UNIT OPERATIONS OF CHEMICAL ENGINEERING.pdf

06-361 page 1 Spring, 2001 Chapter 1. Course Introduction WHAT ARE "UNIT OPERATIONS"? A typical process which a chemical engineer might work with is the production oI gasoline Irom crude oil process ÷ sequence oI 'unit operations¨ (physical changes) ¹ chemical reactors Any process can be subdivided into a number oI steps which are perIormed in sequence to go Irom some initial starting material (crude oil, in this case) to some Iinal material (gasoline). Eor example, we might start by heating the crude oil to lower its viscosity, then pump the oil to the distillation column, where we then separate various components oI the crude. A unit operation is one oI the steps oI this sequence. Usually the term reIers to steps intended primarily to perIorm some physical transIormation (as opposed to chemical transIormation) oI the input stream. Examples oI unit operations: - heat exchange (change temperature oI a stream) - Iluid Ilow (transportation) - distillation (separation oI mixture into multiple streams which are richer in some components that original) - evaporation (remove water Irom liquid) - humidiIication (increase water content oI gas) - gas absorption (remove one component oI a gas mixture) - sedimentation (separate solid Irom liquid) - classiIication (divide mixture oI particles into diIIerent "classes" on the basis oI size) In this course, we will apply the principles learned in the engineering science courses you have had the last three semester to the design oI equipment Ior physical transIormations. We might divide up the courses in the curriculum as Iollows: 06-361 page 2 Spring, 2001 Course Objective Scale Thermo 1) conservation oI energy 2) criteria Ior phase/chemical equilibria molecular Iluids, heat/mass transIer rates oI physical processes microscopic unit operations design equipment Ior physical transIormations macroscopic kinetics 1) rate oI chemical processes 2) design equipment Ior chemical transIormations molecular macroscopic process design divide process into sequence oI steps (synthesize Ilowsheet) supramacro 06-361 page 3 Spring, 2001 Chapter 2. Heat Exchange Equipment DOUBLE-PIPE HXER One particularly simple heat-exchange device is the double-pipe heat exchanger, shown in the Iigure at right. It is called this because the basic component is constructed Irom two pipes, one inside the other. One Iluid (Eluid B) Ilows inside the inner pipe while the second Iluid (Eluid A) Ilows in the annular space between the two pipes. OI course, one Iluid is hotter than the other and heat Ilows through the wall oI the inner pipe Irom the hot Iluid to the cold Iluid. To obtain larger heat exchange area, several pipes are arranged side-by-side and Iittings are attached to allow the Iluids to contact the pipes in series so the overall Ilow is the same as iI only one very long pipe were used rather than several shorter ones. SHELL-AND-TUBE HXER II very large heat exchange areas are required, the double-pipe heat exchanger design becomes impractical. Instead, the shell-and-tube heat exchanger design is used. Shown above is a bundle oI small-diameter tubes which are arranged parallel to each other and reside inside a much larger-diameter tube called the 'shell¨, much like strands oI uncooked spagetti come in a tube-shaped container. The tubes are all maniIolded together at either end so 06-361 page 4 Spring, 2001 that the 'tube Iluid¨ enters the leIt side and is distributed equally among all the tubes. At the right side, the Iluid exits Irom each tube, is mixed together in a second maniIold, then leaves as a single stream. The second Iluid, called the 'shell Iluid¨, Ilows in the space in between the outside oI tubes. BaIIle plates inside the shell Iorce the shell Iluid to Ilow across the tubes repeatedly as the Iluid moves along the length oI the shell. In the design above (called a 1-1 hxer), the Ilow is Irom leIt to right inside the tubes and Irom right to leIt Ior the shell-side Iluid. Another common design is shown below (called a 1-2 hxer). HalI oI the tubes have Ilow Irom leIt to right while the other halI have Ilow in the opposite direction. 06-361 page 5 Spring, 2001 Chapter 3. Fundamentals of Heat Transfer: Review LOG-MEAN DRIVING FORCE The design equation used Ior sizing heat exchangers generally takes the Iorm oI q U A T = A (1) where q is the heat duty oI the exchanger (rate oI energy transIer) U is some heat-transIer coeIIicient, A is the area oI the heat exchange surIace, and AT is the driving Iorce Ior heat transIer. In any real heat exchanger, the local driving Iorce AT will vary along the length oI the exchanger: say it varies between AT 1 at one end oI the exchanger to AT 2 at the other end. Then the question arises: which value oI AT do you use in (1)? The answer turns out to be: use the log mean oI AT 1 and AT 2 : A A A A A T T T T T L = ÷ 1 2 1 2 ln This turns out to be correct in nearly every situation. We will prove it now Ior one particular case and leave some other cases to be proved in the homework (HWK #1, Probl. 1) Proof: When the driving Iorce varies with position along the exchanger, the usual approach to math modelling is to divide the length oI the hxer (L) into a large number oI thin slices, such that the thickness oI each slice is thin enough so that the temperature oI either stream is essentially constant over the slice, leaving the driving Iorce constant over the slice: AT ÷ T´(x) T(x) Depending on which x we choose, we get a diIIerent driving Iorce. The rate oI heat transIer Irom the hot Iluid to the cold Iluid in any slice is given by a local heat transfer coefficient U(x): dq ÷ U(x) AT(x) dA (2) 2 2 1 1 A x÷0 x x¹dx x÷L one slice 06-361 page 6 Spring, 2001 where q(x) is the rate oI heat transIerred Irom the hot Iluid to the cold Iluid in the length oI hxer Irom x÷0 and x÷x; and dq is q(x¹dx) ÷ q(x) ¹ dq and dA is the heat exchange area in this slice oI length dx. II the inner pipe (oI a double-pipe hxer) is a cylinder oI diameter D, then the heat exchange area oI this slice is dA ÷ tD dx Integrating (2) as a Iunction oI x is diIIicult since we`d have to predict AT(x). There is a easier way (but a 'trick¨ is required). Consider steady-state counter-current Ilow through a double-pipe heat exchanger (shown at right). A steady-state energy balance on the hot Iluid gives 1 rate in rate out m H m H q = ' ' ' ' = + (3) A similar balance on the cold Iluid gives 1 rate in rate out mH q mH = + = (4) Solving (3) and (4) Ior q: 1 1 rate of energy rate of energy gained by lost by cold stream hot stream ( ) m H x H q x m H x H ' ' ' ÷ = = ÷ ( ( ¸ ¸ ¸ ¸ ¸¸¸_¸¸¸ ¸¸¸_¸¸¸ (5) where m is the mass Ilowrate oI the stream (units are Kg/sec), H is the enthalpy per unit mass (units are J/Kg), the subscripts and primes have the Iollowing meaning: - unprimed reIers to the cold stream, - prime reIers to the hot stream, - '1¨ reIers to the leIt end oI hxer, and - '2¨ reIers to the right end oI hxer. Now, let`s try to relate each enthalpy to the appropriate temperature. In general, enthalpy changes can have both latent heat (caused by change in phase) and sensible heat (caused by change in temperature) contributions. x÷0 x÷x x÷L pipe wall hot Iluid cold Iluid q x ( ) T´ 1 T´ T´ 2 T 1 T T 2 06-361 page 7 Spring, 2001 dH c dT dv p = + sensible heat latent heat ¸_¸ ì ¸ where - dH is the change in enthalpy (energy/mass) - c p is the heat capacity (energy/mass/degree) oI the Iluid, also known as 'speciIic heat¨ - dT is the change in temperature (degree) - ì is the heat oI vaporization (energy/mass) - dv is the mass Iraction oI Iluid vaporized When the Iluid consists oI two phases (liquid and vapor), which have diIIerent heat capacities c pL and c pJ , we use weighed average oI the two heat capacities c p ÷ (1-v)c pL ¹ v c pJ where v is the mass Iraction oI vapor present and 1-v is the mass Iraction oI liquid present. Eor the time being, let`s assume: Assumption #1: no phase changes occur in either Iluid Now we only have to worry about contributions Irom changes in the temperature (sensible heat): dH ÷ c p dT Integrating over the temperature change allows us to evaluate the change in enthalpy AH c dT c T T p T T p b a a b = = ÷ using assumption #1 using assumption #2 In general, the speciIic heat (c p ) depends on temperature. But iI the temperature change is not too large say 10-20° then the speciIic heat doesn't change too much. Let's assume we can neglect these small changes: Assumption #2: constant speciIic heat Ior both streams Then we can Iactor c p out oI the integral and the integration becomes trivial: AH ÷ c p AT 06-361 page 8 Spring, 2001 In particular, H-H 1 ÷ c p (T-T 1 ) and ' ÷ ' = ' ' ÷ ' H H c T T p 1 1 where c p and c p are the speciIic heats oI the cold and hot streams. Thus (5) can be rewritten as: at x÷x: mc T x T m c T x T q x p p ( ) ÷ = ' ' ' ( ) ÷ ' = ( ) 1 1 , (6) Eor example, at x÷0: mc T T m c T T q p p 1 1 1 1 0 0 ÷ = ' ' ' ÷ ' = ( ) = , In particular, we are interested in the total rate oI heat transIer over the entire length L oI the exchanger: at x÷L: mc T T m c T T q L q p p T 2 1 2 1 ÷ = ' ' ' ÷ ' = ( ) ÷ , Now let me show you the "trick": using (6), I can show that the temperature oI either stream |i.e. T(x) or T(x)| is a linear Iunction oI the amount oI heat transIered, q(x): T x T mc q x p ( ) = + ( ) 1 1 ' ( ) = ' + ' ' ( ) T x T m c q x p 1 1 Since both temperatures are linear Iunctions oI q, their diIIerence is also a linear Iunction oI q. A straight line has the very nice property that its slope is the same at every point: d T dq T T q T A A A = = ÷ ÷ const 2 1 0 (7) I can easily evaluate the driving Iorce (AT) at the two ends oI the heat exchanger, and I know the total heat duty (q T ), so I can calculate this slope. The dq and d(AT) which appear in (7) represent the incremental rate oI heat transIer and the change in driving Iorce AT which occurred over a diIIerent length dx oI the heat exchanger(Irom x to x¹dx). They are related by dq ÷ U dA AT (8) where dA is the heat exchange area in that slice oI the heat exchanger Irom x to x¹dx. Treating the heat exchange surIace as a cylinder oI radius R: 06-361 page 9 Spring, 2001 dA ÷ 2tR dx (8) into (7): d T U T dA T T q T A A A A = ÷ 2 1 or d T T T T q U dA T A A A A = ÷ 2 1 Next, we integrate over the length oI the heat exchanger Irom x÷0 (where AT ÷ AT 1 ) to x÷L (where AT ÷ AT 2 ): d T T T T q U x Rdx T T T x x L dA A A A A A A 1 2 2 1 0 2 = ÷ = = t ¸_¸ The integral on the right-hand-side becomes trivial to evaluate iI we can make one additional assumption: Assumption #3: U is constant Then we can Iactor U outside the integral and integrate the remainder: 2 2 2 0 0 t t t Rdx R dx RL A x x L L T = = = = = which is just the total heat exchange area. Thus we have ln A A A A T T T T q UA T T 2 1 2 1 = ÷ Solving Ior the total heat duty q UA T T T T T T = ÷ A A A A 2 1 2 1 ln or q UA T 1 1 L = A where A A A A A T T T T T L = ÷ 2 1 2 1 ln is called the log-mean temperature difference (LMTD). It represents the average driving Iorce Ior heat transIer. This result provides us with the main design equation Ior heat exchangers. 06-361 page 10 Spring, 2001 Summary oI assumptions: - U and c p 's are constant - AH oI neither stream has contributions Irom both sensible and latent heats (i.e. AH is either all sensible heat or all latent heat) Actually, we assumed in the analysis above that AH was all sensible heat Ior both streams. In the homework, you will show that AH can also be all latent heat and the LMTD is still the appropriate average driving Iorce. Eor example, iI you use condensing steam as your hot Iluid, as long as the steam remains saturated: Ior saturated steam: T(x) ÷ const ÷ T bp We also assumed in the analysis above, that the Ilow was counter-current. You will also show in the Iirst homework problem that the same design equation applies to co-current Ilow. RESISTANCES IN SERIES Consider heat transIer through the inner pipe oI a double-pipe heat exchanger. Let`s take a slice out oI the heat exchanger and examine the temperature proIile along a radius to the pipe at one particular axial position. oblique view of slice 06-361 page 11 Spring, 2001 One can identiIy three regions in which the temperature varies sharply. These correspond to the three resistances to heat transIer: the pipe wall itselI, the Iluid Iilm inside the pipe and the Iluid Iilm outside the pipe. These resistances act in series. The total resistance is related to the resistance oI each Iilm by: 1 1 1 U h x k D D h D D o o w w o L i o i total resistance resistance of outer film resistance of pipe wall resistance of inner film ¸ ¸ = + + ¸_ ¸ ¸ ¸ ¸_¸ That resistances add in this way is a very important concept, so I`m going to take the time to prove it to you again. This will also give us an opportunity to review the deIinitions oI the quantities. Consider the temperature proIile along some line drawn perpendicular to the pipe wall through the center oI the pipe. These temperature variations drive heat transIer down the temperature hill. The rate dq oI heat transIered through some small element oI the inner surIace oI the pipe having area dA i is dq h dA T T i i D dL h wh i = ÷ t ¸ (9) where h i is a single-phase heat-transfer coefficient Ior the inner Iilm oI Iluid next to the pipe and dL is the length oI the pipe segment. At steady-state, this same rate oI heat transIer must occur through the liquid Iilm outside the pipe: dq h dA T T o o D dL wc c o = ÷ t ¸ (10) At steady-state, this same rate oI heat transIer must also occur through the pipe wall itselI: dq k x d A T T w w L wh wc = ÷ (11) D i D o T a T c T g T h T wh T wc end view oI slice r r i r o r r e s i s t a n c e # 1 r e s i s t a n c e # 2 r e s i s t a n c e # 3 dq D i D o 06-361 page 12 Spring, 2001 where k w is the thermal conductivity oI the pipe wall and x w is the wall thickness. Since the inner and outer areas are somewhat diIIerent, we use an average oI them in (11): where d A dA dA dA dA D D dL D D D dL L o i o i o i o i L = ÷ = ÷ = ln ln t t is the log-mean oI dA o and dA i . Now we deIine the overall heat-transfer coefficient U o as dq U dA T T o o h c ÷ ÷ (12) This is called 'overall¨ because we are using the overall driving Iorce in the deIinition. Solving (12) Ior the overall temperature diIIerence, then writing this overall as the sum oI the individual temperature diIIerences dq U dA T T T T T T T T o o h c h hw dq h dA hw cw dq k x d A cw c dq h dA i i w w L o o = ÷ = ÷ + ÷ + ÷ ¸_ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸_ ¸ ¸ ¸ Substituting (9), (10), and (11): dq U dA dq h dA dq k x d A dq h dA o o i i w w L o o = + + Dividing through by dq dA o leaves: 1 1 1 U h D D x k D D h o i o i w w o L o = + + aIter substituting the expressions relating the dA`s to D`s and dL. Heat transIer coeIIicients (either U or h) are like conductances in electricity: the heat Ilux is proportional to them (like electrical current through a resistor is proportional to its conductance). The reciprocal oI the heat transIer coeIIicients (1/U or 1/h) is like an electrical resistance. The Iormula above (relating 1/U to 1/h´s) is like that relating the overall resistance oI a circuit composed oI three resistors in series: the total resistance is the sum oI each oI the three individual resistances. 06-361 page 13 Spring, 2001 FOULING FACTORS The useIulness oI analogies is that we can extend our knowledge in one area using knowledge in other. Eor example, iI there are additional layers through which heat must be conducted, then we can just add their resistances to get the total. A situation involving additional layers is: fouling deposition oI dirt, scale or any solid deposit on one or both sides oI the tube Eouling generally reduces heat transIer by imposing an additional resistance. It's eIIect is accounted Ior by use oI Iouling Iactors: h k x d d d = There may be a Iouling Iactor Ior both the inside and the outside oI the pipe. Adding these resistances to the others: 1 1 1 1 1 U h h x k D D h D D h D D o o do w w o L di o i i o i = + + + + fouling outside pipe fouling inside pipe ¸ ¸_ ¸ ¸ ¸ By means oI the electrical analogy, we are able to guess the Iorm oI the correction without having to go through the analysis. CORRELATIONS TO ESTIMATE `S In all the problems we have worked so Iar, we have been given the values oI the 1-phase heat transIer coeIIicients (the h i and h o ). Now we are going to turn our attention to how these can be estimated iI they are not otherwise known. The empirical correlations appropriate Ior a given situation usually depend on three questions: 1. What is the geometry? (e.g Ilow inside pipe, Ilow across bank oI tubes, Ilow around sphere) 06-361 page 14 Spring, 2001 2. Is the Ilow turbulent or laminar? 3. Does a phase change occur? Our book presents a number oI common correlations which cover the most common situations: Sect. 4.5, 4.6 - without phase change laminar Ilow inside tubes p238 (not covered in class) turbulent Ilow inside tubes 239-244+ Sect. 4.8 - with phase change (not covered in class) Eor Ilow inside long tubes, the criteria Ior turbulent Ilow is simple: just calculate the Reynolds number: N v D GD i i Re = = µ µ µ where µ ÷ Iluid density µ ÷ Iluid viscosity v is the cross-sectional average Iluid velocity The latter quantity is deIined as i Q v S = where Q ÷ volumetric Ilowrate through tube (units oI volume per time) S i ÷ tD i 2 /4 ÷ cross-section area oI tube Another term which is sometimes used to calculate Reynolds number is the mass velocity: i i Q m G v S S µ ÷ µ = = ` (units oI mass per area per time) The mass velocity is oIten used in problems involving gas Ilow where both density and velocity depend on pressure, but their product is independent oI pressure (since it represents mass Ilowrate and at steady state, a mass balance requires the mass Ilowrate to be the same at all axial positions). However Reynolds number is calculated, the criteria Ior turbulent Ilow is as Iollows: N Re ~ 6000 ¬ Ilow is turbulent N Re · 2100 ¬ Ilow is laminar 2100 · N Re · 6000 ¬ Ilow is in transition + You should make a special effort to read these pages before working the HWK problems, since I don`t have time in class to present all the details. 06-361 page 15 Spring, 2001 h for 1urbulent Flow In Long Pipes Eor turbulent heat transIer in long pipes, the individual Iilm coeIIicient inside a pipe can be estimated using the Sieder-1ate equation |eq. (4.5-8) in G|: + h D k v D c k i i N i N p N w Nu ¸ = 0 023 0 8 1 3 014 . Re Pr . / . µ µ µ µ µ ¸ _ ¸ ¸ ¸ ¸_¸ where k ÷ thermal conductivity oI Iluid inside tube c p ÷ heat capacity (speciIic heat) oI Iluid inside tube µ ÷ viscosity oI Iluid inside tube Unless speciIically indicated, all Iluid properties are evaluated at the mixing-cup temperature oI the Iluid: µ, k, c p evaluated at T The subscript 'w¨ indicates the property is to evaluated at the wall temperature: µ w evaluated at T w Experimentally, one Iinds that heat transIer coeIIicients can be diIIerent Ior heating and cooling over the same temperature range. The reason, is that the viscosity at the wall is very important in determining transport and µ w can be very diIIerent Irom µ. This is accounted Ior in the µ/µ w Iactor. Which 1 do you use for property evaluations? When the inlet and outlet Ts diIIer, what T do you use to evaluate the properties at? Use arithmetic mean values Ior T and T w : T T T a b = + 2 and T T T w wa wb = + 2 where the subscript 'a¨ reIers to the inlet and 'b¨ reIers to the outlet. Since the outlet T and the outlet T w depend on the heat transIer coeIIicient, this makes the determination oI h a trial-and- error (see Example 4.5-2). + The coefficient of 0.023 in this equation is the value which appears in McCabe, Smith & Harriott. Geankoplis gives a value of 0.027 instead. In homeworks and exam, you can use either value. The Notes use 0.023 because most of the homework problems using this equation were taken from MS&H. 06-361 page 16 Spring, 2001 Chapter 4. Shell-and-Tube Heat Exchangers The value oI h o Ior shell-and-tube heat exchangers depends on arrangement oI tubes and baIIles inside the shell. We will Iirst describe typical arrangements and then give a common correlation Ior h o . TUBE PITCH The bundle oI tubes in a shell-and-tube heat exchanger can be stacked in one oI two diIIerent ways The use oI triangular pitch allows the tubes to be more tightly packed -- more tubes and thereIore more area per unit volume oI shell. This makes the shell cheaper. On the other hand, the square pitch has the advantage that it is easier to clean. You can get a long brush in between two rows or between two columns. OI course, the shell has to be pulled oII oI the tube bundle beIore you can get at it in either case. Regardless oI which geometry is used, the center-to- center distances between adjacent tubes is called the tube pitch. BAFFLES To some extent, it is advantageous to reduce the cross-sectional area available Ior Ilow. Eor the same volumetric Ilowrate, this will increase the mass Ilux or the mass velocity oI the Iluid Ilow, which will increase h o . OI course, this also increases the pressure drop and the pumping cost. 06-361 page 17 Spring, 2001 A common technique Ior thus increasing h o is to install baIIle plates, which partially block the cross-sectional area on the shell side. A baIIle plate is a disk whose diameter is the inside diameter oI the shell: D s ÷ diameter oI shell with holes drilled through it Ior the tubes to pass through. Part oI the disk is cut oII to Iorm what is called the baffle window, which is the cross section available to the shell-side oI the Iluid. When the height oI the window is 1/4 oI the shell diameter, we call the plate a 25" baffle. A number oI these plates is placed along the length oI the exchanger with the location oI the baIIle window alternating between the top and bottom oI the plate. A second advantage oI these plates is that they prevent the Iormation oI large stagnant regions (having low heat transIer rates) which usually accompany 'channelling¨ oI the Iluid in a direction path connecting inlet to outlet. The spacing between the baIIle plates is called the baffle pitch: P ÷ baIIle pitch Typical baIIle pitch is a Iraction oI the shell diameter: 0.2 · P/D s · 1 Besides reducing the area available Ior Ilow on the shell side, the baIIles also Iorce the Iluid across all oI the tubes, which normally will enhance the heat transIer coeIIicient. FOR SHELL-AND-TUBE EXCHANGER ReIerence: McCabe, Smith & Harriott, 5 th ed., p432-436 06-361 page 18 Spring, 2001 An approximate but generally useIul correlation Ior estimating the shell-side coeIIicient in a shell-and-tube heat exchanger is the Donohue equation: h D k D G c k o o o e p w = 0 2 0 6 0 33 014 . . . . µ µ µ µ (13) where µ, k and c p are properties oI the shell-side Iluid. The Iorm oI this equation is very similar to the Sieder-Tate equation, except the exponents and coeIIicient are diIIerent. Another diIIerence is that the mass velocity G e is a weighted average G G G e b c = (geometric mean) oI the mass velocity through the baIIle window G m S b b = ` and the mass velocity Ior crossIlow perpendicular to the tubes G m S c c = ` where S b is the area available Ior Ilow oI the shell-side Iluid through the baIIle window and S c is the intersticial area available Ior crossIlow perpendicular to the bank oI tubes at the widest point in the shell. S f D N D b b s b o area of blue region ¸ = ÷ t t 2 2 4 4 (14) where D s ÷ diameter oI entire shell N b ÷ number oI tubes in baIIle window f b ÷ area Iraction oI baIIle plate that is window Eor a 25° baIIle plate, we have f b ÷ 0.1955 (15) The second area is calculated Irom 06-361 page 19 Spring, 2001 S PD D p c s o = ÷ 1 p ÷ tube pitch P ÷ baIIle pitch Appendix: Derivation for S and S In case you are curious as to how these Iormulas we obtained, I have put their derivation in this appendix. D o D s p P widest row oI tubes area Ior cross-Ilow (shaded) S c is the area open to Iluid Ilow across the widest row oI tubes in the shell and between two adjacent baIIle plates. Eirst the total area (ignoring tubes) is total area ÷ PD s where P is the baIIle pitch. To calculate the Iraction oI PD s which is open. let's subdivide D s into repeating unit cells: length oI unit cell ÷ p (tube pitch) length which is open ÷ p-D o Iraction open ÷ (p-D o )/p ÷ 1 - D o /p Multiplying this Iraction by the total area gives the area open: S PD D p c s o = ÷ 1 (16) 06-361 page 20 Spring, 2001 OI course, we have ignored some end eIIects which could be signiIicant iI the number oI tubes is not large. Likewise the open area oI a baIIle window is the total area oI the window minus the area blocked oII by tubes: S f D N D b b s b o = ÷ t t 2 2 4 4 where f b ÷ 0.1955 (Ior a 25° baIIle) and N b ÷ no. oI tubes thru window To get this value oI f b , we have to do a little geometry. Eirst, recall that the area oI a circle oI radius r is tr 2 . A r circle = t 2 A 'sector¨ is a portion oI a circle (see Iigure at right): A r sector = 1 2 2 2u ¸ total angle subtended by sector where 2u is the total angle subtended by the arc oI the sector. Note that iI we substitute 2t Ior the 2u, we get the A sector ÷ A circle . The area oI the triangle whose sides are x,v, and r equals A bh xv triangle = = 1 2 1 2 Einally, the area oI the segment is calculated Irom A segment ÷ A sector - 2A triangle Eor a 25° baIIle, the height oI the segment, r-x, will be 25° oI the diameter -- or 50° oI the radius: x r v r x r = ¬ = = ° = ¬ = ÷ = ÷ 05 05 60 3 3 2 1 2 2 . cos . u t A r r r r triangle = = = 1 2 2 3 2 3 8 0 217 2 2 . 06-361 page 21 Spring, 2001 A r r tor sec = = u t 2 2 3 A r r r r segment = ÷ = ÷ = t t 3 2 3 8 3 3 4 0 61418 2 2 2 2 . Now f b is the Iraction oI the total area oI the circle which is represented by this segment: f A A r r b segment circle = = ÷ = = t t 3 3 4 0 61418 31416 01955 2 2 . . . MULTIPASS CONSTRUCTION Just as we can improve the heat transIer coeIIicient on the shell side by reducing the baIIle pitch, we can improve the heat transIer coeIIicient on the tube side by restricting the Ilow to halI oI the tubes. Both have the eIIect oI reducing the area available Ior Ilow, thereby increasing the Iluid velocity and the heat transIer coeIIicient. Suppose we partition the maniIold in such a way that the inlet Iluid can only enter halI oI the tubes. The other halI oI the tubes are used Ior return Ilow. We say that such an exchanger has two tube passes. Although two tube passes improves h i , one oI the passes will be co-current. A 2-4 exchanger improves both h i and h o , and in addition, the Ilow is more nearly countercurrent. We could also partition the shell side with a horizontal baIIle in addition to the vertical baIIles, producing two shell passes. The most commonly used designs are: shell-side Ilow tube side Ilow tube sheets shell-side Ilow tube side Ilow tube sheets 06-361 page 22 Spring, 2001 - 1-2 exchanger - 2-4 exchanger where the Iirst number is the number oI shell passes and the second number is the number oI tube passes. An even number oI tube passes is desireable because then input and output oI the tube-side Iluid occur on the same end oI the exchanger. This allows the other end oI the exchanger to "Iloat" in response to thermal expansion oI the metal, which can otherwise be a serious problem when the shell changes temperature during startup or shutdown. "TRUE" MEAN TEMPERATURE DIFFERENCE Eor multipass construction, the Ilow is neither counter-current nor co-current. As a consequence, the average driving Iorce is no longer the LMTD. As it turns out, what is called the true mean temperature difference (MTD) is somewhat less than the log-mean: AT m ÷ F T AT L (17) This is the AT, which when multiplied by UA gives the total rate oI heat transIer Ior a shell-and- tube exchanger: q T ÷ U o A o AT m F T is a correction Iactor which is unity Ior pure co-current or pure counter-current Ilow: F T ÷ 1 Eor pure crossIlow to a bank oI tubes, or Ior a mixture oI crossIlow and counter-current Ilow, the Iactor is less than unity: F T ÷ F T (Y, Z, type oI Ilow) s 1 where Y and Z are ratios oI temperature diIIerences: Y T T T T cb ca ha cb = ÷ ÷ and Z T T T T ha hb cb ca = ÷ ÷ where the subscript 'c¨ reIers to the cold Iluid while the subscript 'h¨ reIers to the hot Iluid and the subscript 'a¨ reIers to the inlet while 'b¨ reIers to the outlet. 06-361 page 23 Spring, 2001 F T is usually plotted as a Iunction oI Y Ior diIIerent values oI Z. Such a plot is sketched at right. More accurate plots can be Iound in Geankoplis`s book: 1-2 HX ÷ Eig. 4.9-4a 2-4 HX ÷ Eig. 4.9-4b Note that as Z÷0, F T ÷ 1 over the entire range oI Y·1. This special case is applicable Ior condensing vapors. As long as the vapor is condensing, its temperature stays at the boiling point: T ha ÷ T hb ÷ T bp so that Z ÷ 0 and F T ÷ 1 Ior condensing steam as the hot Iluid. Similarly iI we have boiling water as the 'cold¨ Iluid, its temperature doesn`t change (i.e. T ca ÷ T cb ÷ T bp ) and then we have Y ÷ 0 and F T ÷ 1 Although the Ilow in the exchanger is complex, we calculate AT L in (17) as iI the Ilow were counter-current. The deviation caused by not being counter-current is included in the values oI the empirical correction Iactor F T . Example: A steel mill is considering whether a spare heat-exchanger will be able to cool oI 3000 gal/hr oI a dilute acid Irom 250°E to 180°E using 5000 gal/hr oI 80°E river water. You are asked to estimate: a) the true-mean driving Iorce, AT m , and b) the shell-side coeIIicient, h o . Description of available heat exchanger: type: 1-2 tubes: 158 1-inch, 14BWG stainless tubes, 16 It long, stacked in 1/ inch square pitch baIIles: 25° with 6 inch pitch shell: 20 inch inside diameter, mild steel Solution: Eirst we have to determine the outlet temperature oI the cold stream. The rate oI energy lost by the hot stream is the rate oI energy gained by the cold stream q m c T T m c T T T h ph ha hb c pc cb ca = ÷ = ÷ ` ` 06-361 page 24 Spring, 2001 Assuming the physical properties oI dilute acid and river water are the same as pure water, we can take c ph ÷ c pc so that heat capacity can be cancelled out oI the second equation above, leaving T T T T m c m c m m cb ca ha hb h ph c pc h c ÷ ÷ = ~ = = ` ` ` ` . 3000 5000 0 6 gal hr gal hr (18) Erom the problem statement we have T ha ÷ 250°E T hb ÷ 180°E T ca ÷ 80°E Substituting these into (18) and solving Ior T cb , we obtain T cb ÷ 122°E + Next, we calculate the two quantities needed to evaluate the true-mean driving Iorce: Z T T T T ha hb cb ca = ÷ ÷ = = 1 0 6 167 . . and Y T T T T cb ca ha ca = ÷ ÷ = 0 247 . Since this is a 1-2 exchanger, we read F T Irom Eig. Eig. 4.9-4a: F T ÷ 0.97 Although the Ilow in the exchanger is complex, we calculate AT L as iI the Ilow were counter- current: AT 1 ÷ T ha -T cb ÷ 128°E and AT 2 ÷ T hb -T ca ÷ 100°E A A A A A T T T T T L = ÷ = ° 1 2 1 2 113 ln E Einally 110 E L m 1 T F T A = A = ° The shell-side coeIIicient (h o ) can be estimated using Donohue`s correlation (13). Since acid is corrosive, we would be well-advised to place it inside the stainless tubes, leaving river water on the shell side (which is mild steel). With this choice, the mass Ilowrate is + Such a high temperature in the river could be harmful to fish and other wildlife. Engineers need to be mindful of the impact of their designs on the environment as well as on the health and safety of people in the vicinity. Eaced with this result, the engineer should suggest altering the design (e.g. increase the flowrate of cooling water to obtain a lower increase in temperatue). Eor the purpose of this academic example, we will ignore this objection and proceed through the analysis. 06-361 page 25 Spring, 2001 ` . . m = = × 5000 8 34 417 10 4 gal hr lb gal lb hr Eor crossIlow, the maximum available area Ior Ilow can be calculated Irom (16): S PD D p c s o = ÷ = 1 0167 . It 2 using P ÷ 6 inch, D s ÷ 20 inch, D o ÷ 1 inch (see Appendix A.5-2), and p ÷ 1.25 inch. The mass velocity Ior cross-Ilow is G m S c c = = × ` . 2 5 10 5 lb It - hr 2 Let`s assume that the Iraction oI tubes which pass through the baIIle window equals the Iraction oI the shell cross-section represented by the baIIle window. In other words: N b ÷ f b ×158 ÷ 0.1955×158 ÷ 30.8 where 158 was given as the total number oI tubes in the shell and f b ÷ 0.1955 applies Ior 25° baIIle windows (see (15) and Appendix on page 20). Now everything in (14) is known: S f D N D b b s b o = ÷ = t t 2 2 4 4 0 258 . It 2 The mass velocity through the baIIle window is G m S b b = = × ` . 162 10 5 lb It - hr 2 and the eIIective mass velocity is G G G e b c = = × 2 01 10 5 . lb It - hr 2 BeIore we can estimate h o Irom (13), we need some Iluid properties. Since we have no idea what the wall temperature is, let`s just get a rough estimate oI h o by setting µ/µ w ÷ 1. The other properties (besides µ w ) are evaluated at the average temperature oI the shell-side Iluid |T ÷ 0.5(80°E ¹ 122°E) ÷ 101°E (38.3°C)|. Using Appendix A.2: µ ÷ 0.681 cP, k ÷ 0.363 BTU/It-hr-°E, µ ÷ 62.0 lb/It 3 , c p ÷ 0.999 BTU/lb-°E (13) yields h o ÷ 363 BTU/It 2 -hr-°E 06-361 page 26 Spring, 2001 Once we have h o and h i we can estimate T w and iterate until the area required no longer changes. 06-361 page 27 Spring, 2001 Chapter 5. Evaporation evaporation -- concentrating a solution containing a nonvolatile solute by boiling away the solvent (usually water) Examples: - production oI orange juice concentrate - production oI concentrated H 2 SO 4 - production oI distilled water Usually the product is a concentrated solution (called the 'liquor¨) and the vapor is condensed and discarded. liquor - the concentrated solution An important exception is the production oI distilled water Irom tap water. Here the condensed vapor is the product and the more concentrated solution oI minerals or (in the case oI desalination) salts is discarded. Evaporation is not the only unit operation that involves boiling. Some others that involve boiling include: - distillation - involves two or more volatile components (evaporation has just one) - drying - product is a solid (in evaporation, product is usually a more concentrated liquid solution) - crystallization - product is a slurry oI crystals precipitated Irom solution (not a simple solution) Usually these diIIerences are suIIicient to require diIIerent equipment to perIorm the operation. EQUIPMENT FOR EVAPORATION There are several types oI evaporators, but they all have the Iollowing parts: Part #1: HX (Ior adding the latent heat oI vaporization to the liquid Ieed). Usually HX is a shell-and-tube with condensing steam as the hot Iluid on the shell side (because steam is Iree oI minerals which Iorm scale). Part #2: vapor space or vapor head (larger chamber in which liquid, entrained in the vapor as droplets or Ioam, can be separated Irom the vapor, usually by impinging stream onto a plate.) You also need something to move the liquid through the heat exchanger. Sometimes a pump is used and sometimes it's just leIt to gravity, either in the Iorm oI a Ialling Iilm on the outside oI the HX tubes or the entrainment oI liquid by rising bubbles. 06-361 page 28 Spring, 2001 Part #3: liquid mover - pump - gravity - Ialling Iilm - gravity - entrainment by rising bubbles Long-1ube Jertical Evaporator When the heat exchanger tubes are oriented vertically, the bubbles produced by boiling rise and carry liquid with them. Tubes are typically 1 or 2 inches in diameter and 12 to 32 Ieet in length. The liquid entrained in the vapor coming out oI the HX is richer in solute than that entering the HX. Vapor and Iine liquid droplets are separated in the vapor head by placing a baffle plate in the path oI the two-phase Iluid coming out oI the HXer. Notice the vent at the top oI the shell-side oI the HX. Usually a very small Iraction oI the steam Ilow is allowed to bleed out oI the shell. This is done to allow air which is oIten mixed with the steam to escape. Otherwise non-condensable gases will accumulate in the shell, causing the pressure to rise and decreasing the steam Ilowrate. Long-tube vertical evaporators are especially eIIective in concentrating liquids that tend to Ioam. Eoam is broken when the two-phase mixture hits the baIIle plate. 06-361 page 29 Spring, 2001 Forced-Circulation Evaporator Many concentrated solutions are highly viscous. Eor example, a concentrated sugar solution can have a viscosity that is 10 6 times larger than water. II the concentrated solution is highly viscous, the velocity oI rising bubbles will be very small and the heat transIer coeIIicient will also suIIer. Eor very viscous liquids, a pump is oIten used to improve the circulation. OI course a pump also raises the pressure signiIicantly which can prevent the liquid Irom boiling. In a Iorced-circulation evaporator, boiling usually occurs near the end oI the HX where the pressure is lower or in the pipe leading Irom the HX to the vapor space. HX tubes are now oriented horizontally so they are easier to clean. Again a baIIle plate is used to separate vapor Irom entrained liquid. EVAPORATOR PERFORMANCE Regardless oI which type oI evaporator we use, there are two standard measures oI perIormance: capacity pounds water evaporated per hour economy lb water evaporated per lb steam used (usually somewhat less than unity) EXAMPLE: SIMPLE EVAPORATOR PROBLEM Suppose we are trying to concentrate a solution oI orange juice Irom 10wt° solids to 50wt° in a long-tube vertical evaporator. Our past experience has shown that a reasonable heat transIer coeI. is Given: U ÷ 500 Btu/h-It 2 -°E To keep the boiling T low enough, we keep the vapor space under vacuum: p ÷ 4 inHg 06-361 page 30 Spring, 2001 II the Ieed is 55,000lb/hr at 70°E, what area oI heat exchanger is required and what Ilowrate oI 15 psig steam? Eind: ` m s , A Eirst we do a mass balance to determine the Ilowrates. A mass balance on the solids yields: 10°(55,000 lb/hr) ÷ 50°( ` m) so ` m ÷ 11,000 lb/hr (liquor) is the Ilowrate oI the liquor. The corresponding Ilowrate oI the vapor must be (Irom an total mass balance): ` m f - ` m ÷ 44,000 lb/hr (vapor) The heat duty oI the HXer is just the heat gained by the Ieed. In terms oI enthalpies oI the streams: q ÷ vapor ¹ liquor - Ieed q ÷ ( ` m f - ` m)H v ¹ ` mH - ` m f H f (19) Eor high-molecular-weight substances like orange juice and other 'organic colloids¨ v we can neglect boiling-point elevation and heat of dilution. Then the enthalpies oI the Ieed and the liquor are essentially the properties oI pure water under those conditions. The enthalpy oI liquid water and water vapor can be read Irom the Steam Tables Iound in Appendices A.2-9 and A.2-10 (pages 857-861). + v 'Organic colloids¨ are solutions of high molecular-weight polymers or dispersions of fine particles. Then the molar concentration of these solutions is negligibly small and their properties are close to that of pure water: see example on page 35. An example of an organic colloid is orange juice. + Eor consistency with later examples in which heat of dilution effects are important, we read all the enthapies from the steam tables, which have a reference state (where H is defined to be zero btu/lb) corresponding to saturated liquid water at its freezing point. In Example 8.4-1 on p498, Geankoplis works a very similar problem in which all enthalpies are calculated using the conditions of the liquor as the reference state. Then (by definition) H in (19) is zero; the enthalpy of the feed H f differs from that of the liquor just in sensible heat, which can be estimated as (8.4- 9); and the enthalpy of the vapor H v differs from that of the liquor just in latent heat (i.e. H v is just the heat of vaporization at the T,p of the vapor space). See the discussion on 'steam economy¨ on page 33. Try re-working Example 8.4-1 evaluating the enthalpies appearing in (8.4-7) using the steam tables (as above): you should get very nearly the same answer. 06-361 page 31 Spring, 2001 at 70 °E: + H f ÷ 38.09 btu/lb m But what is the temperature oI the liquor? Well, since the liquor is in equilibrium with the vapor, their temperature and pressure should be the same. Since the vapor is just saturated steam at 4 inHg, we look up its properties in the steam table: p v ÷ 4 inHg ÷ 1.97 psia Irom App. A.2-9 (p858): T ÷ 125.1 °E Interpolating the tables, + we obtain at 125.1 °E: H v ÷ 1115.7 btu/lb m and H ÷ 93.1 btu/lb m Substituting back into (19): q lb hr btu lb lb hr btu lb lb hr btu lb btu hr = + ÷ = × + × ÷ × = × 44 000 11157 11 000 931 55 000 38 09 491 10 010 10 0 21 10 480 10 7 7 7 7 , . , . , . . . . . Now, using the design equation, we can calculate the required area: q ÷ UAAT m (20) In a typical evaporator problem, steam is condensing on the shell side while water is evaporating on the tube side. When a phase change occurs with a single component (water in this case), the temperature stays at the boiling point. Thus the driving Iorce AT is practically constant along the length oI the heat exchanger. + Strictly speaking, the enthalpies tabulated in A.2-9 are for saturated liquid, meaning that the pressure on the liquid equals the vapor pressure. Here the pressure on the feed is probably 1 atm (or 14.7 psia) rather than 0.3622 psia (the vapor pressure). Pressure has a very weak effect on the enthalpy of liquid water and the effect of pressure usually neglected. Eor example, the effect of a change on pressure equal to 1atm on the enthapy of liquid water can be estimated as cm btu 1 atm 1 0.044 lb H p J g | | A = A ~ = ] \ ¹ which is equivalent to a change in temperature of only 0.044 °E. As long as we are dealing with much larger sensible temperature changes, we can neglect the effect of pressure on the enthalpy of liquid water. + In working the homework assignments, you will have frequent need to perform this interpolation. A Mathcad document has been prepared to assist you: http://www.andrew.cmu.edu/course/06-202/Steamtab.mcd This file reads the steam table from http://www.andrew.cmu.edu/course/06-202/STEAMTAB.PRN. Links to both files can be found at our homepage. 06-361 page 32 Spring, 2001 However, some changes might still occur at either end oI the heat exchanger. Eor example, steam might enter superheated, meaning that it`s temperature is above the boiling point. Until the temperature drops to the boiling point, condensation does not occur. Similarly, so much heat might be removed Irom the steam, that it completely condenses and is then subcooled below the boiling point. Einally, the Ieed solution has to be heated up to its boiling point beIore evaporation occurs. v The temperature proIile at right illustrates the most general situation. Eor a preliminary design (which is all we will do in this course), it is customary to approximate the 'true mean¨ MDT as: AT m ~ T s - T where T s ÷ T oI sat`d steam at p in steam chest T ÷ T bp oI conc'd liquor at p in vapor space Knowing the pressure oI the steam, we can Iind its temperature in the steam tables: at 15 psig ÷ 29.7 psia: T s ÷ 250°E The heat oI vaporization ì s is the diIIerence between the enthalpy oI saturated vapor H s and the enthalpy oI the condensate H c leaving (assumed here to be saturated liquid): ì s ÷ H s H c ÷ 1164.2 btu/lb - 218.6 btu/lb ÷ 945.6 btu/lb The driving Iorce Ior heat transIer is the diIIerence in temperature oI saturated steam and saturated vapor: AT m ~ T s -T ÷ 250-125 ÷ 125°E Now everything in (20) is known except Ior A: 4 77 10 500 125 7 . × = ÷ ÷° ° Btu h Btu h It E E 2 A v The heat required to preheat the feed from 70°E (the feed conditions) to 125°E (the boiling point at the pressure in the vapor space) is lb btu btu btu 55000 93.1 38.09 0.30 10 hr lb lb hr | || | ÷ = × ] ] \ ¹\ ¹ This is only 6.3% of the heat duty q. We are willing to tolerate this size of error in our preliminary design. 0 Position in Hxer L T e m p e r a t u r e Steam enters superheated Steam leaves subcooled Eeed solution enters subcooled T s T 06-361 page 33 Spring, 2001 or A ÷ 768 It 2 The amount oI steam required is just the heat duty divided by the latent heat oI the steam: 7 Btu 4.80 10 lb h 50, 800 Btu hr 945.6 lb s s q m × = = = ì ` Recall that our two main perIormance indicators were 'capacity¨ and 'economy¨. Capacity is just the rate oI evaporation, which will be the Ilowrate oI the vapor. Eor this example, capacity = ` m lb hr f The economy is the mass oI vapor produced per mass oI steam consumed: economy = ÷ = = ` ` ` , , . m m m lb hr lb hr f s 44 000 50 800 087 Steam Economy Why is the economy less than one? The answer becomes clear, iI (instead oI looking up the enthalpies oI water and its vapor in a table) we estimate them as sensible and latent heat. Eirst, let`s select the liquor as the reIerence state Ior the calculations oI the enthalpies. Then choose liquor as reIerence: H ÷ 0 The vapor is at the same temperature as the liquor so there are not sensible heat changes, but the vapor has a much higher enthalpy owing to its heat oI vaporization: H v ÷ ì Since the Ieed is also a liquid (like the liquor), its enthalpy diIIers Irom the liquor only in sensible heat: H f ÷ c p (T f - T) Substituting these enthalpies in (21): f f p f s s m m m c T T m ÷ ì ÷ ÷ = ì ` ` ` ` 06-361 page 34 Spring, 2001 or 1 f f s p f s s m m m c T T m m < ÷ ì = ÷ ÷ ì ` ` ` ` ` Generally the heat oI vaporization oI water decreases with increases in temperature. Since the steam must be hotter than the water being boiled, this means that ì s · ì. So the Iirst term in the equation Ior economy is already less than one. The second term decreases it Iurther; this term represents the energy required to preheat the Ieed up to the temperature oI the liquor where boiling occurs. So a pound oI steam boils less than one pound oI water because some oI the energy oI the steam is required to preheat the Ieed. However, even iI the Ieed were at the same temperature as the liquor, a pound oI steam would boil less than one pound oI water because the latent heat oI the hot steam is less than the latent heat oI the water being boiled. OVERVIEW OF EVAPORATOR DESIGN The design equation Ior evaporators can be summarized as heat duty: ` ` ` ` ` m m H m H mH q m H H f v f f s s c ÷ ÷ + = = ÷ ¸ vapor feed liquor steam ¸ _ ¸¸ ¸ ¸¸ ¸_¸ ¸ _ ¸¸ ¸ ¸¸ (21) HXer area: q ÷ U A AT m Eor preliminary designs, we can approximate AT m ~ const. ÷ T s - T where T is the boiling point oI the liquor at the pressure in the vapor space. This is approximate when there is signiIicant BPE because T will vary along the length oI the heat exchanger tube as the concentration oI solute also changes. - Evaluation oI enthalpies with negligible BPE and negligible heat oI dilution (example on page 29 and also Example 8.4-1 on p498). Assuming the solvent is water, we can read the enthalpies oI all three streams Irom the steam tables. read T, T s , H, H f and H v Irom steam tables - Evaluation oI enthalpies with complications (Example 8.4-3 on p501): read H, H f Irom special table or chart read T Irom Dühring chart; T s , H v Irom steam tables + + Since enthalpies are being taken from more than one source, we should take care that the reference state is the same for all tables and charts used in a given problem. The steam tables in Geankoplis use saturated liquid water at 32°E and 1 atm as the reference state. The enthalpy chart for NaOH solutions in Geankoplis also use 0 wt% NaOH at 32°E and 1 atm as the reference state. 06-361 page 35 Spring, 2001 COMPLICATIONS Boiling-Point Elevation Some complications arise in other evaporator problems because the enthalpy and partial pressure oI a liquid more generally depends on its concentration as well as its temperature. Whenever a solute is added to water, the partial pressure oI the water is reduced. Recall Raoult's law Irom thermodynamics. Roault`s law: p i (x i ,T ) ÷ x i p i o (T ) where p i ÷ partial pressure oI component i x i ÷ mole Iraction component i p i o ÷ vapor pressure oI pure i at some T Note that adding any solute with water causes x w ·1 so that p w · p w o In general, this implies that I must heat the solution up to a higher temperature (i.e. higher than the b.p. oI pure solvent) beIore it will boil. This increase in boiling point due to addition oI solute is called boiling-point elevation (BPE) - increase in b.p. due to addition oI solute This eIIect can be quite signiIicant Ior concentrated solutions oI inorganic salts. Eor example, Ior at a pressure oI 1 atm, the boiling point oI pure water is 1 atm, 0° NaOH in water: T bp ÷ 212°E II the pressure is other than 1 atm, the temperature can be read Irom the steam tables. Given this boiling point Ior pure water, we can look up the boiling point oI solutions oI NaOH Irom a Dühring chart (see p500 oI Geonkoplis): 50wt° NaOH in water: T bp ÷ 283°E at 1atm Note that the boiling point oI the solution is signiIicantly greater than the boiling point oI pure solvent. BPE is a colligative property, which means colligative property -- depends on molar (not mass) concentration Example: Use Raoult`s law to estimate the partial pressure oI water at 100°C oI a) a 50wt° NaOH solution and b) a 50wt° suspension oI 'organic colloids¨ (use MW average ÷ 1000 g/mol). Estimate the BPE Ior each. 06-361 page 36 Spring, 2001 Solution: To calculate the mole Iraction oI water in the solution, take 1 gram oI solution as a basis. This 1 gram mixture consists oI 0.5 g oI water (MW ÷ 18) and 0.5 g oI NaOH (MW ÷ 40). We calculate the corresponding number oI moles oI each and the mole Iraction: NaOH (MW÷40) Organic Colloids (MW÷1000) x w 05 05 05 0 690 . . . . g 18 g mol g 18 g mol g 40 g mol + = 05 05 05 1000 0 982 . . . . g 18 g mol g 18 g mol g g mol + = At 100°C (the b.p. oI pure water), the vapor pressure oI pure water is 1 atm (÷ p w o ). Using Raoult`s law, the partial pressure is estimated as p w 0.690(1 atm) ÷ 0.690 atm 0.982(1 atm) ÷ 0.982 atm To cause the solution to boil, the partial pressure must be raised to 1 atm. Eor the same mole Iraction oI NaOH, this means raising the vapor pressure to p w o 1 1450 atm 0.690 atm = . 1 1018 atm 0.982 atm = . Using the steam tables (Appendix A.2-9), we Iind that the temperature has to be T 110.0°C 100.8°C Thus using Raoult`s law, the BPE is estimated to be BPE Irom Raoult`s law 110.0°C - 100°C ÷ 10.0°C 100.8°C - 100°C ÷ 0.8°C OI course, this is only an estimate. Raoult`s law is only asymptotically correct as x w ÷1. The actual BPE is 39°C. BPE determined Irom Eig. 8.4-2 283°E - 212°E ÷ 71°E ÷ 39°C Although BPE is signiIicant Ior a low-molecular-weight solute like NaOH, the same 50wt° concentration oI an "organic colloid" (e.g. orange juice) may have a negligible BPE because the molecular weight is much higher. Thus the same wt° gives a much smaller molar concentration oI a high-molecular-weight solute and thereIore a much smaller BPE. 06-361 page 37 Spring, 2001 Heat of Dilution II you dilute concentrated sulIuric acid by adding water, large quantities oI heat can be released. II the mixing is adiabatic and both Iluids are initially at room temperature, the Iinal temperature aIter mixing can by high enough to causing boiling. You were probably warned to use great caution when diluting sulIuric acid in chemistry lab. The problem is that the enthalpy oI a solution depends not only on its temperature, but also on its concentration: H ÷ H(T,c) which means you might need more than just a heat capacity to calculate sensible heat changes. Eig. 8.4-3 gives the enthalpy-concentration diagram Ior NaOH solutions. Example: Suppose we dilute 1-lb oI 50wt° NaOH solution at 70°E with 1-lb oI pure water, also at 70°E. What is the Iinal temperature iI the mixing is adiabatic? Solution: read the enthapies oI the two solutions Irom 8.4-3: H(50°, 70°E) ÷ 123 BTU/lb H(0°, 70°E) ÷ 40 BTU/lb Multiplying each by the corresponding weight and adding, we obtain the Iinal enthalpy oI the mixture ('adiabatic¨ just means that no energy is lost during mixing): (123 BTU/lb)(1 lb) ¹ (40 BTU/lb)(1 lb) ÷ 163 BTU Dividing by the mass oI the Iinal mixture, we obtain the enthalpy per pound: 163 815 25°, BTU 2 lb BTU lb = = = . ? H T This represents the speciIic enthalpy oI the Iinal mixture, which we also know is 25wt°. Again using Eig. 8.4-3, to get this H, we need to choose the temperature as T ÷ 123°E Thus the Iinal solution is warmer than the two initial solutions, due to the 'heat oI dilution.¨ Question: when diluting sulIuric acid (which also exhibits a signiIicant heat oI dilution), why is is wiser to add (concentrated) acid to water, rather than water to (concentrated) acid? We will now look at two methods to improve steam economy oI an evaporator: 1. multiple 'eIIects¨ 2. vapor recompression 06-361 page 38 Spring, 2001 MULTIPLE EFFECTS In the simple evaporators we have dealt with so Iar, one oI the product streams is water vapor also known as steam. Why can`t this vapor be used in place oI steam? The answer is simple: iI the vapor replaced the steam, there would be no driving Iorce Ior heat transIer because the 'steam¨ and the vapor would have the same temperature. To get a driving Iorce in the right direction, we have to use steam with a higher pressure. One method oI making use oI the latent heat oI the vapor is the compress it. We will look at that option below, but Iirst let`s examine another option. One way to improve economy is to divide the heat duty among several evaporators which are piped together in such a way that the vapor produced by one can be used to boil the liquid in another. This reduces the steam usage and improves economy. You can have any number oI evaporators connnected together in this way. Each evaporator in this sequence is called an effect, and the entire sequence is called a multiple-effect evaporator. In the Iigure above, we show a double-eIIect evaporator. In order to maintain a driving Iorce Ior heat transIer in each eIIect, we must have T 2 · T 1 · T s This can be accomplished by having + p 2 · p 1 · p s This requires a vacuum pump or compressor: + Assuming we have saturated liquid and vapor in each effect, the pressure in any effect is just the vapor pressure at the temperature for that effect. Recall that vapor pressure increases with temperature. J L 06-361 page 39 Spring, 2001 p 2 · p 1 + AT 2 ÷ T 1 - T 2 ~ 0 q 2 ÷ U 2 A 2 AT 2 (22) OI course, we still have to supply steam to the Iirst eIIect AT 1 ÷ T s - T 1 ~ 0 q 1 ÷ U 1 A 1 AT 1 (23) Eor simple designs where BPE and AH dil are negligible and where the sensible heat needed to preheat the Ieed up to its boiling point is negligible, the same latent heat added to the vapor in the Iirst eIIect is recovered by condensing in the second eIIect: q 1 ~ q 2 (24) Usually the pressure in the last eIIect and the pressure oI the steam are known. Using the steam tables, we can convert these pressures into temperatures. The unknowns are the temperatures and pressure in the other eIIects. Example: given T s and T 2 , Iind T 1 . Solution: (22) and (23) into (24) yields U 1 A 1 (T s T 1 ) ÷ U 2 A 2 (T 1 T 2 ) This represents one equation in one unknown (T 1 ), assuming that the U`s and A`s are known. II instead, we have N eIIects, the analog oI (22), (23) and (24) yield N-1 equations in N-1 unknowns. These equations are strongly coupled. A simpler approach is to try and solve Ior the AT i `s instead oI the T i `s. We will illustrate this by re-solving the last example. Erom (22) and (23), we have AT q U A 1 1 1 1 = (25) and AT q U A 2 2 2 2 = (26) Adding (14) and (26): AT q U A q U A i i = = + 1 2 1 1 1 2 2 2 (27) Dividing (25) by (27): + The expression for T which follows assumes that the vapor produced in the first effect condenses in the steam chest of the second effect at the same temperature at which it boiled. If BPE is significant, the vapor will condense at a lower temperature than it boiled (even if the pressure is the same). Why? 06-361 page 40 Spring, 2001 A A T T q U A q U A q U A U A U A U A q q 1 1 1 1 1 1 1 2 2 2 1 1 1 1 2 2 1 2 1 1 1 = + ¬ + = (28) Since LAT can be easily calculated Irom the inIo given in the problem statement: A A A T T T T T T T T T s s = + = ÷ + ÷ = ÷ 1 2 1 1 2 2 (29) we can easily calculate AT 1 Irom (28): AT U A U A U A T T s 1 1 1 1 1 2 2 2 1 1 1 = + ÷ The analog oI (24) - (29) Ior N eIIects are (25): AT q U A i i i i = Ior i ÷ 1,2, ... N (30) A A T T q U A q U A q U A q U A i i i i N N N = + + + 1 1 1 2 2 2 . Applying the analog oI (24) (q 1 ÷ q 2 ÷ ... ÷ q N ÷ q), we can cancel out the q i , leaving: A A T T U A U A U A U A i i i N N = + + + 1 1 1 1 1 1 2 2 . (31) Now the sum oI the driving Iorces can also be expressed as A A A A T T T T T T T T T T T T T T N s N N N N s N = + + + = ÷ + ÷ + + ÷ + ÷ = ÷ ÷ ÷ ÷ 1 2 1 1 2 2 1 1 . . (32) (32) into (31): AT T T U A U A U A U A i s N i i N N = ÷ + + + 1 1 1 1 1 1 2 2 . Ior i ÷ 1,2, ... N 06-361 page 41 Spring, 2001 In the simple case in which all the A`s are equal and all the U `s are equal, this last result reduces to AT N T T i s N = ÷ 1 Ior i ÷ 1,2, ... N In short, the total driving Iorce is divided equally among each oI the N eIIects. Performance Each eIIect results in the evaporation oI water at a rate equal to q/ì i . Summing the rate oI evaporation in each eIIect, we obtain the total rate oI evaporation: ` m q J i i i N = = ì 1 Steam is used only in the Iirst eIIect. The rate oI steam consumption in the Iirst eIIect is ` m q s s = 1 ì The steam economy can then be calculated as ` ` m m q q N J s i i i N s q q q s i i N N = ¬ ~ = = = = = ì ì ì ì 1 1 1 1 2 1 II we neglect changes in the latent heat with temperature and pressure, then all the ì`s are equal and the sum reduces to N. II changes in the latent heat with temperature are considered, the sum will be less than N. Summary: Eor the same capacity, steam requirements are signiIicantly less Ior multiple-eIIect evaporators, than Ior single-eIIect evaporators: generally, the economy increases with the number oI eIIects. The price that is paid Ior this increased steam economy is greater equipment costs: generally the cost oI the equipment is proportional to the number oI eIIects. Eor a single eIIect: economy · 1 Eor N eIIects: economy · N The reason that the economy increased is that we need to supply steam only to the Iirst eIIect. The source oI heat Ior 2 nd , 3 rd , ... N th eIIects is supplied by condensing the vapor produced in the previous eIIect. 06-361 page 42 Spring, 2001 VAPOR RECOMPRESSION Multiple eIIects recover some oI the latent heat oI the overhead vapor by using it in place oI steam to boil the liquor in downstream eIIects. Is there any way we can use the vapor in place oI steam in a single-eIIect evaporator? The problem with using the vapor in place oI steam is that the temperature oI the vapor is the same as the temperature oI the liquor: the liquor and vapor leaving the same eIIect are already in thermal equilibrium with each other. In other words, there is no driving Iorce Ior heat transIer. without compression: AT ÷ T s - T ÷ 0 One way to increase the temperature oI the vapor is to compress it. Eig. 1 shows how adiabatic compression oI water vapor (treated as an ideal gas) increases its temperature. + Note that the temperature rises Iaster with increases in pressure than the boiling point increases. Thus adiabatic compression oI a saturated vapor nearly always produces a superheated vapor. When heat is removed Irom this compressed vapor, it will eventually condense at its boiling point Ior that pressure. Increasing the pressure oI the vapor also increases its boiling point, so that as the vapor condenses, it stays at a high temperature T s so we always have some driving Iorce. With compression: AT ÷ T s - T ~ 0 UnIortunately, raising the temperature and pressure oI the vapor actually lowers its latent heat per pound ì s · ì (because T s ~ T) + Temperatures calculated from (7.30) in Smith & van Ness (4 ed., p237). Pressure (psia) 1 10 100 T e m p e r a t u r e ( d e g R ) 1000 Boiling Point Fig. 1 Colored curves show how temperature increases with adiabatic compression, starting with saturated vapor at 1 psia (green), 14.7 psia (red) or 400 psia (yellow). For comparison, the boiling point is shown by the black curve. 06-361 page 43 Spring, 2001 so the vapor no longer has enough latent heat to boil the same mass oI liquid. Thus make-up steam is required, although generally much less steam is needed than iI the vapor`s latent heat is not recycled. The reduction in steam is not without cost. We still have to pay the utilities (electricity or steam turbine) to operate the compressor. Generally, compressors are expensive to buy, maintain, and operate; so you have to do some economic calculations to see iI using a compressor is economical. At the usual conditions oI an evaporator, the vapor can be treated as an ideal gas. Then the work required Ior adiabatic compression can be estimated Irom + W p p p pr in in out in = ÷ ÷ , = ÷ ¸ ¸ µ 1 1 1 1 | | Btu lb (33) where 'out¨ and 'in¨ reIer to the outlet and inlet oI the compressor and where ¸ is the heat capacity ratio, which is for ideal gas 1.324 p p v p c c R c c M ¸ ÷ = ~ ÷ ¸¸_¸¸ Ior water vapor using c p ÷ 0.45 Btu/lb-°E, R ÷ ideal gas constant ÷ 1.986 Btu/lbmol-R ÷ 1.986 cal/mol-K M ÷ mol. wgt. ÷ 18 lb/lbmol ÷ 18 g/mol Now (33) gives the work needed to compress a unit mass oI gas, under ideal conditions (no irreversible losses oI energy). To get the power requirements, we must multiply by the mass Ilow rate and divide by the mechanical eIIiciency (q) oI the compressor: P m m W f pr = ÷ = ` ` | | q |÷| Btu hr kW By compressing the vapor to raise its temperature, we are lowering its latent heat oI vaporization. This means that, even iI we completely condense all the vapor, we still won`t have enough latent heat to boil the liquor at its lower temperature. The diIIerence must come Irom make-up steam. To estimate the make-up steam requirements, let`s return to the design equation (19) Ior the total heat duty: + See Smith & van Ness, 4 ed., footnote on page 238. 06-361 page 44 Spring, 2001 q m m H m H mH f v f f = ÷ ÷ + ` ` ` ` ¸ vapor feed liquor ¸ _ ¸¸ ¸ ¸¸ ¸_¸ In the simple case where BPE and AH dil are negligible (i.e. Ior inorganic colloids), H f and H diIIer Irom the enthalpy oI saturated liquid only in sensible heat. To the extent that sensible heat is negligible, + we can substitute the enthalpy oI saturated liquid Ior H f and H: q m m H m H mH m m H m m H m m H H m m f v f L L f v f L f v L f = ÷ ÷ + ~ ÷ ÷ ÷ = ÷ ÷ = ÷ ` ` ` ` ` ` ` ` ` ` ` ` ì heat to boil vapor: q m m f ~ ÷ ` ` ì (34) This heat duty is partially oIIset by the latent heat recovered by condensing the compressed vapor. The remainder comes Irom make up steam: q m m m f s s s = ÷ + ` ` ` ì ì compressed vapor make- up steam ¸ _ ¸ ¸ ¸ ¸_¸ (35) where we are assuming we are compressing the vapor to the same pressure as the make-up steam. Eliminating q between (34) and (35), we can solve Ior the steam requirements: ` ` ` m m m s f s ~ ÷ ÷ ì ì 1 Notice that any compression allows us to recover all the latent heat oI the vapor, thus signiIicantly reducing steam consumption without compression. However greater compression actually increases steam consumption (because ì s decreases with pressure). Rearranging this expression, we can calculate the economy: 1 f s s s m m m ÷ ì ~ >> ì ÷ ì ` ` ` + Sensible heat is often negligible compared to latent heat. To see why, note that the latent heat of vaporization of water at 1atm, 212°E is 970.3 btu/lb, where 1 btu is the amount of sensible heat required to raise the temperature of one lb of water 1°E. This means that, for sensible heat to equal latent heat, I need to raise the temperature of water 970°E. A temperature rise of only 30°E, say, represents a sensible heat change of 30/970 or 3% of the latent heat of water. This is an estimate of the relative error made in this estimate of m . 06-361 page 45 Spring, 2001 which is now very much greater than unity. However, this way oI computing economy is unIair because the reduction in steam consumption is at the cost oI operating the compressor. In computing the 'economy¨ Ior an evaporator with vapor recompression, it is customary to add an equivalent steam rate Ior the compressor to the make-up steam rate beIore dividing into the capacity. - II the plant uses a steam-powered turbine or iI the plant uses steam generators to make its own electricity, then the apparent steam usage is btu hr lb | | | | btu lb hr s s app s s P m m = + = = ' ì q ` ` where P is the electrical power supplied to the compressor, s ' ì is the latent heat oI the steam used (the prime is added because the pressure oI the steam used here might be diIIerent Irom that used in the evaporator) and q s is the eIIiciency oI electricity generation Irom steam (typically 0.35). - On the other hand, iI electricity is purchased Irom a utility to drive the compressor, then it makes sense to convert the electricity usage into an equivalent amount oI steam (on a Iinancial basis): ` ` $ m m P s app s = + × $ kW- hr lb oI stm II P is in kW then the last term yields lb/hr oI steam. Regardless oI which method oI converting the power usage is appropriate, the economy is computed as: economy ÷ ÷ ` ` ` m m m f s app In Hwk #3, Prob. 5, we will consider the trade oII oI operating costs (direct steam usage plus electricity Ior compressor) and capital costs (heat exchanger area) as a Iunction oI the output pressure oI the compressor. The results are summarized in the graph at right, which shows that an optimum pressure exists at which total costs are minimized. Notice that both power requirements Ior the compressor as well as steam usage increase with pressure. Only when the cost oI the heat exchangers is considered does an optimum exit. $0 $50,000 $100,000 $150,000 $200,000 $250,000 $300,000 15 20 25 30 Pressure of compressed vapor (psia) C o s t s ( $ / y e a r ) Hxer area electricity steam total 06-361 page 46 Spring, 2001 Chapter 6. Vapor-Liquid Equilibrium ReIerence: Smith & Van Ness, 4th Ed., Sect. 10.5 Today we switch Irom heat-transIer operations (heat exchange, evaporation, condensation) to mass-transIer operations (distillation, extraction, absorption, humidiIication). Instead oI sizing equipment to exchange heat between two phases we will size equipment to exchange mass between two phases. NO. OF PHASES PRESENT Suppose I have an apparatus like that shown in the Iigure at right in which I can place an arbitrary mixture oI components and control the temperature and pressure. How many phases will be present at equilibrium? One Component Eor example, suppose I put just water in the piston at 1 atm and room temperature (say 70 °E) Just one component no air! Is the water liquid, solid, or vapor? We know Irom our own experience that water is a liquid near room temperature. Thus we have just one phase present at equilibrium. + II I repeated this experiment at 250 °E, we would have the water present as all vapor again I have just one phase present. Only iI the temperature corresponds to the boiling point (212 °E) will I have two phases present. II I repeated the experiment at some other pressure there will be only one temperature at which two phases are present. The relationship between temperature and pressure Ior which two phases co-exist at equilibrium is called the vapor pressure curve. This diagram summarizes all the vapor-liquid phase behavior Ior a one-component system. 1wo Components Now suppose I put two components in the pot: let`s say a mixture oI n-heptane (C 7 ) and n-octane + If air were present as a second component, we would also have a gas phase. 06-361 page 47 Spring, 2001 (C 8 ), which are important components in gasoline. The vapor-liquid phase behavior oI two- component systems is a little more complicated, because I have an additional variable composition as well as T and P. Instead oI a single temperature at which both liquid and vapor can co-exist, there is a range oI temperatures. At a given pressure, we can summarize the phase behavior by means oI a Txv diagram like that shown at right. To understand the meaning oI this diagram, consider what happens iI we charge the piston with a mixture oI heptane and octane having a particular composition. II the temperature is low enough, we will have just one phase: liquid. As we heat this mixture to the temperature corresponding to the lower curve, we would begin to see bubbles oI vapor Iormed in the liquid as more heat is added. This temperature at which the bubbles Iirst Iorm is called the bubble point when heating a subcooled liquid, the temperature at which the Iirst bubble Iorms The composition oI this Iirst bubble can also be read Irom the Txy diagram. In general, the vapor will be richer than the liquid in one component and leaner in the other. This diIIerence in composition oI the vapor and liquid can serve as the basis Ior separating two or more components oI a mixture. By completely condensing the vapor, I produce a liquid which is richer in one component than the original liquid. Thus I have achieved a degree oI separation oI the two components. Eor a binary system having two phases, we denote the composition oI the liquid and vapor using x and v: x ÷ mole Iraction oI more volatile component + in liquid phase v ÷ mole Iraction oI more volatile component in vapor phase Keep in mind that liquid and vapor must have the same temperature at equilibrium. The line connecting the two points representing vapor and liquid is called a tie line. It`s horizontal on a Txy diagram. + By 'more volatile component¨ we mean that component which has either 1) the higher vapor pressure at a given temperature, or 2) the lower boiling point for a given pressure. Example: for a mixture of C 7 (heptane) and C 8 (octane), heptane has the lower normal boiling point and so heptane is the more volatile component. 06-361 page 48 Spring, 2001 As we continue to heat the mixture, the liquid becomes less rich in the more volatile component since the more oI the more volatile component is leaving Ior the vapor phase. The compositions oI the liquid and vapor move along the heavy solid line shown in the Iigure at right. The highest temperature at which liquid remains is called the dew point when cooling a superheated vapor, the temperature at which the Iirst drop oI dew Iorms Eventually, as we continue to heat, all the mixture ends up in the vapor. + IDEAL SOLUTIONS OBEY RAOULT`S LAW ReIerence: Smith & Van Ness, Example 10.1 (p305) Now let`s try to predict the Txy diagram Ior the simplest case: an ideal solution in equilibrium with an ideal gas. So let`s suppose we are at some temperature Ior which both phases co-exist at equilibrium. Let P i ´ ÷ vapor pressure oI pure i P ÷ total pressure x i ,v i ÷ mole Iractions in liq or vapor at equil. p i ÷ v i P ÷ partial pressure oI component i Raoult's law: p i ÷ x i P i ´ IDEAL BINARY MIXTURES What we have said so Iar about ideal solutions applies to any number oI components. Let's now restrict our attention to systems with only two components, which we will call A and B. The mole Iractions must sum to one Ior each phase, so we can express the mole Iraction oI B in terms oI that Ior A: x B ÷ 1-x A + The vertical line in left margin indicates that this material was not covered during class. I do however expect you to read it and you will be responsible for this material on the exams. v i x i 06-361 page 49 Spring, 2001 v B ÷ 1-v A Applying Raoult's law to each component: p A ÷ v A P ÷ x A P A ´ (36) p B ÷ (1-v A )P ÷ (1-x A )P B ´ (37) Adding the partial pressures Irom (36) and (37) we obtain the total pressure: P ÷ p A ¹ p B ÷ x A P A ´(T) ¹ (1-x A )P B ´(T) Solving Ior the mole Iraction x A : x A ÷ (P-P B ´)/(P A ´-P B ´) (38) Irom (36): v A ÷ x A P A ´/P (39) Keep in might that the vapor pressures in these equations depend on temperature. So iI you speciIy a T and P, I can determine the vapor pressures and calculate the composition oI the vapor and liquid which will be at equilibrium. Given: T,P Look up: P A ´(T) and P B ´(T) Calculate: x A ,v A Irom (38) and (39) Thus I can construct a complete phase diagram Ior the mixture using only vapor-pressure data Ior the pure components. Example. v n-heptane and n-octane Iorm a nearly ideal mixture. Use Raoult's law to construct a phase diagram at constant pressure (1 atm). Solution. We Iirst look up the normal boiling points oI the two pure components. component 1 bp (° °° °C) denoted by n-heptane 98.4 A n-octane 125.6 B v Illustration 9.1 from Treybal. A similar example is 10.1 from S&VN (p305). 06-361 page 50 Spring, 2001 We choose the "low boiler" as component A since it has the larger vapor pressure over this range oI temperatures. Next, we choose a temperature between the two boiling points, look up vapor pressures oI the pure components and substitute into (38) and (39). The results are summarized in the table at right and the graph below. 98 100 102 104 106 108 110 112 114 116 118 120 122 124 126 0 0.2 0.4 0.6 0.8 1 mole fraction of heptane T e m p e r a t u r e ( d e g C ) Liquid Vapor L&V 1 (°C) P' (mmHg) P' (mmHg) x y o oo o 98.4 760 333 1.000 1.000 2.28 105 940 417 0.656 0.811 2.25 110 1050 484 0.488 0.674 2.17 115 1200 561 0.311 0.492 2.14 120 1350 650 0.157 0.279 2.08 125.6 1540 760 0.000 0.000 2.03 06-361 page 51 Spring, 2001 XY AND TXY DIAGRAMS The Txy diagram is a convenient graphical method Ior summarizing vapor-liquid equilibrium (VLE) data at a particular pressure. Eor binary mixtures, we usually denote the mole Iraction oI the more volatile component as x in the liquid phase as as v in the vapor phase. Eor this particular choice oI x and v, the Txy diagram will always slope downward as shown at right. Erom the Txy diagram, one can construct an 'xy¨ diagram, which is the one we will use in distillation calculations. The xy diagram is shown at right. Eor xy diagrams which resemble is one, the relationship between x and v can be sometimes be represented by a single number: the relative volatility, which is deIined as i if f K K o ÷ where the K `s are called the distribution coefficient, which in turn are deIined as: i i i v K x ÷ Since this term is also useIul in mixtures having more than two components, we will use subscripts to denote components. Generally the most volatile component will have the largest distribution coeIIicient and will have its concentration increased (in the vapor compared to the liquid) by the largest percentage. The ratio oI the distribution coeIIicients oI two components is the relative volatility. T mole Iraction oI A T bp,A T bp,B x x v v 06-361 page 52 Spring, 2001 NONIDEAL BEHAVIOR Fig. 2. Txy and xy diagrams for nonideal systems. On left is carbon disulfide in acetone at one atmosphere pressure. On right is acetone in chloroform. Taken from Treybal, 2 ed., pages 290 and 295. Examples Txy and xy diagrams Ior binary mixtures which display signiIicant nonideal behavior are illustrated in Eig. 2. In particular notice that the curve on the xy-diagram can cross the 45° line. This point oI crossing is called an azeotrope: azeotrope - vapor and liquid have the exactly the same composition at equilibrium (i.e. v ÷ x) The occurance oI azeotropes can pose diIIiculty Ior distillation since producing a new phase by condensation or by evaporation produces no separation oI components. 06-361 page 53 Spring, 2001 Chapter 7. Flash Distillation Now let`s apply these concepts about vapor-liquid equilibria to distillation. The simplest device Ior continuous distillation oI some liquid Ieed is the 'Ilash unit.¨ This is just a HXer to add enthapy oI the Ieed and a chamber to allow the vapor and liquid to separate. The schematic at right is the same as that used Ior evaporation, however the vapor now contains all oI the components not just the 'solvent¨. Given the Ieed composition and the amount oI heat added to the Ieed, we would like to determine the composition and Ilowrates oI the distillate and bottoms products. Given: F, x F and q Eind: D, B, v D and x B The mole and enthalpy balances are: F ÷ D ¹ B (40) x F F ÷ v D D ¹ x B B (41) H F F ¹ q ÷ H D D ¹ H B B (42) where F,D,B are total molar Ilowrates oI each stream and x and v are mole Iractions oI the more volatile component. The remaining equation comes Irom requiring the vapor and liquid products to be at equilibrium: (x B , v D ) lies on equil.curve (43) (40)-(43) represent Iour equations in the Iour unknowns. APPROXIMATE SOLUTION The solution to these Iour equations can be obtained in a number oI ways. Let's start with the simplest method. In many cases, the molar heat oI vaporization is nearly the same Ior both components. (Eor this reason, we tend to solve all distillation problems using mole Iraction rather than mass Iractions, and express enthalpies per unit mole rather than enthalpy per unit mass.) II you can assume that the molar heat oI vaporization oI the mixture is completely independent oI its composition, then solution is quite easy: - Assume H D -H B ÷ const, say AH v |÷| Btu/mol (independent oI v D ) - neglect sensible heat compared to latent heat J L F F x H D D x H B B x H q 06-361 page 54 Spring, 2001 Let's choose saturated liquid having the composition oI the Ieed as our reIerence state Ior enthalpies. Then H F ÷ H B ÷ 0 and H D ÷ AH v Even iI the liquid is not saturated liquid, its enthapy diIIers Irom that oI saturated liquid only through sensible heat, which is neglected here. The Iirst equality is the result oI neglecting sensible heats Then (42) reduces to: q ~ AH v D (44) Now we can deIine and solve Ior the Iraction oI the Ieed that is vaporized: f D F q H F v ÷ ~ A which can be considered known. We will now express the molar Ilowrates in terms oI this f: Dividing (40) by F and rearranging: B F D F f = ÷ = ÷ 1 1 (45) Dividing (41) by F: x v f x f F D B = + ÷ 1 (46) Solving Ior v D : v f f x x f D B F = ÷ + 1 (47) which is a straight line on the xv diagram. To plot this line, we need to know the slope and one point which lies on the line. slope ÷ f f ÷1 To obtain one point, we can substitute an arbitrary value in Ior x B , say we substitute x B ÷x F . Then (47) becomes v f f x x f x D F F F = ÷ + = 1 Thus one point on the line given by (47) is one point: (x B , v D ) ÷ (x F , x F ) 06-361 page 55 Spring, 2001 Knowing one point and the slope, we can draw the line. In eIIect, this line represents the relationship between v D and x B which is imposed by the mass and energy balances Ior a given q (or a given f) and Ior the given assumptions on calculating enthalpy. We call this line the operating line: relationship between v D and x B which is imposed by the mass and energy balances To Iind which point on this line is the solution Ior a given q, we need also to satisIy (43), which represents phase equilibrium. This relationship is a curve on the xv diagram, known as the equilibrium curve: relationship between v D and x B which is imposed by vapor- liquid equilibria The intersection oI operating line and the equilibrium curve satisIies all Iour oI the original equations and yields the answer. Inverse Lever-Arm Rule on 1xy Diagram There is also a simple graphical procedure Ior estimating the Iraction vaporized on a Txy diagram. Suppose you are given a Ieed composition and (instead oI speciIying the Iraction evaporat- ed) the Iinal temperature oI the mixture at equilibrium. Then you can locate a point (x F , T) on the Txy diagram. A horizontal line through this point intersects the vapor and liquid curves at the composition oI those phases: (x B , T) and (v D , T). As it turns out, the amount oI vapor produced is inversely proportional to the distance between the Ieed point (x F , T) and the vapor point (v D , T). Similarly, the amount oI liquid remaining is inversely proportional to the distance between the Ieed point (x F , T) and the liquid point (x B , T). More precisely, the ratio oI vapor D to liquid B Ilowrates in an continuous Ilash unit is given by D B x x v x F B D F = ÷ ÷ (48) where D, B and F are proportional to the distances indicated on the Txy diagram above. Because oI the inverse nature oI this relationship, this is called the inverse lever-arm rule. As a useIul analogy, think oI the Ilowrates as being like masses being balanced on a beam, resting on a Iulcrum (see sketch at right). The Iurther the mass is Irom the pivot, the small the mass can be. At (mechanical) equilibruim: T x B v D x F D B F l 1 l 2 M 1 M 2 06-361 page 56 Spring, 2001 1 1 2 2 M l M l = or 1 2 2 1 M l M l = Notice the inverse relationship between the mass and the distance Irom the pivot. Now we will try to prove (48): Proof: Solving (46) Ior f: f D F x x v x F B D B ÷ = ÷ ÷ (49) Substituting into (45): B F D F x x v x v x v x F B D B D F D B = ÷ = ÷ ÷ ÷ = ÷ ÷ 1 1 (50) Dividing (49) by (50), we obtain (48). RIGOROUS SOLUTION Although the approximate solution works reasonable well Ior some mixtures, Ior others it might not be good enough. Then you will need a lot more thermodynamic data besides latent heat and heat capacity. You will need an Hxy diagram. An Hxv diagram is at least superIicially similar to a Txv diagram in that it gives you some idea oI the thermal condition oI the mixture as a Iunction oI composition, except that instead oI T on the v-axis, we have H. At right we show the Hxv diagram which would apply iI 1) the molar heats oI vaporization oI the two components are equal, and 2) sensible heat is negligible compared to the latent heat: two parallel, horizontal lines. Below it is an Hxv diagram Ior a real mixture. Unlike a Txv diagram, the saturated liquid and saturated vapor curves do not connect at the ends or at any other composition. The vertical distance between these two curves represents the latent heat oI vaporization at a particular composition, which is usually substantial Ior all compositions. Unlike temperature, enthalpy oI liquid and vapor will be quite diIIerent, so the tie lines are not horizontal l I ti A B enthalpy per mole A B ideal case sat`d vapor sat`d liquid y 06-361 page 57 Spring, 2001 like on a Txy diagram. In Iact, tie lines tend to be more nearly vertical. As we shall soon see, we will need to determine the tie line through a particular point. No matter how many tie lines your graph has, it would be a real stroke oI luck iI one passes through the particular point speciIied. So we need some way to interpolate tie lines. This is accomplished by means oI: auxiliary line -- a curve on an Hxy diagram Ior interpolating tie lines II we Iorm a right triangle out oI each tie line, then the locus oI the points where the right angle is located is the auxiliary line. OI course, the ends oI the auxiliary line coincides with the ends oI the saturated vapor line. By reversing this procedure, we can locate more tie lines. Now let's see how this diagram can be use do solve our Ilash distillation problems. Eirst, I will give you the receipe and then later I will show why the receipe works. receipe: - locate the Ieed point (x F ,H F ) - locate a second point q/F above the Ieed point (x F ,H F ¹q/F) - Iind the tie line which passes through this second point - ends oI the tie line are (x B ,H B ) and (v D ,H D ). - f can be determined with the help oI (49): f : x v x F B D B = ÷ ÷ Proof: now let's show that this procedure really works. The main idea which we need to prove is that the point (x F , H F ¹q/F) |call this point 'E¨| lies on the tie line. We know that (x B , H B ) |call this point 'B¨| and (x D , H D ) |call this point 'D¨| lie on the tie line since the two streams leaving the Ilash unit must be in phase equilibrium with each other. Our approach will be show that the slope oI the line drawn Irom B to E is the same as the slope oI the tie line drawn Irom E to D. Substituting (40) into (41): x F (D¹B) ÷ v D D ¹ x B B Dividing by D and solve Ior B/D: B D v x x x D F F B = ÷ ÷ (51) 06-361 page 58 Spring, 2001 Divide (42) by F: H q F D F H B F H F D B + = + Using (40) to eliminate D: H q F B F H B F H H B F H H F D B D D B + = ÷ + = ÷ ÷ 1 Solving Ior B/F: B F H H q F H H D F D B = ÷ + ÷ (52) II instead, we had used (40) to eliminate B: D F H q F H H H F B D B = + ÷ ÷ (53) (52) ÷ (53): B D H H q F H q F H D F F B = ÷ + + ÷ (54) Equating (51) and (54), we could show that: H H q F v x H q F H x x D F D F F B F B ÷ + ÷ = + ÷ ÷ Note that both sides oI this equation have the Iorm oI rise/run on an Hxv diagram; thus each side represents the slope oI a line drawn between two points on an Hxv diagram. These two slopes involve three distinct points: (x B ,H B ); (x F ,H F ¹q/F); and (v D ,H D ) The equality oI these two slopes means that these three points must lie on the same line. That line connecting B and D must be a tie line. The conclusion is that (x F ,H F ¹q/F) lies on tie line This is the basis Ior the receipe. 06-361 page 59 Spring, 2001 Inverse Lever-Arm Rule on Hxy Diagram (54) provides a basis Ior an inverse lever-arm rule which is similar to (48). The total diIIerence in enthalpy between saturated vapor and saturated liquid is H D -H B . + This diIIerence can be partitioned into (H F ¹q/F) - H B and H D - (H F ¹q/F). Thus the red tie line can be divided into two parts which are proportional to this diIIerence in enthalpies. (54) says that ratio oI the lengths oI these two parts oI the tie line equal the molar Ilowrate ratio B/D. BATCH (DIFFERENTIAL) DISTILLATION The oldest device Ior distillation consisted oI a kettle into which a liquid mixture was placed. By heating the kettle either by Iire or steam or electrically, the liquid is partially vaporized. AIter condensing the vapor, you have a product which is richer in one oI the components. II both oI the components are volatile (e.g. ethanol and water), it wouldn't make sense to vaporize all oI the liquid in the kettle. A mass balance can tell you that, iI you vaporize the entire charge, the cumulative condensate composition will be essentially the same as the initial charge. Oh, perhaps you have removed some nonvolatile minerals which might have been in the water, but you won't enrich the ethanol concentration in this way. And enriching the composition oI the more volatile component is the objective oI distillation. The solution is to evaporate only partially, because the Iirst vapors which come oII will be the richest in the more volatile component. Vapors produced later will only dilute the condensate. This process is called batch distillation because we process one batch at a time rather than continously processing as with Ilash distillation. Let's now try to predict how the composition oI the liquid and vapor change as we evaporate more and more oI the charge. Let n A (t), n B (t) ÷ moles oI component A or component B leIt in liq. n(t) ÷ n A (t) ¹ n B (t) ÷ total moles oI liq. leIt in kettle at time t + H D -H B is not the 'heat of vaporization¨, which is the heat to totally vaporize one mole of feed. On the Hxy diagram, heat of vaporization is the length of a vertical line between the two curves. steam kettle (still) condenser condensate collector cooling water n, n , x A v cond x 06-361 page 60 Spring, 2001 A n t x t n t = ÷ mole Iraction oI A in liq. remaining (55) v(t) ÷ instantaneous mole Iraction oI vapor produced Suppose we evaporate a small amount oI liquid at time t over a time interval dt: n(t¹dt) - n(t) ÷ dn · 0 -dn ÷ total moles evaporated in time dt A Iraction oI this is component A: -dn A ÷ v(-dn) (56) (55) into (56): -d(xn) ÷ v(-dn) xdn ¹ ndx ÷ vdn ndx ÷ (v-x)dn dn n dx v x = ÷ Integrating Irom t÷0, at which time x,n have their initial values: x(0) ÷ x 0 and n(0) ÷ n 0 to some later time, at which time x,n have values: x(t) and n(t) we obtain dn n dx v x n n t x x t 0 0 ( ) ( ) = ÷ Integrating: ln n n t dx v x x t x 0 0 ( ) = ÷ ( ) Solving Ior n(t): 0 0 exp x x t dx n t n v x ( ( = ÷ ( ÷ ¸ ¸ 06-361 page 61 Spring, 2001 which is called Rayleigh's equation. This serves as the design equation Ior batch distillation. To use this equation, we must recall that the vapor coming out oI the kettle at any time is in equilibrium with the remaining liquid: at equilibrium: v ÷ f(x) Eor each x between x(0) and x(t), we can look up the corresponding v(t) and compute the integrand in Rayleigh's equation 1/(v-x). The integral is the area under the curve between these two limits. Example: A 50mol° mixture oI heptane and octane is placed in a batch and heated. Plot the composition oI the remaining liquid as a Iunction oI the Iraction oI the initial charge which remains. At the pressure in the still, the vapor-liquid equilibrium can be reasonably represented by a constant relative volatility oI o ÷ 1.7 (Ior a discussion on when this might arise, see p105). Solution: Eor binary solutions, it's customary to deIine x and v to be the mole Iraction oI the more volatile component (heptane in this case); this choice yields relative volatilities greater than unity. We will have to guess values oI x(t). Below we go through the procedure Ior one particular guess, say x÷0.2. Step 1:Guess x. We expect the vapor to be richer in the more volatile component Step 2:Calculate the corresponding v. Recalling the deIinition oI relative volatility (see page 51): v x 1 v x x t ( ) x 0 dx v x x t x 06-361 page 62 Spring, 2001 1 1 i i i if f f f K v x v x K v x v x o = = = ÷ ÷ Dropping the subscripts and solving Ior v: 1 1 x v x o = o ÷ + This Iunction is evaluated Ior o÷1.7 and plotted in the Iigure at right. Eor x÷0.2, this yields, v÷0.298. Step 3:We evaluate n/n 0 Irom Rayleigh's equation: 0.5 0 0.2 exp n t dx n v x ( ( = ÷ ÷ ( ¸ ¸ The integral is evaluated by assuming x's between 0.2 · x · 0.5. The integrand is plotted in the Iigure at right. The area under the curve between the two dotted lines is 0.5 0.2 2.45 dx v x = ÷ Rayleigh's equation yields 0.5 0 0.2 exp 0.086 n t dx n v x ( ( = ÷ = ÷ ( ¸ ¸ This represents the Iraction oI the initial charge which remains in the still as liquid. The Iraction oI the initial charge which has been distilled is just unity minus this 0 1 0.914 n t n ÷ = 0 0.5 1 0 0.5 1 x 1.7 0 0.5 1 0 5 10 15 1 y i x i 0.2 0.5 x i 06-361 page 63 Spring, 2001 Step 4:Calculate the condensate concentration II all the condensate is collected in a container and mixed together, the average concentration oI the condensate will be 0 0 0 moles oI A depleted total moles depleted 0.528 cond x x n xn n n = ÷ = = ÷ The graph at right shows what typically happens as the initial liquid is more and more distilled. Notice that the vapor is initially richer than the liquid. Since the leaving vapor is richer, the remaining liquid becomes leaner. Thus both x and v become leaner as distillation proceeds. 0 0.5 1 0 0.2 0.4 0.6 0.8 x(t) y(t) xbar¸condensate x(0) fraction of initial charge distilled = [n(0)-n(t)]/n(0) 06-361 page 64 Spring, 2001 Chapter 8. Multistage Operations In both batch distillation and Ilash distillation, the separation oI the two components is limited by equilibrium. Eor example, suppose I have a 50-50 mixture oI hexane and octane. II I vaporize halI oI this mixture, I produce a vapor having: x F ÷0.5, f÷D/E÷0.5 ÷ v D ÷0.55, x B ÷0.45 which we read oII oI an xy diagram. Suppose you want a 95° pure product? II I vaporized a smaller Iraction oI the Ieed, that would increase the composition oI the distillate, but the best I can do is: x F ÷0.5, f~0 ÷ v D ÷0.58, x B ÷0.5 This helps a little but it's still a long way Irom 0.95 and you don't generate much product with such a small Iraction vaporized. One solution is to take the distillate product oI the Iirst Ilash and partially condense it. This will produce a liquid phase which will be leaner in the more volatile component, leaving the remaining vapor richer. This could be repeated many times until the vapor composition reaches or exceeds the desired value. Although we have produced a stream oI the desired composition, it turns out that we are "throwing out" alot oI the desired component in the "waste" streams -- the bottoms. Erom the Iirst Ilash unit alone, we are losing 25° oI component A in the initial Ieed |(1-f)x 1 ~ 0.25|. II this component A is valuable, we don't want to throw it away. - success at achieving high v D , but low recovery (recovery is ° oI A in Ieed which ends up on Iinal product) 06-361 page 65 Spring, 2001 - requires n¹1 HX'ers Another problem is that each Ilash unit requires a heat exchanger and then we will need one additional heat exchanger to condense the product Irom the Iinal Ilash unit. COUNTER-CURRENT CASCADE Both oI these drawbacks can be removed by a slight modiIication: we recycle the bottoms oI one Ilash unit into the Ieed oI the previous Ilash unit. Instead oI n¹1 HXers, we now need only two. To condense the vapor, we need to transIer some oI its latent heat to the liquid recycled Irom upstream. Eor this to work, the temperature oI the vapor must be greater than that oI the liquid. Let`s see: generally, the vapor will get richer in the more volatile component as we move leIt to right: v 1 · v 2 · ... · v N Assuming that the vapor and liquid streams leaving each Ilash are in equilibrium, then the corresponding inequalities Ior temperature can be deduced Irom the typical Txy diagram at right: T 1 ~ T 2 ~ ... ~ T N So the recycled liquid will be cooler than the vapor it's mixed with. Thus the liquid will absorb some oI the heat oI the vapor. This condenses some oI the vapor and vaporizes some oI the liquid. Now that we have shown that the heat exchangers might be eliminated, let`s return to a discussion oI the countercurrent cascade. The product recovery Ior a countercurrent cascade is much better than Ior the series oI Ilash units without recycle. Eirst, we have only one waste stream (B) -- instead oI one Irom each Ilash. Second, the composition oI this waste stream will usually turn out to be leaner in the valuable component -- because we are enriching the Ieed to the Iirst Ilash by recycling Irom upstream. 06-361 page 66 Spring, 2001 Note that we also recycled some oI the product liquid to condense the vapor Ieed to the last stage. This recycling oI the product is called reflux -- that part oI the product stream recycled to the cascade reflux ratio -- L a /D (ratio oI molar Ilowrates) stage -- one unit oI a cascade (e.g. one Ilash unit) ideal stage -- a stage whose exit streams are at equilibrium 06-361 page 67 Spring, 2001 Chapter 9. Equipment for Tray Towers In our schematic oI a countercurrent cascade, we had the streams moving horizontally. Generally, a pump would then be required between each stage to overcome Iriction. With vapor and liquid, the density diIIerence is suIIiciently large, that we can use gravity to move the Iluids iI we stack the stages vertically rather than horizontally. A vertical stack oI stages is called a tray tower. At right we see a schematic oI a typical tray tower (Treybal, p129). Each stage or Ilash unit is represented by one tray. Eive trays are evident in Eig. 3. Liquid Ilows down Irom the tray above and Ilows across the tray where in comes into contact with vapor Ilowing upward Irom the tray below. AIter mixing, the liquid and vapor are separated, with the new liquid Ilow down to the tray below and the vapor Ilowing to the tray above. Through the act oI mixing, the composition oI the liquid and vapor streams has changed. Fig. 4. An expanded view of a single °bubble cap¨ The contacting oI vapor and liquid on any given tray is Iacilitated by bubble caps, one oI which is shown in more detail in Eig. 4. Each tray contains many bubble caps arranged in a array covering the central region oI the tray (see Eig. 5). Fig. 3. A typical tray tower. 06-361 page 68 Spring, 2001 In this particular design shown in Eig. 4, vapor penetrates the tray through a circular hole 2.47 inches in diameter, upward through a 3-inch length oI pipe beIore being diverted downward by the cap and Iorced through the liquid outside the pipe. The Ilow pattern under both normal and abnormal operation is shown in Eig. 6. Fig. 6. Flow patterns under normal and abnormal operating of a bubble cap Fig. 5. Top view of arrangement of bubble caps on a tray. 06-361 page 69 Spring, 2001 An alternative tray design is the sieve-plate tray (see Eig. 7). Sieve-plate trays are much cheaper to Iabricate compared to bubble-cap trays since the tray consists oI a circular plate with holes drilled through Ior the passage oI vapor. These holes are generally much smaller in diameter than the bubble caps. A disadvantage oI sieve plates is that they are much more prone to weaping and Ilooding (conditions corresponding to poor contact oI liquid and vapor). Fig. 7. Side view of a °sieve plate¨ tray. 06-361 page 70 Spring, 2001 Chapter 10. Performance of a C-C Cascade ANALYSIS OF A C-C CASCADE At right is a schematic representation oI a countercurrent cascade. It ignores some oI the details oI plumbing inside the stages and, oI course, in reality there are no pipes to carry the liquid and vapor between the adjacent stages. But these are unimportant details as Iar as mass and energy balances are concerned. The schematic preserves the essential Ieature -- which is that the Ilow oI liquid and vapor occurs countercurrently. Since we have so many streams to deal with, some convention Ior numbering the stages and labelling the streams will be helpIul. McCabe, Smith & Harriott use the Iollowing conventions: tray numbers: increase in direction oI liquid Ilow (1>i>N) stream labels: subscripts denote stream - L i , J i are molar Ilowrates oI total (A¹B) liquid and vapor leaving the ith stage. - L a , J a ÷ J 1 are the Ilowrates at the top oI the cascade while L N , J b are the Ilows at the bottom oI the cascade ("b" Ior bottom). * - x i , v i are the mole Iraction oI the more volatile component in the liquid and vapor leaving the ith stage. Suppose we know everything about the two input streams: their Ilowrate, composition, and enthalpy. Suppose we also have all the thermodynamic data: Hxv diagram and xv diagram. Given: L a ,x a ,J b ,v b ; Hxv and xv diagrams We want to determine the Ilowrate, composition and enthalpy oI the two output streams and oI all the intermediate streams: Eind: L i ,x i ,J i ,v i Ior i÷1,...,N ( ÷ 4N unknowns) I haven't bothered to list the enthalpies or temperatures here, because iI I know the mole Iraction and pressure, I can look up the enthalpy and temperature oI saturated liquid or vapor. * In heat transfer, the subscripts "a" and "b" were used for entering and leaving streams, respectively. That convention is abandoned here. 06-361 page 71 Spring, 2001 As our system, let's select the Iirst "i" stages. An overall mole balance Ior this system requires: total: L a ¹ J i¹1 ÷ L i ¹ J a , (57) while a mole balance on the more volatile component requires: component A: L a x a ¹ J i¹1 v i¹1 ÷ L i x i ¹ J a v a . (58) Einally, we can also perIorm an enthalpy balance: enthalpy: L a H x,a ¹ J i¹1 H v,i¹1 ÷ L i H x,i ¹ J a H v,a . (59) Do I have enough inIormation to calculate the result? How many independent equations do I have? equations: (57),(58),(59) × N ( ÷ 3N eqns) x i and v i at equil. ( ÷ N eqns) So, at least when I am dealing with ideal stages, I have as many equations as unknowns. Many distillation towers have N÷20 stages or more, so I could easily be talking about 80 equations -- a Iormidible problem. Eortunately, there are assumptions which can make this problem tractable so that it can be solved easily with pencil and paper. CONSTANT MOLAL OVERFLOW In our discussion oI Ilash distillation, we noted that many pure components have the same heat oI vaporization per mole. II you can assume that the molar heat oI vaporization oI the mixture is completely independent oI its composition, then solution is greatly simpliIied. Assumptions: 1) H v,i -H x,i ÷ ì Ior all i. 2) neglect sensible heat compared to latent heat Let's choose saturated liquid having the composition oI the Ieed as our reIerence state Ior enthalpies. Since the only diIIerence between the various liquid streams is sensible heat, we take them all to be equal -- equal to zero: H x,a ÷ H x,i ÷ 0 Ior all i and H v,i¹1 ÷ H v,a ÷ ì 06-361 page 72 Spring, 2001 With these assumptions, the enthalpy balance about the top i stages L a H x,a ¹ J i¹1 H v,i¹1 ÷ L i H x,i ¹ J a H v,a (60) simpliIies to: J i¹1 ÷ J a Ior all i (61) and the total mole balance about the top i stages L a ¹ J i¹1 ÷ L i ¹ J a (62) Subtracting (16) Irom (15): L a ÷ L i Ior all i In other words, all oI the vapor streams have the same molar Ilowrate within the cascade (although their composition and mass Ilowrate will change) and all the liquid streams have the same molar Ilowrate: J a ÷J 1 ÷J 2 ÷...÷J N ÷J b ( ÷ J, say) L a ÷L 1 ÷L 2 ÷...÷L N ÷L b ( ÷ L, say) which is called the state oI constant molal overflow. Since the input Ilowrates are known, we know all the Ilowrates. Thus we have just halved the number oI unknowns Irom 4N to 2N. We can then drop the subscripts on the Ilowrates and Iorget the enthalpy balance (60) and the total mole balance (62). This leaves just the component mole balance L a x a ¹ J i¹1 v i¹1 ÷ L i x i ¹ J a v a . which can be rearranged to: v L J x v L J x i i a a + = + ÷ 1 (63) or v i¹1 ÷ mx i ¹ b which is a straight line on an xv diagram. This line is called the: operating line v i¹1 vs. x i ; a graphical representation oI the mass balance. Another important curve on an xv diagram is the: equilibrium curve v i vs. x i ; a graphical representation oI VLE In the next section, we will see how these two curves can be used to determine the number oI equilibrium stages required to achieve some desired degree oI separation. 06-361 page 73 Spring, 2001 A RECTIFYING CASCADE In a typical distillation tower, the liquid Ied to the top oI the enriching cascade is a portion oI the vapor product which has been condensed and reIluxed back into the tower, as shown at right. The ratio oI the amount reIluxed to the amount not reIluxed is called the reflux ratio L R D ÷ (64) Since all oI the vapor stream leaving the top oI the cascade is condensed to Iorm the distillate product and the liquid Ied to the top, the three streams must have the same composition: x D ÷ x a ÷ v a (65) Then the operating line (63) can be simpliIied to + v L J x L J x L J x D J x i i J L J D J D i D + ÷ = = + ÷ = + 1 1 ¸_ ¸ ¸ ¸ (66) The ratio oI the Ilowrates can be expressed in terms oI the reIlux ratio R using (64): L J L L D RD RD D R R = + = + = +1 (67) 1 1 D D D J L D RD D R = = = + + + The operating line (66) can be rewritten as 1 1 1 1 i i D R v x x R R + = + + + (68) Thus the slope oI the operating line (a plot oI v i¹1 versus x i ) is R/(R¹1). + Alternatively, we could 'start over¨ and perform a component mole balance on the top i stages of the cascade (indicated by the blue line in the sketch). L,x a D,x D ÷0.95 F,x F ÷0.5 L,x 1 L,x N-1 L,x 2 B÷L,x ÷x N b J,v ÷v 1 a J,v 2 J,v N J,v 3 J,v b 1 2 N 06-361 page 74 Spring, 2001 Now, one point on the operating line is x i ÷ v i¹1 ÷ x D . In other words, iI we substitute x D Ior both x i and v i¹1 , (68) is automatically satisIied. Example. Suppose I have a Ieed stream containing 50° alcohol in water which I want to enrich to 95° alcohol. How many stages do I need iI I use a reIlux ratio oI unity and what will be the Ilowrate oI this product? Given: x F , x D , F and R Eind: N and D Solution: The Iirst step is to draw the operating line on the xy- diagram. One point is (x D ,x D ) or (0.95, 0.95). And the slope is R R + ( ) 1 ÷ 1/2 Ior R÷1. Armed with one point and the slope, we could draw the line. But it is usually easier to construct a line iI we know two points. A second convenient point corresponds to the y-intercept. Substituting x i ÷ 0 into (68) yields x i ÷ 0: 1 1 D i x v R + = + which can be conveniently calculated Irom what`s given. Thus 0 1 , x R D + is a second point on our operating line. Knowing the slope and one point, or two points, we can draw the operating line. * Once the operating line is plotted on the xv diagram (which is the equilibrium curve), we can determine the composition oI other streams as Iollows. The two streams leaving a given ideal stage are at equilibrium, thus the compositions of the two streams leaving a given stage must lie on the equil. curve. Since we already know the composition oI the vapor stream leaving stage 1 (v 1 ÷ x D ÷ 0.95), we can determine x 1 by moving horizontally Irom one point on the operating line (x D , x D ) to one point on the equilibrium curve and then read oII the value oI x 1 : (x 1 ,v 1 ) on equil. curve: gives x 1 * The point (x D ,x D ) and slope of 1/2 are not drawn to scale in the sketches. Eor that matter, neither is the equilibrium curve. However, given the equilibrium curve and operating line as drawn, the method of stepping off the number of stages is correct. x D x D x R D 1 06-361 page 75 Spring, 2001 On the other hand, the operating line represents a component mole balance about an arbitrary number oI stages measured Irom the top. In particular, Ior just the top stage, the operating line (66) gives me a relationship between x 1 and v 2 : (x 1 ,v 2 ) on oper. line: gives v 2 Knowing x 1 , I can determine v 2 . In general, the compositions of the two streams between a given pair of adjacent stages must lie on the operating line. Now I can repeat the process: (x 2 ,v 2 ) on equil. curve: gives x 2 (x 2 ,v 3 ) on oper. line: gives v 3 and so on. Continuing to step oII stages in this manner until the vapor mole Iraction equals or drops below x F ÷ v b , requires Iour steps: answer: N ÷ 4 Comment #1: Although the Ieed might be a liquid, it`s the vapor mole Iraction which must be at or below the Ieed composition, because the actual Ieed to be bottom oI the cascade is a vapor (see stream whose composition is labelled v b in the Ilowsheet on page 73). The horizontal line v ÷ x F (÷ v b ) on the xy-diagram is called the feed line. Comment #2: AIter taking 4 steps (or any integer number), we don`t end up exactly at the Ieed composition (except by some extreme stroke oI luck). Instead we end up stepping below the Ieed composition. II the trays really are perIectly ideal, and the Ieed composition is 0.5, the distillate composition would turn out to be slightly richer than 0.95. Although any real column has an integer number oI trays, it is customary to report the answer as a Iraction; Ior example, to go Irom a Ieed oI 0.50 to a distillate oI 0.95, we require 3.7 ideal stages (or 3 ¹ 0.7, where 0.7 is the Iraction oI the last step which corresponds to the Ieed composition) in the current problem. Comment #3: Suppose that our Ieed composition had been a little lower, say v b ´. No matter how many steps we took, we could never get a vapor composition lower than that at the intersection oI the operating and equilibrium curves. In that case, any solution Ior the speciIied reIlux ratio is impossible. Instead the reIlux ratio would have to be increased, which will rotate the operating line counter-clockwise about the speciIied point. This will lower the composition at the intersection point and make the separation possible. It turns out there exists a minimum value oI x a x F v b v´ b v a 1 2 3 4 Ieed line Ior sat`d vapor 06-361 page 76 Spring, 2001 the reIlux ratio which allows a particular distillate composition x D to be produced Irom a given Ieed composition x F . This called the minimum reflux ratio. Let`s Iinish this problem by calculating the Ilowrates. Since the vapor Ilowrate leaving each tray is the same, and because we are creating the vapor to the Iirst tray by completely vaporizing the Ieed, we can say that J ÷ F The ratio oI liquid to vapor Ilowrate inside the cascade is given in terms oI the known reIlux ratio by (67): L J R R R = + = = 1 1 2 1 or L J F = = 1 2 1 2 Since the bottoms stream B is one oI the liquid streams B L F = = 1 2 Einally, the distillate Ilowrate can be calculated Irom the deIinition oI the reIlux ratio: recall (64): D L R L F R = = = =1 1 2 A STRIPPING CASCADE Use oI a single cascade with 3.7 stages allows us to convert a Ieed oI 50mol° alcohol into one containing 95mol°. But a signiIicant amount oI the alcohol is being lost in the second product stream. It`s possible to strip the alcohol Irom this stream by Ieeding it to a second cascade. In a stripping cascade, liquid is Ied into the top. Liquid out the bottom oI the cascade is partially vaporized and the vapor is recycled into the bottom oI the cascade. The vapor stream at the top represents the second product stream. Because the bottoms product is only partially vaporized to Iorm the vapor stream which is recycled to the cascade, its composition is not the same as the bottoms product. In other words x N = v N¹1 = x B So no analogy can be made to (65). However we still have a L,x a B,x B F,x F ÷0.5 L,x 1 L,x N-1 L,x 2 L,x N J,v ÷v 1 a J,v 2 J,v N J,v 3 J,v N¹1 1 2 N 06-361 page 77 Spring, 2001 operating-line relationship like (63) or (66) which is imposed by mole balances. Erom a balance over the entire stripping cascade on the more volatile component, we know that Lx Fx Bx Jv a F B a = = + in out ¸ _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ so Jv a - Lx a ÷ -Bx B Substituting this result into (63), the operating line Ior a stripping cascade becomes + v L J x B J x i i B + = ÷ 1 (69) Like all the operating lines we`ve encountered so Iar, this one has a slope oI L/J. One point on this operating line is (x B ,x B ) since substituting v i¹1 ÷ x i ÷ x B automatically satisIies (69) provided L - B ÷ J which is always true since this represents a total mole balance. Another point is (x F ,v 1 ) iI we happen to know or can somehow calculate v 1 . Then we have two points and can draw the operating line. Otherwise, we use one point and the slope. Example. Suppose I have a Ieed stream containing 50° alcohol in water which I want to strip the alcohol Irom to produce a stream containing only 5° alcohol. How many stages do I need iI I boil halI oI the liquid leaving the bottom oI the cascade and return it as vapor? Given: x F , x B and J/L Eind: N Solution: Once again, vapor and liquid streams leaving the same stage are assumed to at equilibrium; thus (x i ,v i ) lies on the equilibrium curve. On the other hand, the composition oI adjacent streams (x i ,v i¹1 ) lies on an operating line Ior this cascade. One point on this operating line is x i ÷ v i¹1 ÷ x B . II we 'boil halI the liquid leaving the bottom oI the cascade and return it as vapor¨ we know that J ÷ L/2, so the slope oI the operating line is: L/J ÷ 2 Knowing the slope and one point, we can immediately draw + Alternatively, we could 'start over¨ and perform a component mole balance on the bottom N-i stages of the cascade (indicated by the blue line in the sketch). x B x B x F v 1 L J x B x N v N¹1 06-361 page 78 Spring, 2001 the operating line. The liquid product being drawn oII at x B is in equilibrium with the vapor being sent back into the cascade; thus (x B , v N¹1 ) lies on the equibrium curve. On the other hand, (x N , v N¹1 ) are adjacent streams in the cascade and so their compositions must lie on the operating line. Similarly, (x N , v N ) lies on the equilibrium curve, while (x N-1 , v N ) lies on the operating line. We continue to step oII stages in this manner until we step past a liquid composition equal to the Ieed x F ; in other words, we continue stepping until we cross the blue vertical Ieed line. In this case, three steps are required. But one oI these steps is the reboiler; only two additional equilibrium trays are required: answer: N ÷2 plus reboiler Comment #1: Notice that the 'Ieed line¨ Ior a stripping cascade is vertical, whereas the Ieed line Ior a rectiIying cascade was horizontal (see page 75). This is because the Ieed into the bottom oI a rectiIying cascade is a vapor, whereas the Ieed to a stripping cascade is a liquid. Comment #2: Suppose that our Ieed composition had been a little higher, say x F ´. No matter how many steps we took, we could never get a liquid composition higher than that at the intersection oI the operating and equilibrium curves. In that case, any solution Ior the speciIied L/J ratio is impossible. Instead the L/J ratio would have to be decreased, which will rotate the operating line clockwise about the speciIied point. This will raise the liquid composition at the intersection point and make the separation possible. It turns out there exists maximum L/J, or a minimum vapor rate J which allows a particular bottoms composition x B to be produced Irom a given Ieed composition x F and liquid Ilowrate L. This minimum J/L is called the minimum boil-up ratio. Let's Iinish the problem by determining the Ilowrates. Once again, all the Ilowrates can be deduced Irom the equimolal overIlow assumption and the reboil Iraction. All the liquid Ilowrates correspond to the Ilowrate oI the liquid Ied to the top: L ÷ F Since I`m reboiling halI oI this liquid to return to the cascade as vapor, all the vapor Ilowrates are J ÷ L/2 ÷ F/2 The bottoms Ilowrate is what remains aIter halI oI the liquid to the reboiler is boiled: B ÷ L J ÷ F/2 x B x N x F x F x N-1 v N¹1 v N 1 2 3 Ieed line Ior a sat`d liq 06-361 page 79 Spring, 2001 STEAM DISTILLATION Instead oI using a heat exchanger to partially boil the liquid stream leaving the bottom stage to produce the vapor entering the bottom stage, we could directly inject steam as the vapor. This has the advantage that it avoids the cost oI a heat exchanger and you are getting direct contact between the hot and cold Iluids. OI course, use oI steam as the vapor is only useIul iI water is one oI the two components being separated (e.g. alcohol/water) and Iurther, it must be the less volatile component. II instead you are trying to separate heptane Irom octane, injecting steam would be produce a second liquid phase (since oil and water are not miscible). This produces the operating line shown at right. Note that one point on the operating line is (x B , 0) since the 'vapor¨ entering the bottom is pure steam (water). OI course, part oI the operating line lies below the 45° line (x÷v). But there are no restrictions that v must be larger than x. As with the stripping cascade, we continue to step oII stages between the operating line and the equilibrium curve until our liquid composition exceeds x F . Comment: Suppose that our Ieed composition had been a little higher, say x F ´. No matter how many steps we took, we could never get a liquid composition higher than that at the intersection oI the operating and equilibrium curves. In that case, any solution Ior the speciIied L/J ratio is impossible. Instead the L/J ratio would have to be decreased, which will rotate the operating line clockwise about the speciIied point. This will raise the composition at the intersection point and make the separation possible. It turns out there exists maximum L/J, or a minimum steam rate J which allows a particular bottoms composition x B to be produced Irom a given Ieed composition x F and liquid Ilowrate L. ÷0.5 ÷ 0 (live steam 1 2 x B x F v 1 L J 0 x F 06-361 page 80 Spring, 2001 Chapter 11. McCabe-Thiele Method A typical distillation tower will consist oI at least two cascades -- one to enrich the Ieed in the more volatile component and a second one to strip the more volatile component out oI the bottoms stream beIore disposing oI it. rectification -- enrichment oI vapor by contact with liquid reIlux With enough trays, we can usually make D as pure as we like (an exception occurs Ior an azeotropic mixture). However, the liquid stream at the bottom oI the cascade might still contain a signiIicant Iraction oI the more volatile component. This is where a second cascade is useIul: stripping -- removal oI the more volatile component Irom a liquid by contact with reboled bottoms (or direct injection oI steam) A typical problem in designing a distillation column is the Iollowing: Given: : F ,x D ,x B ,L/D,H F Eind: N and N where N and N are the number oI stages in enriching and stripping sections. Having two cascades is a little tricker than one, so let me Iirst tell you how to solve the problem by giving you the receipe, and later we will rationalize some oI the steps in this receipe. This receipe is called the McCabe-1hiele Method and it works when you have only two components and you can assume equal molal overIlow. Step 1) Plot equilibrium data on an xv diagram. Step 2) Locate x B ,: F , and x D on 45° line We use ": F " to denote the mole Iraction ignoring which phase the Ieed is in: : F ÷ moles oI A in liquid ¹ vapor total moles in liquid ¹ vapor condenser F : H F F D x H D D B x B L L J J enriching cascade stripping cascade Ieed tray 06-361 page 81 Spring, 2001 Step 3) Calculate q (a dimensionless variable representing the thermal condition oI the Ieed): J F J L H H q H H ÷ ÷ ÷ where H F ÷ enthalpy oI Ieed H J ÷ H oI sat'd vapor with v÷: F H L ÷ H oI sat'd liq. with x÷: F Step 4) Plot q-line, which is a line given by the Iollowing equation (Q.L.): v q q x : q F = ÷ + ÷ 1 1 OI course, the name "q-line" is given because this line depends on q the thermal condition oI the Ieed. Eor example, suppose the Ieed is ... superheated vapor: H F ~H J ÷ q·0 ÷ 0 1 1 q q < < ÷ sat'd vapor: H F ÷H J ÷ q÷0 ÷ 0 1 q q = ÷ (horizontal line) 2-phases mixture: H L ·H F ·H J ÷ 0·q·1 ÷ 0 1 q q < ÷ In this case, q represents the Iraction oI the Ieed which is liquid. sat'd liquid: H F ÷H L ÷ q÷1 ÷ 1 q q = · ÷ (vertical line) subcooled liquid: H F ·H L ÷ q~1 ÷ 1 1 q q < < +· ÷ As it turns out, the two operating lines Ior the two cascades intersect somewhere along this line. sat`d L L&V sat`d V subcooled superheated : F 06-361 page 82 Spring, 2001 one point: (: F ,: F ) slope: q/(q-1) Stop 5) Plot the operating line Ior the rectiIication section (R.O.L.): v R R x x R D = + + + 1 1 one point: (x D ,x D ) another point: (0,x D /(R¹1)) Step ô) Plot the operating line Ior the stripping section (S.O.L): B Bx L v x J J = ÷ one point: (x B ,x B ) another point: intersection oI R.O.L. and Q.L. Step 7) Step oII stages You could start at either end (x÷v÷x D or x÷v÷x B ), although it is customary to start at the top. In stepping oII the stages, you eventually have a choice as to which operating line to drop down to. LOCATION OF FEED PLATE At some point during the stepping process, you will have a choice oI either operating line to step down to. Either operating line can be used, but once you switch operating lines, you cannot go back to the original line: switching operating lines corresponds to switching Irom one cascade to the other. The stage at which we switch operating lines is the feed plate: stage at which you switch operating lines 06-361 page 83 Spring, 2001 Either oI the two stepping schemes above is OK, but one scheme leads to Iewer total trays than the other. optimum feed plate: to get the Iewest number oI total stages, you switch operating lines when that allows you to take larger steps In other words, iI you have a choice oI which operating line to use, use that one which takes you Iurthest Irom the equilibrium curve. This choice gives the biggest step and Iewest total stages. CONVERGENCE OF OPERATING LINE AND EQUILIBRIUM CURVE In some oI the homework problems, the distance between the operating line and equilibrium curve becomes very small near the point oI their intersection. When this happens, a very large number oI steps might be required to make any progress on separation. Rule: an infinite number oI steps is required to get to an intersection oI the operating line and equilibrium curve. Proof: when the size oI your steps are no longer large compared to the width oI the pen or pencil you are using to draw them, you should "blow up" the drawing to reduce drawing errors. To demonstrate this graphically, let me present the results oI stepping oII stages in a particular example: Example: given the xy-diagram Ior ethanol and water at 1 atm pressure (has an azeotrope at x ÷ v ÷ 0.8), step oII stages Ior the Iollowing conditions: Given: : F ÷ 0.25, x D ÷ 0.7999, x B ÷ 0.01, q ÷ 0.3 Eind: N and N Ior the optimum Ieed location 06-361 page 84 Spring, 2001 0 0.2 0.4 0.6 0.8 0 0.2 0.4 0.6 0.8 equilibrium curve y÷x line Rectifying section oper. line Q line Stripping section oper. line stepping off stages yprime xprime Enter values: 1 Murphree plate efficiency x D 0.7999 Mol frac in distillate x E 0.25 Mol frac in feed x B 0.01 Mol frac in bottoms q 0.3 Thermal conditions of feed (liquid fraction) R 8.618 Reflux ratio (to obtain total reflux, use R=0) R min 4.483 After N steps, this is actual xB: xB calc 5.57 10 3 Nrect 60 Feed tray N 67 Total number of stages Solution: This is a particularly diIIicult separation because the distillation composition x D ÷ 0.7999 is very close to the azeotrope. We anticipate a large number oI steps. Available online is a Mathcad document |McCabe-Thiele (Pro).mcd | which steps oII stages automatically using the optimum Ieed location. The results are shown in the Iigure at right. The program reports (not shown) the total number oI steps and the Ieed location. In this Iigure, there are a total oI 67 steps with the Ieed introduced on the 60 th Irom the top. In the upper right corner, the steps are becoming so small as to be invisible. The below on the leIt is a blow-up oI the region Irom 0.7 to 0.8. Even on this blow-up the steps are becoming invisible in the upper right corner. So the second graph below on the right is another blow-up: this Irom 0.79 to 0.80. Still the steps are becoming vanishingly small. The red (equilibrium) curve does not show up on the second graph below on the right. However, the equilibrium curve is just the locus oI points Irom the upper-leIt corner oI each step. Similarly, the lower right corner oI each step is the operating line. Notice that in these blow-ups, both the equilibrium curve and the operating line very linear. 06-361 page 85 Spring, 2001 0.7 0.72 0.74 0.76 0.78 0.7 0.72 0.74 0.76 0.78 0.79 0.792 0.794 0.796 0.798 0.79 0.792 0.794 0.796 0.798 Over a narrow range, any continuous curve becomes virtually linear. When both the operating line and equilibrium curve are straight lines, an analytic expression exists Ior calculating the number oI steps: N v v v v v v v v a a b b a b a b = ÷ ÷ , ÷ ÷ , log log * * * * , , , (70) which is called Kremser's equation. The subscripts a and b denote streams at the top and bottom oI the cascade, respectively; the v* are vapor compositions in equilibrium with the liquid at the top and bottom oI the cascade. Proof: the linear equilibrium curve can be expressed as v i ÷ mx i ¹ b (71) A mole balance about the top i stages oI the cascade yields x b x a v b v a v * b v * a E C O L 06-361 page 86 Spring, 2001 1 i i a a L L v x v x J J + | | = + ÷ | \ . Using (71) to eliminate x i : 1 i i a a v b L L v v x J m J + ÷ | | = + ÷ | \ . Now we deIine a new parameter, which represents the ratio oI the slope oI the operating line (L/J) to the slope oI the equilibrium curve (m): L A mJ ÷ In terms oI A, our mole balance becomes 1 * a i i a a i a a v v A v b v Amx Av v A mx b + = ÷ + ÷ = + ÷ + ¸¸_¸¸ Erom (71), we see that mx a ¹b is just the vapor concentration in equilibrium with liquid at concentration x a , which we denote at v a *. AIter minor rearrangement, the above equation becomes 1 * i i a a v Av Av v + = ÷ + (72) This serves as a recursion Iormula, which we can applied repeatedly to obtain the concentration on successive stages, starting at the top. Eor i÷1, recalling v 1 ÷ v a , (72) yields i ÷ 1: 2 1 * 1 * a a a a a v v A v Av v v A Av = ÷ + = + ÷ i ÷ 2: 3 2 2 2 * 1 * * 1 * a a a a a a a a v Av Av v A v A Av Av v v A A v A A = ÷ + = + ÷ ( ÷ + ¸ ¸ = + + ÷ + where the second equality above results Irom substituting v 2 Irom the previous step. Now we begin to see a pattern developing: i ÷ n: 2 2 1 1 * n n n a a v v A A A v A A A + = + + + + ÷ + + + . . (73) The sums represent partial sums oI geometric series, whose sum is known analytically: 1 2 i i¹1 N x a x i x ÷x N b v v ÷ a v i v v N b ÷ 06-361 page 87 Spring, 2001 1 2 1 1 1 n n A A A A A + ÷ + + + + = ÷ . Substituting this result into (73): 1 1 1 1 * 1 1 n n n a a A A v v v A A A + + ÷ ÷ = ÷ ÷ ÷ Eor the bottom stage oI the cascade, n÷N and v N¹1 ÷ v b : 1 1 1 * 1 1 N N b a a A A v v v A A A + ÷ ÷ = ÷ ÷ ÷ Multiplying through by 1-A and collecting terms multiplied by A or by A N¹1 : 1 * * N b a a b a a v v A v v A v v + ÷ + ÷ = ÷ (74) This can be simpliIied. Recall that A is deIined as the ratio oI the slopes oI the operating line to the equilibrium curve. This ratio can be calculated Irom the concentration deIined on the xy- diagram above: slope oI OL * * slope oI EC * * a b a b a b a b a b a b v v x x v v A v v v v x x ÷ ÷ ÷ ÷ = = ÷ ÷ ÷ (75) We can use this result to obtain v a - v b ÷ A(v a *-v b *), which will now be substituted into (74): 1 * * * * * * * * b a a b b b N b a a b a a A v v v v A v v A v v A v v A v v + ÷ + ÷ = ÷ ÷ + ÷ = ÷ ¸¸¸¸¸¸_¸¸¸¸¸¸ AIter dividing through by A and solving Ior A N : * * N b b a a v v A v v ÷ = ÷ or * log * log b b a a v v v v N A | | ÷ | ÷ \ . = Taking the log oI both sides, recalling that log(A N ) ÷ N logA, then solving Ior N we obtain the second expression above. To obtain the Iorm oI Kremser's equation given as (70), we substitute (75) Ior A. Comment #1: In the xy-diagram shown next to (70), the equilibrium curve lies above the operating line. In other words, v a * ~ v a and v b * ~ v b . To obtain N~0, you must have v a ~ v b . II 06-361 page 88 Spring, 2001 you inadvertently (but consistently) interchange the subscripts a and b, you will obtain negative values Ior N. Kremser's equation can also be applied to situations in which the operating line lies above the equilibrium curve; then to obtain N~0, the inequalities above must be reversed. Comment #2: II the operating line and equilibrium curve should cross (which is physically impossible), when evaluating (70) you will Iind yourselI trying evaluate the log oI a negative number. Comment #3: The situation in which the operating line and equilibrium curve are parallel (i.e. A ÷ 1) is a special case: (70) becomes indeterminant. However, since the driving Iorce v*-v is constant, the number oI steps is just the total change in concentration divided by the driving Iorce: Ior A ÷ 1: * * a b a b v v N v v ÷ = ÷ Now we will return to discuss the motivation Ior Kremser's equation: approaching the intersection oI OL and EC: Notice that, as x b approaches the point oI intersection (see Iigure on p83), v v b b * ÷ and N = , = · ÷ · log log log log 0 Thus vou can never reach the point of intersection Ior any Iinite number oI steps. In particular, a very large number oI stages is required as we increase the purity oI either oI our two product streams. Eor example, as x D ÷1, the end-point (x D , x D ) oI our operating line approaches (1,1), which is always one point on the equilibrium curve: as x D ÷1, N÷·. CHOOSING THE REFLUX RATIO An important part oI the design oI any distillation column is the choice oI reIlux ratio. There are a Iew qualitative considerations which must be kept in mind. 06-361 page 89 Spring, 2001 1) Existence of R The Iirst thing you should know is that there is a minimum value that the reIlux can assume and still achieve the desired separation. Recall that the slope oI the ROL is slope oI ROL ÷ R R +1 Notice that, as R is reduced, the intersection oI the ROL and the Q-line moves toward the equilibrium curve. As it does so, the minimum number oI stages required approaches inIinity Ior the reasons mentioned above: you can never get to the intersection oI the operating line and the equilibrium curve. Usually this minimum reflux ratio can be determined using the point oI intersection between the q-line and the equilibrium curve. On some peculiar xv diagrams the R.O.L. or S.O.L. constructed in this way might cross the equilibrium curve at points other than where the Q-line crosses. Then the reIlux ratio must be increased until neither operating line crosses equilibrium curve beIore reaching the q-line. OI course, the operating line can be tangent to the equilibrium curve. This point oI tangency is called a pinch point. In stepping oII the stages, it will take an inIinite number just to reach the pinch, but iI the reIlux ratio is increased every so slightly, then the separation can be achieved, although the number oI stages might be large. minimum reflux ratio -- smallest value oI R Ior which the desired separation can be achieved with a Iinite number oI stages. The pinch point might also occur on the S.O.L. as shown at right. Regardless oI where the pinch occurs, once you have drawn the q- line and operating lines, the minimum reIlux ratio (thus determined graphically) can be calculated Irom (11.4-24) in Geankoplis: R x v v x D min = ÷ ' ' ÷ ' (76) where (x´,v´) is any point on the R.O.L. (x´,v´) can be chosen Ior convenience: Ior example, you might want to choose the y-intercept. 06-361 page 90 Spring, 2001 To determine the minimum reIlux ratio using the McCabe- Thiele method, we might imagine drawing the ROL and SOL on the xv-diagram starting with a high reIlux ratio (where ROL and SOL coincide with 45° line) and then decrease the reIlux ratio until any one oI three things happens: 1. the ROL becomes tangent to the equilibrium curve 2. the SOL becomes tangent to the equilibrium curve 3. the intersection oI the ROL and SOL occurs on the equilibrium curve An important thing to notice about the minimum reIlux ratio is that, as we approach it, the number oI stages approaches inIinity: as R ÷ R min , N ÷ · Clearly you don't want to operate near the minimum reIlux ratio because it would cost too much to build a column with so many stages. Generally as you increase the reIlux ratio, both operating lines move away Irom the equilibrium curve so that the steps become larger. Then Iewer stages will be needed. as R|, N! 2) 1otal Reflux: R÷ ÷÷ ÷· ·· · gives Minimum Stages In the limit that R approaches inIinity, the v-intercept oI the R.O.L. becomes zero. The R.O.L. becomes the 45° line. OI course, the S.O.L. and R.O.L. always intersect along the q-line, which means that the S.O.L. also becomes the 45° line. This is the Iurthest the operating lines can ever be Irom the equilibrium curve, so this condition gives the minimum number oI stages: as R÷·: R.O.L. ÷ S.O.L. ÷ 45°-line and N N min Although it might appear that this is the best value oI the reIlux ratio, it turns out that there is a "dark side" to this solution. The problem is what is happening to Ilowrates inside the column. Recall R L/D Now to get inIinite reIlux ratio, one oI two things has to happen: R ÷ ·: L÷· or D÷0 06-361 page 91 Spring, 2001 In most designs, you are interested in processing a Iixed amount oI material, this implies that F,D, and B are Iixed, which eliminates the second alternative above. A simple mass balance about the split oI L and D tells us that the vapor Ilowrate must also get very large: L ÷ ·: J ÷ ·, q c ÷ ·, q r ÷ ·, tower diameter ÷ · This means that the heat exchanger Ior both the condenser and the reboiler are getting very large as are the cost oI utilities (steam and cooling water). Thus the costs oI building the tower becomes large either near the minumum reIlux ratio (where the height oI the tower becomes large) or near total reIlux (where the diameter oI the column becomes large). The costs oI utilities increase about linearly with reIlux ratio. The total costs will experience a minimum at some reIlux ratio, which we call the optimum reflux ratio: minimizes total costs/year Typically the optimum occurs Ior: 1.1 · R opt /R min · 1.5 However most plants actually operate at slightly larger R since the additional cost is not very much and it buys you Ilexibility in case you decide later you want to change x D or in case the Ieed composition changes, you have some room to maneuver: 1.2 · R/R min · 2.0 $ vear 06-361 page 92 Spring, 2001 CALCULATING FLOWRATES AND CONDENSER/REBOILER HEAT DUTIES To compute the heat duties oI the condenser and reboiler, we will need to know the Ilowrates in various parts oI the column. We will start at the top oI the column and work our way down. Eirst we use the deIinition oI reIlux ratio: L ÷ RD and J ÷ (R¹1)D (77) Given F or D and the composition oI the three streams (x D , : F and x B ), we can determine the two unknown Ilowrates just by doing a total mole balance and a component mole balance over the entire distillation column (the blue system in and Iigure at right). F ÷ D ¹ B : F F ÷ x D D ¹ x B B Solving these two equations Ior the two unknown Ilowrates, we obtain F x x : x D D B F B = ÷ ÷ (78) and B x x x x F x x : x D D F D B D F F B = ÷ ÷ = ÷ ÷ (79) Given F, x D , : F and x B , we can thus calculate D, J and L. In a total condenser, we are condensing all oI the overhead vapor. Assuming the condenser output is a saturated liquid, the condenser heat duty is just q J c = ì To calculate the reboiler heat duty, we need to evaluate J . A look at the streams entering the Ieed plate (see Iigure at right), we realize that we have at least two unknowns L and J . A total mole balance gives: J F L J L + + = + which represents only one relationship in two unknowns. The remaining relationship is obtained Irom an enthalpy balance about the Ieed tray: F L J L J FH LH JH LH JH + + = + 06-361 page 93 Spring, 2001 Our analysis oI countercurrent cascades is based on the equimolar overIlow assumption, which contains some assumptions about enthalpy: see "Constant Molar OverIlow" on p71. In particular, we neglect sensible heat and assume that all components have the same latent heat oI vaporization, then we can evaluate the enthalpies oI all saturated liquid or saturated vapor streams as: H L ÷ 0, H J ÷ ì Recall the deIinition oI q Irom p81. This allows us to express the enthalpy oI the Ieed in terms oI q: J F F J L H H H q H H ÷ ì ÷ ÷ = ÷ ì or H F ÷ (1-q)ì Substituting the values Ior enthalpy into the enthalpy balance about the Ieed plate gives: F(1-q)ì ¹ J ì ÷ Jì or 1 J J q F ÷ = ÷ (80) Substituting this into the total mole balance and dividing by F: 1 1 q L L J J F F ÷ ÷ ÷ = + ¸_¸ or L L q F ÷ = (81) Thus introducing the Ieed onto a plate causes a discontinuity in the Ilowrates oI liquid and vapor in the column. Although the Ilowrates are the same Ior each stage in the cascade above the Ieed, and the same Ior each stage in the cascade below the Ieed, the Ilowrates are diIIerent in the two cascades. Eor example, Ior a saturated liquid Ieed, q÷1, and the liquid leaving the Ieed tray has a higher Ilowrate than than entering by an amount equal to the Ieed Ilow rate: Ior q÷1: L -L ÷ qF ÷ F and J-J ÷ (1-q)F ÷ 0 whereas the vapor Ilow is unchanged. On the hand, iI the Ieed is saturated vapor, then the vapor Ilowrate leaving the stage will be higher than than entering by an amount equal to the Ieed rate. Ior q÷0: L -L ÷ 0 and J-J ÷ F The Ilows Ior other thermal conditions oI the Ieed are summarized graphically below: 06-361 page 94 Spring, 2001 Knowing all the Ilowrates, we can now calculated the reboiler heat duty. The reboiler has to boil all the vapor send into the bottom cascade. Eor a partial reboiler (where the bottoms product is withdrawn Irom the liquid remaining in the kettle), there can be no superheating oI the vapor (it must be a saturated vapor). Then the reboiler heat duty is q J r = ì ROLE OF Q-LINE Recall (see 'Step 4¨ on p81) that we said one point oI the SOL is the intersection oI the q-line (QL) and the ROL. In this section, we present the arguments supporting this claim. The equations oI the operating lines are obtained by perIorming a mole balance on component A between some pair oI streams in the middle oI the column and the end oI the column away Irom the Ieed (balances use the systems denoted by the yellow lines): R.O.L.: Jv ÷ Lx ¹ Dx D (82) S.O.L.: J v ÷ L x - Bx B (83) We claim that the q-line represents the intersection oI these two lines. Think oI these as two equations in two unknowns (x,v). The point oI intersection oI these two lines is that point (x,v) which simultaneously satisIies both equations. Now we are not going to try to solve these two equations Ior (x,v). But we can Iind other linear relationships between x and v by combining these two equations. One linear combination is particularly convenient in that it yields an equation which we can plot without knowing any oI the Ilowrates that appear in (82) and (83): 06-361 page 95 Spring, 2001 (83) minus (82): (J -J)v ÷ ( L -L)x - (Bx B ¹Dx D ) Notice that the last term in this equation represents the total rate oI component A leaving the tower. This must equal the total rate at which A enters the tower. So the above becomes: (J -J)v ÷ ( L -L)x - F: F Dividing by F gives: 1 F q q J J L L v x : F F ÷ ÷ ÷ = ÷ ¸_¸ Substituting (80) and (81): (q-1)v ÷ qx - : F or 1 1 F : q v x q q | | = + | ÷ ÷ \ . which is the equation oI our q-line. This equation represents a linear combination oI the two equations which are satisIied at the intersection oI the two operating lines. Thus this equation must also be satisIied at that point. It turns out that that QL represents the locus oI intersections oI the ROL and SOL Ior diIIerent reIlux ratios. PLATE EFFICIENCY So Iar we have always assumed that the vapor and liquid streams leaving a given stage are at equilibrium. In stepping oII the stages, we have always gone all the way Irom the operating line to the equilibrium curve. II the contacting oI the two phases on the tray is good (that is, iI the area oI contact is large and the time oI contact is long) we can approach this ideal state, but we will never quite reach it. Eor a nonequilibrium stage, our horizontal step will be a little shorter than Ior an ideal stage. OI course, we still end up at a point on the operating line, because this represents a mole balance which must still be satisIied. Because the real steps are shorter, it will take more oI them to achieve the same separation. The ratio oI the number oI ideal to real stages required will be less than one and is called the: overall (column) efficiency -- N ideal /N real ÷ E oc · 1 06-361 page 96 Spring, 2001 While this eIIiciency is easy to use, the value oI the eIIiciency is hard to determine. A more Iundamental quantity is the Murphree (tray) efficiency -- v v v v E n n n n Mv ÷ ÷ ÷ + + 1 1 * (84) where v n ÷ actual v leaving stage n v n * ÷ v in equil with x n On an xv diagram, this represents the height oI a real step divided by the height oI an ideal step Ior the same x n . Alternatively, one could also deIine eIIiciency based on the change in liquid composition (rather than vapor composition): E x x x x Mx n n n n ÷ ÷ ÷ ÷ ÷ 1 1 * (85) This deIinition is based on the width oI a step (involves changes in x) rather than based on the height oI a step (changes in v) as in (84). Some typical values oI plate eIIiciencies are given in Table 18-4 (taken Irom Perry`s 6 th edition). x n v n * v n v n1 x* n x n-1 EC OL 06-361 page 97 Spring, 2001 06-361 page 98 Spring, 2001 DETERMINING NUMBER OF REAL STAGES Suppose we have a value Ior E Mx or E Mv , how do we use it to determine the number oI real stages required? The procedure is similar to that Ior ideal stages: Steps 1-ô) same as Ior ideal stages We locate the two operating lines and the q- line, just as beIore. But beIore we step oII the stages we use the eIIiciency to determine a new curve between the operating lines and the equilibrium curve. Step 7) Plot psuedo-equilibrium curve using either (84) or (85). The blue curve in the xy-diagram at right was drawn using (84). At several x-values between x B and x D , we draw a vertical line between the lowest operating line and the equilibrium curve. On each oI these line segments, we locate a point which is a Iraction oI the distance up the line segment Irom the bottom, where that Iraction equals the plate eIIicients. The locus oI these points is the psuedo-equilibrium curve. Einally, we step oII the stages as beIore, but using the psuedo-equilibrium curve in place oI the true equilibrium curve. Step 8) Step oII stages between operating lines and pseudo-equilibrium curve. Since we used deIinition (84) based on the vapor mole Iractions, we should start our stepping process at the bottom oI the column, rather than the top. In other words, we start at (x B , x B ) and step upwards until either the vapor or the liquid mole Iraction exceeds x D . By stepping Irom the bottom up, the step Irom the operating line toward the equilibrium curve involves changes in v, whereas when we step Irom the top down, the step Irom the operating line toward the equilibrium curve involves changes in x instead oI v. The latter is appropriate when the eIIiciency is deIined in terms oI x instead oI v. v n * v n v n+1 06-361 page 99 Spring, 2001 Chapter 12. Multi-Component Distillation The objective oI distillation is the separation oI the Ieed into streams oI nearly pure products. In binary distillation, we have two components and two product streams. With more than two components in the Ieed, there is no way to achieve this with a single column (a single column might have a third product stream, but it`s composition would never be nearly pure). To separate a mixture oI three components into three nearly pure streams, we will need at least two separate distillation columns to get three streams. In general, iI there are N c components, we will need N c -1 columns. N c components N c -1 columns This assumes that each column can achieve a sharp separation between two components oI adjacent volatility. Alternatively, the designer might try to Iirst split oII component C in the Iirst column rather then A. The choice between these two alternatives is a matter oI experience. One 'rule oI thumb¨ is to perIorm the easiest separations Iirst. In this course, we will Iocus instead on the design oI the individual columns aIter this decision has been made. The Iirst step in the design oI a single column is to speciIy the product distribution. Let A,B,C,... denote components ranked in order oI decreasing volatility. sharp split -- distillate contains only components A-M, while bottoms contains only components N-Z In any real distillation, all components will be Iound in all product streams in at least trace amounts. When we say 'only¨ components A-M are Iound in the distillate, we usually mean that the mole Iractions oI the remaining components (i.e. N-Z) can be neglected when summing the mole Iractions to unity. Sharp separations are not always possible, but we will Iocus on 06-361 page 100 Spring, 2001 them in this introduction, Ior the sake oI simplicity. In the description above, components M and N play a special role and are called key components -- M,N (Ior sharp split) The more volatile oI the two is called the light key, while the other is the heavy key: light key -- M heavy key N PRODUCT DISTRIBUTION FOR A SHARP SPLIT EXAMPLE #1: A mixture with 33° hexane, 37° heptane and 30° octane is to be distilled to give a distillate product oI 1° heptane and a bottoms product oI 1° hexane. Eind D/F, B/F, and x Di , x Bi Ior the other components. Solution: Eirst we arrange the list oI components in order oI relative volatility. In this case, that is easy since we have a homologous series* oI alkanes. Since vapor pressure general drops with molecular weight in any homologous series, we don't need to consult tables oI boiling points or vapor pressures to order the components. Component Formula hexane C 6 H 14 most volatile heptane C 7 H 16 octane C 8 H 18 least volatile Let's take a basis oI 100 moles oI Ieed (F÷100) and try to Iill out the Iollowing table: Component F: Fi : Fi Dx Di x Di Bx Bi x Bi hexane 33 0.33 -- -- -- ô.ô1 light key heptane 37 0.37 -- ô.ô1 -- -- heavy key octane 30 0.30 -- -- -- -- totals ÷ 100 1 D 1 B 1 Comment: Generally the more volatile components end up in the distillate stream and the less volatile components in the bottoms. Being the most volatile, most oI the hexane is expected to end up in the distillate; we are told that the bottoms actually contains a little hexane (x B6 ÷ 0.01). Heptane is oI intermediate volatility and so might end up in either or both streams. Erom the * A 'homologous series¨ is a set of very similar items, which differ in some property. In this case, the series is a set of components of similar molecular structure (the alkanes) which differ only in the number of CH 2 groups in the chain. 06-361 page 101 Spring, 2001 problem statement, we also know that the distillate contains a little heptane (x B7 ÷ 0.01) and so presumably most oI the heptane ends up in the bottoms. Being less volatile than heptane, we can reasonably assume that all oI the octane ends up in the bottoms. In essence, we are then making a sharp split between hexane and heptane with the '1°¨ values speciIying the sharpness oI the split desired. In other words, hexane has been choosen as the light key and heptane as the heavy key. II we assume that there is virtually no octane in the distillate (octane is heavier than the heavy key), then we can Iinish speciIying the composition oI the distillate stream. Component F: Fi : Fi Dx Di x Di Bx Bi x Bi hexane 33 0.33 -- ô.99 -- 0.01 heptane 37 0.37 -- 0.01 -- -- octane 30 0.30 ô ô 3ô -- totals ÷ 100 1 D 1 B 1 Knowing the concentration oI hexane in both oI the product streams allows us to determine the Ilowrates Irom mole balances: component C 6 : 0.99D ¹ 0.01B ÷ 33 total: D ¹ B ÷ 100 Solving simultaneously: D ÷ 32.65, B ÷ 67.35 Component F: Fi : Fi Dx Di x Di Bx Bi x Bi hexane 33 0.33 32.32 0.99 ô.ô7 0.01 heptane 37 0.37 0.01 -- octane 30 0.30 0 0 30 -- totals ÷ 100 1 32.ô5 1 ô7.35 1 Now using the Iact that the balance oI the distillate and bottoms streams must be heptane, we can Iill out the table: C , C , C C , C C 06-361 page 102 Spring, 2001 Component F: Fi : Fi Dx Di x Di Bx Bi x Bi hexane 33 0.33 32.32 0.99 0.67 0.01 heptane 37 0.37 ô.33 0.01 3ô.ô7 ô.5445 octane 30 0.30 0 0 30 ô.4454 totals ÷ 100 1 32.65 1 67.35 1 Comment #1: In this example we have completed the speciIications oI the two product streams Ior a 3-component mixture in which the sharpness of split was speciIied by giving the mole Iraction oI the heavy key in the distillate stream and the mole Iraction oI the light key in the bottoms. Alternatively, we might have been given the recovery + oI the two key components in either product stream. Comment #2: The only additional inIormation that is needed iI the number oI components is larger than 3 is the mole Iraction oI those components in the Ieed. The initial guess Ior the distribution oI additional components in the product streams is determined by the Iollowing rules: Rule 1) Components heavier than the heavy key are assumed to end up in entirely in the bottoms Rule 2) Components lighter than the light key are assumed to end up in entirely in the distillate MINIMUM NUMBER OF TRAYS Now let's see iI we can get some idea oI how many trays we will need. This is a diIIicult problem to do precisely. What can be done relatively easily is to estimate the number oI trays needed at total reflux -- this is the minimum number oI trays. Some gross assumptions: total reIlux: D~0, L~J equal molal overIlow: L÷J÷const Ior all trays constant relative volatilities: o if ÷const Ior all trays Recall o if i f K K ÷ (86) is the relative volatility oI component i compared to f, where + Eor example, '5% of the heavy key is recovered in the distillate¨ means Dx D,HK ÷ 0.05×F: F,HK . 06-361 page 103 Spring, 2001 K v x i i i ÷ (87) is the distribution coefficient Ior component i (or f). With these assumptions, we can start at the top oI the column and work out the compositions oI streams leaving every stage. Eor a total condenser: v 1i ÷ x 0i ÷ x Di (88) Ior i÷1,...,N c . The Iirst subscript is the stage number (as in binary mixtures) and the second subscript is the component. With all the compositions oI the vapor stream leaving stage 1 known, we can determine the compositions oI the exit liquid stream Irom the known values oI the relative volatility: o if i f i i f f K K v x v x ÷ = 1 1 1 1 or x x v v v v i f if i f fi i f 1 1 1 1 1 1 1 = = o o (89) Substituting (88): x x x x i f fi Di Df 1 1 = o (90) Ior each component i. Notice that iI i÷f, (1) yields Ior i÷f: x x x x i i ii Di Di 1 1 1 1 1 ¸ ¸ ¸ = o or 1 ÷ 1 which doesn`t tell us anything new; thus we have only N c -1 independent relations oI this type instead oI N c . Choosing one element as the reIerence (say f÷N c ), we can write N c -1 relations Ior the N c -1 independent mole Iractions. The missing equation is x 1i ÷ 1. Thus the composition oI the liquid leaving tray 1 has been completely determined. Next we determine v 2i by mole balances. component i: Jv 2i ÷ Dx Di ¹ Lx 1i but Ior total reIlux D~0 so that L~J, leaving: 06-361 page 104 Spring, 2001 v 2i ÷ x 1i (91) Ior i÷1,...,N c . Knowing the composition oI the vapor leaving stage 2, we can calculate the composition oI the liquid Irom the known volatilities. Rewriting (89), replacing the stage number 1 by 2: x x v v i f fi i f 2 2 2 2 = o Irom (91): x x x x i f fi i f 2 2 1 1 = o Using (90): x x x x i f fi Di Df 2 2 2 = o (92) Erom (90) and (92), we begin to see a trend. The generalization oI this trend yields: x x x x ni nf fi n Di Df = o In particular, we are interested in the number oI steps required to get Irom the distillate to the bottoms: x x x x Bi Bf fi N Di Df = o min The exponent represents the minimum number oI steps on the xv diagram. One oI these steps is the partial reboiler. Solving Ior N min : N x x x x Di Bi Df Bf if min ln ln = o which is called Fenske's equation. Now in any real system, the relative volatility will not be constant. One way to patch things up is to use some kind oI average volatility: o o o if Dif Bif = where o Dif and o Bif are evaluated at the bubble-points oI the distillate or bottoms, respectively. In this case, an appropriate mean is the geometric mean oI the volatilities evaluated at the conditions oI the distillate and bottoms compositions. We use the bubble-point temperature 06-361 page 105 Spring, 2001 rather than the dew-point temperature because the distillate and bottoms streams are usually saturated liquids. II they were saturated vapors, then we would use the dew-point instead. Example #2. Calculate the N min Ior the separation in Example #1 iI the column is operated at 1.2 atm. Solution: Evaluating the relative volatilities We start by evaluating the relative volatilities in the Ieed, distillate and bottoms streams knowing their composition and thermal state. Time out: Vapor pressure data Ior pure components is available in the CRC handbook. C6, C7 and C8 Iorm a nearly ideal mixture and Raoult`s law applies: p i ÷ x i P i o where p i is the partial pressure and P i o is the vapor pressure. Then the distribution coeIIicient can be calculated Irom K v x p P x x P x P P P i i i i i i i o i i o = = = = (93) and the relative volatility Irom o if i f i o f o i o f o K K P P P P P P = = = (94) Vapor pressure depends very strongly on temperature: recall the Clausius-Clapeyron equation (S&VN, p182): d P dT RT i o i ln = ì 2 or d P d T R i o i ln 1 = ì where ì i is the molar heat oI vaporization oI the pure component. Integrating this with respect to T (ignoring the weak temperature dependence oI ì i ) yields ln P A RT i i i 0 = ÷ ì (95) or P T A RT i o i i ( ) = ' ÷ exp ì (96) 06-361 page 106 Spring, 2001 where A i is just some integration constant. Note the exponential sensitivity oI vapor pressure to temperature. Vapor pressure data is oIten summarized in handbooks (e.g. Langes Handbook of Chemistrv, 13 th edition, pages 10-29 to 10-54) using Antoine`s equation: ln P A B T C i i i i 0 = ÷ + (97) which is a semi-empirical generalization oI (95) that accounts Ior the variation oI ì i with T. Other handbooks (e.g. CRC Handbook of Chemistrv & Phvsics, 57 th edition, pages D-183 to D- 215) give vapor pressures at several temperatures and leave you to develop your own interpolation scheme. Since vapor pressure depends strongly on temperature, so does the partition coeIIicient K i given by (93). However, the relative volatility o is much less dependent on temperature. The reason becomes clear when we substitute (96) into (94): o ì ì if i o f o i f i f P P A A RT ÷ = ' ' ÷ ÷ exp Note that when the two molar heats oI vaporization are equal, the temperature-dependence vanishes. Even iI they are only approximately equal, the temperature-dependence oI the relative volatility will be much less than oI the vapor pressures. Eor this reason, the assumption oI constant relative volatility is Irequently used. Time in: We took the values oI the vapor pressure oI each component given in CRC handbook and Iit them to (97) to Iind the constants A i , B i and C i Ior each component. Now the distillate is a saturated liquid containing nearly pure hexane, so its temperature is near the boiling point oI hexane at 1.2 atm (the pressure throughout the column) or 75°C. Component x D P i o (atm) K P P i i o = i HK i HK , C6 0.990 1.200 1.000 2.574 C7 0.010 0.466 0.388 1.000 C8 0.000 0.187 0.156 0.401 The bubble point oI the bottoms is 115°C (see Geankoplis p681-2 Ior how to determine bubble points Ior multicomponent mixtures): 0.33 0.37 0.30 0.990 0.010 0.000 0.010 0.545 0.445 06-361 page 107 Spring, 2001 Calculate Minimum Stages To calculate N min , Eenske`s equation is usually applied to the heavy key and the light keys. So we are interested in o LK,HK o 67 2 574 2 236 2 399 = × = . . . Eenske`s equation is N x x x x Di Bi Df Bf if min ln ln ln . . . . ln . . = = = o 099 0 01 0 01 0544 2399 982 So at an absolute minimum, we need 10 ideal steps, or 9 ideal stages plus the reboiler. Knowing the number oI trays, we can also use Eenske`s equation to check iI indeed the octane Iraction in the distillate is negligible as we assumed in Example #1. o 87 0 401 0 469 0 434 = × = . . . N x D min ln . . . ln . = = 10 0 446 0 01 0544 0 434 8 Solving Ior x D8 , we obtain x D8 ÷ 1.93×10 -6 which indeed is negligible compared to the other mole Iractions in the distillate. MINIMUM REFLUX RATIO Method 1) Psuedo-Binary Method Assuming a saturated liquid Ieed (q÷1) and constant relative volatility, the minimum reIlux ratio turns out to be Component x B P i o (atm) K P P i i o = i HK i HK , C6 0.010 3.448 2.873 2.236 C7 0.544 1.542 1.285 1.000 C8 0.446 0.723 0.603 0.469 06-361 page 108 Spring, 2001 R x : x : D F AB D F AB min = ÷ ÷ ÷ ÷ o o 1 1 1 (98) where x D and : F are the moleIraction oI the light key in the psuedo-binary mixture and 1-x D and 1-: F are the moleIraction oI the heavy key in the psuedo-binary mixture: : : : : F F LK F LK F HK = + , , , and 1÷ = + : : : : F F HK F LK F HK , , , where : F,LK and : F,HK are the mole Iractions in the multi- component mixture. * Proof: We can obtain the minimum reIlux ratio Irom the point oI intersection oI the q-line and the equilibrium curve. Recall (76) Irom page 89: R x v v x D min = ÷ ' ' ÷ ' (99) which was derived using two points on the ROL to determine the slope and knowing that the slope is R/(R¹1). Eor a saturated liquid Ieed, the q-line is vertical so that ' = x : F (100) The v-coordinate oI this point is determined knowing that ( ' ' x v , ) lies on the equilibrium curve. Eor constant relative volatility, we can express the equilibrium curve as a simple equation using the deIinition oI relative volatility: o AB A B A A B B K K v x v x v v x x ÷ = = ÷ ÷ 1 1 Solving Ior v: v x x = + ÷ o o 1 1 Substituting x x : F = ' = , we can calculate ' = + ÷ v : : AB F AB F o o 1 1 (101) * To derive the relations above, take a basis of 1 mole of feed. then : F,LK and : F,HK represent the number of moles of LK and HK in the feed. 06-361 page 109 Spring, 2001 Substituting (101) and (100) into (99) leads to (98). Method 2) Underwood's Method This method makes similar assumptions to the psuedo-binary method except that all components are considered, not just the two key components. In particular, the assumptions made by Underwood`s method are: 1) equimolal overIlow 2) constant relative volatility 3) a sharp split with only one 'pinch¨ Eirst, we will just state the method and then we will take a quick look at its prooI. Recipe: Step 1) Eind the root (denoted |) in the interval o | o HK LK < < which satisIies the polynomial (order oI polynomial ÷ number oI components) o o | i Fi i i : q ÷ = ÷ 1 (102) Step 2) Once we have |, we can compute the minimum reIlux ratio Irom R x i Di i i min + = ÷ 1 o o | (103) where the sum is just over those components present in the distillate (i.e. we ignore components heavier than the heavy key). 06-361 page 110 Spring, 2001 ProoI: A mole balance on component i about the top n stages oI the rectiIication section yields: J n¹1 v n¹1,i ÷ L n x n,i ¹ Dx D,i (104) Recall that the distribution coeIIicient is deIined as: K i ÷ v i /x i or K v x n i n i n i , , , = We put the tray subscript on K because its value depends on the temperature, which varies Irom tray to tray. Thus (104) can be rewritten as: J v L v K Dx n n i n n i n i D i + + = + 1 1, , , , Eor constant molal overflow, the Ilowrates will be the same Ior each tray in the rectiIying cascade, just as with binary mixtures. Thus we can drop the subscripts on Ilowrate. Jv L v K Dx n i n i n i D i + = + 1, , , , (105) When we approach the minimum reIlux ratio, we will encounter a "pinch" somewhere in the column. At the pinch, we take an inIinite number oI steps without changing the mole Iractions: at a pinch: v n¹1,i ÷ v n,i (106) (106) into (105) and solving Ior v i : v D J x L J K i Di i = ÷ · · · · 1 , (107) We have put a subscript '·¨ on each Ilow rate to emphasize that this corresponds to the minimum reIlux, which requires an inIinite number oI trays. II we require (107) to be satisIied Ior each component and sum over all components: v D J x L J K i i Di i i = = ÷ · · · · 1 1 , (108) Now we can write K i in terms oI the relative volatility: 06-361 page 111 Spring, 2001 o i i ref K K = While the Ilowrate ratios can be related to the reIlux ratio: L J R R · · = + min min 1 and D J R · = + 1 1 min (108) becomes: o o | i Di i i x R ÷ = + min 1 (103) where | ÷ + R K R ref min min 1 II K ref were known, then R min could be determined as the root oI this equation. UnIortunately, to calculate K ref we need to know the temperature at the pinch, which is not known. Generally there is also a pinch in the stripping section oI the tower. Eor sharp splits, the temperature at this second pinch will be the same. We can then derive a second equation like (103) by perIorming our balances about the bottom n stages oI the column to obtain a second equation. v B J x L J K i i Bi i i = = ÷ · · · · 1 1 , where the overbars denotes Ilowrates in the stripping section. Then we will have two equations and two unknowns: K ref and R min . Their solution is the value oI | which satisIies the Iollowing equation: o o | i F i i i : q , ÷ = ÷ 1 where : F,i is the composition oI the Ieed and q is its thermal condition. There are several |'s which satisIy this equation. It turns out that we want: o HK · | · o LK Once | is determined, we can calculate R min Irom (103). This is called Underwood's method. Example #3. Calculate the R min Ior the separation in Examples #1 and #2 iI the Ieed is 60° vapor (q ÷ 0.4). 06-361 page 112 Spring, 2001 Solution: Recall the Ieed and product compositions (see sketch at right). Using Underwood`s method, we Iirst determine the root oI (102). o o | i Fi i i : q ÷ = ÷ 1 Substituting into known values: 2 399 0 33 2 399 1 0 37 1 0 434 0 30 0 434 1 0 4 . . . . . . . . × ÷ + × ÷ + × ÷ = ÷ | | | | f ¸ _ ¸¸¸¸¸¸¸ ¸ ¸¸¸¸¸¸¸ which is a cubic equation. The leIt-hand side f(|) is plotted at right. This equation has three singular points at | ÷ o 1 , o 2 and o 3 These are shown as the 3 vertical lines in the plot. Notice that there are also 3 diIIerent |`s Ior which f(|) ÷ 1-q ÷ 0.6 We are interested in the root in the interval: o | o HK LK < < or 1 · | · 2.399 which turns to be | ÷ 1.739. Once we have the value oI |, we calculate the reIlux ratio Irom (103): R x i Di i i min + = ÷ 1 o o | where the sum only includes those components present in signiIicant quantities in the distillate. In this case, octane (which is heavier than the heavy key) was neglected in the distillate Ior Example #1. It should also be neglected in this sum, although in this case, it`s inclusion does not change the result. R min . . . . . . . + = × ÷ + × ÷ = 1 2 399 0 99 2 399 1739 1 0 09 1 1739 359 ANSWER: R min ÷ 2.59 0.33 0.37 0.30 0.990 0.010 0.000 0.010 0.545 0.445 1 q 2 0 2 4 2 0 2 3 2 1 f ( 06-361 page 113 Spring, 2001 NUMBER OF IDEAL PLATES AT OPERATING REFLUX ReI: MSH p588-609 Method 1) Underwood's method Underwood's method can be extended to calculate the number oI plates analytically when the equal molal overIlow assumption and constant relativity are appropriate. But even then, the calculations are suIIiciently involved that a computer would be helpIul. II we are going to use a computer, we might as well do the rigorous tray-by-tray calculations (see Method 3); thus we will not extend Underwood`s method Iurther in these Notes. Method 2) Gilliland correlation An empirical but much easier to use method is the Gilliland correlation. This is an empirical relation between the number oI ideal stages and the operating reIlux ratio. You also need to know the minimum number oI ideal stages (at total reIlux) and the minimum reIlux ratio. Because oI its simplicity, this correlation is widely used Ior preliminary estimates. A second type oI empirical correlation is Erbar & Maddox (see Eig. 11.7-3 in Geankoplis): N N f R R R R min min min , = + + 1 1 Method 3) Rigorous Solution of MESH Equations ReIerence: Henley & Seader, Equilibrium-Stage Separation Operations in Chemical Engineering, Chpt. 15. The Iinal design oI multistage equipment to multicomponent distillation usually requires rigorous determination oI temperatures, Ilow rates, and compositions at each stage. This must be done numerically by computer. We will use a computer program called CHEMSEP to look at a couple oI typical columns. Although the program is designed to be used like a "black-box", I'd like to give you a little peek at what goes on inside. 06-361 page 114 Spring, 2001 A rigorous mathematical description oI a multistage cascade involves material and energy balances and some representation oI the vapor-liquid equilibrium. The resulting set oI algebraic equations is called the MESH Equations, Ior reasons which will soon become apparent. Consider a general stage in the cascade, which we will label stage n. It has the usual liquid and vapor streams entering and leaving. It might also be a Ieed stage, in which case F n = 0. II it is not a Ieed stage, we just set F n ÷ 0. There might also be liquid or vapor product streams U n or W n as well as a heat removal Q n . The MESH equations result Irom balances done about this arbitrary stage. M Equations are Mole balances Ior each component i (N c equations Ior each stage): L x J v F : L U x J W v n i n n i n n i n n n i n n n i n + + = + + + 1 1 1 1 , , , , , Note that the Iirst subscript is the component i and the second subscript Ior the stage n. E Equations describe the liquid-vapor Equilibrium relations among the two outlet streams, which are assumed to be at equilibrium (N c equations Ior each stage): v K x i n i n i n , , , = where K i,n is the distribution coeIIicients. S Equations are the mole Iraction Summations (2 equations Ior each stage): x i n i , = 1 and v i n i , = 1 H Equations is the energy balance Ior each stage (1 equation Ior each stage): L H J H F H L U H J W H Q n L n n J n n F n n n L n n n J n n + + = + + + + 1 1 1 1 , , , , , II there are a total oI N s stages in the column and N c components, then the number oI equations is number oI equations ÷ N s (2N c ¹3) Now we consider the Iollowing to be unknowns: x i,n , v i,n , L n , J n , T n Ior i ÷ 1,...,N c and n ÷ 1,...N s where T n is the temperature oI the liquid and vapor streams leaving stage n. number oI unknowns ÷ N s (2N c ¹3) L L U J W J F Q stage n 06-361 page 115 Spring, 2001 OI course, there are more variables involved in the equations. We assume that the Ieeds and any side drawoII Ilows are speciIied: given: F n , : i,n , H F,n , U n , W n , Q n Also, we assume that the thermodynamic relations exists so that we can compute the enthalpy oI saturated liquid and saturated vapor, given the composition, pressure and temperature: known Iunctions: H H T P x x L n L n n n n N n , , , , , , = 1 , H H T P v v J n J n n n n N n , , , , , , = 1 , K K T P x x i n n n n N n , , , , ; = 1 , 06-361 page 116 Spring, 2001 Chapter 13. Computer Simulation of Distillation We will be using a computer program called 'ChemSep¨ to simulate distillation. Today I'd like to introduce this program. Compared with industrial soItware, it is user-Iriendly and written Ior student use. Below is the welcoming screen. Arno Haket and Harry Kooijman are Irom the University oI Technology in DelIt (The Netherlands), while Ross Taylor is a ProIessor oI Chemical Engineering at Clarkson University (New York state). AIter a copyright screen, we reach the main menu: 06-361 page 117 Spring, 2001 SOME PRELIMINARY STEPS Although a mouse can be used with the program, the deIault is to use the cursor keys to move around the screen. When the desired menu item is highlighted, press ENTER to select it. You can also select menu items by typing the Iirst letter oI the desired item. Below is the result oI choosing the File menu. File Input Results Options Help e ep F Help F st elp F o F e F nits F Options F In o F in enu File Input Results Options uit e ep o e e i e to ie it op Ren e se O s ell it F1:Help F2:Last help F3:Load F4:Save F5:Units F6:Options F9:Info F10:Main menu 06-361 page 118 Spring, 2001 Notice that a sub-menu has appeared. You can probably guess what most oI these commands do. BeIore going any Iurther, you should make sure that any Iiles we create will be placed on the Eloppy Disk drive A:\. This can be accomplished by changing the deIault directory to A:\. Select the Directory command. You will then be presented with the current directory name which can be edited to read: You should also save your conIiguration Iile on your Iloppy disk. This can be accomplished by going to the Options menu and selecting the Save options command. You should edit the Iile name to read 'A:\ChemSep.cnf¨: I11e lnpuf kesu1fs Opf1ons ue1p Chem5ep v3.50 +-----------+ ¦ Load ¦ ¦ NeW ¦ ¦ 5ave ¦ ¦ D1¡ecfo¡y ¦ +------------------------------------------------+ ¦ Enfe¡ neW d1¡ecfo¡y A:\" ¦ +------------------------------------------------+ ¦ kename ¦ ¦ E¡ase ¦ ¦ DO5 she11 ¦ ¦-----------¦ ¦ Ex1f ¦ +-----------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu I11e lnpuf kesu1fs Opf1ons ue1p Chem5ep v3.50 +---------------+ ¦ 5o1ve opf1ons ¦ ¦ un1fs ¦ ¦ Mac¡os ¦ ¦ lnfe¡face ¦ ¦ D1¡ecfo¡1es ¦ ¦ v1deo dev1ce ¦ ¦ Oufpuf dev1ce ¦ ¦ 5ave opf1ons ¦ +---------------------------------------------+ ¦ Opf1ons f11ename A:\Chem5ep.cnf¸¸¸¸¸¸¸¸¸¸¸¸ ¦ +---------------------------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 119 Spring, 2001 Now any changes you make in the conIiguration can be saved in this Iile and re-loaded when you return to ChemSep at some later time. LOADING A FILE (PROBLEM) We are going to create a new Iile which simulates the solution to Prob. 3 on Hwk. #6. We start by selecting the New command Irom the File menu. You will then prompted Ior a Iile name: type 'A:\HWK6P3.SEP¨. The 'A:\¨ will Iorce this Iile to be stored in the root directory oI the Iloppy drive, which is the recommended location Ior your personal Iiles. The Iirst step in inputing the problem description is to select the components. This particular homework problem is based on MS&H5 Prob. 18-3 in which we are to separate a mixture oI acetone and methanol. Under the Input menu, select Components, Iollowed by several 'Enter`s¨ until you are prompted Ior a name (see below): I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ ¦ P¡ope¡f1es ¦ ¦ 5pec1f1caf1ons ¦ ¦ 5o1ve ¦ ¦ ChemP¡op ¦ +----------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 120 Spring, 2001 Enter the Iirst component name (acetone) and hit 'enter¨. This will bring up another short menu containing 'acetone¨ at the bottom: Select 'acetone¨ and hit 'enter¨. This will add acetone to the list oI components. Next select ->Restart search and type the next component name (methanol). AIter selecting methanol and hitting 'enter¨, we have two components in our list: I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ +-< Componenfs >- + ¦ Componenfs ¦ ¦ None ¦ ¦+-------------+ ¦ +---------------- + ¦¦ 5e1ecf ¦ ¦ ¦¦+-----------+¦ ¦ ¦¦¦ 5ea¡ch by ¦¦ ¦ ¦¦¦-----------¦¦ ¦ +¦¦ Name ¦¦-+ +--------------------------------------------------------------------------+ ¦ {5ubsf¡1ng of} name = ¦ +--------------------------------------------------------------------------+ ¦¦ C1ass ¦¦ +¦ Iam11y ¦+ ¦ P¡ope¡fy ¦ +-----------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ +-< Componenfs >- + ¦ Componenfs ¦ ¦ None ¦ ¦+-------------+ ¦ +---------------- + ¦¦ 5e1ecf ¦ ¦ +---------------------------+ ¦ Name ¦ ¦---------------------------¦ ¦ -> kesfa¡f sea¡ch ¦ ¦ -> kef1ne sea¡ch ¦ ¦ Acefone ¦ +---------------------------+ ¦ kefu¡n ¦ +-------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 121 Spring, 2001 Hit 'Esc¨ twice brings us back to the Input menu. This time, select the Operation command, then the Column subcommand. The cursor is now on the '*¨ on the line with Operation. Hitting the 'enter¨ key leads to the Iollowing screen: I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ +-< Componenfs >- + ¦ Componenfs ¦ ¦ Acefone ¦ ¦+-------------+ ¦ ¦ Mefhano1 ¦ ¦¦ 5e1ecf ¦ ¦ +---------------- + +---------------------------+ ¦ Name ¦ ¦---------------------------¦ ¦ -> kesfa¡f sea¡ch ¦ ¦ -> kef1ne sea¡ch ¦ ¦ Mefhano1 ¦ +---------------------------+ ¦ kefu¡n ¦ +-------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 122 Spring, 2001 Move the cursor to simple distillation and hit 'enter¨. This moves the cursor to the '*¨ on the line with Condenser. Hitting 'enter¨ brings up a menu oI options Ior the operation oI the condenser: OI course, in the homework problem we want to subcool the condensate, so we choose Total (Subcooled product). AIter making a Iew other choices in the remaining items, then selecting Show flowsheet, we have: I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ +-----------------------+ ¦ I1ash ¦ ¦ Co1umn ¦ +-----------------------------------------------+ ¦ Ope¡af1on " ¦ ¦------------------+----------------------------+ ¦ Condense¡ ¦ 51mp1e D1sf111af1on ¦ ¦ kebo11e¡ ¦ Exf¡acf1ve D1sf111af1on ¦ ¦ 5fages ¦ Azeof¡op1c D1sf111af1on ¦ ¦ Ieed sfages ¦ 51mp1e Abso¡be¡/5f¡1ppe¡ ¦ ¦ 51desf¡eam sfages¦ kebo11ed Abso¡be¡/5f¡1ppe¡ ¦ ¦ Pumpa¡ound sfages¦ kef1uxed Abso¡be¡/5f¡1ppe¡ ¦ ¦ ¦ 51ng1e Co1umn 5fage ¦ ¦------------------¦ 51mp1e Exf¡acfo¡ ¦ ¦ ¦ 51ng1e Exf¡acf1on 5fage ¦ +------------------¦ Comp1ex Co1umn ¦ +----------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ +-----------------------+ ¦ I1ash ¦ ¦ Co1umn ¦ +-----------------------------------------------+ ¦ Ope¡af1on 51mp1e D1sf111af1on ¦ ¦-----------------------------------------------¦ ¦ Condense¡ " ¦ ¦ kebo11e¡ +---------------------------+¦ ¦ 5fages ¦ 1ofa1 {L1qu1d p¡oducf} ¦¦ ¦ Ieed sfages ¦ 1ofa1 {5ubcoo1ed p¡oducf} ¦¦ ¦ 51desf¡eam sfages¦ Pa¡f1a1 {vapou¡ p¡oducf} ¦¦ ¦ Pumpa¡ound sfages¦ None ¦¦ ¦ +---------------------------+¦ ¦-----------------------------------------------¦ ¦ kefu¡n ¦ +-----------------------------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 123 Spring, 2001 We next select Return and the cursor moves to the Properties submenu oI Input. Hit 'enter¨ twice to enter the Thermodynamic models dialog box. Since this is not a course in thermodynamics, I will always tell you which thermo models to us. AIter making a Iew choices, the dialog box appears as Iollows: I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ +---------------I1oWsheef--------------- + ¦ Componenfs ¦ ¦ +----+ ¦ ¦ Ope¡af1on ¦ ¦ +------>-----+ ¦ ¦ +-----------------------+ ¦ ¦ ¦1 ¦ ¦ ¦ ¦ I1ash ¦ ¦ +--¦------+ +--¦-+ ¦ ¦ Co1umn ¦ ¦ ¦ ¦ +----- ¦ +--------------------------------------¦ ¦--¦---¦--¦ ¦ ¦ Ope¡af1on 51mp1e D1sf111af1on¦ ¦--¦---¦--¦ ¦ ¦--------------------------------------¦ ¦--¦---¦--¦ ¦ ¦ Condense¡ 1ofa1 {5ubcoo1ed p¡¦ ¦--¦---¦--¦ ¦ ¦ kebo11e¡ Pa¡f1a1 {L1qu1d p¡o¦ ¦--¦---¦--¦ ¦ ¦ 5fages 23 ¦ ----->---17-¦--¦---¦--¦ ¦ ¦ Ieed sfages 17 ¦ ¦--¦---¦--¦ ¦ ¦ 51desf¡eam sfages ¦ ¦--¦---¦--¦ ¦ ¦ Pumpa¡ound sfages ¦ ¦ +---¦--------------->- ¦ ¦ kefu¡n ¦ 23+----+ ¦ +--------------------------------------¦ ¦ ¦ ¦ ¦ ¦ +-------P¡ess any key fo conf1nue------- + I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 124 Spring, 2001 Selecting Return takes us to the Load data submenu. Some thermo models do not require any Iurther input Irom the user, but the cubic EOS does require a value Ior the interaction parameter. AIter selecting Load data, then Cubic EOS leads us to the Iollowing dialog: Move the cursor to Correlation and hit 'enter¨. A value Ior kif has been entered Irom the library, then select Return. Hitting 'enter¨ three more times brings us back to the Input/Specifications submenu. Hit 'enter¨ and choose Pressures. Make the selections shown as Iollows: I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ ¦ P¡ope¡f1es ¦ +----------------------+ ¦ 1he¡modynam1c mode1s ¦ +----------------------------------------------+ ¦ 1he¡modynam1c mode1s: ¦ ¦----------------------------------------------¦ ¦ k-mode1 EO5 ¦ ¦ Acf1v1fy coeff1c1enf ¦ ¦ Equaf1on of 5fafe Peng-kob1nson ¦ ¦ vapou¡ p¡essu¡e ¦ ¦ Enfha1py Excess ¦ ¦----------------------------------------------¦ ¦ kefu¡n ¦ +----------------------------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ ¦ P¡ope¡f1es ¦ +----------------------+ ¦ 1he¡modynam1c mode1s ¦ ¦ Load dafa ¦ ¦--+----------------+--¦ ¦ k¦ Load dafa fo¡: ¦ ¦ +--¦----------------¦--+ ¦ Cub1c EO5 ¦ +----------------------------------------------------------+ ¦ Cub1c EO5: Peng-kob1nson {-} ¦ ¦----------------------------------------------------------¦ ¦ Componenf 1 Componenf ¸ k1¸ ¦ ¦ Acefone Mefhano1 " ¦ ¦----------------------------------------------------------¦ ¦ kesef L1b¡a¡y Co¡¡e1af1on kefu¡n ¦ +----------------------------------------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 125 Spring, 2001 The units Ior pressure might not show up as atm. II not, you can change the units Ior entering any quantities hitting the Iunction key F5. AIter setting the pressure, choose Return which moves the cursor to Heaters/Coolers. No heat is to be added or removed except Ior the condenser and reboiler, so we move the cursor on down to Efficiencies. Here we enter the Murphree tray eIIiciency. We enter '1¨ as the default to obtain ideal trays throughout the column. Using the Insert command, we could speciIy diIIerent eIIiciencies Ior each tray. Choosing Return moves the cursor to the Input/Specifications/Feeds command. This is where we I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ ¦ P¡ope¡f1es ¦ ¦ 5pec1f1caf1ons ¦ +-----------------+ ¦ Ana1ys1s ¦ ¦ P¡essu¡es ¦ +---------------------------------------------------+ ¦ P¡essu¡es {afm} ¦ ¦---------------------------------------------------¦ ¦ Condense¡ p¡essu¡e 1.000 ¦ ¦ Co1umn p¡essu¡e Consfanf p¡essu¡e ¦ ¦ 1op p¡essu¡e 1.000 ¦ ¦ P¡essu¡e D¡op {afm/sfage} ¦ ¦ 8offom p¡essu¡e ¦ ¦---------------------------------------------------¦ ¦ kefu¡n ¦ +---------------------------------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ ¦ P¡ope¡f1es ¦ ¦ 5pec1f1caf1ons ¦ +-----------------+ ¦ Ana1ys1s ¦ ¦ P¡essu¡es ¦ ¦ ueafe¡s/Coo1e¡s ¦ ¦ Eff1c1enc1es ¦ +-------------------------+ ¦ Mu¡ph¡ee Eff1c1enc1es ¦ ¦-------------------------¦ ¦ 5fage Eff1c1ency {-} ¦ ¦ Defau1f 1.000 ¦ ¦-------------------------¦ ¦ lnse¡f De1efe kefu¡n ¦ +-------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 126 Spring, 2001 speciIy the thermal condition, composition and Ilowrate oI the Ieed. The thermal condition can be speciIied either as temperature and pressure or as vapor Iraction and pressure. Since our Ieed is 70° vapor, we use the latter option. AIter entering the values appropriate to our problem, we have: A second Ieed could be added using the Insert command. Selecting Return, brings us to the Input/Specifications/Operation command. This is where we enter speciIications regarding the two remaining degrees oI Ireedom in the column design. Hitting 'enter¨ twice brings up a range oI choices Ior speciIying the Iirst degree oI Ireedom. I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ ¦ P¡ope¡f1es ¦ ¦ 5pec1f1caf1ons ¦ +---------------------------------+ ¦ Ieed 1 ¦ ¦---------------------------------¦ ¦ Ieed sfage 17 ¦ ¦ 5fafe: P & v ¦ ¦ P¡essu¡e {afm} 1.000 ¦ ¦ vapou¡ f¡acf1on {-} 0.7000 ¦ ¦ 1empe¡afu¡e {´C} ¦ ¦---------------------------------¦ ¦ I1oW¡afes {mo1/s} ¦ ¦ Acefone 25.00 ¦ ¦ Mefhano1 75.00 ¦ ¦---------------------------------¦ ¦ 1ofa1 f1oW¡afe 100.0 ¦ ¦---------------------------------¦ ¦ lnse¡f De1efe kefu¡n ¦ +---------------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 127 Spring, 2001 We can choose only one oI these. II we speciIy the mole Iraction oI acetone to be 78° (as in our problem statement), then the program adjusts the reIlux ratio in an eIIort to obtain this composition. II we speciIy the reIlux ratio, then the program determines the composition oI the distillate product. Here we will choose the reIlux ratio (to be 8.62) and see what happens. Since we speciIied a subcooled reIlux, we next get to choose its temperature or the number oI degrees oI subcooling. We speciIy the temperature as 25°C. Hitting 'enter¨ when the cursor is over the only remaining '*¨ brings up the Iollowing list: I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ ¦ P¡ope¡f1es ¦ ¦ 5pec1f1caf1ons ¦ +-----------------+ ¦ Ana1ys1s ¦ ¦ P¡essu¡es ¦ ¦ ueafe¡s/Coo1e¡s ¦ ¦ Eff1c1enc1es ¦ +--------------------------------------+ ¦ kef1ux ¡af1o ¦ +--¦ ueaf dufy of condense¡ ¦------------------------------------ + ¦ "¦ D1sf111afe f1oW ¡afe ¦ ¦ ¦ D¦ kef1ux f1oW ¡afe ¦ " ¦ ¦ "¦ Componenf f1oW ¦ ¦ ¦--¦ Mo1e f¡acf1on of a componenf ¦------------------------------------ ¦ ¦ ¦ Componenf ¡ecove¡y ¦ kefu¡n ¦ +--¦ I¡acf1on of comb1ned feeds ¡ecove¡ed ¦------------------------------------ + ¦ 5p11f befWeen fWo componenfs ¦ ¦ I1ex1b1e ¦ +--------------------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 128 Spring, 2001 Now we get to speciIy one property oI the bottom stream. We choose to speciIy the mole Iraction oI the bottoms product. Now the Operation speciIication is complete. I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ ¦ P¡ope¡f1es ¦ ¦ 5pec1f1caf1ons ¦ +-----------------+ ¦ Ana1ys1s ¦ ¦ P¡essu¡es ¦ ¦ ueafe¡s/Coo1e¡s ¦ +--------------------------------------+ ¦ 8o11up ¡af1o ¦ ¦ ueaf dufy of ¡ebo11e¡ ¦ +--¦ 1empe¡afu¡e of ¡ebo11e¡ ¦------------------------------------ + ¦ k¦ 8offom p¡oducf f1oW ¡afe ¦ 8.620 ¦ ¦ 1¦ kebo11ed vapou¡ f1oW ¦ 25.00 ¦ ¦ "¦ Componenf f1oW ¦ ¦ ¦--¦ Mo1e f¡acf1on of a componenf ¦------------------------------------ ¦ ¦ ¦ Componenf ¡ecove¡y ¦ kefu¡n ¦ +--¦ I¡acf1on of comb1ned feeds ¡ecove¡ed ¦------------------------------------ + ¦ 5p11f befWeen fWo componenfs ¦ ¦ I1ex1b1e ¦ +--------------------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 129 Spring, 2001 Choosing Return twice places the cursor on the Input/Solve command. Hitting 'enter¨, the program saves the description and attempts to solve the MESH equations. II all the data has been entered correctly, the next screen appears: I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP {NeW p¡ob1em} Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ ¦ P¡ope¡f1es ¦ ¦ 5pec1f1caf1ons ¦ +-----------------+ ¦ Ana1ys1s ¦ ¦ P¡essu¡es ¦ ¦ ueafe¡s/Coo1e¡s ¦ ¦ Eff1c1enc1es ¦ ¦ Ieeds ¦ ¦ Ope¡af1on ¦ +------------------------------------------------------------------------------ + ¦ kef1ux ¡af1o {-} 8.620 ¦ ¦ 1empe¡afu¡e of ¡ef1ux {´C} 25.00 ¦ ¦ Mo1e f¡acf1on of a componenf {-} Acefone 0.010000 ¦ ¦------------------------------------------------------------------------------ ¦ ¦ kefu¡n ¦ +------------------------------------------------------------------------------ + I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP 2 seconds 5 1fe¡af1ons Chem5ep v3.50 +-------------+ ¦ G¡aphs ¦ ¦ 1ab1es ¦ ¦ 5p¡eadsheef ¦ +-------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 130 Spring, 2001 Now we are ready to display the results. Choosing Results/Graphs gives us a menu oI choices: PLOTTING THE MCCABE-THIELE DIAGRAM Let`s Iirst check out the McCabe-Thiele diagram: The diagram at right seems to resemble that obtained in Hwkkey #6, except that the azeotrope occurs at a mole Iraction oI about 0.76 instead oI 0.80. This diIIerence might occur because the thermodynamic model we chose is not the best one. In any case, it means that the distillate composition will never reach the 0.78 required in Prob. 3. Let`s take a closer look at the proIile. To get a better look at what's happening at the top oI the column, we might want to expand that section oI the McCabe-Thiele diagram. The scale as well as the Iormat oI any graph in ChemSep can be customized by hitting the SPACE key when the graph is displayed. The Iollowing dialog box appears: I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP 2 seconds 5 1fe¡af1ons Chem5ep v3.50 +-------------+ ¦ G¡aphs ¦ +-----------------------------------+ ¦ L1qu1d phase compos1f1on p¡of11es ¦ ¦ vapou¡ phase compos1f1on p¡of11es ¦ ¦ k-va1ue p¡of11es ¦ ¦ 1empe¡afu¡e p¡of11e ¦ ¦ P¡essu¡e p¡of11e ¦ ¦ I1oW p¡of11es ¦ ¦ Mass f¡ansfe¡ ¡afes ¦ ¦ 5f¡1pp1ng facfo¡s ¦ ¦ key ¡af1o p¡of11es ¦ ¦ ke1af1ve vo1af111f1es ¦ ¦ McCabe-1h1e1e d1ag¡am ¦ ¦ Ponchon 5ava¡1f ¦ ¦-----------------------------------¦ ¦ kefu¡n ¦ +-----------------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 131 Spring, 2001 I have shaded the two lines which can be edited to expand the scale. Changing the range on both x and v Irom the interval (0, 1) to the interval (0.6, 0.8) and then hitting the ESC key yields the second McCabe-Thiele diagram below at the leIt. A second diagram appears below at right in which we have expanded the lower end oI the diagram. I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP 2 seconds 5 1fe¡af1ons Chem5ep v3.50 +-------------+ +-----------------------------------------+ ¦ McCabe-1h1e1e d1ag¡am ¦ ¦-----------------------------------------¦ ¦ 11f1e McCabe-1h1e1e d1ag¡am ¦ ¦ 1sf key componenf Acefone ¦ ¦ 2nd key componenf Mefhano1 ¦ ¦ I11p No ¦ ¦ M1n1mum/Max1mum X 0 1 ¦ ¦ X 11c 1nfe¡va1 0.2 0 ¦ ¦ X Labe1 X Acefone/{Acefone+Me ¦ ¦ M1n1mum/Max1mum Y 0 1 ¦ ¦ Y 11c 1nfe¡va1 0.2 0 ¦ ¦ Y Labe1 Y Acefone/{Acefone+Me ¦ ¦ G¡1d Off ¦ ¦ Ax1s co1o¡ 1G¡ay ¦ ¦ 5fages 11nes 1-81ue :Doffed ¦ ¦ Equ111b¡1um L1ne 1-ked :5o11d ¦ ¦ Ope¡af1ng 11nes Ye11oW :5o11d ¦ ¦ q 11nes Wh1fe :5o11d ¦ ¦ P¡1nfe¡ 5efup P¡1nf ¦ +-----------------------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 132 Spring, 2001 PRINTING THE GRAPHS Once you get the desired graph on the screen, you probably want to make a hard copy. To print, hit the SPACE key to bring up the graph customization dialog box shown above. Notice the "Print" and "Printer Setup" items on the bottom line oI the dialog. To print, you move the cursor over the "Print" item and hit the ENTER key. ChemSep will send the graph to the printer identiIied by the "Printer Setup" menu. To get the graphs into this document, I instructed ChemSep to write a Windows bitmap into a Iile which I later imported into MS Word. I edited the "Printer Setup" (by moving the cursor over this item and hitting ENTER) to read as Iollows: AIter you have Output Device Setup reading like that shown above, hit ESC to return to the graph customization dialog, select Print and hit ENTER. You will be asked to supply a Iilename. Use a diIIerent name Ior each graph. When you "Quit" the ChemSep program, you can run Word. Bitmap Iiles can be imported into Word using the Insert/Picture command. The picture can be cropped (eliminate unwanted portions oI the picture around edges) and resized using the Eormat/Picture command. You can also use the mouse to do this. - In Word6 (OIIice 95), "Cropping" is done with the mouse by dragging one oI the six "handles" while holding down the SHIET key. "Scaling" is done with the mouse by dragging one oI the handles without using any keys. II you use one oI the Iour corner handles Ior scaling, the ratio oI height to width will be preserved. I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP 2 seconds 5 1fe¡af1ons Chem5ep v3.50 +-------------+ +-----------------------------------------+ ¦ McCabe-1h1e1e d1ag¡am ¦ ¦-----------------------------------------¦ ¦ 11f1e McCabe-1h1e1e d1ag¡am ¦ ¦ 1sf key componenf Acefone ¦ ¦ 2nd key componenf Mefhano1 ¦ ¦ I11p No ¦ ¦ M1n1mum/Max1mum X 0 0.1 ¦ ¦ X 11c 1nfe¡va1 2.0E-02 0 ¦ ¦ X Labe1 X Acefone/{Acefone+Me ¦ ¦ M1n1mum/Max1mum Y 0 0.2 ¦ ¦ Y 11c 1nfe¡va1 5.0E-02 0 ¦ ¦ Y Labe1 Y Acefone/{Acefone+Me ¦ +--------------------------------+ ¦ ¦ Oufpuf Dev1ce 5efup ¦ ¦ ¦--------------------------------¦ :Doffed ¦ ¦ Dev1ce 8MP ¦ :5o11d ¦ ¦ Po¡f A:\1E51.8MP ¦ :5o11d ¦ ¦ Mode ua1f ¦ :5o11d ¦ ¦ Wo¡k pafh ¦ ¦ +--------------------------------+----------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 133 Spring, 2001 - In Word7 (OIIice 97), "Cropping" is done with the mouse using the 'Crop Tool¨ which is shown by the icon at right. The Crop Tool can usually be Iound on the Picture toolbar, which can be displayed with the View/Toolbar command. Einally, you can position the graph anywhere on the page: - In Word6, you can enclose the picture in a Irame and the move the Irame. Make sure the picture is selected (a single click oI the mouse) and the six handles appear around it. Use the "Insert Erame" command Irom the "Insert" menu. The program might ask iI you want to switch to "Page View". Answer "yes". Once in pageview, you can drag the picture to the desired position with the mouse. You can also use the "Eormat Erame" command to position the picture and to control "text wrapping" etc. - Word7 uses Textboxes instead oI Irames. When inserting the picture Irom Iile, check the 'Eloat over text¨ option oI the Insert Picture dialog box. This puts the picture inside a Textbox which can be dragged anywhere on the page and text will wrap around it. II you have already imported the picture and did not check this box, you can still do so by selecting the picture with a single click, then use the Eormat/Object command. In the Eormat Object dialog, click on the Position tab and check the 'Eloat over text¨ option. You can add text, equations, drawings, and more bitmaps. Once you have the document looking the way you want, print it using the "Eile/Print" command. TESTING THE MCCABE-THIELE METHOD Now that we have some idea how to run ChemSep, let's use it to test the validity oI the assumptions oI the McCabe-Thiele method. The main assumption is "equimolal overIlow." We showed that iI 1) you neglect sensible heat changes compared to latent heat and 2) you assume that the latent heat oI vaporization is independent oI composition, then the molar Ilowrates oI liquid and vapor streams do not vary Irom stage to stage. ProIiles oI liquid and vapor Ilowrates can be plotted using the Results/Graphs/Flow Profiles command. Eor the description above, we obtain the graph below on the leIt: 06-361 page 134 Spring, 2001 The discontinuity oI Ilowrates on the 17 th stage occurs because this is the Ieed tray. Because the Ieed is 30° liquid in the bottom cascade is higher than in the top; because the Ieed is 70° vapor Ilowrate in the top cascade is higher than in the bottom. There is also a drop in both liquid and vapor Ilows at the top oI the column. This drop occurs because oI the subcooled reIlux. Beyond the expected changes in Ilowrates at the Ieed tray and at the top, there is a continuous drop in Ilowrate oI both streams as we move down either the rectiIying cascade or the stripping cascade. These changes in Ilowrates within a cascade are not consistent with the equimolal overflow assumption. The second graph on the right above is what would be predicted by making the equimolal overIlow assumption. We Iorced Chemsep to make the equimolal overIlow assumption by altering slightly the thermodynamic model. Navigating back to the Input/Properties/Thermo-dynamic models menu, we change the Enthalpy setting to None: 06-361 page 135 Spring, 2001 Selecting Return and re-Solveing the model leads to the Ilowrate proIiles shown on the right side. These variations in Ilowrate within a cascade stem Irom the Iact that the enthalpy oI saturated liquid is not independent oI temperature or acetone content as is required Ior "equimolal overIlow." The plot at right shows the liquid and vapor enthalpies, and the mole Iraction oI acetone in the liquid, versus stage number. + Eor equimolal overIlow, the enthalpy proIiles should be two vertical lines. Clearly, they are not vertical lines. The departure arises because the heat oI vaporization oI acetone and methanol are not quite equal. The Iollowing values were read Irom the ChemSep library: ì acetone ÷ 29.5 kJ/mol ì methanol ÷ 35.3 kJ/mol + To obtain these plots, I customized the "liquid phase composition profile" graph -- deleting the methanol composition, and adding the enthalpies using the 'top axis¨ for enthalpy. I11e lnpuf kesu1fs Opf1ons ue1p A:\uWk6P3.5EP 5 1fe¡af1ons Chem5ep v3.50 +----------------+ ¦ Componenfs ¦ ¦ Ope¡af1on ¦ ¦ P¡ope¡f1es ¦ +----------------------+ ¦ 1he¡modynam1c mode1s ¦ +----------------------------------------------+ ¦ 1he¡modynam1c mode1s: ¦ ¦----------------------------------------------¦ ¦ k-mode1 EO5 ¦ ¦ Acf1v1fy coeff1c1enf ¦ ¦ Equaf1on of 5fafe Peng-kob1nson ¦ ¦ vapou¡ p¡essu¡e ¦ ¦ Enfha1py None ¦ ¦----------------------------------------------¦ ¦ kefu¡n ¦ +----------------------------------------------+ I1:ue1p I2:Lasf he1p I3:Load I4:5ave I5:un1fs I6:Opf1ons I9:lnfo I10:Ma1n menu 06-361 page 136 Spring, 2001 How important are these enthalpy diIIerences? On the McCabe-Thiele diagram, these diIIerences show up as curvature in the operating lines. The plot at right shows points taken Irom the operating lines in the rigorous ChemSep separation and compares them with the ROL and SOL calculated using the McCabe-Thiele method. + These calculated operating lines are the straight, solid lines on the plot. There is very little diIIerence. A closer examination reveals that the maximum diIIerence in mole Iraction is 0.00473, which occurs on stage 19. Put another way, the main prediction Ior this problem is what the distillate composition and Ilowrate will be Enthalpy Model x D D/F Excess 0.7161 0.3399 None 0.7142 0.3408 The diIIerence between the two is insigniIicant. This insigniIicant diIIerence is typically the case when the molar heats oI vaporization diIIer by less than 20°. + This graph was produced by EXCEL after importing the WK1 file produced by "Results Spreadsheet" menu of ChemSep. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Acetone in Liquid A c e t o n e i n V a p o r Excess None 45-deg Equilib 06-361 page 137 Spring, 2001 THE TEMPERATURE PROFILE Another interesting proIile which can be plotted by ChemSep is temperature. The plot at right is obtained Ior the current problem. Notice that the hotter temperature occur at the bottom oI the column. This is because the bottoms contain the less volatile component which has the higher boiling point. DEGREES OF FREEDOM ChemSep lets you speciIy a number oI attributes oI the distillate and bottoms streams. We saw this list when we speciIied the condenser and the reboiler. In particular, there were 10 diIIerent kinds oI speciIications which can be made regarding the condenser, product stream or reIlux (see page 127): 1. ReIlux ratio 2. Heat duty oI condenser 3. Temperature oI condensate 4. Distillate Ilow rate 5. ReIlux Ilow rate 6. Component Ilow 7. Mole Iraction oI a component 8. Component recovery (-) 9. Eraction oI combined Ieeds recovered (-) 10.Split between two components There are also 10 diIIerent speciIications which can also be made about the reboiler (see page 128). 1. Boilup ratio 2. Heat duty oI reboiler 3. Temperature oI reboiler 4. Bottom product Ilow rate 5. Reboiled vapour Ilow 6. Component Ilow 7. Mole Iraction oI a component 06-361 page 138 Spring, 2001 8. Component recovery 9. Eraction oI combined Ieeds recovered 10.Split between two components How many oI these are independent? In ChemSep, you must speciIy the number oI stages in each section, the pressures everywhere, and a complete description oI all Ieeds. Then it turns out that there are really only two degrees oI Ireedom leIt: there are only two additional aspects oI the solution which you can speciIy. We conclude by comparing the list oI inputs and outputs Ior the two techniques we have examined. The comparison assumes we have ideal stages, constant pressure, a total consender producing a saturated liquid product, and a partial reboiler. Solution by McCabe-Thiele Graphical Method Given: 1. x D 2. desired x B 3. R ~ R min 4. N E or use oI optimum Ieed Eind: 1. N S 2. (actual x B ) Solution by Chemsep Given: 1. x D 2. x B 3. N E 4. N S Eind: 1. R 06-361 page 139 Spring, 2001 Chapter 14. Gas Absorption/Stripping gas absorption -- 2 components oI a gas are separated by contact with a liquid (in which one component is preIerentially soluble) stripping -- 2 components oI a liquid are separated by contact with a gas An example oI gas absorption is the removal oI ammonia Irom air by contact with liquid water. Ammonia is very soluble in water whereas air is only slightly soluble. Both gas absorption and stripping involve at least three components. Usually only one oI these components crosses the phase boundary. In the example oI ammonia and air, ammonia is the component whose molar Ilowrate changes by the largest percentage oI the inlet value. Although some air will also dissolve in water, and some water will evaporate into the air, the molar Ilowrates oI air and water change by negligible Iractions: their Ilows can usually be considered constant. These are the two main diIIerences between them and distillation: 1) at least 3 components, and 2) oIten only "one transIerrable component". By constrast, in distillation, all oI the components are present in both phases. EQUIPMENT FOR ABSORPTION/STRIPPING Although liquid and gas streams Ior absorption or stripping could be contacted using a tray column (like that used in distillation), tray columns are seldom used. The reason is that tray eIIiciencies are generally much lower Ior absorption and stripping than Ior distillation (perhaps only 5° instead oI 50°). Because oI this very low eIIiciency, very large numbers oI trays are required -- perhaps 100's or 1000's. Eabrication costs just become prohibitively expensive. Eortunately, a viable alternative exists: the packed tower. 06-361 page 140 Spring, 2001 A packed tower is simply a tube or pipe, which is Iilled with some sort oI "packing." The packing typically consists oI particles around an inch in diameter. In commercial packed towers, the usual choice are particles with one oI three diIIerent shapes: - raschig ring (which is just a piece oI pipe which has been cut into segments, whose length and diameter are about the same) L D ~ ~ 1 2 to 1 inches 1 2 - Berl saddle - Pall ring Although, in a pinch, almost anything you have laying around would do -- ping-pong balls, golI balls, etc. The purpose oI the packing is to promote good contact between the liquid and vapor streams which are being brought together to permit interIacial mass transIer. The liquid stream is usually Ied into the top oI the tower while the vapor is Ied into the bottom. Thus we have countercurrent Ilow oI the two streams, which has the same advantages Ior mass transIer as it did Ior heat transIer. The packing promotes good contact between the phases by dividing the two Ieed streams into many parallel interconnected paths. Ideally, you would like the liquid to Ilow downward as a thin Iilm over the surIace oI the packing. This would give the maximum surIace area oI contact between the gas and liquid. II you just pour the liquid Irom the end oI the pipe onto the top oI the packing in tower having much larger diameter than the pipe, most oI the packing will not even be wet. Only some oI the channels will be carrying Ilow. This is called: 06-361 page 141 Spring, 2001 channeling -- maldistribution oI liquid Ilow So some sort oI device to distribute the Ilow over the entire cross section oI the tower is needed. This device is called a distributor. Even iI the Ilow is evenly distributed at the top oI tower, channeling might still develop as the Iluid trickles down. When two thin Iilms converge they tend Ior Iorm a thick Iilm and a dry patch, which results in a reduction in contact area. So redistributors are placed every 10-15 Ieet along the length oI the tower. A TYPICAL ABSORBER DESIGN PROBLEM To motivate the next Iew lectures, let`s pose a typical design problem. Eor this example, I'm going to take Prob. 22-1 Irom McCabe, Smith & Harriott. Problem: treat 500 SCEM oI air containing 14 mol° acetone to remove 95° oI the acetone by absorption in liquid water in a packed bed operating at 80°E and 1 atm, with 1-inch raschig rings. The Ieed water contains 0.02° acetone and the Ilowrate is 1.1 times the minimum. The partial pressure oI acetone over an aqueous solution at 80°E can be calculated Irom p P x x A A o A A A = = ÷ ( ) ¸ ¸ where ln . 195 1 2 where P A o ÷ 0.33 atm is the vapor pressure oI acetone at 80°E. As the designer, you must select the Iollowing: 500 SCEM air containing 14 mol° acetone water with 0.02° acetone ÷ 1.1 L L min 80° E 1 atm recover 95° oI acetone 06-361 page 142 Spring, 2001 - Ilowrate oI water - diameter oI tower - height oI packing Solution Overview: Choice of liquid flowrate: L L = 1.1 to 1.5 times min Just as with distillation, there is a minimum L/J required to achieve the desired separation. In typical operation oI an absorber, the liquid rate is chosen to be just above the minimum. Tower diameter: Ap Z T = 0.25 to 0.5 inch H 0 It oI packing 2 The diameter oI the tower is usually chosen on the basis oI pressure drop. Generally, Ior a particular L and J, the smaller the column diameter, the larger the mass velocities will be, and the larger the pressure drop. Generally, large mass velocities are desirable because they give high mass transIer coeIIicients, but too large mass velocities cause 'Ilooding¨ which severely decreases mass tranIer rates. The pressure drop above is about optimum. Tower height: Z J S K a dv v v T v H v v N Oy a b Oy = ÷ ¸ * ¸ _ ¸ ¸ ¸ The height oI the tower is determined by mass transIer rates. Basically, the gas and liquid phases need to be in contact Ior a certain time Ior the acetone to diIIuse Irom the gas phase into the liquid. The equation above is called the "design equation". H J S K a Ov v ÷ is called the height of one transfer unit, where K y a is the overall mass transIer coeIIicient times the interIacial area per unit volumn oI packing, and N dv v v Ov v v a b ÷ ÷ * is called the number of transfer units, * where v is the mole Iraction oI the transIerable component and v-v* is the local driving Iorce Ior mass transIer. * This particular formula is for dilute solutions, which is probably not really applicable to the current example, but it is applicable in the homework problems. 06-361 page 143 Spring, 2001 Detailed Solution: Eirst oI all, what is 'SCEM¨? This stands Ior 'standard cubic Ieet per minute¨ SCEM ÷ standard (0°C, 1 atm) cubic Ieet per minute where 'standard¨ means that the volume is evaluated at standard conditons. We can convert SCEM into moles. Starting with the ideal gas law, PJ nRT = or J n RT P R K atm = = ° = = 273 1 22 4 359 . liters gmol It lbmol 3 depending oI which set oI units are used to express R. Recall that a gmol is the quantity oI material whose mass equals the molecular weight in grams. Similarly, a lbmol is the quantity oI material whose mass equals the molecular weight in pounds. Eor example, air has a molecular weight oI 29: MW oI air g gmol lb lbmol = = 29 29 Thus we can convert 500 SCEM into a molar Ilowrate: 500 359 139 It min It lbmol lbmol min 3 3 = = . J Now let`s move on to determine the liquid Ilowrate required. Recall that in distillation the slope oI the ROL is given by slope oI ROL = = + L J R R 1 You should also recall that there exists a minimum value oI the reIlux ratio Ior any given combination oI product and Ieed speciIications. This minimum value oI R translates into a minimum allowable L/J. There is a similar minimum L/J Ior gas absorption. To Iind the minimum reIlux ratio in distillation, we needed to plot an operating line and an equilibrium curve. EQUILIBRIUM CURVE In Prob. 22-1, McCabe gives us equilibrium data in the Iorm oI partial pressure oI acetone in the gas phase, p A , as a Iunction oI the mole Iraction oI acetone in the liquid, x: 06-361 page 144 Spring, 2001 p P x A A o A = ¸ (109) P A o ÷ vapor pressure oI acetone at 80°E ÷ 0.33 atm ¸ A ÷ activity coeIIicient Ior liquid mixture Activity coeIIicient is a measure oI nonideality oI the liquid phase. Eor ideal solutions the activity coeIIicient is unity: ¸ A ÷ 1 (ideal solution) Eor ideal solutions, (109) reduces to Raoult`s law. Our solution oI acetone in water is not ideal, but McCabe kindly gives us a model ln . ¸ A x = ÷ 195 1 2 Note that ¸ A ÷ 1 (i.e. ln¸ A ÷ 0) as x ÷ 1. This is a general rule: pure components also behave ideally since nonideal behavior results Irom interactions between diIIerent molecules; there are no diIIerent molecules when the Iluid is pure. We can obtain an expression Ior the mole Iraction oI acetone in the gas just by dividing the partial pressure by the total pressure: v p P A = where we use P ÷ 1 atm. Repeating this Ior diIIerent x`s to obtain v`s up to 0.14 (the Ieed concentration), we obtain the curve at right, where x ÷ mole Iraction oI acetone in the liquid (2 nd component is water) v ÷ mole Iraction oI acetone in the gas (2 nd component is air) OPERATING LINE Suppose we are trying to absorb acetone Irom a mixture oI air and acetone by contacting the air mixture with water. Let x v 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0 0.02 0.04 0.06 0.08 0.1 0.12 06-361 page 145 Spring, 2001 L,J ÷ total molar Ilowrates x,v ÷ mol.Irac. oI transI. comp. where the transIerrable component is acetone in this example. PerIorming a component balance on the transIerrable component about the top section oI the tower yields: L a x a ¹ Jv ÷ Lx ¹ J a v a (110) Solving Ior v: v L J x J v L x J a a a a = + ÷ This relationship v(x) between the mole Iractions in the liquid and gas streams is again known as the operating line, since the relationship is imposed by a component mole balance. However as acetone is transIerred Irom the gas to liquid, L and J change. In particular, their ratio changes. Thus the operating line is not straight. This makes it diIIicult to determine the minimum water Ilowrate There are two limiting cases where we can easily predict the change in L and J. Approx. #1: Only One 1ransferable Component It is oIten possible to assume that only one component is undergoing tranIer between the liquid and gas streams. Eor example, in our problem acetone is being transIerred Irom the air to the water: transIerable: acetone non-transIerable: air, water Although some oI the water will evaporate when it contacts the air and some oI the air will dissolve in the water, the molar rates oI transIer oI these components can oIten (but not always) be neglected compared to rate oI acetone transIer. When you can neglect the molar transIer rate oI all but one component, then signiIicant simpliIication can be made. II we can neglect evaporation oI water and dissolution oI air, then the moles oI water (the nontransIerable component) in the liquid stream must be the same at all elevations: (1-x)L ÷ L´ ÷ const so L L x = ' ÷ 1 (111) E C OL Ior L min x a v a v b OL`s Ior inc L 06-361 page 146 Spring, 2001 Similarly, the moles oI air in the gas stream must be the same at all elevations: (1-v)J ÷ J´ ÷ const so J J v = ' ÷ 1 (112) (111) and (112) into (110): ' ÷ + ' ÷ = ' ÷ + ' ÷ L x x J v v L x x J v v a a a a 1 1 1 1 where L´,J´ are constants. II we now express concentrations in terms oI mole ratios instead oI mole Iraction: X x x ÷ ÷ = 1 moles oI A in liquid moles oI non - A in liquid Y v v ÷ ÷ = 1 moles oI A in gas moles oI non - A in gas then this equation becomes very simple: L´X a ¹ J´Y ÷ L´X ¹ J´Y a or Y L J X Y L J X a a = ' ' + ÷ ' ' (113) Since L´/J´ ÷ const, this is the equation oI a straight line. Thus Ior the special case oI one-transIerable component, the operating line is straight on mole ratio coordinates. Note that mole ratios need not be smaller than unity: 0 s x s 1 but 0 s X s · although they are in this example. Let`s apply this 'one transIerable component¨ approximation to Prob. 22-1. One point on the operating line is (X a ,Y a ). We are told that the inlet water contains a small amount oI acetone: x a ÷ 0.0002 ÷ X x x a a a = ÷ = 1 0 0002 . 06-361 page 147 Spring, 2001 The concentration oI acetone in the gas phase is determined Irom the speciIication that we want to remove 95° oI the acetone Irom the Ieed: v J v J a a b b = 0 05 . v J v v J v a a b b ' ÷ = ' ÷ 1 0 05 1 . Since J´ (the molar air Ilowrate) is the same at either end oI the column, we can cancel it out, leaving Y Y a b = = ÷ = 0 05 0 00814 014 1 014 01628 . . . . . ¸ The second point on the operating line must lie somewhere along the line Y ÷ Y b ÷ 0.1628 When you plot this up, you Iind that the equilibrium curve is below the operating line. This is generally true Ior gas absorption and makes sense when you realize that the gas must be richer in the ammonia than at equilibrium otherwise the acetone would not spontaneously absorb into the liquid. Note that there is a minimum slope which the operating line can have beIore a 'pinch¨ developes with the equilibrium curve. This minimum slope represents the minimum value oI ' ' = ÷ ÷ = L J min . . . . . 01628 0 00814 0 081 0 0002 191 ' = ' = × = L J min . . . . 191 191 139 2 66 lbmol min lbmol min Multiplying this by the molecular weight oI water (18 lb/lbmol) and dividing by the density oI water (8.33 lb/gal) yields a minimum water Ilowrate oI 5.75 gal/min. X Y 0 0.04 0.08 0.12 0.16 0 0.02 0.04 0.06 0.08 0.1 06-361 page 148 Spring, 2001 Approx. #2: Dilute Solution In other problems, the solutions might be very dilute. II the solutions are suIIiciently dilute, then mole ratios are virtually equal to mole Iraction: iI x ·· 1 then 1-x ~ 1 and X x x x ÷ ÷ ~ 1 When this is also true Ior the gas stream, then (113) can be approximated as v L J x v L J x a a = ' ' + ÷ ' ' In Iact, we can also drop the prime, since L´ ~ L: v L J x v L J x a a = + ÷ Thus the operating line will also be linear on mole Iraction coordinates iI the solutions are suIIiciently dilute. This linear operating line on mole Iraction coordinates will also apply even iI we have more than one transIerrable component, provided both phases are sufficientlv dilute in all the transferrable components. INTERFACIAL MASS TRANSFER: REVIEW An important design parameter is the depth oI packing required. In distillation, the tower height was determined by the number oI plates required times the plate separation (which is usually 1-2 Ieet). The number oI plates required Ior a given separation is determined by the operating and equilibrium lines. Eor packed towers, the height also depends on the operating and equilibrium lines, but in addition it is inversely proportional to the mass transfer coefficient, which in distillation plays only a minor role in determining plate eIIiciency. Now that we have established the importance oI interIacial mass transIer in packed towers, let's talk about modelling mass transIer across a phase boundary. 06-361 page 149 Spring, 2001 Let's start by recalling the driving Iorce Ior heat transIer across an interIace. Suppose I contact a hot gas with a cold liquid. The temperature proIile near the interIace will look something like that show at right. There are two characteristics oI this sketch which are important: 1. heat Ilows Irom high to low temperature 2. temperature is continuous across the interIace InterIacial mass transIer is similar to interIacial heat transIer, but it is also diIIerent. 1. mass transIer occurs in the direction Irom high to low chemical potential (not necessarily Irom high to low concentration * ) 2. chemical potential is continuous across interIace (concentration is generally not continuous) The main diIIerence is evident in the sketch oI the concentration proIile near the interIace. Note the discontinuitv in the mole fraction at the interface. + v i = x i T vi ÷ T xi ÷ T i The reason Ior this discontinuity in concentration across the interIace has to do with thermodynamic criteria Ior phase equilibrium. Recall: phase equilibrium: µ f J ÷ µ f L Ior f÷1,...,N c thermal equilibrium: T J ÷ T L where µ j is called the chemical potential which plays the role oI temperature in mass transIer. UnIortunately, there exists no 'thermometer¨ Ior measuring chemical potential. Instead, we are Iorced to measure chemical concentration. While chemical potential usually * Within a single phase, transport is usually from high to low concentration. + Instead of component i, the subscript 'i ¨ will be used to denote quantities evaluated at the interface between two phases. In place of i, we will use f to denote components. 06-361 page 150 Spring, 2001 increases with concentration within any given phase (this is why diIIusion oI a solute occurs Irom high to low concentration), when comparing chemical potentials between two phases, there is no general correlation between chemical potential and concentration. To illustrate this, recall the simplest case oI VLE: an ideal gas mixture in equilibrium with an ideal solution. This leads to Raoult's law: v P x P f f f o = Note that: v x P P f f f o = = 1 Thus v f = x f Ior VLE even in the simplest case oI vapor-liquid equilibrium, the mole Iractions are not equal, except in the trivial case when you have only one component and the total pressure is the vapor pressure. Definitions of 1ransfer Coefficients Recall Irom the Iirst part oI this course, the local heat Ilux through the interIace can be related to the local temperatures using any one oI three types oI local heat transIer coeIIicients: rate oI heat transIer interIacial area = ÷ = ÷ = ÷ h T T h T T U T T x h i v i c h c (114) where h x ,h v ÷ one-phase local heat transIer coeIIicients U ÷ overall local coeIIicient. Basically these equations say that the rate oI heat transIer is proportional to the driving Iorce (which is the departure Irom equilibrium) and the proportionality constant is the heat transIer coeIIicient. Similar deIinitions oI transport coeIIicients can be made Ior mass transIer: molar rate oI transIer interIacial area = ÷ = ÷ = ÷ = ÷ k x x k v v K x x K v v x i v i x v ( ) ( ) ( * ) ( *) (115) where k x ,k v ÷ one-phase local mass transIer coeIIicients K x ,K v ÷ overall local mass transIer coeIIicients Comparing the mass transport expressions in (115) with their heat-transIer analogues in (114), there is a good deal oI similarity especially in the one-phase relations. Eor a single phase, the 06-361 page 151 Spring, 2001 driving Iorce is the diIIerence between the concentration or temperature in the bulk and in the concentration or temperature at the interIace. But the driving Iorce Ior the overall coeIIicients look a little diIIerent. The overall driving Iorce Ior heat transIer is just the diIIerence between the temperature oI the hot and cold Iluid T h - T c By simple analogy, you might expect the overall driving Iorce Ior mass transIer to be the diIIerence in concentration oI the two phases: v - x = overall driving Iorce Instead v - v* and x* - x appear in (115) as the overall driving Iorce. The *'s are deIined as shown on the Iigure at right. Basically v - v* and x* - x are two diIIerent measures oI the distance between the operating line and the equilibrium curve at that elevation in the packed bed where the liquid and gas concentrations are (x,v). x-v does not represent the overall driving Iorce Ior mass transIer, because the transport rate does not go to zero when x-v ÷ 0. The tranport rate is zero only at equilibrium, and x-v = 0 at equilibrium. Determining the Interfacial Concentrations: (x ,y ) Although overall mass-transIer coeIIicients are the most easy to use in design, correlations are sometimes more available in the Iorm oI single-phase coeIIicients. We may have a need to calculate overall coeIIicients Irom the single-phase coeIIicients. Eor example, in Prob. 22-1 we are given the values oI H x and H v . II we had instead the value oI H Ov we could use the value oI N Ov we computed last time to calculate the height oI packing. In what Iollows, we derive a relationship among H Ov , H x and H v . Consider once again, the rate oI interIacial transport at some arbitrary elevation in the absorber, denoted by the dotted line in the Iigure at right. The proIile oI concentration oI the tranIerable component near the interIace at this location in the absorbed looks like the sketch below at leIt. At steady state, the Ilux oI ammonia through the gas Iilm must equal the Ilux oI ammonia through the liquid 06-361 page 152 Spring, 2001 Iilm: N A ÷ k x (x i -x) ÷ k v (v-v i ) (116) II the Iluxes were not equal, we would have ammonia building up at the interIace. We will denote this interIacial Ilux by N A . Suppose the bulk compositions x,v are known, together with the two single-phase mass transIer coeIIicients. To determine the Ilux, we must also determine the interIacial compositions x i ,v i . Think oI these two interIacial concentrations as two unknowns. We need two equations. One equation is provided by the requirement that (x i ,v i ) must lie on the equilibrium curve. The second relation is (116), which can be re-written as a linear relationship between v i and x i : v k k x v k k x i x v i x v = ÷ + + slope intercept ¸ ¸_ ¸ ¸ ¸ (117) The intersection oI (117) and the equilibrium curve gives the interIacial concentrations. Example: In Prob. 22-1, we are given values oI the heights oI both single-phase coeIIicients (H x and H v ) and need to evaluate the height oI the overall gas-phase coeIIicient (H Ov ). Solution: Consider the Iollowing problem: Given: k x and k v Eind: K v Using the deIinitions oI the k´s and oI K v , N A ÷ k x (x i -x) ÷ k v (v-v i ) ÷ K v (v-v*) (118) The relationship among the concentrations is shown in the Iigure at right. Adding and subtracting v i Irom the overall driving Iorce v- v*: * * A A y y i i N N K k v v v v v v ÷ = ÷ + ÷ ¸_¸ ¸_¸ (119) Using (118) we can assign meaning to two oI the three diIIerences appear above, leaving: slope = ÷ k k x v slope = ÷ k k x v 06-361 page 153 Spring, 2001 * i A A i v v x x m N N v v K k ÷ = + ÷ ¸¸_¸¸ (120) The remaining diIIerence v i -v* can be related to x i -x and the local slope oI the equilibrium curve m: * A i i x N v v x x m m k ÷ = ÷ = (121) In the second equality above, x i -x is expressed in term oI the remaining single-phase mass transIer coeIIicient using (116). Substituting (121) into (120) and dividing by N A : 1 1 K k m k v v x = + (122) where * i i v v m x x ÷ = ÷ ÷ avg. slope oI E.C. OI course, iI the equilibrium curve is straight (as it will be in dilute solutions), then m is its slope. This is similar to the expression Ior overall heat transIer coeIIicients Ior a double-pipe heat exchanger: 1 1 1 U h h x v = + (123) We said that 1/U represented the total resistance to heat transIer through the two phases, which is just the sum oI the resistances oI each phase. The main diIIerence between (122) and (123) is the appearance oI m in (122). Eor heat transIer, the slope oI the equilibrium line is unity (m ÷ 1) because at thermal equilibrium T v ÷ m T x ÷ T x . In short, (122) states that the total resistance to mass transIer equals the sum oI the resistances oI the gas and liquid phases. We could also have showed, in a similar Iashion, that 1 1 1 K mk k x v x = + (124) where now: * i i v v m x x ÷ = ÷ II the operating line is straight, these two slopes would be the same and we could relate the two overall mass transIer coeIIicient: 06-361 page 154 Spring, 2001 straight E.C.: 1 v x m K K = Equimolar Counter-Diffusion vs. Diffusion through Stagnant Fluid In the Iorm most analogous to Eourier's law oI heat conduction, Fick's law oI diIIusion oI a binary mixture oI components A and B, the Ilux (moles/area/time) oI component A in the ¹z direction is given by * A A: AB dc J D d: = ÷ where c A is the molar concentration oI component A and D AB is the diIIusion coeIIicient. This gives the Ilux oI A relative to a reIerence Irame which moves with the mole-average velocity (i.e. the average oI the species velocity oI each component) oI the mixture (A¹B). More useIul in calculations is the molar Ilux oI A relative to a stationary reIerence Irame is * mol average velocity A: A A: B: A: AB A v J dc N N N D c d: c = + = ÷ + ¸¸_¸¸ ¸¸_¸¸ (125) where c ÷ c A ¹c B is the total molar concentration. Case I: II we use a capillary tube to connect two gas reservoirs having the same total pressure but diIIerent amounts oI gases A and B (see Eig. 6.2-1), we will obtain equimolar counter- diffusion oI A and B (i.e. N A: ÷ -N B: ). In this way, the mole-average velocity v ÷ 0 and the total pressure in both reservoirs remains constant. Eor N A: ÷ -N B: , (125) gives * A A: AB A: dc N D J d: = ÷ = (126) Case II: II instead, we have diIIusion oI benzene vapor above its liquid through air (which is virtually insoluble in the liquid), the air must remain stagnant (i.e. N B: ÷ 0) since it cannot enter the liquid (see Eig. 6.2-2a). This is diIIusion oI A through stagnant B. Eor N B: ÷ 0, (125) gives A A A A: AB A: v dc c N D N d: c = ÷ + or * 1 A: A A A: AB J dc v N D d: ÷ = ÷ ¸¸_¸¸ Einally we have * 1 A: A: A J N v = ÷ (127) 06-361 page 155 Spring, 2001 Eor the same driving Iorce (concentration gradient), * A: J is the same Ior both cases, but N A: are diIIerent. Since 1-v A is always less than one, we see that equimolar counter-diIIusion (126) is slower than diIIusion through a stagnant Iluid (127). This can be qualitatively understood as Iollows. Suppose that to get to class, you need to walk down a corridor that's crowded with other students. II everyone else was standing almost still (stagnant Iluid), it would be easier to walk around them than iI everyone is walking toward you (counter diIIusion). All oI the Ilux expressions above apply locally at every point in the Iluid. Eor 1-D steady equimolar counter-diIIusion (Irom a reservoir having c A1 to a second reservoir having c A2 ), (126) integrates to * 1 2 1 2 1 2 2 1 2 1 AB AB A: A: A A A A v A A D cD N J c c v v k v v : : : : ' = = ÷ = ÷ = ÷ ÷ ÷ (128) where z 2 -z 1 is the length oI the capillary through which diIIusion is occuring and k´ v is the single-phase (gas) mass transIer coeIIicient to be used with mole Iractions and equimolar-counter diIIusion. The analogous expression Ior diIIusion through stagnant Iilm oI B is (see Section 6.2C): 1 2 1 2 2 1 1 AB A: A A v A A A M cD N v v k v v : : v = ÷ = ÷ ÷ ÷ (129) where 1 2 1 2 1 1 1 1 ln 1 A A A M A A v v v v v ÷ ÷ ÷ ÷ = ÷ ÷ is the log-mean oI v B evaluated at either end oI the diIIusion path. Comparing (128) and (129), we see that 1 v v A M k k v ' = ÷ Similar relations exist Ior diIIusion in the liquid phase: 1 x x A M k k x ' = ÷ and between the overall mass-transIer coeIIicients Ior equimolar counter-diIIusion and diIIusion through a stagnant Iilm: * 1 v v A M K K v ' = ÷ and * 1 x x A M K K x ' = ÷ 06-361 page 156 Spring, 2001 where * * * 1 1 1 1 ln 1 A AG A M A AG v v v v v ÷ ÷ ÷ ÷ = ÷ ÷ and * * * 1 1 1 1 ln 1 A AL A M A AL x x x x x ÷ ÷ ÷ ÷ = ÷ ÷ and where v AG is the mole Iraction in the bulk oI the gas, * A v is the mole Iraction which would be in equilibrium with the bulk liquid having a mole Iraction oI x AL , and * A x is mole Iraction in the bulk oI the liquid which would be in equilibrium with the bulk gas having a mole Iraction oI v AG . HEIGHT OF A PACKED TOWER The analysis which Iollows has as its goal the determination oI the height oI packing required. The approach is similar to that used in the design oI double-pipe heat exchangers in which the goal is the determination oI the area oI heat-exchange surIace required. The main complication arises Irom the Iact that the mass Ilux is not proportional to v-x; instead, the mass Ilux is proportional to v-v*. We have to chop up the tower into pieces which are small enough so that the driving Iorce is virtually uniIorm throughout each piece. Since the compositions change only with :, we chop up the tower in such a way that we produce pieces which have :~const, which is a thin horizontal slice. Now let's take a closer look at what happens inside this slice oI the tower. The slice contains solid packing as well as liquid and gas streams. In what Iollows, we will ignore the solid and treat the liquid and gas phases as iI they were completely separated, rather than interspersed in each other. double-pipe hxer analogy: Let's do a mass balance on the acetone in the gas phase only contained within our slice oI column. Besides the liquid streams entering and leaving the slice, we have acetone crossing the 06-361 page 157 Spring, 2001 interIace between the gas and liquid phases. The rate oI absorption per unit area can be expressed in terms oI the local overall mass transIer coeIIicient: molar rate oI transIer interIacial area = ÷ K v v v ( *) (130) U T T o h c ( ) ÷ We could also have used one oI the other expressions in (115); this one proves later to be more convenient than some oI the others. Just like the overall heat transIer coeIIicient U o depends on the Ilowrates oI both Iluids being contacted in the heat exchanger, the overall mass transIer coeIIicient K v depends on both Ilowrates K K L J y y = , To obtain the rate oI transIer across the interIace in our slice oI column, we need to multiply (130) by the interfacial area in that slice. In a heat exchanger, the heat transIer surIace is Iixed by the geometry oI the equipment selected: it is just the area oI the pipe wall or the tubes. In particular, the heat transIer area does not depend on the Ilowrates oI the hot and cold streams. On the other hand, the boundary between liquid and gas in a packed bed is very complex and very hard to measure directly. Most importantly, the area also depends on the Ilowrates oI the gas and liquid streams (L,J). The interIacial area is usually expressed as a, the interIacial area per unit volume oI packing: interIacial area volume oI packing ÷ = a a L J ( , ) (131) t t D d: D d: D o o o 2 4 2 = One empirical correlation relating area to Ilowrates is the Schulman Correlation (Table 6.3 oI Treybal): a mG G x v = o | where G x and G v are the mass Ilowrates oI the liquid or gas stream divided by the cross-sectional area oI the tower 06-361 page 158 Spring, 2001 G D ÷ = mass time mass velocity t 2 4 and where m, o, | are constants which depend on the type and size oI packing used. Multiplying (130) by (131): rate oI transIer interI area interI area vol. oI packing × = ÷ × K v v a v ( *) r K a v v v ÷ = ÷ molar rate oI transIer volume oI packing ( *) (132) 4U D T T o o h c ÷ II we now multiply by the volume oI this slice oI the packed column, we will obtain the rate at which acetone crosses the interIace in this slice: rate oI transIer ÷ r S A: ÷ dN A dq U dA T T o o D d: h c o = ÷ t ¸ where S ÷ empty tower cross-section ÷ tD T 2 4 S A: ÷ volume oI slice At steady state, the rate oI transport oI acetone into the vapor must equal the rate out: in ÷ out (Jv) : ÷ (Jv) :¹: ¹ rSA: (133) ` ` m c T : m c T : : dq p p µ µ ( ) = + ( ) + A Dividing by -A: and letting A: ÷ 0, (133) becomes: d Jv d: rS = ÷ (134) ` m c dT d: dq d: U D T T p o o h c µ t = ÷ = ÷ 06-361 page 159 Spring, 2001 Now the total molar Ilowrate (J) will change as the gas phase loses acetone, but the Ilowrate oI acetone-Iree air (J´ ) does not change with Z: J J v = ' ÷ 1 ÷ Jv J v v = ' ÷ 1 d Jv d J v v J d v v J dv v J dv v = ' ÷ = ' ÷ = = ' ÷ = ÷ 1 1 1 1 2 (134) becomes: J v dv d: rS K a v v S v 1÷ = ÷ = ÷ ÷ * (135) The second equation is obtained by substituting our expression Ior the rate oI absorption r Irom (132). We can now solve this equation Ior the height dZ oI the slice: d: Jdv v v v K a S v = ÷ ÷ ÷ 1 * So iI we know the local mole Iractions v and v* and the change in the mole Iraction dv oI the gas phase which occurred between the top and bottom oI this slice, we can calculate the height oI the slice. The total height oI packing is the sum oI the height oI each slice: Z d: Jdv v v v K a S T Z v v v T b a = = ÷ ÷ ÷ 0 1 * Eactoring out the S, which is constant along the entire height oI the packing, and changing the order oI integration (since v b ~v a ): Z S Jdv v v v K a T v v v a b = ÷ ÷ 1 1 * Now K v depends on G x and G v , which in turn depend on gas composition. Eor the case oI a dilute gas stream, we know that v v J J K a v