Master & Slave casting Pot

[preachinpilot]

First off i'm no electrition. I want to build a casting pot with a master pot to melt up to 100lbs of lead at one time. Next I want to have a slave pot which is fed by the master pot. I want to controll the setup with two PID temperature controllers and I want to mount them so that the slave pot top is flush with the top of my bench. The idea is that I can pre melt lead in sufficient quantity that I do not have to wait for my lead to get upo to temperature. I am making a business of casting large bullets with a ladel (from car movers) and am fairly fast with a usual rate of 500/hr 45 cal, 260 grn bullets and can pour 850 40 cal bullets an hour. I need to put a bottom pour type valve in the master pot so as to fill the slave pot. I could use two lee pots and let one recover while using the other. I buy my lead in 63lb pigs and could save time by being able to feed 30 lbs of the pig at a time.

I am thinking of a 230v/2500w band heater element for the master pot and a 500w/230v band heater on the 5"x3" slave pot. I have two PID conrtrollers. One auburn instruments, and one from a cast boolit poster in Alaska , which is awsum. If I use a 440 band and hook it to 230 it will only produce 1/2 the rated wattage, correct?

[shadowcaster]

For your master pot here is a link to a large bottom pour pot that I designed and built. I fire it with propane, but it could be adapted to electric.

http://castboolits.gunloads.com/show...r+smelting+pot

Shad

[Catshooter]

[QUOTE=preachinpilot;If I use a 440 band and hook it to 230 it will only produce 1/2 the rated wattage, correct?[/QUOTE]

Am I correct in thinking that your question is that if you have 440 volt rated 2500 watt resistance heater and you apply 230 volts to it you'll get 1250 watts of heat? The answer is yes. And it will work fine on that voltage.

Welcome to the forum too.


Cat

[earthling121757]

I think your wattage will be aprox. 1/4 of the rating. The heater is basically a resistor. The formula for power is P=V^2/R or P=I^2*R
The square of .5 is .25. Of course going from 440 to 230 isn't exactly half, but if you measure your mains voltage you should be able to calculate the result.

[Zymurgy50]

No squared in Ohm's law.
I=V/R
I= current
V=voltage
R=resistance
If you cut the voltage in half while maintaining the same resistance the current will also be halved.

[dilly]

Okay I can't help you at all with this project, but I hope someone else does and that you will get it working and show us all how it's done when you get it there. This sounds pretty neat.

[Catshooter]

earthling,

Welcome to the site.

Sorry but that turns out to not be the case. As zymurgy points out, there's no squared in Ohms law. It's direct. Cutting the voltage cuts the current the same amount. And vice versa.

Get out your Amprobe and check it . . .


Cat

[earthling121757]

Yes, that's true about ohms law... but preachinpilot was asking about power, not current.

Power = Voltage * Current. So if you drop the voltage to half, the current also drops to half, thus the power drops to 1/4.

For example, suppose I have a resistance of 10 ohms. I apply 100 volts across that resistor. I have a current of 10 amps. P=V*I so P=100*10 or 1000 watts.

Now, I lower the voltage to 50 volts. 50 volts across 10 ohms will produce a current of 5 amps. So now I have a power of P=50*5 or 250 watts.

At half voltage, the POWER is 1/4 the value it would be at full voltage.

Ohms law does not have a squared term, but the formula for power does (when referenced to a constant resistance).

[GabbyM]

If it were me I'd copy the Magma pot used on there big casting machines.

What works good for me currently. Is to use my Magma 40 lb bottom pour and a second pot to melt ingots. Feeding the casting pot with a large soup ladle.

[kitsap]

Earthling121757, Zymurgy50,

The typical configuration for a dual voltage heating element is to have two segments with three connection points. The segments can be connected in series or parallel to accommodate high or low voltage and end up with the same power output.

DougF

[preachinpilot]

Thanks. I just know I want to be able to dip faster, lower to the table and mold and have a lot of lead hot with no down time. I usually cast for 2 hrs and get tired of having to stop. To make money you need production with the least change in temperature and the least number of missfits. With the rowell ladel im probably 95% efficient. I have looked at the Magma and so am stuck between a band heater and running 3/1000w rods in a 3/4" pot bottom.

[Frozone]

No squared in Ohm's law.
I=V/R
I= current
V=voltage
R=resistance
If you cut the voltage in half while maintaining the same resistance the current will also be halved.

Actually there IS. It's just hidden in the equation you quote. P = I²R is the correct equation.

You quote V like it is an atomic thing, but in reality V = IR.
So if you plug (IR) into the equation in place of V and simplify, you will find a Squared value. that leads to the square root of the current in the power problem

[D Crockett]

first off i'm no electrition. I want to build a casting pot with a master pot to melt up to 100lbs of lead at one time. Next i want to have a slave pot which is fed by the master pot. I want to controll the setup with two pid temperature controllers and i want to mount them so that the slave pot top is flush with the top of my bench. The idea is that i can pre melt lead in sufficient quantity that i do not have to wait for my lead to get upo to temperature. I am making a business of casting large bullets with a ladel (from car movers) and am fairly fast with a usual rate of 500/hr 45 cal, 260 grn bullets and can pour 850 40 cal bullets an hour. I need to put a bottom pour type valve in the master pot so as to fill the slave pot. I could use two lee pots and let one recover while using the other. I buy my lead in 63lb pigs and could save time by being able to feed 30 lbs of the pig at a time.

I am thinking of a 230v/2500w band heater element for the master pot and a 500w/230v band heater on the 5"x3" slave pot. I have two pid conrtrollers. One auburn instruments, and one from a cast boolit poster in alaska , which is awsum. If i use a 440 band and hook it to 230 it will only produce 1/2 the rated wattage, correct?

instead of going threw all the trouble and expense get a magma 90 lb pot and you can add 5 lb ingots to it and never run out of casting miteral if you have any questions pm me d crockett