Limiting condition McCabe Thiele Graphical Equilibrium-Stage When analzing or designing a process, it is useful to look at limiting cases to assess the possible values of process parameters. In distillation analsis, separation of a pair of components can be improved b increasing the number of stages while holding reflu constant, or b increasing the reflu flow for a given number of stages. This tradeoff sets up two limiting cases: 1. Total Reflu (minimum ideal stages) 2. Minimum Reflu (infinite ideal stages) The design tradeoff between reflu and stages is the standard economic optimization problem chemical engineers alwas face -- balancing capital costs (the number of tras to be built) vs. the operating cost (the amount of reflu to be recirculated). good design will operate near a cost optimum reflu ratio. ChE 334: Separation rocesses r Saad l-shahrani
McCabe Thiele Graphical Equilibrium-Stage a) Minimum number of plates: L If the reflu ratio ( R ) is increased to ver large value, the operating lines become the 45 o line. The infinite reflu ratio occurs in real life when the column is operated under what are called (total reflu) condition Under these conditions, no feed is added to the column (F=0) and no products are withdrawn (=0, =0), but the vapor is raised up and condensed to the column. So the column is just circulating vapor and liquid up and down. Most columns are started up under total reflu conditions. ChE 334: Separation rocesses r Saad l-shahrani
istillation of inar Miture Since the liquid flow rate in the column is same as the vapor flow rate, L V V L 1.0 The operating line and n1 L L n L n, m1 L L m L m The composition in the base of the column under total reflu =, and the composition of the liquid in the reflu drum = In this case the number of ideal plates is minimum. ChE 334: Separation rocesses r Saad l-shahrani
inar Multistage istillation The minimum number of ideal plates can be done b: a) Graphicall as shown in the figure Minimum number of plates = 3+reboiler 2 1 X 1 Composition of liquid in reflu drum 1 = 2 = 1 3 Operating lines as total reflu 3 = 2 X 2 4 = 3 4 = X X 3 Composition of liquid in re-boiler X X F X ChE 334: Separation rocesses r Saad l-shahrani
inar Multistage istillation b) nalticall (using Fenske Equation) This equation gives the number of plates required under total reflu at constant. It is applicable to multi-component sstem as well as binar sstem (= constant, total reflu, ideal sstem). It is ver useful for getting quick estimates of the size of a column. erivation of Fenske Equation Consider two component (,) forming ideal solution K K / / / / mole ratio in top product mole ration in bottomproduct (1) ChE 334: Separation rocesses r Saad l-shahrani
r Saad l-shahrani ChE 334: Separation rocesses n ideal miture follows Raoult s law and = vapor product ratio inar Multistage istillation K K / / / / / does not change much over the range of temperature encountered, constant 1, 1 (2)
inar Multistage istillation Substitute (2) in (1) 1 1 For plate n+1 n1 n1 1 n 1 1 n 1 Since = 0 (total reflu), L / V= 1.0, n1 L L n L zero Then n+1 = n and n n1 1 n 1 n 1 ChE 334: Separation rocesses r Saad l-shahrani
inar Multistage istillation t the top of the column, if a total condenser is used 1 =, n = 0 Substitute in (2) 1 1 2 1 1 n ChE 334: Separation rocesses 1 For plate (1) 1 1. 1. 2.......... n1 1 n 1 1 n 1 1 n n For plate (2) For plate (n) For re-boiler plate 1 n-1 n n-1 r n L b, b water 0 1 2 2 3 3 4 steam Re-boiler r Saad l-shahrani V b b
inar Multistage istillation If all equations are multiplied together and all the intermediate terms canceled, n ( ) 1 1 1 ( ) N min 1 1 Where n= Nmin + reboiler N min 1 ln[( /1 ln ) /( /1 )] ln[(mole ration) /(mole ration) or N min1 ln( / ln ln ] ) /( / ) ChE 334: Separation rocesses r Saad l-shahrani
McCabe Thiele Graphical Equilibrium-Stage Eample: Calculate the minimum number of tras required to achieve a separate from 5 mole % bottoms to 90 moles % distillate in a binar column with =2.5 solution = 0.05, = 0.9 N min 1 ln[( /1 ln ) /( /1 )] N min ln[(0.9/1 0.9) /(0.05/1 0.05)], ln 2.5 1 N min 5.14 0.9163 1 4.61 ChE 334: Separation rocesses r Saad l-shahrani
McCabe Thiele Graphical Equilibrium-Stage Eample: in a miture to be fed to a continuous distillation column, the mole fraction of phenol is 0.35, of o-cresol 0.15, of m-cresol 0.3 and of lenes 0.2. it is hoped to obtain a product with a mole fraction of phenol 0.952, of o-cresol 0.0474, of m-cresol 0.0.0006. if p-o = 1.26, m-o =0.7, estimate how man theoretical plates would be required at total reflu. ssume no phenol in the bottoms. Solution: light component (o-cresol) heav component (m-cresol) Total balance 100= + = zero For phenol 100*0.35=*0.952+*,p = 36.8 Kmol, = 63.2 Kmol For o cresol 100*0.15=0.0474*36.8+,o *63.2,o =0.21 ChE 334: Separation rocesses r Saad l-shahrani
McCabe Thiele Graphical Equilibrium-Stage For m cresol 100*0.3=0.0006*36.8+,m *63.2,m =0.474,X =0.316 component Feed top ottms phenol 0.35 0.952 0 p-o = 1.26 o-cresol 0.15 0.0474 0.21 o-o = 1.0 m-cresol 0.3 0.0006 0.474 m-o =0.7 lenes 0.2 0 0.316 ln[(0.0474 / 0.474)] o-m = 1/0.7=1.43 N 1 0.0006) /(0.21/ min N min 13. 5 ln1.43 ChE 334: Separation rocesses r Saad l-shahrani
McCabe Thiele Graphical Equilibrium-Stage b) Minimum Reflu Ratio The net figure shows how changing the reflu ratio affects the operating lines: the lower the reflu ratio, the closer the operating line moves toward the equilibrium curve, and the larger the number of plates. If the reflu ratio finall reduced to the point where either operating line intersects or becomes tangent to the VLE curve, an infinite number of plates will be required and the reflu ratio is minimum. ChE 334: Separation rocesses r Saad l-shahrani
McCabe Thiele Graphical Equilibrium-Stage To obtain the R min n1 R R 1 n R 1 (, ) ab intercept R min 1 R a min 1 b ` ChE 334: Separation rocesses r Saad l-shahrani
McCabe Thiele Graphical Equilibrium-Stage If the equilibrium curve has a cavit upward, e.g., the curve for waterethanol shown in the figure in this case the minimum reflu ratio must be computed from the slope of the operating line (ac) that is tangent to the equilibrium R a min 1 ` Feed line Non-ideal Line VLE c b ` ChE 334: Separation rocesses r Saad l-shahrani
McCabe Thiele Graphical Equilibrium-Stage Eamole. continuous fractionating column is to be design to separate 30,000 kg/h of a miture of 40 percent benzene and 60 percent toluene into an overhead product containing 97 percent benzene and a bottom product containing 98 percent toluene. These percentages are b weight. reflu ratio of 3.5 mol to 1 mol of product is to be used. The molal latent heats of benzene and toluene are 7,360 and 7,960 cal/ gmol, respectivel. enzene and toluene form an ideal sstem with a relative volatilit of about 2.5. The feed has a boiling point of 95 o C at a pressure of 1 atm. (a) Calculate the moles of overhead product and bottom product per hour. (b) etermine the number of deal plate and the position of the feed plate (i) if the feed is liquid and at its boiling point. ii)if the feed is liquid and at 20 o C (specific heat 0.44 cal/ g- o C) (iii) if the feed is a miture of two-thirds and one-third liquid. (c) If steam at 20 Ib,/in 2 (1.36 atm) gauge is used for heating, how much steam is required per hour for each of the above three cases, neglecting heat losses and assuming the reflu is a urated liquid? (d) If cooling water enters the condenser at 25 C and leaves at 40 C, how much cooling water a required, in gallons per minute? ChE 334: Separation rocesses r Saad l-shahrani
1 0.9 3 2 1 0.8 0.7 0.6 Feed line 6 5 4 0.5 7 R 1 0.4 0.3 0.2 0.1 0. 10 8 9 R F =0.44 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 =0.0235 =0.974