So - picking up were I left off Tuesday ------------
I have two furnaces - one that takes a #4 crucible which is used for melting pure metal &/or small assay smelts
The other one will take a #40 crucible & is used for actual larger smelting loads - the smallest crucible I have for this furnace is a #20 crucible which is what I used to determine how much - ore - flux - & collector metal I could put in that crucible - as if I was going to do the smelt on his ore
His furnace will only take a #10 crucible so his smelts will only be half of what my #20 crucible will do
I used sand from a bag of "play sand" I have --- play sand is sand that has the "dirt" washed out of it - so it is clean (fine) rock & therefore a very good representation of crushed ore
I sifted the sand through my 80 mesh screen - this is the largest mesh size you want for smelting other wise it will take a VERY extend smelt time for the flux to dissolve the gangue in the ore & slag the gangue off --- finer (100 - 150 mesh) is better but 80 mesh is just fine proved there is not a lot of it that big
The sand I have is mostly finer then 80 mesh - like only 15 -20% 80 mesh with the other 85 - 80% being 100 mesh & smaller so it will smelt just fine
So - as I posted the other day --------
so in order to figure out how much ore you need to recover 1 gram of gold - you need to divide the grams in a ton by the grams of gold in a ton --- so ----------
907,185 (grams in a ton) divided by 871 grams gold in a ton = 1,041 grams of ore needed to recover 1 gram gold
Per the bold print - the question becomes - how much volume (how many cups) of ore does it take to get that 1,041 grams of ore - AND - can you put that much - or more - or less - that much ore in the crucible for the smelt --- keeping in mind that besides the ore you also need flux & collector metal in the crucible for the smelt
The ore + the flux + the collector metal = your "smelt load" --- that is the max combination of ore/flux/collector you can put in the crucible for the smelt --- keeping in mind that you need to leave head room in the crucible so that you don't have a "boil over" (the "foaming up that happens) when the smelt comes up to smelt temp
Due to the chemical reactions that take place in the smelting process there is
A LOT of foaming for a period of time until those chemical reactions start to die down --- therefore you can only fill the crucible about half full - to
two thirds full "at most" --- other wise you will most certainly lose part of you smelt to the boil over
Making the "smelt load" --- for every 1 part of ore you need
"at least" 2 parts of flux - & when trying to collect 1 gram of gold from a kilo of ore the smelt load needs to have
"at least" 30% of the load to be collector metal
If you do not have enough collector metal - when the smelt gets molten - the metals will not collect together properly & settle as a molten pool of metal in the bottom of the crucible --- instead (without enough collector metal) little beads of metal will form in the flux/slag - get hung up there - thereby loosing metal to the slag instead of pooling to the bottom of the crucible
So - back to the math - how much
volume is the 1,041 grams ore (to recover 1 gram of PMs) & how much of a load can you put in a crucible
I started with filling a 1 cup measuring cup with my sifted sand & weighed it --- it weighed 352 gram
So 3 cups of ore = 1,059 grams (so very close to the 1,041 grams)
Next we need to know how much of an actual "smelt load" will fit in a crucible
So next I filled a coffee can with sand - one cup at a time - it took 9 cups to fill the coffee can
I then poured the sand into my #20 crucible to see if that was enough for a load - not enough for a load - or more then a load --- the result was that it actually filled the crucible to it's max load level --- there is no way I would put more then a coffee can of ore/flux/collector metal in a #20 crucible & in fact at that much of a load you are at the high point of risking a boil over --- so max 9 cups of ore/flux/collect all combined for the smelt load
Let's now do the math to make that load
The load needs to be "at least" 30% collector metal - lets call it 33% (or one third) to be on the safe side of good collecting
So - 1/3 of 9 cups is 3 cups
fine chopped copper needed for the load --- that leaves 6 cups for the ore & flux
Side note here; - there is a reason why I put fine chopped copper in bold print - for proper collection you can not use large pieces of copper &/or wadded up copper wire --- if you need me to explain why I will be more then glad to --- fine chopped - shavings - or powder (copper) insures best collection
so - back to the 6 cups remaining for the ore/flux part of the of the load
You need at least 2 parts flux for every 1 part ore - so to make the ore/flux part of the load that comes to 2 cups ore + 4 cups flux = 6 cups ore/flux + 3 cups fine chopped copper = the 9 cups max smelt load that will fit in a #20 crucible
Your furnace only takes a #10 crucible so cut that in half which means you can only smelt 1 cup (350 grams) of ore at a time
As we have already determined - we need 1,041 grams to recover 1 gram of precious metals from your ore which means at best with your furnace you can recover only about (plus/minus) 1/3 gram of PMs
per smelt load
Side note again; - in the event you have questions like --- ya but Kurt you are only talking about gold - what about PGMs &/or silver - simply ask & will be more them glad to better explain
Anyway - now that we have figured out how much ore you can process "per smelt load" with your 10K furnace we next need to figure how many smelt loads we can do in a day
That will be my next post - which I will
try to do tomorrow
Kurt