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8. Multi-component distillation Prof. S. Scott, ChE 128, UC Santa Barbara How many distillation columns are required? One distillation column can be optimized to separate one pair of volatile components. D, xD,1 F zF,1 zF,2 We can specify two fractional recoveries: FR1 = (DxD,1)/(FzF,1) = 0.95 FR2 = (BxB,2)/(FzF,2) = 0.98 If the feed contains more than two volatile components, we cannot specify the recoveries of the additional components. However, we can add more distillation columns. B, xB,1 Distillation of multicomponent mixtures 1 Alternative? 1 1, 2 F II 2 2,3,4 F zF,1 zF,2 zF,3 zF,4 3 3,4 2 I 4 3 3, 4 III Separation of C components requires (C1) distillation columns. 4 Key components • Each column is designed to separate two components of adjacent relative volatility. These components are the keys. • All other components are non-keys. design for separation component a 1 1.5 Light non-key (LNK) 2 1.4 Light key (LK) assume exclusively in distillate specify recovery in distillate 3 1.3 Heavy key (HK) specify recovery in bottoms 4 1.2 Heavy non-key (HNK) 5 1.0 Heavy non-key (HNK) designation assume exclusively in bottoms • Distributions in the distillate and bottoms streams are specified for the two key components. • If we assume that the non-keys do not distribute, the overall mass balance is easily solved. Overall mass balance with non-distribution of non-keys If non-keys do not distribute, then LNK: Dxi,D = Fzi,F HNK: Bxi,B = Fzi,F Define fi, di, and bi as molar flow rates of component i in feed, distillate and bottoms, respectively. Ex.: 100 mol/h of an equimolar mixture of four components. Assume 80 % recovery of LK in D, and 80 % recovery of HK in B. component i zF,i fi = F zF,i di xD,i = di/D bi xB,i = bi/B LNK 0.25 25 25 0.50 0 0 LK 0.25 25 20 0.40 5 0.10 HK 0.25 25 5 0.10 20 0.40 HNK 0.25 25 0 0 25 0.50 Σdi = D = 50 ΣxD,i = 1 Σbi = B = 50 ΣxB,i = 1 Stage-by-stage calculations For a specified external reflux ratio R = L0/D and feed quality q, and assuming constant molal overflow (CMO), first perform the overall column balance to obtain D. Then, L = (L0/D) D V=L+D ⇒ L/V L̅ = L + q F V̅ = V + (1 – q) F ⇒ L̅/V̅ Algebraic solution, starting at top of column: 1. yi,1 = xi,D 2. xi,1 = yi,1 / Ki,1 3. yi,2 = (L/V) xi,1 + (1 – L/V) xD section) (total condensor) (VLE) (mass balance, rectifying Repeat steps (2) and (3) down to feed stage. Then, change mass balance equation to yi,k+1 = (L̅/V̅ ) xi,k + (1 – L̅/V̅ ) xB We just need to find Ki for each stage. Obtaining Ki(T) for each stage • Consider a ternary system with components A, B, C • Assume constant relative volatility • Choose B as the reference compound a AB K A y A / xA = = K B y B / xB xA,j = K B,j = y A,j K A,j y A,j a AB = + a BB = 1 y A,j a ABK B,j y B,j a BB + yC,j a CB xA,j + xB,j + xC,j = æ y i,j ö = åç a iB ÷ø i è a CB = y A,j a ABK B,j + KC KB y B,j a BBK B,j y A,j xA,j = + y C,j a CBK B,j a AB æ y i,j ö å çè a ÷ø iB i IF relative volatility is not constant, use geometric average, (a1aN)½. Or, find Tj from dewpoint calculation (starting with xD,i), then find Ki(Tj) using VLE. =1 Trial-and-error solution Perform stage-by-stage analysis, from top to bottom of column. xHK,N+1 ≥ xHK,B xLK,N+1 ≤ xLK,B must be satisfied simultaneously • if not true, choose different starting concentrations at the top of the column, and repeat • cannot assume xD,HNK = 0, but no way to guess • works best if there are no HNKs. If there is no LNK, can start at bottom of column and work to the top • not obvious where the optimum feed stage is • very tedious, convergence is difficult Approximate shortcut methods for multicomponent distillation • Suitable for preliminary designs • Three-part F-U-G method: 1. Use Fenske equation to estimate Nmin (at total reflux) 2. Use Underwood equations (2) to estimate Rmin (with N = ∞) 3. Use Gilliland correlation to estimate Nactual and Nfeed (can also use Kirkbride equation) Operating at total reflux • designate the keys A and B, and assume constant aAB • according to the definition of equilibrium: yi,1 1 xi,0 é yA ù é xA ù ê ú = a AB ê ú ë y B ûN+1 ë xB ûN+1 (even though there is no liquid stream with composition xi,N+1) • at total reflux, the operating line is y = x yi,j+1 = xi,j so yi,N+1 = xi,N yi,N+1 N xi,N N+1 é xA ù é xA ù ê ú = a AB ê ú ë xB ûN ë xB ûN+1 • move up to stage N: é yA ù é xA ù é xA ù 2 ê ú = a AB ê ú = a AB ê ú ë y B ûN ë xB ûN ë xB ûN+1 • continue to the top of the column Estimating Nmin • at the top of the column Fenske equation • solve for Nmin é xA ù é y A ù Nmin é xA ù ê ú = ê ú = a AB ê ú ë xB û0 ë y B û1 ë xB ûN+1 Nmin éæ x ö æ xA ö ù A ln êç ÷ ú ç ÷ êëè xB ø 0 è xB ø N+1 úû = lna AB • can also write in terms of fractional recoveries (FR) Nmin éæ FR ö ù æ 1- FRB ö A ln êç ú ÷ ç FR ÷ 1FR è êëè A ø distillate B ø bottoms ú û = ln a AB Other uses of the Fenske equation Once Nmin is known, • check non-distribution of non-keys, or estimate their distribution: FRC,distillate = N a CB min æ FRB ö Nmin + a CB ç 1- FR ÷ è ø B bottoms • estimate optimum feed stage at total reflux: NF,min éæ x ö æ zA ö ù A ln êç ÷ çz ÷ ú x êëè B ø 0 è B ø F úû = lna AB Pinch points in binary distillation Recall for binary distillation: Rmin Binary composition profile •xD xD • • • • pinch point • x zF• •zF • • • • • plateau of • constant • composition • • • xB Using Rmin generates a pinch point at the feed stage xB• • condensor • • feed • reboiler stage number • near the pinch point, composition changes little from stage to stage - passing streams are very close to equilibrium - no change in temperature between stages Behavior in ternary systems • in the presence of a third component whose composition does change through the feed stage, the pinch point moves Ternary composition profile LK-HK-HNK 1 stripping rectifying Consider LK • LK is the most volatile component in the system on every stage • there is almost no LK at the reboiler stage xi Consider HNK • HNK is the least volatile component in the system on every stage • almost all HNK is found at the reboiler stage • finite HNK at feed stage, drops off rapidly above 0 C F stage number R Behavior of HK Consider HK: Ternary composition profile LK-HK-HNK 1 pinch point • HK behavior is most complex (HK-LK), with HK the less volatile component xi • near reboiler, distillation is almost binary ~binary ~binary • above feed stage, distillation is almost binary (HNK-HK), but HK is the more volatile component • HNK goes through a maximum in the stripping section, creating a pinch point there 0 C F stage number R Pinch points in multicomponent distillation pinch F LK, HK, HNK pinch F LNK, LK, HK pinch F LNK, LK, HK, HNK pinch • LK-HK-HNK system • LNK-LK-HK system • LNK-LK-HK-HNK has a pinch point in has a pinch point in has pinch points in the stripping section the rectifying section both sections Graphical methods are not helpful in finding these pinch points. Second Underwood equation CMB in rectifying section at minimum reflux: at the pinch point: xi,j = xi,j +1 VLE: y i,j +1 = K i xi,j +1 relative volatility: a i = K i Kref Lmin Vmin y i,j +1 = y i,j +1 + Dxi,D a i Kref Vmin y i,j +1 = a i Dxi,D Lmin ai VminK ref Vmin y i,j +1 = Lmin xi,j + Dxi,D y i,j +1 = a i Kref xi,j +1 æ Lmin ö Vmin y i,j +1 ç 1= Dxi,D ÷ è a iVminK ref ø define: f = a i Dxi,D 2nd Underwood equation: Vmin = åVmin y i,j +1 = å ai - f i i Lmin VminK ref First Underwood equation in the stripping section: assume a i Bxi,B -Vmin = åVmin y i,k +1 = å i i ai - f f =f DVfeed é a i Dxi,D a i Bxi,B ù = Vmin -Vmin = å ê + ú a f a f i ê úû i ë i é a i Fzi,F ù = åê ú i ê ë a i - f úû 1st Underwood equation: DVfeed feed quality definition: DVfeed = F(1- q) searching forφ: with C components, the 1st Underwood equation has C roots, one between each adjacent pair of a-values, and one between 0 and the smallest a-value. We need the root aHK < φ < aLK • binary system: use quadratic equation to solve forφ • trial-and-error: use ½(aHK + aLK) as initial guess Estimating Rmin Onceφis known, obtain Vmin using the 2nd Underwood equation. a i Dxi,D Vmin = åVmin y i,j +1 = å ai - f i i To get xi,D values for non-keys: • assume non-distribution OR • assume distribution is the same at minimum reflux and total reflux i.e., use Fenske equation Lmin = Vmin – D Rmin = Lmin/D Gilliland correlation We have Nmin (Fenske) and Rmin (Underwood). R is often specified as a multiple of Rmin, e.g., R = 1.5 Rmin y= y = 1- e éæ 1+54.4 x ö æ x-1ö ù êç ÷ø çè 0.5 ÷ø ú 11+117.2x è x ë û x= Figure 7-3 Gilliland correlation as modified by Liddle (1968); Reprinted with permission from Chemical Engineering, 75(23), 137 (1968), copyright 1968, McGraw-Hill. From Separation Process Engineering, Third Edition by Phillip C. Wankat (ISBN: 0131382276) Copyright © 2012 Pearson Education, Inc. All rights reserved. Optimum feed stage • obtain Nmin and NF,min from Fenske eqn • obtain N from Gilliland correlation D Assume relative position of the feed stage is the same as it was at total reflux: NF,min Nmin = NF N D F F NF NF,min Remember! Nmin and N include partial reboiler (and partial condensor). Works well for symmetrical columns (feed stage close to middle). B B Kirkbride equation Better feed stage estimation for unsymmetrical columns: Nrectifying Nstripping éæ z ö æ x ö 2 æ ö ù B = êç HK ÷ ç LK ,B ÷ ç ÷ ú êè zLK ø è xHK ,D ø è D ø ú F ë û F D 0.206 D where Nrect and Nstrip do not include the feed stage: B F N = Nrect + Nstrip + 1 B Sandwich components At the beginning of this section, we specified that the key components should have adjacent relative volatilities. What if there is a non-key with intermediate volatility (i.e., between the two key components)? This component is called a “sandwich” component. component a 1 1.5 Light non-key (LNK) 2 1.4 Light key (LK) 3 1.3 Sandwich non-key 4 1.2 Heavy key (HK) 5 1.0 Heavy non-key (HNK) designation Sandwich components tend to concentrate in the middle of the column, and can cause flooding even when present at trace concentrations. If unavoidable, use side-streams to withdraw them. D F S B