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Calculations CharteLYONDELL (Sheet _____ Of ______
File_____________
SUbject __C__~--f__p_r_0_k_laquo_b_1-__ Oate____________
~====~----~======================~--~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
~--~--------
V I
IRect S~tl(V ~yjH L V Xi 1- ( - ~)Y--)
(ZI ~ Lshy __~V~~~- __S(oP~ ) (l-r )V
(-L ~---~-----~---
if )x~
EC03010A (902) wwwlyondellcom
------ --- ---
r
f -0 O Cf r
2- = J-r
~LYONDELL Calculations Chart
Sheet Z- Of ___3c--_ File________
Subject _________________ Oate________
By
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JN~flJ -0 J _shye--middot-----middotmiddot- --
--lt)12 - o~s~j ~- b ~
EC03010A (902) wwwlyondelLcom
~LYONDELL Calculations Chart
Sheet ---3=-_ of_-=3=---_ Flle________
Subject________________ Oate________
By________
C-s-o$it_ -- 03 -lt
~=bK-tlt
q ( I - 1) = - J
- -- llt L- Crt I-Ilt
EC03010A (902) wwwlyondellcom
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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
=
Calculations CharteLYONDELL (Sheet _____ Of ______
File_____________
SUbject __C__~--f__p_r_0_k_laquo_b_1-__ Oate____________
~====~----~======================~--~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
~--~--------
V I
IRect S~tl(V ~yjH L V Xi 1- ( - ~)Y--)
(ZI ~ Lshy __~V~~~- __S(oP~ ) (l-r )V
(-L ~---~-----~---
if )x~
EC03010A (902) wwwlyondellcom
------ --- ---
r
f -0 O Cf r
2- = J-r
~LYONDELL Calculations Chart
Sheet Z- Of ___3c--_ File________
Subject _________________ Oate________
By
--Ar til Ir-_A A -2-L (-1 D
u - l
)ltl -0 I]
Lell io 00 5 I
-L
O~ ~ C~jJ--t shyV V l5 S(i)e~
-
6( _ ltg08 - -- shyAgt ta
FL~ TllPXS =0 G(S- C= cA) -vb) Xq 013
j -- I 1) So cJi) ~ g ~=~ V
SJ i N1e~lt~(lf of S fitl PINf c Pole iJ E- ~ Of)) ie
JN~flJ -0 J _shye--middot-----middotmiddot- --
--lt)12 - o~s~j ~- b ~
EC03010A (902) wwwlyondelLcom
~LYONDELL Calculations Chart
Sheet ---3=-_ of_-=3=---_ Flle________
Subject________________ Oate________
By________
C-s-o$it_ -- 03 -lt
~=bK-tlt
q ( I - 1) = - J
- -- llt L- Crt I-Ilt
EC03010A (902) wwwlyondellcom
~~~~~~~~~~~~~-~~~~~~~~~-~-~~~~~~~-~~~ ~
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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
bull
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-----------t
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
=
------ --- ---
r
f -0 O Cf r
2- = J-r
~LYONDELL Calculations Chart
Sheet Z- Of ___3c--_ File________
Subject _________________ Oate________
By
--Ar til Ir-_A A -2-L (-1 D
u - l
)ltl -0 I]
Lell io 00 5 I
-L
O~ ~ C~jJ--t shyV V l5 S(i)e~
-
6( _ ltg08 - -- shyAgt ta
FL~ TllPXS =0 G(S- C= cA) -vb) Xq 013
j -- I 1) So cJi) ~ g ~=~ V
SJ i N1e~lt~(lf of S fitl PINf c Pole iJ E- ~ Of)) ie
JN~flJ -0 J _shye--middot-----middotmiddot- --
--lt)12 - o~s~j ~- b ~
EC03010A (902) wwwlyondelLcom
~LYONDELL Calculations Chart
Sheet ---3=-_ of_-=3=---_ Flle________
Subject________________ Oate________
By________
C-s-o$it_ -- 03 -lt
~=bK-tlt
q ( I - 1) = - J
- -- llt L- Crt I-Ilt
EC03010A (902) wwwlyondellcom
~~~~~~~~~~~~~-~~~~~~~~~-~-~~~~~~~-~~~ ~
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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
bull
I I
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-----------t
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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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
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== 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
bull
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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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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
bull
I I
L ~
bull
-----------t
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
bull
I I
L ~
bull
-----------t
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
=
I I
L ~
bull
-----------t
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
=