Lecture 6: Separation process design, sizing & selection Chapters 4, 7 (Textbook) plus additional material* • Part-1: Introduction • Part-II: Two-phase (equilibrium) separations • Part-III: Separation of azeotropic mixtures • Part-IV: Other separation processes • See also ”Separation Process Principles” by Henley, Seader & Roper Chapters 7, 11 & 13 of book for distillation & absorption, and chapter 14 for membrane based separation Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 1 Downstream Separations: The main issues • When is separation necessary? • What are the mechansims of separation? • Which mechanisms and how many need to employed to achieve a desired separation? • How to configure/select/design the separation tasks? • How to verify if the desired separation can be achieved? This lecture will provide a quick overview of the above topics – more details can be found in courses on separation process principles Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 2 When is separation necessary? N2 N2, Argon Crude H2 separation … Reactor NH3 processes Argon H2, Methane Methane Crude Oil Oil products Separating Recovery & different oil purification of products from chemicals crude oil (products) Two Examples Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 3 Downstream Separation in Chemical Process Design - I Chemical processes without reaction Extraction Recovery Purification Finishing • Crude oil based refinery for gasoline, and other fuel products • Seeds and other biomass based edible oil extraction processes • Herbs, plants based extraction of specialty chemicals • ........... Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 4 Downstream Separation in Chemical Process Design - II EL, Pl, M recycle EL, Pl, M EL, Pl, M DEE Reactor Separation W EA Waste Water recycle Identify and design the sequence of separation tasks that need to be performed in order to recover and purify the products, recycle the reactants and purge the inerts that enter the process Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 5 Which mechanisms of separation? Phase 1 Phase 1 Phase 1 Feed Phase Feed Feed Barrier Creation (b) (a) MSA (c) Phase 2 (Solvent) Phase 2 Phase 2 Feed Phase 1 Phase 1 Feed Force field or gradient (d) Phase 2 (e) Phase 2 a) Phase Creation (eg, distillation); b) Agent-based separation (eg, extraction); c) Barrier (eg, pervaporation; filtration); d) Solid agent (eg, adsorption); e) Force field or Gradient (eg, ion-exchange; centrifuge) Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 6 How many mechanisms and which mechanisms? A, B, C, D, E, F • Depends on how many chemicals need to be separated (NS)? A, E, F / C, D, B A / B, C, D, E, F • Minimum number of separation tasks is NS-1 •For each separation task, A, F / F C / D, B find the appropriate separation mechanism • Analyze mixture to be separated (collect data or predict behaviour) • Identify relation between mechanism and system property (separation principles) • Generate and test alternatives (select, design, verify) Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 7 How to configure separation tasks? - Distillation Trains Separation of propane (A), i-butane (B), butane (C), i-pentane (D), pentane (E) A B Secondary Separation Efficiencies at P=1 atm. 0.3 B/C A B 0.25 C D A/B E 0.2 C Propane iButane iButane nButane 0.15 C/D SSE iPentane nButane D nPentane iPentane 0.1 0.05 0 D/E 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 E Liquid composition, mole fractions of the lightest compounds • Order the driving force diagrams in terms of fij| max = abs(yi – xi); • Configure the distillation train in terms of fij| max (easiest separation first – largest driving force equivalent to minimum energy); • Design each distillation column such that it uses the maximum driving force (note: only 4 separation tasks A-B; B-C; C-D; D-E) Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 8 How to configure separation tasks? – Extraction plus recovery Main Features: • Extraction (2-liquid phases); use of solvents • Recovery of solvents & solute (requires energy for VL separation) • Other forms of extraction could be membrane-based, ion- exchange, etc., that does not require the recovery step) 9 Part-II: Two-phase (equilibrium) separations: Design issues Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 10 2-phase VLE based separation: design degrees of freedom? Vapor, V T, P Feed, Z Liquid, L Typical 2-Phase VLE-Model sat sat V Equilibrium relation xi gi ji Pi = yi ji P i = 1,2,….c Condition Si xi = Si yi = Si zi = 1 Mass Balance Z zi = V yi + L xi i = 1,2,….c Degree of Freedom Analysis F = c - p +2 For p = 2, F = c. There are 2c+3 Equations and 3c + 5 sat variables (not counting gi , ji , Pi ). Therefore, c + 2 variables need to be specified. For example, specify Z, T, P and (c-1) z compositions. Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 11 Some Definitions Equilibrium Constant (VLE) Ki = yi/xi sat •Ideal System Ki = Pi (T) / P sat sat V •Non-ideal Liquid & Vapor Ki = gi ji Pi /( ji P) sat •Non-ideal Liquid, ideal vapor Ki = gi Pi /P L V •Any P & T Ki = ji / ji sat sat Relative volatility ij = yj xi / (xj yj) = Pi /Pj Rewriting in terms of yi yi = xi ij/[1+ xi(ij - 1)] Driving Force Dij = yi - xi Rewriting in terms of yi Dij = xi ij/[1+ xi(ij - 1)] - xi Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 12 Binary systems with different volatility 0.30 Distillation, 1 atm. Distillation, 5 atm. Pervaporation 0.25 Extractive distillation i 0.20 0.15 Driving Force,Driving FD 0.10 0.05 0.00 0.0 0.2 0.4 0.6 0.8 1.0 Liquid composition, xi DX-(PT) phase diagram (in XY-(PT) phase diagram (in terms of driving force) – shown terms of relative volatility) for different separation types) Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 13 VLE based separation: Model w.r.t. ij or Dij Vapor, V T, P Feed, Z Liquid, L Typical 2-Phase VLE-Model Equilibrium condition yi = xi ij/[xi (ij - 1) + 1] w.r.t. relative volatility or (eq.1) Dij = xi ij /[xi (ij - 1) + 1] - xi w.r.t. driving force Mass Balance Z zi = V yi + L xi i = 1,2,….c or yi = (R + 1) zi – R xi where R = L/V or (eq. 3) Dij = (R + 1) (zi –xi) Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 14 Graphical calculation technique for V-L based separation Dij = yi - xi = xi ij / (1 + xi (ij - 1)) - xi Dij = (R+1)(zi – xi) When Dij = 0, there is no separation and zi = xi When xi = 0 or L= 0; R= L/V= 0 & Dij = yi - xi = zi (everything is vaporized) When R = specified; xi = Dij /(R + 1) + zi (point of intersection of Eqs. 1 & 3) Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 15 Visualization of 2-phase separations: P-XY & T-XY diagrams P-xy diagram at constant T T-xy diagram at constant P Dew point L A C B V A V & L V & L C B Bubble point V L x1 and y1 x1 and y1 For a feed at C, L/V = CB/AC; at constant T, if you compress a vapor you get a liquid and vice versa. At constant P if you heat a liquid you get a vapor and vice versa. If you have a feed mixture in the 2-phase (V&L) region, it will split into a liquid and a vapor Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 16 VLE - Binary Systems: Azeotropic systems Ethanol - Water Methanol - Toluene y y1 1 x1 x1 XY-PT phase diagrams. At the azeotrope, xi= yi, therefore, Ki=1 and/or ij = 1. Separation beyond the azeotropic point is not possible Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 17 Visualization of 2-phase separations: azeotrope systems T P x , y 1 1 x1, y1 Methanol-Toluene T-xy & P-xy diagrams Generate these diagrams as a function of P to see if the location of the azeotrope can be moved Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 18 Types of VLE-phase separation models & solution approaches Problem Specification Isothermal flash: fast solution Isothermal flash (T, P, F, z) Specified: F, z1, z2, ..zc-1, T, P *Bubble-point T (P, x) Solve for = V/F with fixed Ki *Dew-point T (P, y) sat 1. Calculate Ki = Pi (T)/P *Bubble-point P (T, x) 2. f = Si zi(1-Ki)/[1+ (Ki-1)] = 0 *Dew-point P (T, y) 3. V = F ; L = F - V Adiabatic flash (Q=0, P, F, z) 4. xi = zi / [1+ (Ki-1)] Non-adiabatic flash (Q, P, F, z) 5. yi = ziKi / [1+ (Ki-1)] = xi Ki % Vaporization (V/F, P) 6. Q = hV V + hL L – hF F * Note: Degrees of freedom = c + 1 Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 19 Separation of Binary Mixtures by Distillation: review V1 Component Mass Balance Q Condenser C Overall 1 D F D B Recycle, L0 Distillate, D, x F z = D x + B x V2 L1 Rectifying section Feed VNF LNF-1 F D F, z yn+1= Ln/Vn+1 xn + D/Vn+1x NF y = [R/(R+1)] x + [1/(R+1)]xD VNF+1 LNF Stripping section B VNP LNP-1 ym+1= Lm/Vm+1 xm - B/Vm+1x Vapor boilup VNP+1 NP Reboiler B QR y = [(VB+1)/VB] x - 1/(VB+1)]x Equilibrium relation Bottom product, B, xB yi = xi ij / (1 + xi (ij - 1)) Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 20 •Equilibrium relation •yi = xi ij / (1 + xi (ij - 1)) •Component Mass Balance •Overall •F zF = D xD + B xB •Rectifying section D •yn+1= Ln/Vn+1 xn + D/Vn+1x •y = [R/(R+1)] x + [1/(R+1)]xD •Stripping section B •ym+1= Lm/Vm+1 xm - B/Vm+1x B •y = [(VB+1)/VB] x - 1/(VB+1)]x Course: Process Design Principles & Methods, L6 PSE for SPEED, Rafiqul Gani 21 Driving force based design - I •Given a mixture to be separated into two products in a distillation column with N trays.