sorry, it's been a while. Mitch asked for our equation, so here's our thought process on how we got to it from 30% yield and 70% recovered SM
iteration 1 yield: 1 x 0.3
iteration 2 yield: (1 x 0.7) x 0.3
iteration 3 yield: ((1 x 0.7)0.7) x 0.3
iteration 4 yield: (((1 x 0.7)0.7)0.7) x 0.3
...
thus, total yield for all iterations:
total yield = 0.30[ (1) + (1 x 0.7) + (1 x 0.7
2) + (1 x 0.7
3) ]
total yield = 0.30[ (1 x 0.7
0) + (1 x 0.7
1) + (1 x 0.7
2) + (1 x 0.7
3) ]
total yield = 0.30(0.7
0 + 0.7
1 + 0.7
2 + 0.7
3)
or
Σ
x=0→n y(1-y)
x (where y is the yield for each iteration and n is the number of iterations)
or more generally for any recycling experiment where the mass balance is not 100%
Σ
x=0→n y(z)
x (where y is the yield for each iteration and z is the percent recovered starting material and n is the number of iterations)
There might be a way to convert the summation into a function, but I don't remember my calculus that well.