multistage course chemical engineering

1. MULTISTAGE SEPARATION
PROCESSES
CHE 452
ENG. AMAL MAGDY
SECTION (6)

2. Multistage Distillation
Calculations
• Binary System
• Sorel plate to plate calculations
• Ponchon Sevarit method
• Lewis plate to plate calculations
• McCabe Thiele method

3. McCabe Thiele method
• The graphical representation of Lewis method
• Like Lewis Method the tower is divided into top section
and bottom section
• McCabe Thiele method depends on the x-y diagram only
• Each section has its own operating line
y
x

4. McCabe Thiele method
Calculation steps
• Calculation steps are done through three main
steps:
1. Material Balance:
• OMB  F = D + W
• CMB  𝑭. 𝒙𝒇 = 𝑫. 𝒙𝑫 +𝑾. 𝒙𝒘
Get D & W

5. McCabe Thiele method
Calculation steps
• Calculation steps are done through three main
steps:
2. Operating Line for Top Section:
𝒚𝒏+𝟏 =
𝑳
𝑳+𝑫
𝒙𝒏 +
𝑫
𝑳+𝑫
𝒙𝑫
Or
𝒚𝒏+𝟏 =
𝑹
𝑹 + 𝟏
∗ 𝒙𝒏 +
𝒙𝑫
𝑹 + 𝟏
To draw this line you should have two points:
At xn = 0  yn+1 =
𝒙𝑫
𝑹+𝟏
At xn = xD  yn+1 = xD
xD
xF
xW
1
R
xD


6. McCabe Thiele method
Calculation steps
• Calculation steps are done through three main
steps:
3. Operating Line for Bottom Section:
𝒚𝒎
′ =
𝑳′
𝑽′ 𝒙𝒎+𝟏
′
−
𝑾
𝑽′ 𝒙𝒘
To draw this line you should have two points:
At 𝒙𝒎+𝟏
′
= 0  𝒚𝒎
′
=−
𝑾
𝑽′ 𝒙𝑾
At 𝒙𝒎+𝟏
′
= xw  𝒚𝒎
′ = xw
xD
xF
xW
V'
Wx
- W

7. McCabe Thiele method
Calculation steps
• The diagram of the whole tower will be:
• The number of theoretical stages = No. of stages in
top section + No. of stages in top section + Reboiler
• For Top Section
• NTS = 6 +
𝑎
𝑏
• For Bottom Section
• NTS = 1 + Reboiler
xD
xF
xW
1
R
xD

a
b
V'
Wx
- W

8. McCabe Thiele method
Calculation steps
• Q-line is calculations:
𝒒 =
𝑯𝑽 − 𝒉𝒇
𝑯𝑽 − 𝒉𝑳
=
𝑯𝑽 − 𝒉𝒇
𝝀
Slope of q-line =
𝒒
𝒒 −𝟏
• Q-line direction depends on the phase of feed:
1. Subcooled Feed  q > 1
2. Saturated Liquid Feed  q = 1
3. Partially Vaporized Feed  0 < q < 1
4. Saturated Vapor Feed  q = 0
5. Superheated Feed  q < 1
1
2
3
4
5
xD
xF
xW

9. McCabe Thiele method
Calculation steps
• Minimum reflux ratio:
It is the point that resulted from the extension of
the line plotted from (xD, xD) to the intersection of
q-line with the equilibrium curve
xD
xF
xW
1
R
x
min
D


10. SHEET (4)

11. Example (2)
A continuous fractionating column is to be designed to separate 100 kgmole/h of a
mixture of 60% benzene and 40% toluene into a top product containing 2% toluene
and a bottom product containing 3% benzene. Feed enters the column at 30°C. Pressure
in the column is atmospheric and the reflux ratio is 1.2 the minimum.
Calculate:
a. Flowrates of the top and the bottom products
b. Number of ideal plates
c. Steam consumption if saturated steam at 3 atm is available.
d. Cooling water consumption if cooling water at 30°C are available and left the system at
45°C.

12. Example (2)
Additional Data:
Relative volatility=2.5
Bubble and dew points of feed are 90 and 100°C.
Liquid specific heats of benzene and toluene are 0.43 and 0.45 cal/g.°C respectively
Latent heats of benzene and toluene are 7400 and 8000 cal/gmole respectively
Latent heat of steam at 3 atmosphere is 517 cal/g

13. Solution (2)
Givens:
F = 100 kgmole/h , 𝑃 = 1 𝑎𝑡𝑚 , 𝑇 = 30°𝐶 & 𝑅 = 1.2 𝑅𝑚𝑖𝑛
𝑥𝑓 = 0.6 , 𝑥𝐷 = 0.98 & 𝑥𝑤 = 0.03
Answer:
a. Flowrates of the top and the bottom products
F = D + W
D + W = 100  (1)
𝑭. 𝒙𝒇 = 𝑫. 𝒙𝑫 +𝑾. 𝒙𝒘
100 ∗ 0.6 = 0.98 𝐷 + 0.03 𝑊  (2)
From (1) & (2)
D = 60 kgmole/h , W = 40 kgmole/h

14. Solution (2)
b. Number of ideal plates
• Draw the equilibrium curve
Using the given relative volatility, assume xA to get yA at equilibrium
𝒚𝑨 =
𝜶. 𝒙𝑨
𝟏 + (𝜶 − 𝟏) 𝒙𝑨
xA yA
0 0
0.1 0.217
0.3 0.517
0.5 0.714
0.7 0.854
0.9 0.957
1 1
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
y
A
xA

15. Solution (2)
b. Number of ideal plates and feed location
• Determine Rmin
𝒒 =
𝑯𝑽 − 𝒉𝒇
𝑯𝑽 − 𝒉𝑳
=
𝑯𝑽 − 𝒉𝒇
𝝀
𝜆 = 0.6 ∗ 7400 + 0.4 ∗ 8000 = 7640 cal/gmole
𝐶𝑃𝑚𝑖𝑥
= 𝑥𝑖 ∗ 𝐶𝑃𝑖
= 0.6* 0.43+ 0.4*0.45 = 0.438 cal/g.°C
Mavg = 0.6 * 78 + 0.4 * 92 = 83.6 g/gmole
𝐻𝑉 − ℎ𝑓 = 𝜆 + 𝐶𝑃𝑚𝑖𝑥
∗ 𝑇𝐵𝑢𝑏𝑏𝑙𝑒 − 𝑇𝑓 = 7640 + 0.438 ∗ 90 − 30 ∗ 83.6 = 9837.008 cal/gmole
𝑞 =
𝐻𝑉 − ℎ𝑓
𝜆
=
9837.008
7640
= 1.288

16. Solution (2)
b. Number of ideal plates and feed location
Slope of q-line =
𝒒
𝒒 −𝟏
= 𝟒. 𝟒𝟕𝟐
• Determine Rmin
𝑥𝐷
𝑅𝑚𝑖𝑛 + 1
= 0.51
Rmin = 0.92
R = 1.2 * Rmin = 1.104
𝑥𝐷
𝑅 + 1
= 0.466
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
y
A
xA

17. Solution (2)
b. Number of ideal plates and feed
location
NTS = 17.7 + Reboiler
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
y
A
xA

18. Solution (2)
c. Steam consumption if saturated steam at 3 atm is available.
Q r = mst.lst = V’.lr
lr = yr * lA + (1 - yr ) * lB
lr = 0.07*7400 + 0.93*8000 = 7958 cal/gmole
Slope of op. line in bottom section =
𝐿′
𝑉′ = 1.25
L’ = V’ + W
L’ = 200 kgmole/h & V’ = 160 kgmole/h
Q r = V’.lr = 160 * 7958 *1000 = 1273.28*106 cal/h
lst. = 517*18*1000 = 9306*103 cal/kmole
Q r = mst.lst = 1273.28*106  mst. = 136.82 kmole = 2462.82 kg
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
y
A
xA

19. Solution (2)
d. Cooling water consumption if cooling water at 30°C is available and left the system at
45°C.
Qcond=V1 . lmix =mw . Cpw . ΔT
V1 = Lo + D
Lo = R * D = 1.104* 60 = 66.24 kmole  V1 = 126.24 kmole/h
lmix = (0.98*7400 + 0.02*8000) * 1000 = 7412*103 cal/kmole
Qcond. = 935690*103 cal/h
Cpw = 1 cal/g.°C
ΔT = 15°C
Qcond = mw . Cpw . ΔT  mw = 62379.3 kg/h

20. Special cases in Binary
systems
1. Enriching Section:
• Feed is saturated vapor
• NTS in bottom section = 0
XD
XW XF

21. Special cases in Binary
systems
2. Stripping Section:
• Feed is saturated liquid
• NTS in top section = 0
XD
XW XF

22. Special cases in Binary
systems
3. Complex Feed or Multiple feeds:
• Feed (1) is saturated liquid
• Feed (2) is saturated vapor
• The slope of the middle section operating line
=
𝐿′
𝑉′
• Where,
• L’ = F1 + L
• V’ = V
XD
XF1
XW
1
R
xD

(XD,XD)
(XW,XW
)
XF2
L’/V’

23. Special cases in Binary
systems
4. Top Side Product:
• The top product is saturated liquid
• The middle section operating line is
determined using:
𝑥 = 𝑦 =
𝑆 . 𝑥𝑠 + 𝐷 . 𝑥𝐷
𝑆 + 𝐷
XD
XS
XW
1
R
xD

(XD,XD)
(XW,XW)
XF
x=y

24. Special cases in Binary
systems
5. Bottom Side Product:
• The bottom product is saturated liquid
• The middle section operating line is
determined using:
𝑥 = 𝑦 =
𝑆 . 𝑥𝑠 + 𝑊 . 𝑥𝑊
𝑆 + 𝑊
XW
1
R
xD

(XD,XD)
(XW,XW
)
XS
x=y
XD
XF

25. Special cases in Binary
systems
6. Open Steam:
• The bottom section operating line is
determined by matching (xw, xw) with (x,y)
that calculated using the following equation:
𝑥 = 𝑦 =
𝑊 . 𝑥𝑊
𝑊 − 𝑆
XD
XF
XW
1
R
xD

(XD,XD)
(XW,ys)
S
W
x
W W

.

26. Example (3)
A continuous distillation is supplied with two equimolar feed streams, one is saturated
liquid containing 40% benzene and the other is saturated vapor with 20% benzene.
The required top product is 98% benzene while the bottom product is 2% benzene.
Reflux ratio used is 4.
a. Calculate the required number of theoretical stages and the correct feed location of each
feed.
b. If the two streams are to be mixed before entering the tower, find the required number
of stages.
Equilibrium data are given below:
Mole % of benzene in
liquid phase
0 10 20 30 40 50 60 70 80 90 95 100
Mole % of benzene in
vapor phase
0 20.8 37.2 50.7 61.9 71.3 79.1 85.7 91.2 95.9 98 100

27. Solution (3)
Assume F1 = F2 = 100 mole
F1 + F2 = D + W = 200  (1)
𝐹1 . 𝑥𝑓1
+ 𝐹2 . 𝑥𝑓2
= 𝐷 . 𝑥𝐷 + 𝑊 . 𝑥𝑤  (2)
From (1) & (2)
D = 58.3 mole
W = 141.7 mole
XD
XF1
XW
1
R
xD

(XD,XD)
(XW,XW
)
XF2
L’/V’

28. Solution (3)
𝑥𝐷
𝑅+1
= 0.196  get the top section op. line
L = D * R = 233.2 mole
L’ = F1 + L = 333.2 mole
V’ = V = L + D = 291.5 mole
Slope of the middle section op. line =
𝐿′
𝑉′ = 1.143
 Get the middle section op. line
Then calculate the number of stages
XD
XF1
XW
1
R
xD

(XD,XD)
(XW,XW
)
XF2
L’/V’

29. Solution (3)
The final shape will be:
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
y
A
xA

30. Solution (3)
b. If the two feeds are mixed together:
Assume F1 = F2 = 100 mole
F1 + F2 = F = 200
𝐹1 . 𝑥𝑓1
+ 𝐹2 . 𝑥𝑓2
= 𝐹 . 𝑥𝑓  𝑥𝑓 = 0.3
Then begin to solve as mentioned in example 2 (b)