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~====~----~======================~--~B~Y===============~07 L A distillation column with a total condenser is separating acetone from ethanol A distillate concentration of xD = 090 mole fraction acetone is desired Since CMO is valid LtV =constant If LtV is equal to 08 find the composition of the liquid leavshying the fifth stage below the total condenser

b A distillation column separating acetone and ethanol has a partial reboiler that acts as an equilibrium contact If the bottoms composition is x =013 mole fractionB acetone and the boilup ratio YIB =10 fmd the vapor composition leaving the secshyond stage above the partial reboiler

Co The distillation column in parts a and b is separating acetone from ethanol and has xD =09 xB = 013 LtV = 08 and YIB = 10 If the feed composition is z = 03 (all concentrations are mole fraction of more volatile component) fmd the optimum feed plate location total number of stages and required q value of the feed Equishylibrium data for acetone and ethanol at 1 atm (Perry et al 1963 p 13-4) are

XA 10 15 20 25 30 35 40 50 60 70 80 90 YA 262 348 417 478 524 566 605 674 739 802 865 929

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4D6 Convert from mole fraction to weight fraction Basis 100 kgmoles feed

46 kg 5 kgmoles Ethanol x = 230 kg

kg mole

18 kg 1710 kg95 kgmoles water x == 1

kg mole Tota 1940

EtOH wt frac = 23011940 = 01186

H-h q= F Use Figure 2-4 Approximate feed stage at 97degC H-h

H=550

= 550 - 300 == 0546 q 550-92

hp = - 300

Slope =l=-120 q-l

h=92

== 550-160 =08515Note Using 5 wt gives hF = 160 q 550-92

Slope l == -574 Which are wrong q-l

L4D7 a Plot top op line slope = - =8 x = Y == xD == 9 Step off stages as shown on

V Figure

b Plot bottom op line slope == L == 1+ V= 2 x = Y= xB 013 Step off stagesV fB

(reboiler is an equil stage) Find Y2 = 0515 c Total stages = 8 + reboiler

Optimum feed plate = 7 or 8 from top Plot feed line Goes through x = Y = z = 3 and intersection of two operating lines

slope == - 2 =l gives q = 0692 4 q-l

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4D8 The equilibrium data is plotted and shown in the figure From the Solution to 4D7c

q=0692 and q(q-l)= 94

a total reflux Need 5213 stages (from large graph) 59 from small diagram shown

b (LV) = 9-462 0660 (see figure) mm 9-236

(LID) = (LV)min =1941 mill 1 (LV)min

c In 4D7 LID = LV = =4 act 1 - LV 2

LDacl = (Multiplier) x (LID )mlll Multiplier 411941 = 206

d Operating lines are same as in Problem 4D7 Start at bottom of column Reboiler is

an equilibrium contact Then use EMV AB AC =075 (illustrated for the first real

stage) Stage 1 is the optimum feed stage 11 real stages plus a partial reboiler are

sufficient

=

4D6 Convert from mole fraction to weight fraction Basis 100 kgmoles feed

46 kg 5 kgmoles Ethanol x = 230 kg

kg mole

18 kg 1710 kg95 kgmoles water x == 1

kg mole Tota 1940

EtOH wt frac = 23011940 = 01186

H-h q= F Use Figure 2-4 Approximate feed stage at 97degC H-h

H=550

= 550 - 300 == 0546 q 550-92

hp = - 300

Slope =l=-120 q-l

h=92

== 550-160 =08515Note Using 5 wt gives hF = 160 q 550-92

Slope l == -574 Which are wrong q-l

L4D7 a Plot top op line slope = - =8 x = Y == xD == 9 Step off stages as shown on

V Figure

b Plot bottom op line slope == L == 1+ V= 2 x = Y= xB 013 Step off stagesV fB

(reboiler is an equil stage) Find Y2 = 0515 c Total stages = 8 + reboiler

Optimum feed plate = 7 or 8 from top Plot feed line Goes through x = Y = z = 3 and intersection of two operating lines

slope == - 2 =l gives q = 0692 4 q-l

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4D8 The equilibrium data is plotted and shown in the figure From the Solution to 4D7c

q=0692 and q(q-l)= 94

a total reflux Need 5213 stages (from large graph) 59 from small diagram shown

b (LV) = 9-462 0660 (see figure) mm 9-236

(LID) = (LV)min =1941 mill 1 (LV)min

c In 4D7 LID = LV = =4 act 1 - LV 2

LDacl = (Multiplier) x (LID )mlll Multiplier 411941 = 206

d Operating lines are same as in Problem 4D7 Start at bottom of column Reboiler is

an equilibrium contact Then use EMV AB AC =075 (illustrated for the first real

stage) Stage 1 is the optimum feed stage 11 real stages plus a partial reboiler are

sufficient

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4D8 The equilibrium data is plotted and shown in the figure From the Solution to 4D7c

q=0692 and q(q-l)= 94

a total reflux Need 5213 stages (from large graph) 59 from small diagram shown

b (LV) = 9-462 0660 (see figure) mm 9-236

(LID) = (LV)min =1941 mill 1 (LV)min

c In 4D7 LID = LV = =4 act 1 - LV 2

LDacl = (Multiplier) x (LID )mlll Multiplier 411941 = 206

d Operating lines are same as in Problem 4D7 Start at bottom of column Reboiler is

an equilibrium contact Then use EMV AB AC =075 (illustrated for the first real

stage) Stage 1 is the optimum feed stage 11 real stages plus a partial reboiler are

sufficient

=