We wish to distill 350 kgmol/hr of a saturated liquid feed. The fee is 25 mol% P

Question

We wish to distill 350 kgmol/hr of a saturated liquid feed. The fee is 25 mol% P, 45 mol% Q, and 30 mol% R. The column operates at 1 atm. The column has a total condenser and a partial reboiler. The reflux is a saturated liquid. The optimum feed stage is to be used. A fractional recovery of 95 % P is desired in the distillate and 92% of Q in the bottoms. Determine D. State any assumptions used. D = 95.7 kmol/hr if no distributing Determine the minimum number of stages for the column using the Fenske equation. Given: alpha pQ = 1.5, alpha QR = 2.7, alpha PR = 5.9 N_min ~ 13 Determine the minimum reflux ratio using the Underwood equation using a value of phi = 0.8. Given: alpha p = 1, alpha Q = 0.75, alpha R = 0.25 (L/D)_min ~ 1.4

Explanation / Answer

(a) solve the material balance

Assume no component R and 95% P desired inthe distillate. Take 350kgmol/h of feed as the basis

Overall M.B. F = D + B = 350 Kmol/Hr

Total Feed = 350 KgMol/Hr contains 25 mol% P, 45 mol% and 30 mol% R

so

Total                    P (kmol)                        Q (kmol)                                  R (Kmol)

Feed                  350*0.25=87.5              350*0.45=157.5                          350*0.30=105

Distillate            87.5*0.95=83.123          157*0.08=12.6                                      0

Bottom              87.5*0.05 =4.375              157.5*0.92=144.9                                 105

So here Total Distillate (D) is = 83.125 + 12.6 = 95.725 Kgmol/hr

                              

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