Question

A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and 90 wt% hexane. This slurry is contacted with 100 kg of pure hexane in a single stage so that the value of N for the outlet and underflow is 1.5 kg insoluble solid / kg solution retained. Calculate the amounts and compositions of the overflow V1 and underflow L1 leaving the stage

Answer 1

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

$F_{solution}$ = mass of solvent + mass of oil (i.e A+C)

Feed Phase:

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

 = 25kg

Mass ratio of oil to solution $Y_{F}$ =$\frac{Mass C}{Mass (A+C)}$

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane  in feed = 25 ×  0.9 =22.5kg + 2.5 =25kg

therefore  $Y_{F}$ = $\frac{2.5}{25}$

= 0.1

mass ratio of solid to solution $Y_{A}$  =  $\frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}$

=3

Solvent Phase:

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

Underflow:

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = $\frac{mass B}{mass(A+C)}$

solution in underflow E₁ = Mass (A+C)

Overflow:

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75=  E₁ × N₁

75 =  E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg