John Mark R. Allas Bs Che-3 [PDF]

Citation preview

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. Underwood’s 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 (𝑥𝐴 = 0.35), n-pentane (𝑥𝐵 = 0.30), n-hexane (𝑥𝐶 = 0.20) and n-heptane (𝑥𝐷 = 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 1.2𝑅𝑚, and feed tray location GIVEN: 𝑞 = 0.30 𝑥𝐴 = 0.35 𝑥𝐵 = 0.30 𝑥𝐶 = 0.20 𝑥𝐷 = 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 1.2𝑅𝑚, and feed tray location SOLUTION: a) Amount and composition of products and top and bottom tower temperatures Component (B) 𝑛 − 𝑝𝑒𝑛𝑡𝑎𝑛𝑒 being the light key (L) and Component (C) 𝑛 − ℎ𝑒𝑥𝑎𝑛𝑒 the heavy key (H) Component (B): thru Equation 11.7-16 𝑥𝐵𝐹𝐹 = 0.30(1000) = 300 = 𝑦𝐵𝐷 𝐷 + 𝑥𝐵𝑊𝑊 Since 97% of B is in distillate, 𝑦𝐵𝐷 𝐷 = 0.97(300) = 291. Hence 𝑥𝐵𝑊𝑊 = 0.03(300) = 9 Component (C): thru Equation 11.7-17 𝑥𝐶𝐹𝐹 = 0.20(1000) = 200 = 𝑦𝐶𝐷𝐷 + 𝑥𝐶𝑊𝑊 Since 85% of B is in bottom, 𝑦𝐶𝐷𝐷 = 0.15(200) = 30. Hence 𝑥𝐶𝑊𝑊 = 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 = 𝑦𝐴𝐷𝐷 = 0.35(1000) = 350. Also, moles D in bottoms = 𝑥𝐷𝑊𝑊 = 0.15(1000) = 150. The values tabulated below. Feed, F Distillate, D Bottoms, W Component 𝒙𝑭 𝒙 𝑭𝑭 𝒚𝑫𝑫 𝒚𝑫 = 𝒙𝑫 𝒙 𝑾𝑾 𝒙𝑾 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 𝑭 = 𝟏𝟎𝟎𝟎 𝑫 = 𝟔𝟕𝟏 𝑾 = 𝟑𝟐𝟗



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 65℃ will be estimated for the first trial. The K values are read from Figure 11.7-2 and the 𝛼 values calculated. Using Equation 11.7-7 and 11.7-8, the following values are calculated: 𝒚𝒊𝑫 𝑲𝒊 𝜶𝒊 0.5216 1.75 7 0.4337 0.66 2.64 0.0447 0.25 1 0 0.9 0.36 1 𝐾𝐶 = 0.2835 is equal to 70℃ which will be used for the 2nd Trial Component A B (L) C (H) D



𝒚𝒊/𝜶𝒊 0.0745 0.1643 0.0447 0 0.2835



𝒙𝒊



𝒚𝒊𝑫



𝑲𝒊



𝜶𝒊



𝒚𝒊/𝜶𝒊



𝒙𝒊



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 135℃ will be estimated for the first trial. The K values are read from Figure 11.7-2 and the 𝛼 values calculated. Using Equation 11.7-5 and 11.76, the following values are calculated: Component A B (L) C (H) D 𝐾𝐶 = 0



𝒙𝒊𝑾 0 0.0274 0.5167 0.4559 1



𝑲𝒊 5.10 2.5 1.25 0.65



𝜶𝒊 4.08 2 1 0.52



𝒙𝒊𝜶𝒊 0 0.0548 0.5167 0.2371 0.8086



𝒚𝒊



.= 1.2367 is equal to 134℃ 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 𝛼 values of the light key L (n-pentane) to use in Equation 11.7-13 𝛼𝐿.𝑎𝑣 = √𝛼𝐿𝐷𝛼𝐿𝑊 𝛼𝐿𝐷 = 2.4691 (𝑇 = 70℃ at top column) 𝛼𝐿𝑊 = 2 (𝑇 = 134℃ at bottom column) 𝛼𝐿. 𝛼𝐿.𝑎𝑣 = 2.2222 Then using Equation 11.7-12, to get theoretical stages or steps 𝑁𝑚



𝑁𝑚 = log [(𝑥𝑥𝐻𝐷𝐿𝐷 ×× 𝐷𝐷)(𝑥𝑥𝐻𝑊𝐿𝑊 ×× 𝑊𝑊) log (𝛼𝐿.𝑎𝑣)



𝑥𝐿𝐷 = 0.4337 𝑥𝐻𝐷 = 0.0447 𝑥𝐻𝑊 = 0.5390 𝑥𝐿𝑊 = 0.0274 𝐷 = 671 𝑊 = 329



𝑁𝑚 = 𝑵𝒎 = 𝟔. 𝟓𝟕𝟔𝟖 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 𝛼 value to use is 𝛼𝐴.𝑎𝑣 = √𝛼𝐴𝐷𝛼𝐴𝑊 𝛼𝐴𝐷 = 6.7019, 𝛼𝐴𝑊 = 4.1667 𝛼𝐴. 𝑥𝐴𝐷𝐷



( )𝑁𝑚 𝑥𝐻𝐷𝐷 = 𝛼𝐴.𝑎𝑣 𝑥𝐴𝑊𝑊



𝑥𝐻𝑊𝑊



𝑥𝐻𝐷 = 0.0447,𝑥𝐻𝑊 = 0.5167 𝑥𝐴𝐷𝐷 6.5768 10036.7637 = 𝑥𝐴𝑊𝑊



0.0447(671) = (5.2844) 0.



Making an overall balance on A, 𝑥𝐴𝐹𝐹 = 350 = 𝑥𝐴𝐷𝐷 + 𝑥𝐴𝑊𝑊 Substituting 𝑥𝐴𝐷𝐷 = 10036.7637𝑥𝐴𝑊𝑊 𝑥𝐴𝐹𝐹 = 350 = 10036.7637𝑥𝐴𝑊𝑊 + 𝑥𝐴𝑊𝑊 𝑥𝐴𝑊𝑊 = 0.0349 and 𝑥𝐴𝐷 𝐷 = 349.9651 For the distribution of component D, 𝛼𝐷.𝑎𝑣 = √𝛼𝐷𝐷𝛼𝐷𝑊 𝛼𝐷𝐷 = 0.3880



𝛼𝐷𝑊 = 0.55 𝛼𝐷. 𝑥𝐷𝐷𝐷 = (𝛼𝐷.𝑎𝑣)𝑁𝑚 𝑥𝐷𝑊𝑊



𝑥𝐻𝐷𝐷



𝑥𝐻𝑊𝑊 𝑥𝐷𝐷𝐷 = (



0.0447(671) = 1.0990 × 10−3 0.4620) 𝑥𝐷𝑊𝑊



6.5768



0.5167(329)



Making an overall balance on D, 𝑥𝐷𝐹𝐹 = 150 = 𝑥𝐷𝐷𝐷 + 𝑥𝐷𝑊𝑊 Substituting 𝑥𝐷𝐷𝐷 = 1.0990 × 10−3𝑥𝐷𝑊𝑊 150 = 1.0990 × 10−3𝑥𝐷𝑊𝑊 + 𝑥𝐷𝑊𝑊 𝑥𝐷𝑊𝑊 = 149.8353 𝑎𝑛𝑑 𝑥𝐷𝐷𝐷 = 0.1647 The revised distillate and bottoms compositions are as follows Distillate, D Component



Bottoms, W



𝑥𝐷𝐷



𝑦𝐷 = 𝑥𝐷



𝑥𝐷𝑊



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



𝑾 = 𝟑𝟐𝟖.𝟖𝟕𝟎𝟐



1



TOTAL



𝑫 = 𝟔𝟕𝟏.𝟏𝟐𝟗𝟖



2.4511 × 𝟏𝟎−𝟒 1



𝑥𝑊 1.0612 × 𝟏𝟎−𝟒



c) Minimum reflux ratio, number of stages at 𝟏.𝟐𝑹𝒎, and feed tray location The temperature to use for determining the values of 𝛼𝑖 is the average between the top of 65℃ and the bottom 134℃ and is



=



102℃. The 𝐾𝑖 values obtained from Figure 11.7-2 and the 𝛼𝑖 values and distillates and feed compositions to use in Equation 11.7-19 and 11.7-20 are as follows



COMPONENT



𝑥𝑖𝐹



𝑥𝑖𝐷



𝐾𝑖



𝛼𝑖



A



0.35



0.5215



3.25



4.9242



𝑥𝑖𝑊 1.0612× 10−4



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



2.4511× 10−4



0.29



0.4394



0.4556



1.00 1 Substituting into Equation 11.7-19 with 𝑞 = 0.30 for feed t



1



𝛼𝑖𝑥𝑖𝐹 1−𝑞=∑ 𝛼𝑖 – 𝜃 9242 × 0 .35



1 − 0.30 = +



2 .1212 × 0 .30



1 × 0 .20



0 .4394 × 0 .15 4.



+ + 4.9242 − 𝜃



2.1212 − 𝜃



1−𝜃



0.4394 − 𝜃



This is trial and error, so a value of 𝜃 = 1.5 will be used for the trial. The trials are shown below. 1.7235



0.6364



0.20



0.0659



4.9242 − 𝜃 0.5033 0.4890 0.4849 0.4848 0.4842 0.4836 0.4756



2.1212 − 𝜃 1.0245 0.8824 0.8472 0.8461 0.8410 0.8360 0.7750



1−𝜃 -0.4 -0.5 -0.5405 -0.5420 -0.5487 -0.5556 -0.6667



0.4394 − 𝜃 -0.0621 0.0686 -0.0708 -0.0709 -0.0712 -0.07158 -0.0766



∑𝑠𝑢𝑚



𝜃 1.5 1.4 1.37 1.369 𝟏.𝟑𝟔𝟒𝟓 1.36 1.3



For exact value, 𝟐



.𝟕𝟎𝟓𝟐, 𝟎.𝟕𝟏𝟖−𝟎.𝟔𝟗𝟐𝟒𝟏.𝟑𝟔𝟗−𝟏.𝟑𝟔



The final value of 𝜃 = 1.3645 is substituted into Equation 11.7-20 to solve 𝑅𝑚 𝛼𝑖𝑥𝑖𝐷 𝑅𝑚 + 1 = ∑ 𝛼𝑖 − 𝜃 𝑅𝑚 𝑅𝑚 + 0.3 = 0.7214 + 1.2154 + (−0.1226) + (−1.1642 × 10−4)



1.0657 0.8028 0.7208 0.718 0.7053 0.6924 0.5073



Rm=1.5141 The following values are calculated. 𝑅 = 1.2(𝑅𝑚) = 1.2(1.5141) = 1.8169 𝑅



1.8169



𝑅+1



1.8169 + 1



== 0.6449 𝑅𝑚



1.5141



𝑅𝑚 + 1



== 0.6022 1.5141 + 1



Using Erbar-Maddox Correlation Figure 11.73, plotting these values to get



𝑁𝑚 𝑁



Since 𝑁𝑚 = 6.5768, so 6



.5768 = 𝑁



𝑁𝑚/𝑁



= 0 .5



0.5, 𝑁 = 13.1536



This gives 𝟏𝟑.𝟏𝟓𝟑𝟔 number of stages This 𝟏𝟑.𝟏𝟓𝟑𝟔 − 𝟏. 𝟎 (𝒓𝒆𝒃𝒐𝒊𝒍𝒆𝒓)𝒐𝒓 𝟏𝟐.𝟏𝟓𝟑𝟔 𝒕𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒕𝒓𝒂𝒚𝒔 gives For the location of the feed tray using the Kirkbride, Equation 11.7-21 𝑁𝑒 𝑥𝐻𝐹 𝑊 𝑥𝐿𝑊 2 𝑙𝑜𝑔 = 0.206 log [( ) ( )] 𝑁𝑠 𝑥𝐿𝐹 𝐷 𝑥𝐻𝐷



𝑥𝐻𝐹 = 0.20 𝑥𝐿𝐹 = 0.30 𝑥𝐿𝑊 = 0.0274 𝑥𝐻𝐷 = 0.0447 𝐷 = 671.1298 𝑊 = 328.8702 𝑁𝑒 0.20 328.8702 𝑙𝑜𝑔 = 0.206log [( 𝑁𝑠 0.30 671.1298



2



0.0274 )



(



) ] = −0.1877



0.0447



𝑁𝑒 𝑙𝑜𝑔 = −0.1877 𝑁𝑠 𝑁𝑒 = 0.6491 𝑁𝑠 Also, 𝑁𝑒 + 𝑁𝑆 = 0.6491𝑁𝑆 + 𝑁𝑆 = 𝑁 = 13.1536 Solving, 𝑁𝑆 = 7.9762,𝑁𝑒 = 5.1773 The number of theoretical stages above



𝑵𝒆 is 𝟓.𝟏𝟕𝟕𝟑 and theoretical stages



𝑵𝑺 is 𝟕.𝟗𝟕𝟔𝟐



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 – 37.3 W= 62.7



Bottoms XWW XW 0 0 0.0627 0.001 62.6373 0.999 W=62.7 1



α 4.19 1.58 1



B. Calculate the Rm and the number of stages at 1.25 Rm COMPONENTS



XiF



A B C



αi



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 𝑞 = 1 for feed t 1 − q = ∑



1–1=



4.19 𝑥 0.047 4.19− 𝜃



+



1.58 𝑥 0.072 1.58− 𝜃



+



𝛼𝑖𝑋𝑖𝐹 𝛼𝑖 − 𝜃



1.00 𝑥 0.881 1.00− 𝜃



This is trial and error, so a value of 𝜃 = 0.5 will be used for the trial. The trials are shown below.



𝜃 0.5 0.25 0.05 0.005 0.00075 0.0000080



0=



4.19 𝑥 0.047 4.19− 𝜃



0.1970 4.19 − 𝜃



0.1138 1.58 − 𝜃



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 ∑ 1.0000



1.58 𝑥 0.072 1.58− 𝜃



+



0.881 1.00 − 𝜃



1.762 1.1747 0.9274 0.8854 0.8817 0.8810



1.00 𝑥 0.881 1.00− 𝜃



∑ 1.928 1.3103 1.04 1.00 1.000 1