#### Transcript Distillation

Distillation Prepared by Dr.Nagwa El Mansi Chemical Engineering Department Cairo University Fourth year References:1-Coluson and Richerdson, Chemical Engineering vol , vol II , vol III. 2- Geancoplis, Principles of Unit Operation. 3- Mc-Cabe and Smith, Unit operations for Chemical Engineering. 4- Traybal, Mass Transfer Operations. 5- Sherwood, Mass Transfer. 6-Perry’s , Chemical Engineering. 7- “Separation Process Principles”, 2nd ed, Seader et’al . 8- Site on Google search, Separation Processes. Distillation Distillation:Is a method of separating homogenous mixture based on differences in boiling points. Distillation is done by:a-partial vaporization ( liq → vap) b-partial condensation (vap → liq) c- changing in pressure ( gas → liq) Distillation has a number of applications:1- It is used to separate crude oil into more fractions for specific uses such as transport, power generation and heating. 2-Water is distilled to remove impurities, such as salt from seawater. 3- Air is distilled to separate its components oxygen, nitrogen, and argon for industrial use. Separation in distillation depend on:Relative volatility ( α ) :α = α AB PAo = o PB Where :- A= more volatile component (mvc). B = less volatile component (lvc). PAo = vapor pressure of more volatile component (mvc). PBo = vapor pressure of less volatile component (lvc). When α > 1 or α < 1 separation takes place. But when α = 1 there is no separation. Vapor-liquid Equilibrium:[ When the liquid phase behaves as an ideal solution all molecules have the same size; all intermolecular forces are equal; the properties of the mixture depend only on the properties of the pure components of the mixture]. Phase Diagram (Binary Mixture) In our analysis of phase diagram, we shall consider only Two component mixture, e.g. A (more volatile) and B (less volatile). There are 2 types of phase diagram: constant pressure or constant temperature. Constant Pressure Phase Diagram Figure(1) shows a constant pressure phase diagram for an ideal solution (one that obey ) Raoult's Law A typical equilibrium curve for a binary mixture on x-y diagram. It contains less information than the phase diagram (i.e. temperature is not included), but it is most commonly used. It is useful for graphical design in determining the number of theoretical stages required for a distillation column. Constant Temperature (Isothermal) Phase Diagram:Figure(2)shows the constant temperature phase diagram. The constant pressure phase diagram is more commonly used in the analysis of VLE, but the constant temperature Phase diagram is also useful in the analysis of solution behavior that deviates from Raoult's Law. [ From this Figure (constant temperature phase diagram) we see that the more volatile liquid will have a higher vapor pressure (i.e. pA at xA = 1.0). Note also the regions of vapor-only, liquid-only and vapor-liquid mixture. ] Effects of Increased Pressures:Although most distillations are carried out at atmospheric or near atmospheric pressure, it is not uncommon to distill at other pressures. High pressure distillation (typically 3 - 20 atm). At elevated pressures, the vapor phase deviates from ideal gas behavior, and modifications to the VLE data is required. After pressure P3, the critical pressure of the more volatile component is exceeded, and there is no longer a distinction between vapor and liquid. Distillation is no longer possible beyond this point. The majority of distillations are carried out at pressures below 70% of the critical pressure. Abnormal mixtures:1- Azeotropic Mixture:Very large deviations from ideality lead to a special class of mixtures known as azeotropes , azeotropic mixtures, or constant-boiling mixtures. Azeotrope is a special class of liquid mixture that boils at a constant temperature at a certain composition. At this condition, it behaves as if it was one component with one constant boiling point. A boiling liquid mixture at the azeotropic composition produces a vapor of exactly the same composition, and the Liquid does not change its composition as it evaporates. Two types of azeotropes are known:minimum boiling and Maximum boiling a- Minimum-boiling azeotropes: a- One of the best known minimum-boiling azeotrope is the ethanol-water system which at 1 atm occurs at 89.4 mole% ethanol and 78.2 oC. b- carbon-disulfide - acetone (61.0 mole% CS2, 39.25 oC, 1 atm) c- benzene - water (29.6 mole% H2O, 69.25 oC, 1 atm) b-Maximum-boiling azeotropes:a-Hydrochloric acid - water (11.1 mole% HCl, 110 oC, 1 atm) b-Acetone - chloroform (65.5 mole% chloroform, 64.5 oC, 1 atm) a-Minimum Azeotrope:Figure(5) show the constant pressure phase diagram plus equilibrium curve for a minimum-boiling azeotropic mixture of carbon disulfide (CS2) and acetone. At point L, the concentration in the vapor phase is the same as the concentration in the liquid phase ( y = x ), and a = 1.0. This concentration is known as the azeotropic composition (0.61 mole fraction CS2). At this point , the mixture boils at a constant temperature (39.25 oC under 1 atm) and without change in composition. On the equilibrium diagram, it can be seen that at this point, the equilibrium curve crossed the diagonal b-Maximum boiling Azeotrope:The Figure(6) shows the constant pressure phase diagram plus equilibrium curve for a maximum boiling azeotrope mixture of acetone and chloroform. The Azeotropic composition is 0.345 mole fraction acetone. Point L in the Figures is now a minimum on the constant temperature phase diagram, and a maximum (64.5 oC, under 1 atm) on the constant pressure phase diagram 2- Partial liquid miscbility:Some substances exhibit such large deviations from ideality that they do not dissolve completely in liquid state, e.g., Isobutanol-water 3-Complete immisciblity Mixtures:The mutual solubility of some Liquids is so small that they can be Considered substantially insoluble (Figure(8)) e.g., hydrocarbon and water. The vapor pressure of either component cannot be influenced by the other and each one exerts its vapor pressure at the prevailing temperature. When the sum of the separate vapor pressures equals the total pressure, the mixture boils at a temperature lower than the boiling point of each component:PA + PB = PT , or P◦A + P◦B = PT Calculation of Vapor/Liquid Equilibrium:1- By experimental methods:- (under certain P&T to get y = f(x)). 2-Published data. 3-Prediction of K values. For binary mixture:o i Pi = P . x i ------ (Raoult's law) PT . y i ------- ( Dalton's law) o i P yi = xi = Ki xi PT where Pi :- partial pressure of component i Pio :- vapor pressure of component i Pio Ki = K-value for component i PT x i , yi :- mole fractions of liquid and vapor respectively PT = Pi = o P i . xi For binary mixture ( A & B ):PT = PA + PB PT = PAo . x A + PBo . x B PT = P . x A + (1- x A ) P , where x A + x B = 1 o A o B PT = PAo . x A + PBo - PBo . x A PT - PBo xA = o PA - PBo PAo and y A = x A ----(equilibrium relation) PT Also, another form of equilibrium relation by using relative volatility( ) PAo y A /x A y A /x A y A (1-x A ) α= o = = = PB y B /x B (1-y A )/(1-x A ) x A (1-y A ) α x A - αx A y A = y A - y A x A α x A = (αx A - x A +1) y A α x A = [(α - 1) x A +1] y A yA = α xA ------ (equilibrium relation) [(α - 1) x A +1] For multi-component system:For ideal system:Pio yi = k i . x i = xi PT For non-ideal system:Pio y i = k i . i . x i = . i . xi PT where i is the activation coefficient = measure of non-ideality One Stage Distillation 1-Simple (differential) distillation (ASTM):We will consider a binary mixture of A (more volatile) and B (less volatile). The set-up is as shown in the Figure(9) . The system consist of a batch of liquid feed (F) (fixed quantity) with a composition( xF) inside a kettle (or still) fitted with heating element or steam jacket, and a condenser to condense the vapor produced. The condensed Vapor is known as the distillate (D) with a composition (yD) . The distillate is collected In a condensate receiver. The liquid remaining in the still is known as the residual (W) With a composition ( xW). At any time t, the amount of liquid in the still is L , with mole fraction of A in the liquid being (x). After a small differential time (t + dt) , a small amount of vapor dL is produced, and the composition of A in the vapor is (y) (mole fraction). The vapor is assumed to be in equilibrium with the residue liquid . The amount of liquid in the still is thus reduced from L to (L - dL), while the liquid composition changed from x to (x - dx). See the following Figure (10):- Performing a material balance on A : Initial amount in still Amount left in still Amount vaporized We have, L . x = ( L - dL) ( x - dx ) + dL .y L . x = L . x - dL.x - L.dx - dL.dx + y . dL Neglecting the term dx dL , the equation reduces to : L . dx = y . dL - x dL = ( y - x ) dL By separating variables then integrate: W F x dL W dx = L x F (y-x) F Ln = W xF or x F dL dx W L = x (y-x) W F dx x (y-x) --------(1) W There are two methods to solve equation (1):First method:- (by integration) αx y= ----- (equilibrium relation) (α - 1) x+1 F Ln = W F Ln = W xF dx x (y-x) -------- (1) W xF dx x αx W ( - x) (α - 1) x+1 x F (1 - x W ) 1- x W 1 = Ln + Ln α-1 x W (1 - x F ) 1- x F After performing the integration assume xw, (usually xw unknown) . If left hand side=right hand side your assumption is correct if not repeat your assumption. Second method :- (graphical integration) F Ln = W xF dx x (y-x) -------- (1) W F Ln = Area under the curve (note x W is unknown) W Steps for calculation the correct xw:1-plot x-y diagram. 2- assume xw. 3-costruct the previous table between x and 1/(y*- x). 4 - plot x vs 1/(y*- x) as shown in the last figure. 5- calculate the area under the curve = Ln F/W. 6- if area = Ln F/W , your assumption of xw is correct if not reassume xw. 7- finally calculate the correct xw. Equation(1) can be used to determine (xw) if (F, xF , D) are known. Also the average distillate composition can be Calculated ( yD ) by simple material balance:F = D + W --------(2) F xF = D yD + W xW -------(3) By solving the three equations all the quantities and compositions can be calculated. 2-Flash (equilibrium) distillation:A single-stage continuous operation where a liquid mixture is partially vaporized, then flows through a pressure reducing valve to the separator. In the separator, separation between the vapor and liquid takes place. How much of A is produced in the vapor (and remained in the liquid) depends on the condition of the Feed, (see Figure(13)). Vapor amount (V) with composition (y), and a liquid amount (L) with composition (x) are produced. The two streams leaving the flash drum (y and x )are in equilibrium with each others. Flash distillation is done by:1-Flash vaporization (liquid →heating→throttling valve → Separation ). 2-Flash condensation ( vapor →cooling →separation ). 3- Feed at high pressure → change to low pressure. Some times flash distillation is used as a method for changing conditions of a mixture. A single-stage flash operation can rarely produce the required purity or fractional recoveries. An obvious but inefficient approach is to apply a series of flash separators condensing part of the vapor and boiling part of the liquid products from successive stages. Its preferred to use systems with high relative volatility in flash distillation to obtain considerable separation . Operating line equation for flash distillation:For binary mixture(A &B) Overall Material Balance:F=L+V Component Material Balance:F. x F = L . x + V . y ( L + V ) xF = L . x + V . y L ( x F - x) = V ( y - x F ) (y - x F ) (y - x F ) L L = or = -------(4) (operating line equation) V ( x F - x) V ( x - xF ) This line lies between two points (x,y) and (x F , x F ) L and has a slope ( ) V Figure(14) shows it's the graphical representation. Multi-flash distillation:F=L+V Component Material Balance:F. x Fi = L . x i + V . yi ( L + V ) x Fi = L . x i + V . yi L ( x Fi - x i ) = V ( yi - x Fi ) (y - x Fi ) L = i -------- (5) V ( x Fi - x i ) Equilibrium relation is yi = k i . x i or can be written as x i = yi / k i substitute in equation (5) :(yi - x Fi ) L = V ( x Fi - yi / k i ) L ( x Fi - yi / k i ) = (yi - x Fi ) V L L x Fi (1+ ) = yi (1 + ) V V. k i L ) V --------- (6) L (1 + ) V. k i ( 1+ yi x Fi L ) V L ( ki + ) V ( 1+ or x i x Fi where :- x x i ------- (7) L ) V 1 and , L ( ki + ) V ( 1+ Fi y x i L ) V 1 L (1 + ) V. k i ( 1+ Fi Stepes for calculation of multi-component distillation:Pio 1- Calculate k i values for each component ( k i ) PT at certain temperature. 2- Assume L/V. 3- Calculate 4- If If y y i i y x i L ) V 1 L (1+ ) V ki ( 1+ Fi 1 repeat your assumption of L/V . 1 your assumption is correct then calculate x i 1 ( solution is by trial and error) Temperature calculations inside flash drum:- Figure(16) shows an enthalpy balance for continuous operation. The material and enthalpy balance are:- F=L+V F. x F = L . x + V. y F . h F + Q = L . h L + V. H V -----(8) where :h F = feed enthalpy Q = enthalpy gained by the heater h L = x [cp L,A (T-Tr )] + (1-x) [cp L,B (T-Tr )] + H mix -----(9) H V = y[cp v,A (T-Tr ) +λ A ] + (1-y) [cp v,B (T-Tr ) + λ B ] -----(10) T = operating temperature inside the drum Tr = reference temperature cp L = liquid heat capacity , energy/mol.degree cp V = vapor heat capacity , energy/mol.degree H mix heat of mixing (neglected for ideal system) λ A , λ B = latent heats of vaporization at Tr , energy/mol Steps for calculation T :1- assume T. 2- calculate Ki. 3- assume L/V. 4- calculate yi . 5- check ∑ yi =1. 6- calculate xi . 7- check the enthalpy balance, if (LHS = RHS )→ your assumption is correct, if not reassume T . Usually, T is assumed between the bubble and the dew Points of the mixture. Calculations of dew point and bubble point:A-Bubble point:- yi = k i . x i y i =1= (for the first bubble x i = x F ) k i . x Fi given, x Fi , P assume Tbubble and calculate k i o i o A o B P P P x Fi = x FA + x FB 1 PT PT PT Pio if x Fi = 1 your assumption is correct PT if not repeat your assumption B- Dew point :yi xi = ki xi = 1 = (for the first liquid droplet x i = x Fi ) yi ki given, x Fi , P assume Tdew and calculate k i yA yB xi = 1 = + o PA PBo ( ) ( ) PT PT if yi ki = 1 your assumption is correct if not repeat your assumption Trials to calculate bubble point Trials to calculate dew point 3- Steam distillation:- It is a method for distilling organic compounds which are heat sensitive materials (contaminated by traces of non volatile impurities).This process involves using bubbling steam (which is completely insoluble with the organic compound) through a heated mixture of the raw material to give maximum contact between steam and the compound. By Raoult's law, some of the target compound will vaporize (in accordance with its partial pressure)leaving Impurities and condensed steam in the still. The vapor mixture (carried by steam) is cooled and condensed, usually yielding a layer of organic compound(c) and a layer of water(w). The vapor pressure of each component (c&w) cannot be influenced by the presence of the other and each one exerts its vapor pressure at the Prevailing temperature. When the sum of the separate vapor pressures equal the total pressure, the mixture boils at a lower temperature (lower than boiling points of c & w). The use of steam reduces the partial pressure of feed components and thus permits there vaporization at temperature below there normal boiling point. o W PW = P , yW PW = PT PW = partial pressure of water PWo = vapor pressure of water Also for the component (c):o C PC = P PC , yC = PT PC = partial pressure of water PCo = vapor pressure of water PT = PW + PC (at boiling) At boiling:PC moles of materials n C = = PW moles of water nW PC PC y .P moles of (c) C T PW PT PC y W . PT moles of (w) PC MC moles of (c) × M C mass of (c) × = PT - PC MW moles of (w) × M W mass of (w) mass of (w) = mass of steam used as a carrier. Total amount of steam used = Steam used for heating m cp (Tb mix -Tinital ) + Steam used as a carrier + Steam used for vaporization m Advantage of steam distillation:1-Distillation takes place under low temperature. 2-Prevent decomposition of thermally sensitive materials. 3- Cheap and economic. 4-Avalability of heating medium. Fractional Distillation or Differential Counter-Current Vapor/ Liquid Operations (Multi-stage Distillation):- To obtain high purity use multi-stage arrangement These stages can be combined in one column as shown in Figure(23) Steam Distillation:Steam distillation refers to a process in which live steam is in direct contact with the distilling system. The basis of steam distillation rest on the fact That water forms immiscible mixtures with most organic substances(most commonly employed in petroleum refining operation) and these mixtures will boil at a temperature below that of either water or the other materials. Vaporization takes place at a temp. below its normal boiling point. Steam is widely used because of its energy level, cheapness, and availability. Total and partial condenser arrangements:1-Total condenser:In total condenser all saturated vapors at the top of the distillation column condensed to saturated liquid y1 x0 xD QC = V1 λ mix QC = ( L0 + D) λ mix QC L0 = ( +1) λ mix D D = ( R +1 ) λ mix Where:Q C = condenser duty V1 = vapor flow rate at the top plate in the distillation column y1 = vapor composition at the top plate in the distillation column D = distillate flow rate x D = distillate composition leaving condenser L 0 = liquid flow rate returning back to the top plate x 0 = liquid composition returning back to the top plate L0 R= D = reflux ratio 2- Partial Condenser:Partial condenser acts as one plate with yD in equilibrium with top plate condensate x0 yD = KD x0 and yD = xD Types of Reflux:1- Cold Reflux:- Liquid stream returning back to the Column is less than bubble point of the mixture. 2- Hot Reflux:-Liquid stream returning back to the Column is more than bubble point of the mixture. 3- Recirculating Reflux or pumping around:A side stream is withdrawn from the column to be cooled and return back to the column. Reboilers:1- Internal Reboiler:The tubular heat exchanger built into the bottom of the tower (Figure(33)) provides large surface area (no shell), but Cleaning requires a shut-down of the distillation operation. Liquid level is controlling for small capacity towers and for materials which aren’t corrosive and scaling. 2-External Reboilers:A-Thermosiphon Reboiler:The vertical thermosiphon reboiler as shown in Figure(34) with the heating medium outside the tubes, can be operated so as to vaporize all the liquid entering it to produce a vapor of the same composition as the residue product, in which no enrichment is provided. B- Kettle Reboiler:The kettle reboiler shown in Figure (35), with heating medium inside the tubes, provides a vapor to the tower essentially in equilibrium with the residue product and Thus it behaves like a theoretical stage. It’s large in size , better to control, long residence time , not used for thermally sensitive materials Pipe still heater:Sometimes we use a Furnace or a pipe still heater instead of the reboiler. Advantages:1-Used when no steam is available(heating media is hot oil or hot gases). 2- high temperatures is needed. Disadvantages:1-Expensive. 2- Not suitable for sensitive materials. Distillation Calculations :Overall Material balance (O.M):F=D+W Component material balance(CMB):F xF = D xD + W xW Overall Heat Balance (OHB):F h F + QC = D h D + W h W + Q r where:- QC = condenser duty QC = V λ = V ( y λ + (1-y ) λ ) 1 mix 1 1 1 1 2 V =L +D 1 0 QC =(L + D) λ =m cp (T -T ) 0 mix water water water out water in and Qr = reboiler duty = m λ steam steam Methods for calculation number of stages in distillation (1)Sorel plate to plate calculations:For upper section:OMB Vn+1 = L n + D ----(11) CMB Vn+1 y n+1 = L n x n + D x D --(12) HB Vn+1 H n+1 = L n h n + D h D QC ---(13) Where:H n+1 = f (y n+1 ) and h n = f (x n ) For bottom section:OMB L'm+1 = V'm + W ---(14) CMB L'm+1 x'm+1 = V'm y'm + W x w ---(15) HB L'm+1 h'm+1 + Q r = V'm H'm + W h W ----(16) Where:h'm+1 = f (x'm+1 ) H'm = f (y'm ) Top or rectifying section ( total condenser conditions):First plate ( n = 1 ) V2 = L1 + D ---(OMB) V2 y 2 = L1 x1 + D x D ---(CMB) V2 H 2 = L1 h1 + D h D Q C ---(HB) where:- y1 k1x1 --- (Eq.relation) Calculation steps :1-Assume V2 then get L1 from (OMB) equation. 2- Calculate y2 after calculating x1 ( from Eq.relation → y1 = k1 x1 ) 3- If LHS = RHS (in HB equation) → your assumption is correct if not repeat your assumption. Second plate ( n = 2 ) V3 = L 2 + D ---(OMB) V3 y3 = L 2 x 2 + D x D ---(CMB) V3 H 3 = L 2 h 2 + D h D QC ---(HB) where:- y 2 k 2 x 2 --- (Eq.relation) same previous steps for calculation V3 and L 2 then calculation of y3 , then checking your assumption from heat balance equation. Repeat your calculations till you reach x n = x f or x n x f then calculate the number of stages in the upper section Bottom section (Using partial vaporizer = Reboiler ):Reboiler acts as one theoretical stage outside the distillation column. The two streams leaving it are in equilibrium with each other:yr = kr xw For Reboiler (m = 0 ) L'1 = V'r + W ---(OMB) L'1 x'1 = V'r y'r + W x w ---(CMB) L'1 h'1 + Q r = V'r H'r + W h W ---(HB) Steps for calculations:1-Assume Vr then get L’1 from(OMB) equation. 2- Calculate x’1 after calculating yr ( from Eq.relation → yr = kr xw ) 3-Substituting by Vr , H’r , L’1 , h’1 in the (HB) equation. (after calculating the correct temperature inside the reboiler) 4- If LHS = RHS → your assumption is correct if not repeat your assumption. First plate 1' from the bottom (m = 1 ) L'2 = V'1 + W ---(OMB) L'2 x'2 = V'1 y'1 + W x w ---(CMB) L'2 h'2 + Q r = V'1 H'1 + W h W ---(HB) Where : y'2 = k'2 x 2 Same steps were followed till the vapor composition equal or more than feed composition (y m x F ). All the above depends on the reflux ratio, in which QC and Qr were calculated. 2- Ponchon Savarit method (Graphical representation of Sorel method ) :- F = D + W -----(OMB on the distillation column) Fx F = D x D + W x W -----(CMB) F h F + Q r = D h D + W h W + QC ----(HB) F h F = D h D + W h W + QC - Qr QC Qr F h F = D (h D + ) + W (h W ) D W QC Qr put Q' = ( h D + ) & Q" = ( h W ) D W F h F = D Q' + W Q" ---------(17) Substitute in (CMB eq.) with (OMB eq.): (D + W) x F = D x D + W x W D ( x F - x D ) = W (x W - x F ) (x W - x F ) D = -----(18) W ( xF - xD ) Substitute in eq. (17) with (OMB) eq.:(D + W) h F = D Q' + W Q" D ( h F - Q') = W ( Q" - h F ) ( Q" - h F ) D = -----(19) W ( h F - Q' ) (x W - x F ) ( Q" - h F ) D = ----(20) W ( xF - xD ) ( h F - Q' ) Equation 20 is an operating line equation for the hole column between three points (x D , Q' ) , ( x F , h F ) , ( x W ,Q"). For top section:Vn+1 = L n + D ---(11) Vn+1 y n+1 = L n x n + D x D ---(12) Vn+1 H n+1 = L n h n + D h D QC ---(13) Vn+1 Vn+1 QC H n+1 = L n h n + D ( h D ) D H n+1 = L n h n + D Q' -----(21) By substituting eq.(11) in eq.(12) (L n + D) y n+1 = L n x n + D x D D ( y n+1 x D ) = L n ( x n y n+1 ) ( x n y n+1 ) D = -----(22) Ln ( y n+1 x D ) By substituting eq.(11) in eq.(21) ( L n + D) H n+1 L n h n + D Q' D ( H n+1 Q' ) = L n ( h n H n+1 ) ( h n H n+1 ) D = ------(23) Ln (H n+1 Q' ) ( x n y n+1 ) ( h n H n+1 ) D = = -----(24) Ln ( y n+1 x D ) (H n+1 Q' ) Equation 24 is an operating line equation for the top section between three points (x n , h n ) , (y n+1 , H n+1 ) , (x D , Q' ). For bottom section:L'm+1 = V'm + W ---(14) L'm+1 x'm+1 = V'm y'm + W x w ---(15) L'm+1 h'm+1 + Q r = V'm H'm + W h W ----(16) Qr L'm+1 h'm+1 = V'm H'm + W ( h W ) W L'm+1 h'm+1 = V'm H'm + W Q" ----(25) By substituting eq.(14) in eq.(15) (V'm + W ) x'm+1 = V'm y'm + W x w W ( x'm+1 x w ) = V'm ( y'm x'm+1 ) W ( y'm x'm+1 ) -----(26) V'm ( x'm+1 x w ) By substituting eq.(14) in eq.(25) ( V'm + W) h'm+1 = V'm H'm + W Q" W ( h'm+1 Q") = V'm ( H'm h'm+1 ) W ( H'm h'm+1 ) -----(27) V'm ( h'm+1 Q'') ( H'm h'm+1 ) W ( y'm x'm+1 ) ----(28) V'm ( x'm+1 x w ) ( h'm+1 Q'') Equation (28) is an operating line eq. for the bottom between three points (y'm , H'm ) , ( x'm+1 , h'm+1 ), ( x w , Q'') Enthalpy-composition diagram:- (Ponchon-Savarit Method ) 1-Takes into account latent heats, heats of mixing & sensible heats. 2-No assumption of molal flow rates. 3-Graphical procedure combining enthalpy & material balances. 4- Provides information on condenser & reboiler duties. The diagram at a given constant pressure is based on reference states of liquid and temperature such as 273K. Use enthalpy concentration data:The data that require for enthalpy concentration diagram are: 1-Heat capacity of the liquid, Cp 2-Boiling temperature, Tb 3-Latent heats of vaporization, λ 4-Heat of solution, ∆Hmixing The saturated liquid line in enthalpy h kJ/kg can be obtained from the following :- hL = x [cpL,A (T-Tr )] + (1-x) [cpL,B (T-Tr )] + Hmix -----(9) The saturated vapor line can be obtained from the following equation below:HV = y[cpv,A (T-Tr ) +λA ] + (1-y) [cpv,B (T-Tr ) + λB ] -----(10) x & y = mol or mass fractions of liquid & vapor. T= boiling point of temperature Tr = reference temperature C p,L & C p,v = liquid & vapor heat capacity ∆Hmix = heat of solution at Tr ( neglected for ideal system). Again the difference in enthalpy between the streams passing each other is constant, with the enthalpy of this stream being Q’ = hD + Qc / D , where Q c is the condenser duty in kJ/h. Also, Q C V1 λ = ( L 0 + D ) L0 + 1 ) = D ( R + 1) D Q C = D ( R + 1) λ = D( QC = ( R + 1 ) λ ------(29) D QC If R is assumed or given, calculate Q C then Q' = ( h D + ) D which is the difference point ( or polar ) of the top section. This difference Q’ is a common operating point for all values of yn+1 and xn in enriching section of the distillation tower that having an enthalpy [Q’ = hD + Qc / D ] and a composition of xD . The intersection of y1 is shown in the diagram having the composition y1= xD for a total condenser. The liquid x1 is in equilibrium with y1 and is located by drawing a tie line through y1 intersecting the saturated liquid line at x1. Next, a line is drawn as x1Q’ , which is intersects saturated vapor line at y2. following the operating line equation of the top Section (eq.(24)). Also , the difference point (or polar) in the stripping section is called Q”, having an enthalpy Q”= ( hw -Qr/W) and a composition xw . This point Q” is an operating point just as Q’ for the enriching section. The point Q” is plotted below. Starting at point xw a tie line is drawn through this point intersecting the saturated vapor line at yr, is in equilibrium with xw . Next, the line yr Q” is drawn, which intersects the saturated liquid line at x’1. Hence, x’1 must be on the line between yr and Q”. Next, the tie line x’1y’1 is drawn. This process is continued in stepping off the theoretical plates. Steps for calculation the theoretical number of stages:1-Draw the enthalpy-concentration plot & the x-y equilibrium plot on the same graph. 2-Determine the points xF , xD , xW. 3-Use the following equation to calculate Q’= hD + Qc / D. 4-Locate point Q’ at (xD , Q’) 5-Locate point y1 at (xD , H1) 6-Locate point x0 at (xD , hD) → for total condenser. 7- use the following equation to calculate Q” = hw - Qr/W. 8-Draw the locus of Q” which is a vertical line from xw, { point Q” at (xw , Q”)}. 9-Draw the line from Q’ to intersect hF (enthalpy of saturated liquid at xF and the vertical line of xW). The point of intersection shows Q” . 10-From point y1, draw a line down to 45o line. Then, draw a horizontal tie line to cut the x-y equilibrium curve. Draw a line up to the liquid enthalpy curve. This point is x1. 11-Plot an operating line from x1 to Q’, intersecting the vapor enthalpy curve at y2. From y2, draw a vertical line down to the equilibrium curve. Draw yet another line, go up to obtain x2; repeat the process until you exceed xF 12- Also, from point xw draw a tie line connecting Between xw & yr. 11-Draw an operating line from Q”yr to intersect the liquid enthalpy curve at x’1 . 12-To obtain y’2 repeat the process until you exceed xF 13-The tie lines drawn between x1 to y1, x2 to y2, x3 to y3 and so on = number of stages. Minimum reflux ratio:Determine Q’min from the graph. -From point, (xF ,hF), draw a line down to equilibrium curve. Then, draw a horizontal line to 45o line. Next, draw a line up to saturated vapor line. Let say the point is y. -Then, draw a line from point (xF ,hF) to point y till you reach the vertical line of xD. - Q’min is the point intersect between the line from point (xF ,hF) to point y and vertical line of xD . 3-Lewis plate to plate calculations:Lewis assumes that:1- If the amount of molecules which evaporate and which condensate are the same. This means that within each section the liquid and the vapor flow rates remain constant in the whole section. This can be translated into the following equations: For Top or Rectifying Section:L1 = L2 = L3 = ......... = Ln = constant V1 = V2 = V3 = ......... = Vn = constant For Bottom or Stripping Section:L’1 = L’2 = L’3 = ......... = L’m+1 = constant V’1 = V’2 = V’3 = ......... = V’m = constant 2- The heat of vaporization (λ) of the two components of the feeding mixture must be the same ( λmvc = λLvc ) . 3- There is no losses from the column. 4-No heat of mixing ∆ H mix = 0 (Ideal system). { In this case the saturated liquid and vapor curves becomes two straight parallel lines in the enthalpy composition diagram}. Based on the assumptions made and the identified data and unknows, we can rewrite the mass balances and equilibrium equations for both the sections as Follows:- For upper section:Vn+1 = Ln + D ----(OMB eq.) V = L + D -----(OMB eq. in Lewis method) Vn+1 y n+1 = L n x n + D x D ----(CMB eq.) V y n+1 = L x n + D x D ----(CMB eq. in Lewis method) y n+1 L D = xn + x D -----( 30) V V Also, L D V = L + D y n+1 = xn + xD L+ D L+ D L/D D/D y n+1 = xn + xD L/D + D/D L/D + D/D R 1 y n+1 = xn + x D -----( 31) R+1 R+1 Equation (30) or (31) are operating line equations having slop L/V or R/R+1 and there intersection with y-axis are D/V or xD/ R+1 . For bottom section:L'm+1 = V'm + W ---(OMB eq.) L' = V' + W -----(OMB eq. in Lewis method) CMB L'm+1 x'm+1 = V'm y'm + W x w ---(CMB eq.) L' x'm+1 = V' y'm + W x w ---(CMB eq.in Lewis method ) L' W y'm = x'm+1 x w -----(32) V' V' Equation (32) is an operating line equation has a slope L’/V’ and intersection with y-axis is - Wxw/V’ . Plate to plate calculations using total condenser:Top section:- First plate ( n = 1 ) V = L + D ---(OMB) V y 2 = L x1 + D x D ---(CMB) L D y2 = x1 + xD V V y1 k1x1 --- (Eq.re. at certain T ) Calculation steps:1-For total condenser y1 = xD = x0 , calculate x1 from the Eq. re. → y1 = K1 x1 at certain assumed temperature for the top plate. 2-Substitute in equation (30) or (31) then calculate y2 (notice that the trails only on temperature not on the amount as in Sorel method). 3-After calculating the correct y2 , calculate y3 from:- n=2 L D y3 = x2 + xD V V & y3 = k 3 x 3 4- Continue your calculation and so on till you reach xn ≤ xF . Bottom section:Reboiler ( plate out side the column , m = 0 ) y'r = k r x w L' = V' + W L' W y'r = x'1 xw V' V' calculation steps:1- By assuming the temperature inside the reboiler ,the composition yr is calculated. 2- Calculate x’1 from eq.(32) , for m = 0 First plate inside the column , m =1 y'1 = k1 x1 L' W y'1 = x'2 xw V' V' 3- calculate y’1 from the equilibrium relation, then calculate x’2 4- Continue your calculation and so on till you reach y’m ≥ xF . 4- Mc-Cabe Thiele Method (graphical representation of Lewis method):To obtain the number of theoretical trays using the McCabe-Thiele method, we shall divide the column into 3 sections: rectifying, feed and stripping sections. As these sections are then represented on the (x-y) equilibrium curve for the binary mixture in question and re-combined to make a complete design, as shown in Figure (49). 1- Top section:V = L + D ---(OMB) V y n+1 = L x n + D x D ---(CMB) ( L + D ) y n+1 = L x n + D x D The point of intersection between this operating line equation and 45 line (x = y) is:( L + D ) y = L y + D xD L y + D y = L y + D xD D y D xD y = x = x D the point of intersection between top operating line and 45 line is (x D , x D ) The top operating line is drawn from its upper end at y1 = x0 = xD (on the 45o line), with a slope L/V = R/R+1 and an intercept =(xD /R+1) with y-axis constant for given R and purity of distillate xD . 2- Bottom section:V' = L' - W ---(OMB) V' y'm = L' x'm1 + W x w ---(CMB) ( L' W ) y'm = L' x'm+1 + W x w The point of intersection between this operating line equation and 45 line (x = y) is:( L' W ) y' = L' y' W x w L' y' W y' = L' y' W x w W y' = W x w y' = x' = x w the point of intersection between top operating line and 45 line is (x w ,x w ) The bottom operating line is drawn from its lower end at (xw , xw ) (on the 45o line) with a slope L’/V’ and an intercept = - W xw/ V’ with y-axis . 3-Feed stage consideration:By solving the two operating line equations (top and Bottom) at the point of intersection (xi , yi) ( V yi = L x i + D x D ) (Top) + (V' yi = L' x i W x w ) (Bottom) (V' V) y i = (L' L) x i (D x D +Wx W ) (V' V) yi = (L' L) x i F x F (V' V) (L' L) yi = x i x F ----(33) F F q- line definition:Material balance on feed entrance:F + L + V' = V + L' F + ( V' V) = ( L' L) ( V' V) ( L' L) 1+ = =q F F ( L' L) ( V' V) q & q 1 = F F By substitution in equation (33) with these values:xF q yi = xi -----(34) q 1 q 1 q It is the q line equation , it's slope q 1 Heat balance on feed entrance:F h F + L h L + V' H'V = V H V + L' h'L considering h L h'L & H V H'V F h F + ( V' - V) H V = ( L'- L) h L ( V' - V) ( L'- L) hF + HV = hL F F ( L'- L) ( V' - V) q & q -1 = F F h F + ( q - 1) H V = q h L h F + q HV - HV = q h L HV - h F q= ----(35) HV - h L amount of heat necessary to vaporize one mole of the feed q = latent heat of vaporization For different feed conditions, q has the following numerical limits :L V 1 Saturated liquid feed:F hF = hL HV h F q= HV h L HV h L q= = =1 HV h L q line:q 1 q 1 1 1 q-line is a vertical line bubble point liquid feed 2Saturated vapor feed:- V F hF = hV HV h V q= HV h L L HV h V q= =0 HV h L q line:q 0 0 q 1 0 1 q-line is a horizontal line dew point vapour feed 3 Partial vaporized feed: e.g 0<q< 1 L V F q = 0.7 q line:q 0.7 0.7 = ve q 1 0.7 1 0.3 q-line has a -ve slope partially vaporized feed 4Subcooled feed: hF < hV L V F HV h V q= HV h L m cp ΔT + λ = >1 λ q line:- ( assume q = 1.5 ) q 1.5 +ve q 1 1.5 1 +ve line in the first quarter subcooled liquid feed 5 Superheated feed:- L V h F > HV HV h F ve q= = = ve HV h L +ve q line:q ve ve q 1 ve 1 ve q-line in the third quarter superheated vapour feed Number of theoretical stages using Mc-Cabe Thiele Method The number of theoretical stages required for a given separation is then the number of triangles that can be drawn between these operating lines and the equilibrium curve. The last triangle on the diagram represents the reboiler. Some cases in Mc-Cabe Thiele method:(1) Calculation of no. of stages in complex feed( more than one feed):For first feed:L' L q1 get L' assume sat.liq. F1 L V = L + D get R = D D is calculated from (OMB) calculate L & V . q1 known , F1 known caculate L'. V' V q1 1 get V' F1 For second feed:L" L q2 get L" F2 assume vap.liq. V" V q2 1 get V" F2 Draw operating line for top Section, the draw operating Line for bottom section, then connect between the two Lines as shown in the opposite figure. If the two feeds were mixed together the introduced to The column, how many stages would be required to perform this separation. By mixing:F = F1 + F2 F x F = F1 x F1 + F2 x F2 F q mix = F1 q1 + F2 q 2 No. of stages in second case > No. of stages in the first case. It's prefered to introduce each feed at it's corresponding composition (2) Calculation of no. of stages in case of enriching section:- (3) Calculation of no. of stages in case of stripping section:- (4) Calculation of no. of stages in case of open Bottom section :steam:L' + S = V' + W or V' - L' = S -W---(OMB) L' x'm+1 + S ys = V' y'm+1 + W x W ----(CMB) ys 0 L' x'm+1 = V' y'm+1 + W x W ---(op.line eq.) By solving the operating line equation with y = x line : L' x' = V' x'+ W x W ( L' - V' ) = W x W & - ( S - W ) x = W xW W xW y=x= W-S V' - L' = S - W Open steam presentation Open steam presentation in Mc-Cabe Thiele method in Ponchon Savarit method (5) Calculation of no. of stages in case of side steam:In some applications, a product other than top or Bottom side products are also withdrawn from the column, which is usually a saturated liquid product . Assume the flow rate of intermediate product is S with composition xs. The column is divided into three sections:- top , middle and bottom. Material balance on the top Section:V = L + S + D ---(OMB) (V-L) (S+D) V y n+1 = L x n + S x S + D x D (CMB) The point of intersection between this operating line and 45 line V y = L y + S xS + D x D ( V - L ) y = S xS + D x D S xS + D x D S xS + D x D y=x= (V-L) (S+D) Material balance on the bottom section:L' = V' + S + W ---(OMB) ( L' - V' ) = ( S + W ) L' x'm+1 = V' y'm + S x S + W x w (CMB) The point of intersection between this operating line and 45 line L' y = V' y + S x S + W x w ( L' - V' ) y = S x S + W x w S xS + W x w S xS + W x w y=x= = ( L' - V' ) (S+W) Packed tower for small capacities:- Comparison between Mc-Cabe Thiele method and Ponchon-Savarit method Mc-Cabe Thiele method Ponchon-Savarit method 1- Simple 1- Complex 2-Require (x-y )diagram only 2-Require enthalpy comp. Diagram & (x-y )diagram 3-Less data Much more data 4-Equi-molar flow rates 4-Require more data on each section( M.B, H.B , Eq.re.,…) Operating conditions:(1) Reflux ratio , R :As R decreases, separation becomes difficult, driving force(D.F) decreases, number of stages increase the operating line for the rectification section moves towards the equilibrium line. The ‘pinch’ between operating and equilibrium lines becomes more pronounced and more and more trays are required. In this case R = Rmin and the number of stages reaches infinite ( N=∞)→ Pinch point, and D.F = zero. As R increases, separation is improved, D.F increases, the two operating lines goes down till they coincide With 45◦ line and L/V =1. At this condition R= ∞ = total reflux and number of stages is minimum. Relation between number of stages and reflux ratio: The minimum reflux ratio and the infinite reflux ratio place a constraint on the range of separation operation. Any reflux ratio between Rmin and Total R will produce the desired separation, with the corresponding number of theoretical stages varying from infinity at Rmin to the minimum number (Nmin at Total R). Optimum reflux ratio:A trade off between operating cost and equipment cost is needed. As R increases →less stages , less equipment cost is needed. As R decreases →more stages , more equipment cost is needed, more boiling and condensation. Equipment and operating costs Combine to give total cost( Ropt = 1.1 to 1.5 Rmin) (2) Choice of pressure:Some times high pressure (especially for gases) is used in the distillation column, the equilibrium T/C diagram goes up , temperature increases and the equilibrium curve approaches diagonal (as shown in Figure (4)). This results in decreasing the relative volatility (α) , and The D.F decreases, thus distillation becomes difficult. Number of stages increase(N), and the operating cost Increase ,and (R) increases. Vapor load (V) decreases, as the pressure increases which leads to decrease in column diameter , but column thickness will increase. The high pressure column is tall and thin. Explosion problems must be taken in to consideration. As pressure decreases(e.g vacuum distillation) , α increases, separation is more easier, number of stages decreases, vapor load increases, which leads to big column diameter. The temperature decreases ( which is suitable for highly sensitive materials), but column thickness will increase. The low pressure column is short and fat . Collapse problems must be taken into consideration. The highest temperature and pressure usually at the Bottom of the distillation column. Note:- T bottom < T thermal decomposition P bottom < P critical The atmospheric distillation column operates at 1atm and it is the most economical one. (4) Feed temperature:When hF decreases , Qr increases and Qc decreases. And when hF increases Qr decreases and Qc increases, and R increases. Also the q-line moves from up to down as feed temperature is increased. (5)Feed location:The state of the feed mixture and feed composition affects the operating lines and hence the number of stages required for separation (feed is introduced at the point of intersection of the two operating lines). During operation, if the deviations from design specifications are excessive, then the column may no longer be able handle the separation task. To overcome the problems associated with the feed, some column are designed to have multiple feed points ( two or three nozzles)when the feed is expected to containing varying amounts of components. Batch Distillation:When the quantity of the solution to be separated is small or products at different purities are required, then the rectification operation is carried out batch wise. A plate type batch rectification column is shown in Figure( 56) . Column operates first under total reflux until the top product reaches the desired composition,from there on top product is withdrawn. During the operation, no bottom Product is withdrawn. The purity of top product depends on the number of the plates available in the column and the composition of the liquid in the boiler. The liquid in the boiler becomes poorer in the mvc as the distillation proceeds; as no feed is given to the column and as the vapor formed is always richer in the mvc. As a result of this, top product becomes also poorer in the mvc as the time passes. If the purity of the top product is to be kept constant , then the reflux ratio must be increased steadily . Batch distillation carried out in two different ways:-either under constant reflux ratio or under increases reflux ratio. As the column consists of only enriching section, there is only one operating line which is given as :R 1 y n+1 = xn + x D -----( 31) R+1 R+1 While in operations under constant reflux ratio the slope of this line remains constant but its intercept changes; in operations under increasing reflux ratio, the slope of the line changes but the intercept remains constant. (a)Batch rectification under constant reflux ratio: In this case, xD continuously decreases as the reflux ratio is kept constant. A mvc balance at any moment during the rectification can be written as (ASTM):- F Ln = W xF dx x (x D -x) -------- (1) W F = D + W ------(2) Fx F = D x D + W x w -----(3) With the help of equation (1), (2) and (3), the quantity of top product (D) and it’s average composition (xD) at the end of operation for a given F , xF , R , xw and N can be calculated as follows:1-Draw the operating line from equation(31) starting at point xF ,then locate the given number of stages. 2- By shifting this line up and down, the 2-Select an arbitrary xD value (more than xF) and draw the operating line from this xD value and starting at this point locate the given number of stages and then read The corresponding (x ) value. Repeat this step by shifting this line up and down until reaching or exceeding the given xw as shown in the opposite Figure. 3- By plotting x values, 1/(xD-x) as shown in Figure( 57 ) the area under the curve between the limits of xD and xw gives the value of the integral in equation(1) from which W is then found . 4-Finally from equations (2) and (3) D and xD are computed. The heat to be supplied to provide the reflux is Qc = λRD (b)Batch Rectification at Constant Top Product : If the composition of the top product is to be kept constant, the reflux ratio must be increased steadily. In this type of operation, generally N, F, xF , xD and xw are known. 1-First D and W are calculated from equations (1)and(2). 2-points xF , xD and xW are located on the xy diagram. A Line from point xD is drawn and slope of this line is changed until the given number of stages between point xD and xW fit. 3-The reflux ratio, calculated from the slope of this line gives the reflux ratio at which the operation will be stopped. It is obvious that the reflux ratio will be increased continuously from the initial value to the final value . The total amount of distillate collected D can be determined from material balance (before and after the Distillation process): F=D+W F xF = D xD + W xw F x F = D x D + (F - D) x w F xF = D xD + F xw - D xw F ( xF - xw ) = D ( xD - xw ) F ( xF - xw ) D = ( xD - xw ) R assumed R initial R R R final Xw ,assumed x initial D = F (XF –Xwass )/ (XD – Xwass) D initial ------------ ------------- ------------ -------------- x final D final If the reflux ratio R is assumed to be adjusted Continuously to keep the top product at constant value, then at any moment the reflux ratio given by:R = dL/dD L R final dL = 0 RdD R initial L QR = λ dL = λ 0 R final RdD R initial By graphical integration will give the value of R d D Separation of Difficult Mixtures:Why mixtures are difficult to be separated by Distillation? 1-Low relative volatility of the mixture to be separated. In such operation continuous distillation require high reflux ratio , high reflux ratio, large cross-sectional Diameter and high number of stages. Ex:- separation of n-heptane from methyl cyclohexane The relative volatility ≈ 1.08 , using large number of Stages to achieve the separation. 2-when the relative volatility = 1 → azeotropic mixture , No straight forward distillation (equilibrium curve crosses The diagonal indicating presence of azeotrope. To over come this:1-Use another method for separation. e.g:- Solvent extraction, adsorption , crystallization,… 2-Obvious change in pressure (high vacuum). 3-Addition of third or new material to change α of the two component in which separation becomes easier. Extractive distillation:Extractive distillation refers to those processes in which a high-boiling solvent is added alter the relative volatilities of components in the feed. The boiling point of the solvent is generally much higher than the boiling points of the feed mixture that formation of new azeotropes is impossible. The high boiling point will also ensure that the solvent is will not vaporize in the distillation process . consider the simplified system shown in the Figure (59) for separation of toluene and iso-octane using phenol as the solvent . The separation of toluene (boiling point 110.8 oC) from iso -octane (boiling point 99.3 oC) is difficult using conventional distillation. Addition of phenol(boiling point 181.4 oC) results in the formation of phenol toluene mixture that leaves the extractive distillation column as bottoms, while relatively iso- octane is recovered as overhead product. The phenol-toluene mixture is further separated in a second column (solvent recovery column) whereby toluene appears as distillate and the bottoms product, phenol, is recycled back to the first column. Azeotropic distillation:azeotrope is a special class of liquid mixture that boils at a constant temperature at a certain composition. It behaves as if it were one component with one constant boiling point. Such mixture cannot be separated using conventional distillation methods(α). As an example the azeotrope in the ethanol water binary system has a composition of 89 mole per cent of ethanol The addition of benzene (entrainer) serves to volatilize water to a greater extent than ethanol thus pure ethanol may be obtained. The first tower in Figure (60) gives the ternary azeotrope as an overhead vapor, and nearly pure ethanol as bottom product. The ternary azeotrope is condensed and splits into liquid phases in the decanter. The benzene-rich phase from the decanter serves as reflux, while the water–ethanol rich phase passes to two towers, one for benzene recovery and the other for water removal. The azeotropic overheads from these successive towers are returned to appropriate points in the primary tower.