Question

Question 3 A saturated liquid mixture is fed to a continuous distillation column at 300 kmol/h. The specifications of the operation of the column are provided in Table 2. The column operates at 800 kPa. Table 2: Feed, distillate and bottoms flowrates Component Feed (kmol/h) Bottoms (kmol/h) Distillate (kmol/h) methanol 95 ethanol 33 n-propanol (LK) 67 n-butanol (HK) 45 4 n-pentanol 60 Total 300 Additional information: a. n-propanol is the light key component (LK) and n-butanol is the heavy key component (HK) b. The equilibrium values (K values) are calculated using the equations below: methanol: K = 0.0487e0.03047 ethanol: K = 0.0191e003267 n-propanol: K= 0.0074e003517 butanol: K 0.0038e0.03757 - pentanol: K = 0.0013e0.03917 n Where T C The relative volatilities (relative to the HK) can be determined at the 142°C, the average of the top and bottoms temperature. 3.1. Complete the material balance over the distillation column and the mole fractions of each component in feed, distillate and bottoms streams. Re-draw Table 3 below into your (18) answer book and complete. Table 3: Fill in the molar flows and compositions for the feed, distillate and bottoms products and K and a. Bottoms (kmol/h) XB K Distillate (kmol/h) XD Component XF Feed, (kmol/h) methanol ethanol n-propanol (LK) n-butanol (HK) n-pentanol Total 3.2. Determine the minimum number of stages using the Fenske method. (6) 3.3. Determine the minimum reflux ratio using the Underwood method. (HINT use: 0= 1.05, 1.1 and 1.15) (10) 3.4. Determine the number of ideal stages if the actual reflux is 5 times the minimum reflux using the Gilliland method. (6) 3.5. Determine the number of ideal stages in the rectifying sections and the stripping section. (8) 3.6. Determine the feed location stage. (1) [49 Marks 11

Answer 1

Solution:- Given data A Satunaled liguid mixture is fed to a Conbinuous dishlahon alumn at The column cpenates ad So0kpa. Fobl: Feed, digake and bottems flowrafe udhanel/h Disle kkmatorni (l eneh 300 as methanol ethano pmepanalk -butanol (k) 33 67 -pentano! Total 300 popanol butanal is the he lighty omponeud k) and heavy hey componeud () n n- b. methanolk o.630u 7 - O.OuS7 e ethanol : k OO191e o.0328 T O 0351T 1-ряораnо! K0.00Ue 0.03 75 T butanol k: 0 0038e 0.0391 T n-pentanol

It is qiven that npropanol i 6 kmol in bottom Therefosre n-piopana in dstllate- feed-bottm 61kmal/h kmolh Similoly nbutarol i given a top tharefase n-bedano tn as ishllah.on bottom feed diskllabion in ut kmol lh - assume that all the ethanol and me-thanal n-pentanol wi be n bottom. TOtal lous of distillade Total Bottom lous - 6U[ +60- l07 knollh 3.hematenial balane and Rewnile Re Lable we are in disllode cohruas al the 95+336Q= 193. krel obtain the tollouwing table. It is also mentioned that the avenege demperadure Ltence we use dekomi the column luz °c.Therefore values at that kemaodue -he k and od

molaa Ploos and Composihons fon e fead, dieilale on bottons producs and k Dishllak Componeal me lhFmoh Bottons kmollh) mahonol O316 OO000 3.uqa89 uS9 O.1100 33 O O00 0 2.01281 3 2.53 3 3 0.2233 61 0.316 O 056 1-0810ul.36 renctuu n-betandl US 0.1S00 O.O207 O.3832 Th 0.33slus 0 U2 |npantanal 60 O.2000 total IG3 Q OE

asampton need to checlk whehen Oco we Coe C not AtCxgr the aboue eguolion which disibuh eguakon Componn distlate and bottoms used oflen the Jelalive volabliies ha conslants A and c by subsltuding the values d and bgtey Qnd huavy ay Pixed and knoon wwe Ue in are knousn.we dakomine Hheir distibubion Qus 1-3 A+Co obtain the saluiors whera Awe A -1.0l0 Nous we use -these conslanl to ckkmins the distribukion atio non-ky Componeuds

-1-0107+18.li logu 59 mehand bnhaml dmned 307uu818 307uusi8 3.08 X10 bmuthanal Jegl 53 --1.0l07+ (5.11ux hanol duhano! behanol 3810 33 3810 3 86710 bkanal Inpeudanol dpriendl- pendanal *651X lO 60 x 6 51L0 3.90- dpderel

inhal assunption and ethanol pru seut in the bottom hereoe mehano in his conec. the assumph'on a Pendanol pusud Coect no no in the distale is also n. 3.2 he minimum numben stagej ujing He fenske method. XLK XHK Jog Xd b Minimum og Oumbesages fense mithod Jog/3 amog 354 min detnmined by 0.3820 O.0208d O.055b Nn in5.11stages 3.3 he minimum auflus jaho using Ale undirwood method 1- Stdkraked igi dSo -1.

the valae 0926 Rmis delomine d by fallouing euahon" ixXid -1+ Rm K 133 3 u the numbof ideal stages i he adual flux is s Hmes ke minimum Julluz using tle Gilliland metho d' t whore R-Rm R+ R-5Rm R 6.65 81tti

in the 3.s the numbr idal slages eilying seckons and e ipping Sechon feed Tau Location and the nunben ectiticabion and sipping stages be dekorminud by kisikbaide equalion can 0.206g Mok XLKXduk) 1-23 ** Ns NtNs Yos and : 3. 6 h fead locahion stage the feed tnay coil be 8h stage iom bottarn