Chapter 4 Distillation Design

Chapter 4 Distil ation Design Subject: 1304 332 Unit Operation in Heat transfer Instructor: Chakkrit Umpuch Department of Chemical Engineering Faculty of Engineering Ubon Ratchathani University 1

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What are you going to learn in this chapter? 4.1 Vapor-Liquid Equilibrium Relations 4.2 Single-Stage Equilibrium Contact for Vapor-Liquid System 4.3 Simple Distil ation Methods 4.4 Distil ation with reflux and McCabe-Thiele Method 2

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4.1 Vapor-Liquid Equilibrium Relations 4.1.1 Raoult’s Law An ideal law, Raoult’s law, can be defined for vapor-liquid phases in equilibrium (only ideal solution e.g. benzene-toluene, hexane-heptane etc. p = P x A A A (1) Where p is the partial pressure of component A in the vapor in Pa (atm) A P is the vapor pressure of pure A in Pa (atm) A x is the mole fraction of A in the liquid. A 1 = x Comp + ositi x A B on in liquid: (2) 1 C = y ompo + y A sit B ion in vapor: (3) 3

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4.1.2 Boiling-Point Diagrams and xy Plots Boiling-point diagram for system benzene (A)-toluene (B) at a total pressure of 101.32 kPa. Dew point is the temperature at which the saturated vapour starts to condense. Bubble-point is the temperature at which the liquid starts to boil. The difference between liquid and vapour compositions is the basis for distil ation operations. 4

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4.1.2 Boiling-Point Diagrams and xy Plots Boiling-point diagram for system benzene (A)-toluene (B) at a total pressure of 101.32 kPa. If we start with a cold liquid composition is x = 0.318 (x A1 B1 = 0.682) and heat the mixture, it will start to boil at 98ºC. The first vapor composition in equilibrium is y = 0.532 (y = A1 B1 0.468). Continue boiling, the composition x will move to the A left since y is richer in A. A 5

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4.1.2 Boiling-Point Diagrams and xy Plots The boiling point diagram can be calculated from (1) the pure vapor-pressure data in the table below and (2) the following equations: p + p = P A B (4) P x + P 1 ( − x ) = P A A B A (5) p P x y A A A = = A P P (6) Where p , p are the partial pressure of component A and B in the vapor in Pa (atm) A B P , P are the vapor pressure of pure A and pure B in Pa (atm) A B P is total pressure in Pa (atm) x is the mole fraction of A in the liquid. A 6

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4.1.2 Boiling-Point Diagrams and xy Plots The boiling point diagram can be calculated from (1) the pure vapor-pressure data in the table below and (2) the following equations: 1 7

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Ex 4.1 Use of Raoult’s Law for Boiling-Point Diagram Calculate the vapor and liquid compositions in equilibrium at 95ºC (368.2K) for benzene-toluene using the vapor pressure from the table 1 at 101.32 kPa. Solution: At 95ºC from Table 1 for benzene, P = 155.7 kPa and P = 63.3 A B kPa. Substituting into Eq.(5) and solving, P x + P 1 ( − x ) = P A A B A 155.7(x ) + 63.3(1-x ) = 101.32 kPa (760 mmHg) A A Hence, x = 0.411 and x = 1 – x = 1 – 0.411 = 0.589. Substituting into eqn. A B A (6), p P x 155 A A A .7(0.41 ) 1 yA = = = = 0.632 P P 101 3 . 2 8

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The boiling point diagram can be calculated from the pure vapor-pressure data in the table below and the following equations: 1 9

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A common method of plotting the equilibrium data is shown in Fig. 2 where y is plotted versus x for the benzene-toluene system. The 45º line is A A given to show that y is richer in component A than is x . A A Fig. 2 Equilibrium diagram for system benzene(A) – toluene(B) at 101.32 kPa (1atm). 10

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4.1 Vapor-Liquid Equilibrium Relations 4.1.2 Boiling-Point Diagrams and xy Plots Ideal boiling point diagram Minimum-boiling azeotrope Maximum-boiling azeotrope An azeotrope is a mixture of two or more liquids in such a ratio that its composition cannot be changed by simple distillation. This occurs because, when an azeotrope is boiled, the resulting vapor has the same ratio of constituents as the original mixture. 11

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4.1 Vapor-Liquid Equilibrium Relations4.1.2 Boiling-Point Diagrams and xy Plots 12

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4.2 Single-Stage Equilibrium Contact for Vapor-Liquid System A single equilibrium stage is – the two different phases are brought into intimate contact with each other. – The mixing time is long enough and the components are essentially at equilibrium in the two phases after separation. V V 1 2 Where V , V is a vapor 1 2 L , L is a liquid 0 1 L L 0 1 Total mass balance: L0 + V2 = L1 + V1 = M L x Mass A balance: 0 0 + V y 2 2 = L x 1 1 + V y 1 1 = Mx A A AM In case of constant molal overflow : V = V and L = L 1 2 0 1 13

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Ex 4.2 Equilibrium Contact of Vapor-Liquid Mixture A vapor at the dew point and 101.32 kPa containing a mole fraction of 0.40 benzene (A) and 0.60 toluene (B) and 100 kg mol total is contacted with 110 kg mol of a liquid at the boiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streams are contacted in a single stage, and the outlet streams leave in equilibrium with each other. Assume constant molal overflow. Calculate the amounts and compositions of the exit streams. Solution: The given values are V = 100 kg mol, y = 0.40, L =110 kg 2 A2 0 mol , and x = 0.30. A0 For constant molal overflow, V V 1 2 V = V and L = L . 2 1 0 1 L L 0 1 14

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Material balance on component A, L x 0 0 + V y 2 2 = L x 1 1 + V y 1 1 = Mx A A AM 11 ( 0 . 0 3 ) 0 +10 ( 0 . 0 4 ) 0 = 110x +100y 1 A 1 A To solve equation above, the equilibrium relation between y and x in A1 A1 figure below must be used. First, we assume that x = 0.20 and substitute into equation above to A1 solve for y . A1 11 ( 0 3 . 0 ) 0 +10 ( 0 4 . 0 0) = 11 ( 0 . 0 ) 2 +100y 1A Assuming that x = 0.20 and solving y = 0.51. A1 A1 Next, assuming that x =0.40 and solving, y = 0.29. A1 A1 Next, assuming that x =0.40 and solving, y = 0.29. A1 A1 (These point are plotted on the graph.) At the intersection of this line with the equilibrium curve, 15 y = 0.455 and x = 0.25. Answer A1 A1

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4.3 Simple Distillation Methods 4.3.1 Introduction Distillation is a method used to separate the components of liquid solution, which depends upon the distribution of these various components between a vapor and a liquid phase. The vapor phase is created from the liquid phase by vaporization at the boiling point. Distillation is concerned with solution where all components are appreciably volatile such as in ammonia-water or ethanol-water solutions, where both components will be in the vapor phase. 17

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4.3.2 Relative Volatility of Vapor-Liquid Systems Relative volatility (α ) AB It is a measure of the differences in volatility between 2 components, and hence their boiling points. It indicates how easy or difficult a particular separation wil be. y / x y / x A A A A α = = AB y / x 1 ( − y 1 )( − x ) B B A A Where α is the relative volatility of A with respect to B in the binary system. AB P x P x Raoult’s law: y A A = y B B = A P B P PA α = AB PB α x AB A y = A 1+ (α − ) 1 x AB A when α is above 1.0, a separation is possible. AB 18

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Ex 4.3 Using data from table 1 calculate the relative volatility for the benzene-toluene system at 85ºC (358.2K) and 105ºC (378.2K) Solution: At 85ºC, substituting into equation below for a system following Raoutl’s law, P 116 9 . α = A AB = = 5 . 2 4 P 4 . 6 0 B Similarly at 105ºC, 204 2 . α = = 3 . 2 8 86 0 . The variation in α is about 7%. Answer 19

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4.3.3 Equilibrium or Flush Distillation 4.3.3.1 Introduction to distillation methods Distillation has two main methods in practice. 1. Production of vapor by boiling the liquid mixture to be separated in a single stage and recovering and condensing the vapors. No liquid is allowed to return to the single-stage still to contact the rising vapors. 2. Returning of a portion of the condensate to the still. The vapors rise through a series of stages or trays, and part of the condensate flows downward through the series of stages or trays countercurrently to the vapors (“fractional distillation, distillation with reflux, or rectification”). There are 3 important types of distillation that occur in a single stage or still: Equilibrium or flash distillation, Simple batch or differential distillation and simple steam distillation 20

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4.3.3.2 Equilibrium or Flush Distillation yA Separator heater xA Flash distillation is a single stage separation technique. 1. A liquid mixture is pumped through a heater to raise the temperature and enthalpy of the mixture. 2. It then flows through a valve and the pressure is reduced, causing the liquid to partially vaporize. 3. Once the mixture enters a big enough volume (the “flash drum”), the liquid and vapor separate. 4. Because the vapor and liquid are in such close contact up until the “flash” occurs, the product liquid and vapor phases approach equilibrium. 21

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4.3.3.2 Equilibrium or Flush Distillation Total mass balance: F = V+ L yA Separator Component A balance: Fx = V y + Lx F A A heater where x F, V and L are flow rate of feed, vapor and liquid phases. A x , y and x are mole fraction of component A in feed, vapor and liquid. F A A Material balance for more volatile component : V F V x = ( )y + ( − )x F A A x = f y + 1 ( − f )x F A A F F F Where f = V/F = molal fraction of the feed that is vaporized and withdrawn continuously as vapor. 1-f = one as liquid 22

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Ex 4.4 A mixture of 50% mole normal heptane and 50% normal octane at 30ºC is continuously flash distilled at 1 standard atmosphere so that 60 mol% of the feed is vaporized. What will be the composition of the vapor and liquid products? x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 A y 0.247 0.453 0.607 0.717 0.796 0.853 0.898 0.935 0.968 A Solution: Given: x = 0.5, f = 0.6 F Find: x , y A A Basis: F = 100 mols Applying the mass bala Fnce = V yie + L lds: f = V / F Since V = f F = ,0.6 1 ( 0 ) 0 = 60 L = F −V = 100 − 60 = 40 23

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Material balance for more volatile component, V F V x = ( )y + ( − )x F A A F F F x = f y + 1 ( − f )x F A A Subtituting value of f =0.6 and x =0.5 we get, F 0.5 = 0.6y + 1 ( − 0.6)x A A 0.5 = 0.6y + 0.4x A A Assuming that x = 0.5 and solving y = 0.5. A A Next, assuming that x =0 and solving, y = 0.83. A A (These point are plotted on the graph.) At the intersection of this line with the equilibrium curve, y = 0.58 and x = 0.39. Answer A A 24

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x=0 2nd y-intercept= 0.834 yA xA x =0.39 3th A x =0.5 F 1st y = 0.58 A y = 0.5 F 25

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4.3.4 Simple Batch or Differential Distillation The pot is fil ed with liquid mixture and heated. Vapour flows upwards though the column and condenses at the top. Part of the iquid is returned to the column as reflux, and the remainder withdrawn as distil ate. Nothing is added or withdrawn from the stil until the run is completed. 26

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4.3.4 Simple Batch or Differential Distillation The total moles of component A left in the stil n wil be A n = xn A where n is the moles of liquid left in the stil at a given time y and x is the vapor and liquid compositions If a smal amount of liquid dn is vaporized, the change in the moles of component A is ydn, or dn . A Differentiating equation gives dn = d(xn) = ndx + xdn dn dx A = ndx + xdn = ydn n y − x 27

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4.3.4 Simple Batch or Differential Distillation 1 n 1 x dn dx = n y − x ∫ dn = n ∫ dx = n1 ln y − x n n0 x0 0 dx/(y-x) can be integrated graphically or numerically using tabulated equilibrium data or an equilibrium curve. For ideal mixture: y x A A = α AB y x B B n n A B ln = α ln AB n n dn dn dn n 0 A 0B A A A = = α AB dn dn dn n Integrating 1/ α B B B AB n  n  dn dn B A A B = α =   AB n n n n   A B 0B 0 A 28

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Ex 4.5 A batch of crude pentane contains 15 mole percent n- butane and 85 percent n-pentane. If a simple batch distil ation at atmospheric pressure is used to remove 90 percent of butane, how much pentane wil be removed? What wil be the composition of the remaining liquid? Solution: An average value of 3.5 is used for α . AB Basis: 1 mol feed n n nOB = 8 . 0 5 A = 0 . 0 15 O A = . 0 1 5 (butane) (pentane) From equation: 1/α AB n  n  B A = n  n  0B  0A  n = total mole of B left in still, n = total mole A left in still. B A n = total initial mole of B in still, n = total initial mole A in still. 0B 0A 29

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nB = ( )1. 0 1/ 5. 3 = 51 . 0 8 . 0 85 nB = 5 . 0 1 ( 8 8 . 0 ) 5 = 4 . 0 40 Total mole of liquid left in still: n = 0 4 . 4 + 0 0 . 15 = 4 . 0 55 mol Mole fraction of butane in liquid left: 0 0 . 15 xA = = 0 0 . 33 0 4 . 55 30

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4.3.5 Simple Steam Distillation Note that by steam distillation, as long as water is present, the high- boiling component B vaporizes at a temperature well below its normal boiling point without using a vacuum. The A and B are usually condensed in condenser and the resulting two immiscible liquid phases separated. Disadvantage: large amounts of heat must be used to simultaneously evaporate the water with high-boiling compound. 31

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4.3.5 Simple Steam Distillation When the sum of the separate vapor pressures equals the total pressure, the mixture boils and P + P = P A B Where PA is vapor pressure of pure water A PB is vapor pressure of pure B Then the vapor composition is P P y A = y B = A P B P The ratio moles of B distilled to moles of A distilled is n P B B = n P A A 32

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Ex 4.6 A mixture contains 100 kg of H O and 100 kg of 2 ethyaniline (mol wt = 121.1 kg/kg mol), which is immiscible with water. A very slight amount of nonvolatile impurity is dissolved in the organic. To purify the ethyaniline it is steam-distilled by bubbling saturated steam into the mixture at a total pressure of 101.32 kPa (1 atm). Determine the boiling point of the mixture and the composition of the vapor. The vapor pressure of each of the pure compounds is as fol ows (T1): Temperature P (water) P (ethylaniline) A B (mm Hg) (mm Hg) K ºC 353.8 80.6 48.5 1.33 369.2 96.0 87.7 2.67 372.3 99.15 98.3 3.04 386.4 113.2 163.3 5.33 33

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P + P = P A B Solution: Temperature P P P=P +P A B A B (water) (ethylaniline) (kPa) K ºC (kPa) (kPa) 353.8 80.6 48.5 1.33 49.83 369.2 96.0 87.7 2.67 90.37 372.3 99.15 98.3 3.04 101.34 386.4 113.2 163.3 5.33 169.23 The boiling temperature = 99.15ºC since total pressure in this temperature is equal to atmospheric pressure. The vapor composition are: P 98 3 . P 0 . 3 4 y = A y = B B = = 0 . 0 3 A = kPa = 9. 0 7 P 101 3 . 2kPa P 101 3 . 2 34

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4.4 Distillation with Reflux and McCabe-Thiele Method 4.4.1 Introduction to Distillation with Reflux Rectificaition (fractionation) or stage distillation with reflux is a series of flash-vaporization stages are arranged in a series which the vapor and liquid products from each stage flow countercurrently to each other. The liquid in a stage is conducted or flows to the stage below and the vapor from a stage flow upward to the stage above. V V V 1 2 3 2 n n+1 L L L 0 L 1 1 2 n-1 n A total material balance: V +1 + L −1 = V + L n n n n V y A component balance on A: +1 +1 + L x −1 −1 = V y + L x n n n 35

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4.4.1 Introduction to Distillation with Reflux In a distil ation column the stages (referred to as sieve plates or trays) in a distil ation tower are arranged vertical y, as shown schematical y in figure below. 1. Feed enters the column somewhere in the middle of the column. 2. Feed is liquid, it flows down to a sieve tray or stage. 3. Vapor enters the tray and bubbles through the liquid on this tray as the entering liquid flows across. 4. The vapor and liquid leaving the tray are essential y in equilibrium. 5. The vapor continues up to the next tray or stage, where it is again contacted with a downflowing liquid. 6. The concentration of the more volatile component is being increased in the vapor form each stage going upward and decreased in the liquid from each stage going donwards. 36

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4.4.1 Introduction to Distillation with Reflux In a distil ation column the stages (referred to as sieve plates or trays) in a distil ation tower are arranged vertical y, as shown schematical y in figure below. 7. The final vapor product coming overhead is condensed in a condenser and a portion of the liquid product (distil ate) is removed, which contains a high concentration of A. 8. The remaining liquid from the condenser is returned (refluxed) as a liquid to the top tray. 9. The liquid leaving the bottom tray enters a reboilier, where it partial y vaporized, and the remaining liquid, which is lean in A or rich in B, is withdrawn as liquid product. 10. The vapor from the reboiler is sent back to the bottom stage or trays is much greater. 37

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4.4.3 McCabe-Thiele Method of Calculation for Number of Theoretical Stages4.4.3A) Introduction and assumptions A mathematical – graphical method for determining the number of theoretical trays or stages needed for a given separation of a binary mixture of A and B has been developed by McCabe and Thiele. The method uses material balances around certain parts of the tower, which give operating lines and the xy equilibrium curve for the system. Main assumption 1) Equimolar overflow through the tower between the feed inlet and the top tray and the feed inlet and bottom tray. 2) Liquid and vapor streams enter a tray, are equilibrated, and leave. 38

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A total material balance: V +1 + L −1 = V + L n n n n A component A balance: V y +1 +1 + L x −1 −1 = V y + L x n n n Where V is mol/h of vapor from tray n+1 n+1 L is mol/h liquid from tray n n y is mole fraction of A in V and so on. n+1 n+1 39

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4.4.3B) Equation for enriching section A total material balance: F = D + W (1) A component A balance: Fx = Dx + Wx F D w (2) Where F is the entering feed (mol/h) D is the distillate (mol/h) W is the bottoms (mol/h) 40

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Material balance over dashed-line section: V (3) +1 = L + D n n A balance on component A: V y +1 +1 = L x + D D x n n n n (4) 41

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Solving for y , the enriching-section operating line is n+1 y = L Dx n x n n + D 1 + (5) V V n 1 + n 1 + Si V nce L D L V R R n+ = n + , / n n+ = /( + ) 1 1 1 and equation becomes R x y D n+ = xn + 1 R +1 R +1 (6) where R = L / D n = reflux ratio = constant. The eqn. (1) is a straight line on a plot of vapor composition versus liquid composition. 42

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The slope is L / V or R /( R + ) 1 . It intersects the y=x line (45º diagonal n n 1 + line) at x = x . The intercept of the operating line at x = 0 is y = x D D / ( R + ) 1 . The theoretical stages are determined by starting at x and stepping off the first D plate to x . Then y is the composition of the vapor passing the liquid x . 1 2 1 In a similar manner, the other theoretical trays are stepped off down the tower in the enriching section to the feed tray. 43

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4.4.3C Equation for stripping section Material balance over dashed-line section: V +1 = L −W m m (7) A component A balance: V y +1 +1 = L x − W w x m m m m (8) 44

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Solving for y , the enriching-section operating line is m+1 y = L Wx m x m m − W 1 + (9) V V m 1 + m 1 + Again, since equimolal flow flow is assumed, L = L = constant and m N V +1 = V m N = L constant,V eqn. (2) is a straight line when plotted as y versus x, m = m 1 + with a slope of . It intersects the y = x line at x = x . w y = −Wx /V W m 1 + The intercept at x = 0 is . 45

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The theoretical stages for the stripping section are determined by starting at x , going up to y , and then across to the operating line, etc. W W 46

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4.4.3D Effect of feed conditions The condition of feed stream is represented by the quantity q, which is the mole fraction of liquid in feed. L = L + qF m n (10) V = V + 1 ( − q)F (11) n m The enriching and striping operating-line equations on an xy diagram can be derived as follows: V y = L x + Dx (12) n n D V y = L x −Wx (13) m m w Where the y and x values are the point of intersection of the two operating lines. Subtracting eqn.(12) from eqn.(13), (V −V ) y = (L − L )x − (Dx + Wx ) (14) m n m n D w 47

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4.4.3D Effect of feed conditions Substituting eqn.(2), (10), and (11) into eqn.(14) and rearranging, = q y − x x F (15) q −1 q −1 c (T − T ) c (T − T ) q = 1 pL b F + q = 1 pV F d + λ λ Cold-liquid feed Superheated vapor where C , C = specific heats of liquid and vapor, respectively pL pV T = temperature of feed F T , T = bubble point and dew point of feed respectively b d λ = heat of vaporization 48

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4.4.3E Location of the feed tray in a tower and number of trays. From eqn.(15), the q-line equation and is the locus of the intersection of the two operating lines. Setting y = x in eqn(15), the intersection of the q-line equation with the 45º line is y=x=x , where x is the overall composition of the feed. F F In given below the figure, the q line is plotted for various feed conditions. The slope of the q line is q/(q-1). q = 0 (saturated vapor) q = 1 (saturated liquid) q > 1(subcooled liquid) q < 0 (superheated vapor) 0 < q < 1 (mix of liquid and vapor) 49

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4.4.3D Number of stages and trays 1st point 2nd point 3rd point n = 7 =number of tray + reboiler Number of tray = 6

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4.4.3F Using Operating Lines and the Feed Line in McCabe-Thiele Design Slope = R/(R+1) Slope = q/(1-q) Slope = L/ V

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Ex 4.7 A continuous fractioning column is to be designed to separate 30,000 kg/h of a mixture of 40 percent benzene and 60 percent toluene into an overhead product containing 97 percent benzene and a bottom product containing 98 percent toluene. These percentages are by weight. A reflux ratio of 3.5 mol to 1 mol of product is to be used. The molal latent heats of benzene and toluene are 7,360 and 7,960 cal/g mol, respectively. Benzene and toluene from a nearly ideal system with a relative volatility of about 2.5. The feed has a boiling point of 95ºC at a pressure of 1 atm. a) Calculate the moles of overhead product and bottom product per hour. b) Determine the number of ideal plates and the position of the feed plate (i) if the feed is liquid and at its boiling point; (ii) if the feed is liquid and at 20ºC (specific heat 0.44 cal/g.ºC); (i i) if the feed is a mixture of two-thirds vapor and one-third liquid.

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Solution (a) 40 97 2 78 78 x 78 xD = = 0 9 . 74 xB = = 0 02 . 35 F = = 4 . 0 40 40 60 97 3 + 2 98 + + 78 92 78 92 78 92 100 The average molecular weight of th = 40e fe 6 e 0 d is 85 8 . + 78 92 The average of hea λ t = va4 . 0 p 4 o ( r, 7 iz 36 a 0 t)io + n. 0 i 5 s 6( 9 , 7 60 ) = , 7 696 cal / gmol The feed rate F is 30,000/85.8 = 350 kg mol/h. By an overal benzene balance, using Eq. below  4 . 0 40 − 0 . 0 235  D = 350  =153 4 . kgmol / h  9 . 0 74 − 0 . 0 235  B = 350 1 − 53 4 . =196 6 . kgmol / h

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Solution (b) (i), We determine the number of ideal plates and position of the feed plate. 1) Plot the equilibrium diagram, erect verticals at x , x , and x . D F B 2) Draw the feed line. Here q=1, and the feed line is vertical. 3) Plot the operating lines. The intercept of the rectifying lie on the y axis is, x /(R+1) = 0.974/(3.5+1) = 0.216 (eqn (6)). From the D intersection of the rectifying operating line and the feed line, the stripping line is drawn. 4) Draw the rectangular steps between the two operating lines and the equilibrium curve. The stripping line is at the seventh step. By counting steps it is found that, besides the reboiler, 11 ideal plates are needed and feed should be introduced on the seventh plate from the top.

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Solution (b) (ii), The latent heat of vaporization of the feed λ is 7,696/85.8 = 89.7 cal/g. 4 . 0 4 9 ( 5 −20) q =1+ = . 1 37 89 7 . = q y − x x F q −1 q −1 The slope of the feed line is -1.37/(1-1.37) = 3.70. When steps are drawn for this case, as shown in Fig. below, it is found that a reboiler and 10 ideal plates are needed and that the feed should be introduced on the sixth plate.

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Solution (b) (iii), From the definition of q it follows that for this case q = 1/3 and the slope of the feed line is -0.5. The solution is shown in Fig. below. It cal s for a reboiler and 12 plates, with the feed entering on the seventh plate.

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4.4.4 Total and Minimum Reflux Ratio for McCabe-Thiele Method 4.4.4A Total flux One limiting values of reflux ratio is that of total reflux, or R = ∞. Since R = L /D n and, by eqn.(16). V +1 = L + D n n (16) Then L is very large, as is the vapor flow V . This means that the slope R/(R+1) n n of the enriching operating line becomes 1.0 and the operating lines of both sections of the column coincide with the 45º diagonol line, as shown in Fig below. Minimum number of trays can be obtained by returning all the overhead condensed vapor V from the top of the tower back to the tower as reflux, i.e., total 1 reflux. Also, the liquid in the bottoms is reboiled. 60

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 x 1− x  Minimum number of theoretical steps N D w m log1− x x   D w  when a total condenser is used (α is constant). N = (17) m logαav For small variations in α, α = α α av ( 1 w )1/2 (18) where α is the relative volatility of the overhea d vapor 1 α is the relative volatility of the bottoms liquid. w 61

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4.4.4B Minimum reflux ratio The minimum reflux ratio (R ) will require an infinite number of trays for the given m separation desired of x and x . D W If R is decreased, the slope of the enriching operating line R/(R+1) is decreased, and the intersection of this line and the stripping line with the q line moves farther from the 45º line and closer to the equilibrium line. Two operating lines touch the equilibrium line (“pinch point”) at y’ and x’ (number of steps required becomes infinite). The line passes through the points x’, y’ and x (y=x ): D D R x − y′ m D = (19) R +1 x − x′ m D 62

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4.4.4C Operating and optimum reflux ratio Total reflux = number of plates is a minimum, but the tower diameter is infinite. This corresponds to an infinite cost of tower and steam and cooling water. This is the limit in the tower operation. Minimum reflux = number of trays is infinite, which again gives an infinite cost. These are the two limits in operation of the tower. Actual operating reflux ratio to use is in between these two limits. The optimum reflux ratio to use for lowest total cost per year is between the minimum R and m total reflux (1.2R to 1.5R ). m m 63

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Ex 4.8 What are (a) the minimum reflux ratio and (b) the minimum number of plates for cases (b)(i), (b)(ii), and (b)(i i) of EX 4.7 if α is given as 2.47. av Solution (a) For minimum reflux ratio use eqn. (18). Here x = 0.974. The results D are given in Table below. Case x’ y’ RDm (b)(i) 0.440 0.658 1.45 (b)(ii) 0.521 0.730 1.17 (b)(ii ) 0.300 0.513 2.16 64

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Ex 4.8 What are (a) the minimum reflux ratio and (b) the minimum number of plates for cases (b)(i), (b)(ii), and (b)(i i) of EX 4.7 if α is given as 2.47. av Solution (b) For minimum number of plates, the reflux ratio is infinite, the operating lines coincide with the diagonal, and there are no differences among the three cases. Use the α = 2.47 and equation below we get, av  x 1 x D −  log w 1   − x x D w   0 9 . 74× . 0 976  Nm = = ln  −1 logαav  0 0 . 24× . 0 026  = 8 1 . 05 −1 = 7 The minimum number of ideal plates is 7 plus a reboiler. 65

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Homework No.6 1. The vapor-pressure data are given below for the system hexane-octane. (a) Using Raoult’s law, calculate and plot the xy data at a total pressure of 101.32 kPa. (b) Plot the boiling-point diagram. T(ºF) T(ºC) Vapor Pressure n-Hexane n-Octane kPa mmHg kPa mmHg 155.6 68.7 101.3 760 16.1 121 175 79.4 136.7 1025 23.1 173 200 93.3 197.3 1480 37.1 278 225 107.2 284.0 2130 57.9 434 258.2 125.7 456.0 3420 101.3 760

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Homework No.6 (continue) 2. A mixture of 100 kg mol which contains 60 mol% n-pentane (A) and 40 mol% n-heptane (B) is vaporized at 101.32 kPa pressure under differential conditions until 40 kg mol are distilled. Use vapor-liquid equilibrium data below. If flash distillation is done and 40 kg mol are distil ed, what is the composition of the vapor distil ed and of the liquid left? equilibrium data for n-pentane-n-heptane, x and y are mole fraction of n-pentane x 0.0 0.059 0.145 0.254 0.398 0.594 0.867 1.000y 0.0 0.271 0.521 0.701 0.836 0.925 0.984 1.000

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Homework No.7 1. A mixture containing 50 g mol of benzene and 50 g mol of chlorobenzene is distilled by simple distillation without reflux until 40 percent of the initial charge is taken off as overhead. The system benzene-chlorobenzene may be considered ideal, with an averge relative volatility of 5.3 (a) What are the compositions of overhead and residue after distillation is complete? (b) The overhead from the first distil ation is subjected to a second simple distil ation. Again 40 percent of the charge is taken overhead. What is the composition of the second over head product? What is its mass in grams? How many grams of chlorine does it contain?

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Homework No.7 (continue) 2. A mixture of 50 g mol of liquid benzene and 50 g mole water. Determine the boiling at 101.32 kPa pressure. Liquid benzene is immiscible in water. Determine the boiling point of the mixture and the composition of the vapor. Which component wil first be removed completely from the stil ? Vapor pressure data of the pure components are as follows: Temperature Pwater Pbenzene (mm Hg) (mm Hg) K ºC 308.5 35.3 43 150 325.9 52.7 106 300 345.8 72.6 261 600 353.3 80.1 356 760

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Homework No.8 1. A saturated liquid feed of 200 mol/h at the boiling point containing 42mol% heptane and 58% ethyl benzene is to be fractionated at 101.32 kPa abs to give a distil ate containing 97 mol% heptane and a bottoms containing 1.1 mol% heptane. The reflux ratio used is 2.5:1. Calculate the mol/h distil ate, mol/h bottoms, theoretical number of trays, and the feed tray number. Equilibrium data are given below at 101.32 kPa abs pressure for the mole fraction n-heptane x and y . H H Temperature Temperature K ºC x y K ºC x y H H H H 409.3 136.1 0 0 383.8 110.6 0.485 0.730 402.6 129.4 0.08 0.23 376.0 102.8 0.790 0.904 392.6 119.4 0.25 0.514 371.5 98.3 1.000 1.000

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Homework No.8 (continue) 2. Determine the minimum reflux ration Rm and the minimum number of theoretical plates at total reflux for the rectification of a mixture of heptane and ethyl benzene as given in problem 1. Do this by the graphical mehtods of McCabe-Thiele.

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