16 0 495 KB
John Mark R. Allas BS CHE- 3
Identification: 1. Mccabe Thiele Method 2. Minimum Reflux Ratio 3. q 4. Enriching operating line 5. Stripping operating line 6. Underwoodβs shortcut method 7. Bonus 8. Distillation 9. 10. Heavy Components 11. Light Components 12. Light key 13. Boiling key 14. Dew point 15. Enriching Section 16. Erbar and maddox 17. Pinch point 18. One Tower 19. 20. Feed section analysis
11.7-5. Shortcut Design of Multicomponent Distillation Tower. A feed of part liquid and part vapor (q=0.30) at 405.4 kPa is fed at the rate of 1000 mol/h to a distillation tower. The overall composition of the feed is n-butane (π₯π΄ = 0.35), n-pentane (π₯π΅ = 0.30), n-hexane (π₯πΆ = 0.20) and n-heptane (π₯π· = 0.15). This feed is distilled so that 97% of the n-pentane is recovered in the distillate and 85% of the n-hexane in the bottoms. Calculate the following: a) Amount and composition of products and top and bottom tower temperatures b) Number of stages at total reflux and distribution of other components in the products c) Minimum reflux ratio, number of stages at 1.2π
π, and feed tray location GIVEN: π = 0.30 π₯π΄ = 0.35 π₯π΅ = 0.30 π₯πΆ = 0.20 π₯π· = 0.15 REQUIRED: a) Amount and composition of products and top and bottom tower temperatures b) Number of stages at total reflux and distribution of other components in the products c) Minimum reflux ratio, number of stages at 1.2π
π, and feed tray location SOLUTION: a) Amount and composition of products and top and bottom tower temperatures Component (B) π β ππππ‘πππ being the light key (L) and Component (C) π β βππ₯πππ the heavy key (H) Component (B): thru Equation 11.7-16 π₯π΅πΉπΉ = 0.30(1000) = 300 = π¦π΅π· π· + π₯π΅ππ Since 97% of B is in distillate, π¦π΅π· π· = 0.97(300) = 291. Hence π₯π΅ππ = 0.03(300) = 9 Component (C): thru Equation 11.7-17 π₯πΆπΉπΉ = 0.20(1000) = 200 = π¦πΆπ·π· + π₯πΆππ Since 85% of B is in bottom, π¦πΆπ·π· = 0.15(200) = 30. Hence π₯πΆππ = 0.85(200) = 170
For the first trial, it is assumed that no component D (heavier than the heavy key C) is in distillate and no light A in the bottoms. Hence, moles A in distillate = π¦π΄π·π· = 0.35(1000) = 350. Also, moles D in bottoms = π₯π·ππ = 0.15(1000) = 150. The values tabulated below. Feed, F Distillate, D Bottoms, W Component ππ π ππ ππ«π« ππ« = ππ« π πΎπΎ ππΎ A 0.35 350 350 0.5216 0 0 B (L) 0.30 300 291 0.4337 9 0.0274 C (H) 0.20 200 30 0.0447 170 0.5167 D 0.15 150 0 0 150 0.4559 1 1 π = ππππ π« = πππ πΎ = πππ
F= D+W W= F-D XFF = XDD + XWW XFF = XDD + XW (F-D) 350 = 0.5216 (D) D=671 W= 1000-671 W=329 For the dew point of the distillate (top temperature), a value of 65β will be estimated for the first trial. The K values are read from Figure 11.7-2 and the πΌ values calculated. Using Equation 11.7-7 and 11.7-8, the following values are calculated: πππ« π²π πΆπ 0.5216 1.75 7 0.4337 0.66 2.64 0.0447 0.25 1 0 0.9 0.36 1 πΎπΆ = 0.2835 is equal to 70β which will be used for the 2nd Trial Component A B (L) C (H) D
ππ/πΆπ 0.0745 0.1643 0.0447 0 0.2835
ππ
πππ«
π²π
πΆπ
ππ/πΆπ
ππ
A
0.5216
1.9
6.7019
0.0778
0.2609
B (L)
0.4337
0.7
2.4691
0.1757
0.5892
C (H)
0.0447
0.2835
1
0.0447
0.1499
D
0
0.110
0.3880
0
0
0.2982
1
Component
1
Kc=0.2982 For the bubble point of the bottom, a value of 135β will be estimated for the first trial. The K values are read from Figure 11.7-2 and the πΌ values calculated. Using Equation 11.7-5 and 11.76, the following values are calculated: Component A B (L) C (H) D πΎπΆ = 0
πππΎ 0 0.0274 0.5167 0.4559 1
π²π 5.10 2.5 1.25 0.65
πΆπ 4.08 2 1 0.52
πππΆπ 0 0.0548 0.5167 0.2371 0.8086
ππ
.= 1.2367 is equal to 134β which will be used for the 2nd Trial
b) Number of stages at total reflux and distribution of other components in the products The proper πΌ values of the light key L (n-pentane) to use in Equation 11.7-13 πΌπΏ.ππ£ = βπΌπΏπ·πΌπΏπ πΌπΏπ· = 2.4691 (π = 70β at top column) πΌπΏπ = 2 (π = 134β at bottom column) πΌπΏ. πΌπΏ.ππ£ = 2.2222 Then using Equation 11.7-12, to get theoretical stages or steps ππ
ππ = log [(π₯π₯π»π·πΏπ· ΓΓ π·π·)(π₯π₯π»ππΏπ ΓΓ ππ) log (πΌπΏ.ππ£)
π₯πΏπ· = 0.4337 π₯π»π· = 0.0447 π₯π»π = 0.5390 π₯πΏπ = 0.0274 π· = 671 π = 329
ππ = π΅π = π. ππππ theoretical stages (5.5768 theoretical trays) The distribution or compositions of the other components can be calculated using Equation 11.714. For component A, the average πΌ value to use is πΌπ΄.ππ£ = βπΌπ΄π·πΌπ΄π πΌπ΄π· = 6.7019, πΌπ΄π = 4.1667 πΌπ΄. π₯π΄π·π·
( )ππ π₯π»π·π· = πΌπ΄.ππ£ π₯π΄ππ
π₯π»ππ
π₯π»π· = 0.0447,π₯π»π = 0.5167 π₯π΄π·π· 6.5768 10036.7637 = π₯π΄ππ
0.0447(671) = (5.2844) 0.
Making an overall balance on A, π₯π΄πΉπΉ = 350 = π₯π΄π·π· + π₯π΄ππ Substituting π₯π΄π·π· = 10036.7637π₯π΄ππ π₯π΄πΉπΉ = 350 = 10036.7637π₯π΄ππ + π₯π΄ππ π₯π΄ππ = 0.0349 and π₯π΄π· π· = 349.9651 For the distribution of component D, πΌπ·.ππ£ = βπΌπ·π·πΌπ·π πΌπ·π· = 0.3880
πΌπ·π = 0.55 πΌπ·. π₯π·π·π· = (πΌπ·.ππ£)ππ π₯π·ππ
π₯π»π·π·
π₯π»ππ π₯π·π·π· = (
0.0447(671) = 1.0990 Γ 10β3 0.4620) π₯π·ππ
6.5768
0.5167(329)
Making an overall balance on D, π₯π·πΉπΉ = 150 = π₯π·π·π· + π₯π·ππ Substituting π₯π·π·π· = 1.0990 Γ 10β3π₯π·ππ 150 = 1.0990 Γ 10β3π₯π·ππ + π₯π·ππ π₯π·ππ = 149.8353 πππ π₯π·π·π· = 0.1647 The revised distillate and bottoms compositions are as follows Distillate, D Component
Bottoms, W
π₯π·π·
π¦π· = π₯π·
π₯π·π
A
349.9651
0.5215
0.0349
B
291
0.4336
9
0.0274
C
30
0.0447
170
0.5169
D
0.1647
149.8353
0.4556
πΎ = πππ.ππππ
1
TOTAL
π« = πππ.ππππ
2.4511 Γ ππβπ 1
π₯π 1.0612 Γ ππβπ
c) Minimum reflux ratio, number of stages at π.ππΉπ, and feed tray location The temperature to use for determining the values of πΌπ is the average between the top of 65β and the bottom 134β and is
=
102β. The πΎπ values obtained from Figure 11.7-2 and the πΌπ values and distillates and feed compositions to use in Equation 11.7-19 and 11.7-20 are as follows
COMPONENT
π₯ππΉ
π₯ππ·
πΎπ
πΌπ
A
0.35
0.5215
3.25
4.9242
π₯ππ 1.0612Γ 10β4
B (L)
0.30
0.4336
1.4
2.1212
0.0274
C (H)
0.20
0.0447
0.66
1
0.5169
D
0.15
2.4511Γ 10β4
0.29
0.4394
0.4556
1.00 1 Substituting into Equation 11.7-19 with π = 0.30 for feed t
1
πΌππ₯ππΉ 1βπ=β πΌπ β π 9242 Γ 0 .35
1 β 0.30 = +
2 .1212 Γ 0 .30
1 Γ 0 .20
0 .4394 Γ 0 .15 4.
+ + 4.9242 β π
2.1212 β π
1βπ
0.4394 β π
This is trial and error, so a value of π = 1.5 will be used for the trial. The trials are shown below. 1.7235
0.6364
0.20
0.0659
4.9242 β π 0.5033 0.4890 0.4849 0.4848 0.4842 0.4836 0.4756
2.1212 β π 1.0245 0.8824 0.8472 0.8461 0.8410 0.8360 0.7750
1βπ -0.4 -0.5 -0.5405 -0.5420 -0.5487 -0.5556 -0.6667
0.4394 β π -0.0621 0.0686 -0.0708 -0.0709 -0.0712 -0.07158 -0.0766
βπ π’π
π 1.5 1.4 1.37 1.369 π.ππππ 1.36 1.3
For exact value, π
.ππππ, π.πππβπ.πππππ.πππβπ.ππ
The final value of π = 1.3645 is substituted into Equation 11.7-20 to solve π
π πΌππ₯ππ· π
π + 1 = β πΌπ β π π
π π
π + 0.3 = 0.7214 + 1.2154 + (β0.1226) + (β1.1642 Γ 10β4)
1.0657 0.8028 0.7208 0.718 0.7053 0.6924 0.5073
Rm=1.5141 The following values are calculated. π
= 1.2(π
π) = 1.2(1.5141) = 1.8169 π
1.8169
π
+1
1.8169 + 1
== 0.6449 π
π
1.5141
π
π + 1
== 0.6022 1.5141 + 1
Using Erbar-Maddox Correlation Figure 11.73, plotting these values to get
ππ π
Since ππ = 6.5768, so 6
.5768 = π
ππ/π
= 0 .5
0.5, π = 13.1536
This gives ππ.ππππ number of stages This ππ.ππππ β π. π (ππππππππ)ππ ππ.ππππ πππππππππππ πππππ gives For the location of the feed tray using the Kirkbride, Equation 11.7-21 ππ π₯π»πΉ π π₯πΏπ 2 πππ = 0.206 log [( ) ( )] ππ π₯πΏπΉ π· π₯π»π·
π₯π»πΉ = 0.20 π₯πΏπΉ = 0.30 π₯πΏπ = 0.0274 π₯π»π· = 0.0447 π· = 671.1298 π = 328.8702 ππ 0.20 328.8702 πππ = 0.206log [( ππ 0.30 671.1298
2
0.0274 )
(
) ] = β0.1877
0.0447
ππ πππ = β0.1877 ππ ππ = 0.6491 ππ Also, ππ + ππ = 0.6491ππ + ππ = π = 13.1536 Solving, ππ = 7.9762,ππ = 5.1773 The number of theoretical stages above
π΅π is π.ππππ and theoretical stages
π΅πΊ is π.ππππ
SOLUTION:
A. For a feed rate of 100 mol/h, calculate D and W, number of stages at total reflux and distribution (concentration) of A in the bottoms.
COMPONENT A B C
FEED XFF 4.7 7.2 88.1 100
XF 0.047 0.072 0.881 1
Distillate XDD XD 4.7 0.126 7.1 0.1913 25.5 0.6827 D=37.3 1
F= D+W W= F-D XFF = XDD + XWW XFF = XDD + XW (F-D) 4.7 = 0.1260 (D) D = 37.3 W = 100 β 37.3 W= 62.7
Bottoms XWW XW 0 0 0.0627 0.001 62.6373 0.999 W=62.7 1
Ξ± 4.19 1.58 1
B. Calculate the Rm and the number of stages at 1.25 Rm COMPONENTS
XiF
A B C
Ξ±i
XiD
0.047 0.072 0.881 1
0.126 0.1913 0.6827 1
XiW
4.19 1.58 1
0 0.001 0.999 1
Substituting into Equation 11.7-19 with π = 1 for feed t 1 β q = β
1β1=
4.19 π₯ 0.047 4.19β π
+
1.58 π₯ 0.072 1.58β π
+
πΌππππΉ πΌπ β π
1.00 π₯ 0.881 1.00β π
This is trial and error, so a value of π = 0.5 will be used for the trial. The trials are shown below.
π 0.5 0.25 0.05 0.005 0.00075 0.0000080
0=
4.19 π₯ 0.047 4.19β π
0.1970 4.19 β π
0.1138 1.58 β π
0.0534 0.05 0.0476 0.0471 0.0470 0.0470
0.1054 0.0856 0.0744 0.0723 0.0721 0.0720
+
X= -1.2678x10-12 β 1.0000
1.58 π₯ 0.072 1.58β π
+
0.881 1.00 β π
1.762 1.1747 0.9274 0.8854 0.8817 0.8810
1.00 π₯ 0.881 1.00β π
β 1.928 1.3103 1.04 1.00 1.000 1