Batch Distillation. Gavin Duffy School of Electrical Engineering Systems DIT Kevin St.

Transcription

1 Batch Distillatio Gavi Duffy School of Electrical Egieerig Systems DIT Kevi St.

2 Learig Outcomes After this lecture you should be able to. Describe batch distillatio Defie ad use the Rayleigh equatio Explai batch distillatio with costat product Explai batch distillatio with costat reflux

3 Itroductio I batch distillatio a fixed amout of charge is added to the still. Top product compositio varies with time. It depeds o bottom product compositio, umber of trays ad reflux ratio. There is o steady state compositios are chagig with time At start, top product is rich i MVC. After time, top product becomes less rich i MVC. A batch colum is like the top half of a cotiuous colum it has a rectifyig sectio oly

4 The Batch Colum A portio is retured to the colum as reflux Remaider is removed as Distillate or Top Product Reflux Ratio = Reflux/Distillate Same operatig lie y = R 1 x+ R+ 1 R+ 1 x d 1 V +1 y +1 Reflux L x Codeser D x d Distillate Top Product Bottoms charged to the still at the start

5 Advatages Separatio of small quatities of mixtures, i.e. capacity too small to justify cotiuous separatio Flexibility to hadle differet feedstocks to produce differet products More tha oe product may be obtaied light compoets are removed first. Differet purities of the same compoet ca also be obtaied. Upstream is batch operated ad compositio of feed varies with time Foulig is a serious cocer Seader, J.D. & Heley, E.J. (2006), Separatio Process Priciples, Wiley, p466

6 Mass Balace A overall mass balace for the batch distillatio is as follows: = + 0 D Where 0 = o. of moles i still at start = o. of moles i still at ed D = o. of moles i distillate I all cases, the o. of moles is of all compoets.

7 Rayleigh Equatio Imagie a batch is heated ad the vapour formed is removed immediately from the system without ay reflux a very simple distillatio. This was aalysed by Rayleigh who developed the followig equatio: l 0 = x dx y x Where o = iitial o. of moles i still = o. of moles left i still at time t x = liquid mole fractio of MVC at time t y = vapour mole fractio of MVC at time t x w = liquid mole fractio of MVC i feed (t=0) x f = vapour mole fractio of MVC at time t x w

8 Rayleigh Equatio cotd. Iput to the equatio Iitial umber of moles i the still (of both compoets 0 ) Iitial mole fractio of the MVC (x w ) i the still Fial mole fractio of MVC i the still (x) The results we get are: The total umber of moles left i still () The umber of moles of each compoet left i still The umber of moles of each compoet i distillate

9 To solve Rayleigh 1. We ca use a graphical method of itegratio 2. We ca also use the relatioship betwee y ad x from relative volatility as follows: y αx = 1+ x 1 ( α ) Add this to the equatio ad itegrate to give: l o = 1 l α 1 x o x 1 x + α l 1 x o

10 Activity Use Rayleigh A batch of crude petae cotais 15 mole percet butae ad 85 mole percet petae. It is added to a still ad heated at atmospheric pressure. How may moles are left i the still whe the remaiig charge i the still is 97% petae? Iitial coditios work o the basis of 1 mole, i.e. o = 1 mole. x o = 15% = 0.15 At the ed, x = 3% = What is, the umber of moles left i the still?

11 Alterative to Rayleigh A alterative to the Rayleigh equatio is described i McCabe Smith, 6 th Ed., pp 700 to 701. The followig equatio is derived: 1 B 0B = A 0A Where B = o. of moles of B (LVC) at ed 0B = o. of moles of B (LVC) at start A = o. of moles of A (MVC) at ed 0A = o. of moles of A (MVC) at start α AB = relative volatility See example 21.9 i McCabe Smith. It is similar to the previous problem. Istead of specifyig a fial mol fractio, a fial o. of moles of A are give. α AB

12 Reflux i Batch Distillatio I batch distillatio the top product compositio chages with time. What do we do if we wat a costat top product compositio? There are two optios for reflux i batch distillatio. 1. Icrease the reflux ratio with time to keep the product cocetratio costat. Low reflux iitially; high reflux towards the ed. 2. Use a fixed reflux ratio. Operate the still util the top cocetratio falls below a setpoit. Temperature ca be used to determie whe the top cocetratio has reached the setpoit.

13 Costat Product Variable Reflux To maitai a costat product compositio, the reflux ratio is icreased from a low value iitially to a large value at the ed. High reflux meas that a lot of heat is eeded. The distillatio should be stopped oce a chose reflux ratio is exceeded. The amout of distillate removed ad material left behid is give by the followig: x0 D= 0 xd x x B B x x x D 0 = 0 xd B Where x 0 = mol fractio of MVC i still at the start x B = mol fractio of MVC i still at the ed x D = mol fractio of MVC i distillate (costat!) 0 = o. of moles i still at start D = o. of moles of distillate removed Where x 0 = mol fractio of MVC i still at the start x B = mol fractio of MVC i still at the ed x D = mol fractio of MVC i distillate (costat!) 0 = o. of moles i still at start = o. of moles left i still

14 Desig approach Costat Product Choose the top product compositio, x D, ad the still compositio x 0. Get the x-y graph ready. If you do t have a umber of ideal stages (i.e. ew desig) the choose the iitial reflux ratio, draw the rectifyig operatig lie ad step from x D dow to meet x 0 to give o. of stages, N. If you have N already, the use it to determie R iitial. Choose a upper reflux ratio. Draw aother operatig lie usig the same x D. Step off N stages to give x B, the fial still compositio. The still compositio is chagig less ad less of the MVC. R is icreased with time but o. of stages is fixed so distillatio must be stopped whe bottoms compositio drops to a miimum value Figure out how to measure compositio real time!

15 Costat Product Ethaol Water VLE R 1 R 2 Ya (Eth vapour) x B,ed x B,start Xa (Eth liquid) Must stop at x B,ed or x D will start decreasig x D

16 Costat Reflux Variable Product The other optio is to leave reflux fixed ad work aroud a variable product compositio, x D. Simple solutio. x D varies from x D1 at the start to x D2 at the ed. A equatio very similar to Rayleigh s ca be used: l o = x 0 dx x x x D Where 0 = o. of moles i still at start = o. of moles i still at ed x 0 = mol fractio of MVC i still at the start x B = mol fractio of MVC i still at the ed B x D = mol fractio of MVC i distillate (a variable)

17 Desig approach Costat Reflux Choose the top product compositio, x D, the iitial still compositio x 0, ad the fial still compositio x B. Get the x-y graph ready. If you do t have a umber of ideal stages (i.e. ew desig) the choose the reflux ratio, draw the rectifyig operatig lie ad step from x D1 dow to meet x 0 to give o. of stages, N. If you have N already, the use it to determie R. Redraw the operatig lie with the same slope. The objective is to draw the operatig lie i the right place such that whe N stages are stepped off, the last step hits x B exactly. Where the operatig lie started gives x D2 (where it crosses the x=y lie).

18 Costat Reflux Ethaol Water VLE R 1 Ya (Eth vapour) R x D,ed x D,start x B,ed x B,start Xa (Eth liquid) Must stop at x B,ed or x D will start decreasig

19 Costat Reflux The average mole fractio of the MVC i the distillate is give by: x D, avg = Where 0 = o. of moles i still at start x 0 = mol fractio of MVC i still at start t = o. of moles i still at time t x Bt = mol fractio of MVC i still at time t 0 x 0 0 t t x B t

20 Costat Reflux Distillatio time ca be calculated if boilup rate is kow: time R+ 1 V = 0 ( ) Where R = reflux ratio V = boilup rate (if kmol/hr the time i hr) 0 = o. of moles i still at start t = o. of moles i still at time t t