Problem 7.2.8: Oil is extracted from meal (oil-free solid) using benzene as a solvent. The multi-stage counter-current extraction unit is employed to treat 1000 pounds per hour of meal. The feed meal contains 400 pounds of oil and 25 pounds of benzene. The wash benzene contains 10 pounds of oil dissolved in 655 pounds of benzene. The discharge solids must not contain more than 60 pounds of oil. For multistage counter-current extraction, determine (a) composition of strong solution, (b) the weight of solution leaving with the concentrated meal, (c) amount of strong solution, and (d) number of theoretical stages required.

Solution: Basis of calculation is 1000 pounds of inerts. Associated solution contains 400 pounds of oil and 25 pounds of benzene. Thus quantity of solution in the feed is 425 pounds.

(a) Discharged solids contains all of the inerts 1000 pounds, solute (oil) 60 pounds. Thus solute to inert ratio in the final underfow is 60 over 1000 or 0.06. Using equilibrium data, this corresponds to y_n of 0.118, where y_n is pounds of oil per pounds of solution in the overflow, V_n

Weight of solution in the final underflow ( R_n) is equal to 60 (pounds of oil) over 0.118 (concentration) or 508.5 pounds.

Solvent in the final underflow (R_n) is equal to 508.5 minus 60 or 448.5 pounds.

(a) Solvent in the extract is equal to 655 plus 25 minus 448.5 or 231.5 pounds.

Oil in the extract is equal to 10 plus 400 minus 60 or 350 pounds.

Mass fraction of oil in strong solution, y₁ is equal to 350 overl (350 plus 231.5) or 0.602.

Weight of solution leaving with extracted meal is 60 (oil) plus 448.5 (solvent) or 508.5 pounds per hour

(b) Amount of strong solution, V₁ is equal to 350 (oil) plus 231.5 (benzene) or 581.5 pounds per hour

Locate point R₀ such that X_I is equal to 1000 (inert) over 425 (solution) or 2.353, and X_A is equal to 400 (oil) over 425 (solution) or 0.941.

Locate point R₀ (0.941, 2.353).

Locate point V₁ (0.602, 0).

Draw a line R₀V₁ and extend to a point Δ (difference stream).

Locate point V_n+1 (0.015, 0).

Locate point R_n (0.118) on the underflow curve.

Draw a line V_n+1R_n and extend it to point Δ.

Now number of stages can be found by drawing triangles.

Draw a vertical line from V₁ to intersect the underflow curve at R₁.

Join R₁ with Δ and locate V₂.

Draw a vertical line from V₂ to intersect the underflow curve at R₂.

Join R₂ with Δ and locate V₃. Continue these steps till we reach R_n.