Chapter 4 Total Reflu and Minimum Reflu Ratio a. Total Reflu. In design problems, the desired separation is specified and a column is designed to achieve this separation. In addition to the column pressure, feed conditions, and reflu temperature, four additional variables must be specified. Specified Variables Case A. 2. 3. Eternal reflu ratio L 0 / 4. Use optimum feed plate esigner Calculates, : distillate and bottoms flow rates Q R, Q C : heating and cooling loads : number of stages, feed : optimum feed plate C : column diameter, = mole fraction of more volatile component A in distillate and bottoms, respectively The number of theoretical stages depends on the reflu ratio R = L 0 /. As R increases, the products from the column will reduce. There will be fewer equilibrium stages needed since the operating line will be further away from the equilibrium curve. The upper limit of the reflu ratio is total reflu, or R =. The rectifying operating line is given as y n+ = R R + n + R + When R =, the slop of this line becomes and the operating lines of both sections of the column coincide with the 45 o line. In practice the total reflu condition can be achieved by reducing the flow rates of all the feed and the products to zero. The number of trays required for the specified separation is the minimum which can be obtained by stepping off the trays from the distillate to the bottoms. Figure 4.4-8 Minimum numbers of trays at total reflu. 4-4
y + Total condenser y -, V-, L - 2 y 2 y 0 Total reboiler Figure 4.4-9 istillation column operation at total reflu. The minimum number of equilibrium trays can also be approimated by Fenske equation, m = log ( α ) (4.4-22) In this equation α = (α α ) /2 where α is the relative volatility of the overhead vapor and α is the relative volatility of the bottoms liquid. We can derive Fenske equation using the notation shown in Figure 4.4-9 where stages are numbered from the bottom up. The vapor leaving stage is condensed and returned to stage as reflu. The liquid leaving stage is vaporized and returned to stage as vapor flow. For steady state operation with no heat loss, heat input to the reboiler is equal to the heat output from the condenser. From material balance, vapor and liquid streams passing between any pair of stages h equal flow rates and compositions, for eample, V - = L and y - =. In general, molar vapor and liquid flow rates will change from stage to stage unless the assumption of constant molar overflow is valid. At stage, the equilibrium relation is written as y = K (4.4-23) 4-42
From the material balance y = 2 (4.4-24) Combine Eqs. (4.4-23) and (4.4-24) 2 = K (4.4-25) Similarly for stage 2 y 2 = K 2 2 (4.4-26) Combine Eqs. (4.4-25) and (4.4-26) y 2 = K 2 K (4.4-27) The procedure can be repeated to stage where y = K K - K 2 K (4.4-28) Similarly for the less volatile component i y = K i, K i,- K i,2k i, ( ) (4.4-29) ividing Eq. (4.4-8) by Eq. (4.4-9), we h y y = α α - α 2 α (4.4-30) In this equation α k = K K k i, k Rearranging Eq. (4.4-30) we obtain = relative volatility between the two components on stage k. y = y min αk or + = + k= min αk (4.4-3) k= Since + =, =, and assuming constant relative volatility, Eq. (4.4-3) becomes α min = (4.4-32) Solving for the minimum number of equilibrium trays gives min = log ( α ) (4.4-33) 4-43
Eq. (4.4-33) is the Fenske equation (4.4-22) where α = α = (α α ) /2 m = log ( α ) (4.4-22) b. Minimum reflu ratio. As the reflu ratio is reduced, the distance between the operating line and the equilibrium curve becomes smaller. The minimum reflu ratio R m is the limiting reflu where the operating line either touches the equilibrium curve or intersects the equilibrium curve at the q-line. The minimum reflu ratio will require an infinite number of trays to attain the specified separation of and. Figure 4.4-0 shows an equilibrium plate n with streams L n- and V n+ entering and streams L n and V n le the plate. If the two steams L n- and V n+ are at equilibrium there will be no net mass transfer between the liquid and vapor streams. The equilibrium curve will touch or intersect the operating line at this point. n- L n- V n n L n V n+ n+ Figure 4.4-0 Equilibrium plate n with vapor and liquid streams. Given q, and F, the feed line is fied and the upper operating line depends on the reflu ratio R. At total reflu, the operating line coincides with the 45 o line. As R is decreased, the slope of the enriching operating line R/(R + ) is decreased. The operating line will rotate clockwise around the point ( =, y = ) until it is tangent to the equilibrium curve or it intersects the q-line at the equilibrium point whichever comes first. The location where the operating line touches or intersects the equilibrium curve is called the pinch point. The enriching operating line at minimum reflu is then defined. y n+ = Rm R + n + m R + m The minimum reflu R m can be obtained from either the intercept of the slope of the enriching operating line. The operating flu ratio is between the minimum R m and total reflu. Usual value is between.2r m to.5r m. Figure 4.4- shows the pinch points for case where the operating line intersects the equilibrium curve and case 2 where the operating line touches the equilibrium curve. 4-44
Figure 4.4- The pinch points for minimum reflu. Eample 4.4-4 ---------------------------------------------------------------------------------- A distillation column receives a feed that is 40 mole % n-pentane and 60 mole % n-heane. Feed is saturated liquid with a flow rate of 2,500 lbmol/hr. The column is at atm. A distillate of 90 mole % n-pentane is desired. A total condenser is used. Reflu is a saturated liquid. ottoms from the reboiler is 98 mole % n-heane. etermine the minimum number of equilibrium trays and the minimum reflu ratio. ata: Vapor pressure, P sat, data: ln P sat = A /(T + C), where P sat is in kpa and T is in K. Compound A C n-pentane () 3.9778 2554.6 36.2529 n-heane (2) 4.0568 2825.42 42.7089 Heat of evaporation for n-pentane, λ C5 =,369 tu/lbmol, C pl,c5 = 39.7 tu/lbmol o F Heat of evaporation for n-heane, λ C6 = 3,572 tu/lbmol, C pl,c6 = 5.7 tu/lbmol o F Solution ------------------------------------------------------------------------------------------ (a) Minimum number of equilibrium trays The equilibrium data for n-pentane and n-heane at atm are listed in Table 4.4-4. The data were generated with the Matlab codes listed in Table 4.4-5 assuming ideal solution. 4-45
Table 4.4-4 Equilibrium data for n-pentane and n-heane system at atm. = mole fraction of n-pentane in the liquid y = mole fraction of n-pentane in the vapor = 0.00000, y = 0.00000, T(K) = 342.06 = 0.05000, y = 0.2705, T(K) = 339.40 = 0.0000, y = 0.23699, T(K) = 336.9 = 0.5000, y = 0.33263, T(K) = 334.58 = 0.20000, y = 0.4626, T(K) = 332.39 = 0.25000, y = 0.48975, T(K) = 330.32 = 0.30000, y = 0.55462, T(K) = 328.38 = 0.35000, y = 0.624, T(K) = 326.53 = 0.40000, y = 0.66335, T(K) = 324.79 = 0.45000, y = 0.709, T(K) = 323.4 = 0.50000, y = 0.7506, T(K) = 32.56 = 0.55000, y = 0.787, T(K) = 320.07 = 0.60000, y = 0.82048, T(K) = 38.64 = 0.65000, y = 0.85070, T(K) = 37.28 = 0.70000, y = 0.8786, T(K) = 35.97 = 0.75000, y = 0.9037, T(K) = 34.72 = 0.80000, y = 0.9260, T(K) = 33.53 = 0.85000, y = 0.94692, T(K) = 32.38 = 0.90000, y = 0.9660, T(K) = 3.28 = 0.95000, y = 0.98374, T(K) = 30.22 =.00000, y =.00000, T(K) = 309.20 The minimum number of trays required is obtained by stepping off between the 45 o line and the equilibrium curve from = 0. to = 0.9. The answer is 4. equilibrium trays. Figure E- Minimum numbers of trays at total reflu. 4-46
The minimum number of trays can also be estimated by Fenske equation, m = = log = ( 8) log ( α ) log ( α ) log ( α ) 0.9 0. 0.9 0. In this equation α = (α α ) /2 where α is the relative volatility of the overhead vapor and α is the relative volatility of the bottoms liquid. y + Total condenser y -, V-, L - 2 y 2 y 0 Figure E-2 istillation column operation at total reflu. From the notation in Figure E-2, at the top y = = 0.9 0.9032, = 0.75000. At the bottoms = = 0., y = 0.237 (equilibrium data from Table 4.4-4). Total reboiler y / α = ( y ) / ( ) 0.9032 / 0.7500 = ( 0.9032 ) / ( 0.7500 ) = 3.02 y / α = = ( y ) / ( ) ( 0.237 ) / ( 0.00 ) 0.237 / 0.00 = 2.7955 4-47
α = (α α ) /2 = (3.02 2.7955) /2 = 2.9487 m = ( ) ( α ) log 8 log = ( ) ( ) log 8 log 2.9487 = 4. (b) Minimum reflu ratio. Figure E-3 The pinch point for minimum reflu. Since feed is saturated liquid, the feed-line is vertical and intersects the equilibrium curve at the point = 0.4, y = 0.6633. The enriching operating line for minimum reflu passes through this point ( = 0.4, y = 0.6633) and the point ( = = 0.9, y = 0.9). The slope of the rectifying operating line is given by Rm 0.9000 0.6633 = R + 0.9000 0.4000 m = 0.4734 The minimum reflu ratio is then R m = 0.4734.0000 0.4734 = 0.90 4-48