an atomic mass of 197g/mol
so the molar mass of our gold is 100/197 = 0.5076mol ("Moles")

Gold has a density of 19320kg/m^3
so the volume of our 100g of gold = 0.1 / 19320
= 5.18 x 10^6 m^3
= 5.18cm3 ("Vol")

Calculating the energy requirement:

Heating the gold to melting point:

Energy required = Mass x Specific heat capacity x Change in temperature
Q(heating) = m x c x deltaT
= 0.1 x 129 x (1064 - 20)
= 13442J

Melting the gold:

Energy required = Molar mass x Latent heat of fusion
Q(melting) = Moles x L
= 0.5076 x 64500
=32740.2J

Total energy requirement = Q(heating) + Q(melting)
=46182.2J

Resistance of crucible:
Abandoned

220V @ 5 Amps is close to 1Kw.
1kwh is 3.6 * 10^6 joules.
1kw power for a single minute is 60000 J.
Im guessing I have something to work with here.

 

[haveagojoe]

Im guessing I have something to work with here.

Do the resistance calcs for the crucible and you'll see what I mean. Of course it's possible to deliver enough energy from the supply, we know that from conventional mini-furnaces. The issue is doing it by resistance in something which is not sufficiently resistive.

Graphite has a resistivity of 1 x 10^-5.

Resistance = Resistivity x (path length/cross-sectional area)

 

[Amol Gupta]

Do the resistance calcs for the crucible and you'll see what I mean. Of course it's possible to deliver enough energy from the supply, we know that from conventional mini-furnaces. The issue is doing it by resistance in something which is not sufficiently resistive.

Graphite has a resistivity of 1 x 10^-5.

Resistance = Resistivity x (path length/cross-sectional area)

Ok so Im guessing your concern is my graphite crucible will not have enough resistance to be heated resistively(resistance of the crucible being very low).

I have a voltmeter that can read resistance and a crucible I plan to use, I'm pretty sure it is some sort of graphite clay and not pure graphite.

I'll check the resistance of the crucible using the voltmeter.

 

[Yggdrasil]

Ok so Im guessing your concern is my graphite crucible will not have enough resistance to be heated resistively(resistance of the crucible being very low).

I have a voltmeter that can read resistance and a crucible I plan to use, I'm pretty sure it is some sort of graphite clay and not pure graphite.

I'll check the resistance of the crucible using the voltmeter.

You can always use it as an indirect "target" and use a tight fitting metal band or similar as a susceptor to heat it up.
Heat the Metal band/cup/cover so it heats the crucible.
I still feel an electric furnace is better at this stage.

 

[haveagojoe]

I'll check the resistance of the crucible using the voltmeter.

Yes that would be a good first step, if it's 1 ohm or more then maybe there's a chance... I still think it's a dangerous proposition though, I must say I would advise against it.

 

[Yggdrasil]

Yes that would be a good first step, if it's 1 ohm or more then maybe there's a chance... I still think it's a dangerous proposition though, I must say I would advise against it.

Not sure where you see that danger.

At below 5V you literally need metal to overcome the electric potential.
There will be a lot heat though which is the intention of all furnaces.
I'd be more skeptical on a high Voltage set up

 

[haveagojoe]

Not sure where you see that danger.

At below 5V you literally need metal to overcome the electric potential.
There will be a lot heat though which is the intention of all furnaces.
I'd be more skeptical on a high Voltage set up

It's not the potential it's the current. It won't give you a shock, it will just turn your body into a resistive heater. You might not even realize at the time but it can cook you through. Internal tissue burns can necessitate limb amputation.

 

[Amol Gupta]

It's not the potential it's the current. It won't give you a shock, it will just turn your body into a resistive heater. You might not even realize at the time but it can cook you through. Internal tissue burns can necessitate limb amputation.

I'll surely keep that in mind, running back to the calculations I did with respect to the video you mentioned.

If the numbers shown in the video are correct the resistance on the secondary side of the transformer is 0.025 ohms.

I guess we could agree on that.

 

[Yggdrasil]

It's not the potential it's the current. It won't give you a shock, it will just turn your body into a resistive heater. You might not even realize at the time but it can cook you through. Internal tissue burns can necessitate limb amputation.

Yes but it need to have sufficient voltage to overcome the resistive potential.
If you touch the poles of a 12V car battery nothing will happen unless you have some metal in the hand to let the voltage overcome the resistive potential.
And you are right, the current is the killer. 30-100mA through the hearth will stop it.
But unless something like a high voltage pulse or similar gives the current the push nothing happens.
Static electricity can have 100K to 1M V and is basically harmless due to the low amperage.
But you feel it allright.

 

[haveagojoe]

I'll surely keep that in mind, running back to the calculations I did with respect to the video you mentioned.

If the numbers shown in the video are correct the resistance on the secondary side of the transformer is 0.025 ohms.

I guess we could agree on that.

The important thing is the resistance across the crucible. A bigger crucible will have a lower resistance. A smaller, longer item like the nail he uses will have a higher resistance, hence why the nail burns up but the crucible doesn't. You can imagine the electrons a bit like water flowing through a pipe- it's easy to push water through a short, wide pipe, but if you try to push it though a very long, very thin pipe it's much more difficult. So both the length and the cross-sectional area affect the resistance, and then also the resistivity which is dictated by the composition of the conductor material. As I said above, matters are complicated by the fact it will be heating metal, in your case gold, which itself is a very good conductor, so will dramatically lower the resistance. As you can see with the aluminium, the crucible stops glowing so brightly when the metal melts, because current is then able to flow through the metal as well as the crucible. As a result it doesn't really melt very well. It struggles to deliver enough current when the metal is molten, whereas a conventional furnace would just keep on delivering heat, resulting in very nice liquid metal for pouring. And for gold, the melting temperature is much higher, and it's not just a straight line up, it's exponential.

 

[snail]

I worked in a plant that electroplated steel coils with tin. There were 20 sets of anodes each with a 4500A supply at up to 25 volts.

Busses were uninsulated and no precautions were necessary touching the anodes or busses at that voltage. Current through a resistance depends on resistance. A huge power supply does not increase the current.

 

[haveagojoe]

A huge power supply does not increase the current.

That is plain wrong. The whole point is increasing the current, while reducing the voltage. What you mean is it doesn't increase the power. The power is still there when the voltage is low and the current is high. If you want to depend on the resistance of your skin, more fool you and better hope you havent got sweaty hands or a small cut. I used to work in low voltage electrics (-48V) and have witnessed nasty burns. I am bewildered by people's lack of respect for electricity. I think I'm done with this thread.

 

[Amol Gupta]

That is plain wrong. The whole point is increasing the current, while reducing the voltage. What you mean is it doesn't increase the power. The power is still there when the voltage is low and the current is high. If you want to depend on the resistance of your skin, more fool you and better hope you havent got sweaty hands or a small cut. I used to work in low voltage electrics (-48V) and have witnessed nasty burns. I am bewildered by people's lack of respect for electricity. I think I'm done with this thread.

So here's the thing let us consider a 25V line, the resistance of human body being 1000 ohm(in wet conditions) even if a human is in series connection to the line I'm expecting a shock of 0.625watts will it harm the individual maybe a prolonged exposure will.

 

[Amol Gupta]

This is close to 40 joules of energy

 

[Yggdrasil]

So here's the thing let us consider a 25V line, the resistance of human body being 1000 ohm(in wet conditions) even if a human is in series connection to the line I'm expecting a shock of 0.625watts will it harm the individual maybe a prolonged exposure will.

Not much is needed if it goes through the hearth.
30mA is enough

 

[haveagojoe]

So here's the thing let us consider a 25V line, the resistance of human body being 1000 ohm(in wet conditions) even if a human is in series connection to the line I'm expecting a shock of 0.625watts will it harm the individual maybe a prolonged exposure will.

Impossible to say unless we know the amperage. Yes it's not a shock like with high voltage, its a burn and it can be inside your flesh rather than on the surface, it takes a few seconds or more and you might not even notice it until it starts to hurt later.

 

[Amol Gupta]

Not much is needed if it goes through the hearth.
30mA is enough

Ok that settles the debate, wearing a pair of gloves never hurts and I have no incentive to touch the wires bare hands.

 

[Amol Gupta]

Back to the important stuff
I did check my crucible using a multimeter.
I had a 1.4V battery connected in series with the crucible and used my multimeter to check the current running through the crucible which turned out to be 180 mA, leaving the resistance to be 9 ohms.

Now I'm pretty sure this is some sort of graphite clay, and this is a crucible which is used in the industry for melting gold any thoughts on the recommended voltage to be used.

110V is my best guess.

 

[Amol Gupta]

I will confirm the resistance using a higher voltage battery.

 

[Yggdrasil]

Back to the important stuff
I did check my crucible using a multimeter.
I had a 1.4V battery connected in series with the crucible and used my multimeter to check the current running through the crucible which turned out to be 180 mA, leaving the resistance to be 9 ohms.

Now I'm pretty sure this is some sort of graphite clay, and this is a crucible which is used in the industry for melting gold any thoughts on the recommended voltage to be used.

110V is my best guess.

Most Multi meters have a function to measure resistance directly.