Hey @rattle, I just use your numbers as set assuming that's what...

Hey @rattle, I just use your numbers as set assuming that's what they have...

The problem is with your calculation. Basically, 2,500t of 1.14% ore = 28.5t of tin. Correct, but that's 100% tin. That's a total amount of tin contained in the ore. However, they take the 2,500t of ore and process it to create 35% pre-concentrate. So, they go from 1.14% to 35%, whilst getting rid of a substantial amount of waste. The grade increases ~ 30 times (35%/1.14%=30.7). So, they go from 2,500t to about 81t (2,500/30.7). But then, they also don't have 100% recovery, but only say 75%. Therefore, they lose another 25% of the mass in the process: 81t*0.75=60.7t.
Then, they need to make a, say, 63% concentrate that they send to smelting. 60.7t * (35/63) = 33.7t. But then again, there would be some losses in this process, no? That's a genuine question. Perhaps they can turn pre-concentrate into a concentrate with minimum losses. Let's estimate 90% recovery. 33.7*0.9=30t of 63% sellable (smeltable) concentrate.

That's as for the stockpiles, 30t.

Now, tailings.

1,440t at 0.4% grade to 32% pre-concentrate makes 1,440*(0.4/32) = 18t. Again recovery only 75% makes 13.5t.
Then, 13.5t to 63% concentrate: 13.5*(32/63) = 6.8t. 90% recovery: 6.8t*0.9 = 6.1t. Round up 6t.

All up about 36 tons of a concentrate.

To put it simply, let's look at Granville stage 2.
They say they will process 40,000t of ore to make up to 550t of a concentrate. That means their output will ~1.4% of the mass processed (550/40,000=1.4%)

Considering the grades for stage 1 are equal, by processing 2,500t of ore they will get 35t of a concentrate. The recovery is lower though as they are on a learning curve. So, around 30t is quite reasonable, I think. Then plus the tailings which isn't a big deal.

Here's my brain process. Happy to be corrected or have the figures adjusted.

MiB