Thin steel pot for smelting 500lbs at once?

[birddog]

I have converted a 6gal paint pot to my smelting pot, but also cut it down to hold about 3.5gal. That is all I would care to work with at one time
Charlie

[Oreo]

Running the calculations right now. Looks like a 16" diameter pot, 8" depth of molten lead is not quite 660lbs. At 8" depth lead exerts something like 3psi, if I've calculated correctly. Still working on it...

[jsizemore]

Has anyone tried to melt this much? I have a very large (700lb by my figuring) pot and a 250k burner. A lid is obvious for heat retention, but it has also been suggested to firebrick around the pot to insulate. Anyone with ant experience?

You can use sheetmetal too. I use the sheetmetal jacket from an old hot water heater cuz it was free. I use the bottom or top for a lid. The holes in the lid let me stick my thermometer stem into the melt so I can monitor temp.

[40-82 hiker]

Do smaller batches!!!!! I would never attempt over 100# at a time. No matter what the pot is made of.

I never mix huge batches of alloy because someday I will probably want to change it and I am stuck with it. I mix my alloys in my casting pot from the basic materials to get me to the mix I want using the alloy calc spreadsheet. Works ever time for me!

banger-j

Bangerjim has a lot of wisdom here, along with others.

I can't add much, but am responding to say that you must ask yourself why you need to take such a risk for no added benefit. I too smelt to clean the lead and make ingots, and then make my alloy in my casting pot, which can be done with great consistency (never had any problems with it). A sturdy $50 dollar burner stand with a HP burner, and a 20 pound propane tank properly modified is really all that is needed at largest, IMHO. It's too simple, really. I mean this with great respect, but it appears your making this unsafe by making it too complicated, unnecessarily so. I think this site has hundreds of members and guests who safely smelt lead in the 100 to 200 pound range safely all the time. Think you're out there on a limb with 500 pounds...

[lightman]

I can do about 400# with my set-up. The steel pot and the stand are plenty strong enough for that weight. I like the consistency that I get with the larger volume. I've never felt in danger using my rig. Blending batches smaller than this is plenty good for what we are doing, so mine is probably overkill. Its just what I had at the time.

Oreo, if you were close to me I would go partners with you on this project. I probably either have what we would need or can get it. Good Luck with this project.

[dikman]

I have a 1/2 propane tank that holds about 150 lbs crammed to the gills. I wouldn't consider any more than that, myself, in the interests of safety. If you want to go heavier I would use a pot made of thicker metal.
Don't forget that you're actually heating the bottom of the tank itself too. While it's all very well to say that it's only at the temp. of molten lead, during the heating process you're pouring a lot of heat onto the bottom of the tank and while you won't get it hot enough to melt if you put enough heat on it then it could very well soften a bit. This could apply in particular to a pot with a thin base. Not a good idea if you've got a couple of hundred lbs. of molten lead sitting on top.

As has been said, if in any doubt don't try it.

[theperfessor]

I made up a spreadsheet using several simplifying assumptions that calculates the tensile stress in the walls of a thin wall (t = .032") pot. The equations used are as follows:

Density of lead, constant, 0.4097 (lb/in^3)
Weight of lead, constant, 500 (lbs)
Wall thickness, constant, assumed to be .032 (in)
Volume of lead, calculated, = Weight/Density (in^3)
Diameter of pot, incremented constant, 1 to 16 in .5 increments (in)
Height of lead, calculated, = (Volume x 4)/(pi x Diameter^2) (in)
Pressure at base of pot wall, calculated, = Density x Height (lb/in^2)
Hoop (tensile) stress at base of pot wall, calculated, = Diameter x Pressure/wall thickness (lb/in^2)

What you make of this information is up to you.

[40-82 hiker]

Perfessor,

Thanks for the chart, but how would we calculate the failure point of mild steel at a given temp. based on the hoop stress of your chart? Obviously the strength decreases with temp., but how is that decrease calculated? Also, steel does not transmit heat as efficiently as aluminum (am not advocating the use of Al for smelting - just example), so is it possible to say that the burner side of the steel vessel (using a HP propane regulator) is hotter than the 675 degree inside surface regulated by the lead absorbing the heat? This would weaken the vessel even further than appearance due to lead temp., would it not? SS is an even worse conductor of heat, so the difference there would be even greater.

Thanks.

[badbob454]

my worry is a thin metal pot will get cherry or white hot where the lead is not touching it , over a very hot burner ,
go with a good heavy pot, and support your weight on the burner

[theperfessor]

I don't know of any fundamental way to calculate the strength of a given material at elevated temperatures. If there is an equation to do this it would be a derived equation, i.e. a best-fit curve that would use numerous data points taken from tensile tests across a range of temperatures. There is data for the tensile strength of various materials at elevated temperatures in some engineering handbooks.

The typical design process is to calculate the stress in a certain spot and then multiply that by a safety factor to determine the minimum material strength needed. Most safety factors range from 2 to 10 or more. A safety factor of 2 might be used when sizing a component that wouldn't hurt anyone if it failed, where a safety factor of 10 might be used when there might be a loss of human life involved. If I recall hydraulic hoses use a safety factor of 6 or 8, not sure what sf is applied to elevator cables but I think you get the idea.

Actual stress x safety factor =< material yield strength

If I were doing this I would look up the tensile strength of my material at some point above my melt temp just to be safe. Let's say you find out mild steel (or stainless or whatever) has a tensile strength of 15,000 psi at 900F. Using a safety factor of 10 (melted lead scares me) I would have to be sure the actual stress doesn't exceed 1/10th of 15,000 psi, in other words I would not stress the material above 1,500 psi.

I could change the wall thickness, melt less than 500 lbs, or use a larger diameter pot.

Won't tell anybody else what to do, but I have a propane powered rig that melts 80 lbs or so at a time and once it is fired up and cooking I stay as busy as I ever want to be keeping up with the loading, casting, ingot handling, etc. I personally would probably find 500 lbs at once a little overwhelming.

[40-82 hiker]

I don't know of any fundamental way to calculate the strength of a given material at elevated temperatures. If there is an equation to do this it would be a derived equation, i.e. a best-fit curve that would use numerous data points taken from tensile tests across a range of temperatures. There is data for the tensile strength of various materials at elevated temperatures in some engineering handbooks.

The typical design process is to calculate the stress in a certain spot and then multiply that by a safety factor to determine the minimum material strength needed. Most safety factors range from 2 to 10 or more. A safety factor of 2 might be used when sizing a component that wouldn't hurt anyone if it failed, where a safety factor of 10 might be used when there might be a loss of human life involved. If I recall hydraulic hoses use a safety factor of 6 or 8, not sure what sf is applied to elevator cables but I think you get the idea.

Actual stress x safety factor =< material yield strength

If I were doing this I would look up the tensile strength of my material at some point above my melt temp just to be safe. Let's say you find out mild steel (or stainless or whatever) has a tensile strength of 15,000 psi at 900F. Using a safety factor of 10 (melted lead scares me) I would have to be sure the actual stress doesn't exceed 1/10th of 15,000 psi, in other words I would not stress the material above 1,500 psi.

I could change the wall thickness, melt less than 500 lbs, or use a larger diameter pot.

Won't tell anybody else what to do, but I have a propane powered rig that melts 80 lbs or so at a time and once it is fired up and cooking I stay as busy as I ever want to be keeping up with the loading, casting, ingot handling, etc. I personally would probably find 500 lbs at once a little overwhelming.

Thanks for the post. I was just wondering if there was any way to work this backwards when you posted your chart. I still think it best to use what works safely for so many people, but was hoping there was a quantitative way to determine the stress capable with a given vessel.