This problem looks at the trade-offs between purity measured by \(\mathrm{y}_{\mathrm{I}, \text { out }} / \mathrm{y}_{\text {sugar, out }}\) and sugar recovery in the liquid, \(\left(\mathrm{y}_{\text {sugar,out }} \mathrm{F}_{L, \text { out }}\right) /\left(\mathrm{x}_{\text {sugar,in }} \mathrm{F}_{S, \text { in }}\right)\). A continuous countercurrent leaching system is used to recover sugar from sugar cane with water as the solvent. The effective equilibrium constant \(\mathrm{y}_{\text {sugar }} / \mathrm{x}_{\text {sugar }}=\mathrm{m}_{\mathrm{E}, \text { sugar }}=1.18\) where \(\mathrm{y}\) and \(\mathrm{x}\) are the weight fractions of sugar in liquid and solid, respectively. Feed flow rate is \(\mathrm{F}_{\mathrm{S}, \mathrm{in}}=100 \mathrm{~kg} / \mathrm{h}\) of sugar cane with a sugar concentration of \(\mathrm{x}_{\text {sugar,in }}=5.0 \mathrm{wt} \%\). At the same time, we wish to retain as much of an impurity on the sugar cane as possible. For the impurity, the effective equilibrium constant \(\mathrm{y}_{\mathrm{I}} / \mathrm{x}_{\mathrm{I}}=\mathrm{m}_{E, \mathrm{I}}=0.70\) where \(\mathrm{y}_{\mathrm{I}}\) and \(\mathrm{x}_{\mathrm{I}}\) are the impurity weight fractions. The initial impurity concentration on the sugar cane is \(\mathrm{x}_{\mathrm{I}, \text { in }}=0.4 \mathrm{wt} \%\). Ignore the effect of the impurity on flow rates. Use the average solids flow rate and the average liquid flow rate for the ratio \(\mathrm{F}_{\mathrm{S}} / \mathrm{F}_{\mathrm{L}}\) in the Kremser equation, but use the actual values of \(\mathrm{F}_{S}\) and \(\mathrm{F}_{\mathrm{L}}\) (ignoring impurity) to calculate recovery and weight fractions. Find \(\mathrm{F}_{\mathrm{L}, \text { avg }}, \mathrm{F}_{\mathrm{S}, \text { avg }}\), sugar recovery in liquid, \(\mathrm{y}_{\text {sugar,out }}, \mathrm{y}_{\mathrm{I}, \text { out }}, \mathrm{x}_{\mathrm{I}, \text { out }}\), and \(\mathrm{y}_{\mathrm{I}, \text { out }} / \mathrm{y}_{\text {sugar,out }}\) for the following problems:

a. \(\mathrm{y}_{\mathrm{I}, \text { out }} / \mathrm{y}_{\text {sugar,out }} \leq 0.050, \mathrm{~N}=9\), recovery sugar as high as possible

b. \(\mathrm{y}_{\mathrm{I}, \text { out }} / \mathrm{y}_{\text {sugar,out }} \leq 0.050, \mathrm{~N}=6\), recovery sugar as high as possible

c. \(\mathrm{N}=9\), Recovery \(=95.0 \%\)

d. \(\mathrm{N}=6\), Recovery \(=95.0 \%\)

e. For \(\mathrm{N}=9\), why is the sugar weight fraction higher at \(74.6 \%\) recovery than at \(95 \%\) recovery?