Q4) A distillation column receives a feed that is 40 mole % n-pentane and 60 mol
ID: 702451 • Letter: Q
Question
Q4) A distillation column receives a feed that is 40 mole % n-pentane and 60 mole % n-hexane. Feed flow rate is 2,500 Ibmol/hr and feed temperature is 30°C. The column is at 1 atm. A distillate that is 99.9 mole % n-pentane is desired. A total condenser is used. Reflux is a saturated liquid. The external reflux ratio is Lo/D 3. Bottoms from the partial reboiler is 99.8 mole % n-hexane. Find D, B, QR, Qc. Note: Watch your units on temperature. Data: s11,369 Btu/lbmol 13,572 Btu/lbmol, both A at boiling points. c6 CPLCS39.7 (assume constant) PL-51.7 (assume constant) CPVCS = 27.45 + 0.08 148 T-4.538 × 10-5 T24 10.1 × 10-9 T3 CPC32.850.09763 T- 5.716 x 10- T2+13.78 x 109T where T is in C and Cpy and Cp are cal/(molC) or Btu/(Ibmol F).Explanation / Answer
a)
Distillate and bottoms flow rates
Overall material balance over the entire tower gives
D + B =2,500
Material balance for n-pentane over the entire tower gives
0.97D +0.02B = (0.4)(2,500) = 1,000
? 0.97(2,500?B)+0.02B= 1,000
Solving for B and D
from the above equations we have
B = 1,500 lbmoles/hr and
D = 1000 lbmoles/hr.
(b)
Heating and cooling loads
QC=V1(H1?hD)?V1?Hevap=V1(0.97?C5+ 0.03?C6)V1=L0+D= (L0/D+ 1)D= 4D= (4)(1000) = 4000 lbmoles/hr
QC = (4,000)(0.97×11,369 + 0.03×13,572) =4.574×107Btu/hr
The temperatures of the reflux stream and the reboiler must be known to solve for the heat
load of the reboiler. Since the distillate is almost pure pentane and the bottoms product is
almost pure hexane, the boiling temperatures of pure pentane and hexane at 1 atm are used
for the temperatures of the reflux stream and the reboiler, respectively. Hence
TD?309 K and TB?342 K
(C)
Since the boiling point of a 40 mole % n-pentane is 324.79 K, the feed enters the column at30oC (or 303.15 K) is a subcooled liquid. Making an energy balance over the column, we obtain
FhF+QR=DhD+QC+BhB
Let the reference state be liquid at 30oC, then hF= 0.
hD=CpL,C5(TD?TF) = (39.7)(309?303)(1.8) = 428.8 Btu/lbmol
hB=CpL,C6(TB?TF) = (51.7)(342?303)(1.8) = 3,629.3 Btu/lbmol
QR=DhD+QC+BhB= (1,500)( 3,629.3) + 4.574×107+ (1,000)(428.8) =5.1613×107 Btu/hr
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