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Minimum Reflux Ratio Chapter 4 Total Reflux and Minimum Reflux Ratio a. Total Reflux . 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 reflux temperature, four additional variables must be specified. Specified Variables Designer Calculates Case A D, B: distillate and bottoms flow rates 1. xD QR, QC: heating and cooling loads 2. xB N: number of stages, Nfeed : optimum feed plate 3. External reflux ratio L0/D DC: column diameter 4. Use optimum feed plate xD, xB = mole fraction of more volatile component A in distillate and bottoms, respectively The number of theoretical stages depends on the reflux ratio R = L0/D. 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 reflux ratio is total reflux, or R = ∞. The rectifying operating line is given as R 1 yn+1 = xn + xD R +1 R +1 When R = ∞, the slop of this line becomes 1 and the operating lines of both sections of the column coincide with the 45 o line. In practice the total reflux 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 reflux. 4-41 Total condenser yN xN+1 N yN-1 , V N-1 xN , L N N-1 2 y1 x 2 1 y0 x1 Total reboiler Figure 4.4-9 Distillation column operation at total reflux. The minimum number of equilibrium trays can also be approximated by Fenske equation, x1− x log D B 1− x x N = D B (4.4-22) m ()α log ave 1/2 In this equation αave = ( α1αB) where α1 is the relative volatility of the overhead vapor and αB 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 N is condensed and returned to stage N as reflux. The liquid leaving stage 1 is vaporized and returned to stage 1 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 have equal flow rates and compositions, for example, VN-1 = LN and yN-1 = xN. 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 1, the equilibrium relation is written as y1 = K1x1 (4.4-23) 4-42 From the material balance y1 = x2 (4.4-24) Combine Eqs. (4.4-23) and (4.4-24) x2 = K1x1 (4.4-25) Similarly for stage 2 y2 = K2x2 (4.4-26) Combine Eqs. (4.4-25) and (4.4-26) y2 = K2K1x1 (4.4-27) The procedure can be repeated to stage N where yN = KNKN-1…K 2K1x1 (4.4-28) Similarly for the less volatile component i 1 − yN = Ki,NKi,N-1…K i,2K i,1 (1 − x1) (4.4-29) Dividing Eq. (4.4-8) by Eq. (4.4-9), we have y x N = α α …α α 1 (4.4-30) − N N-1 2 1 − 1 yN 1 x1 Kk In this equation αk = = relative volatility between the two components on stage k. Ki, k Rearranging Eq. (4.4-30) we obtain − Nmin − Nmin yN 1 yN α xN +1 1 xN +1 α = ∏ k or = ∏ k (4.4-31) x 1− x = x 1− x = 1 1 k 1 1 1 k 1 Since xN+1 = xD, x1 = xB, and assuming constant relative volatility, Eq. (4.4-31) becomes x 1− x α Nmin = D B (4.4-32) 1− x x D B Solving for the minimum number of equilibrium trays gives x1− x log D B − 1 xD x B Nmin = (4.4-33) log ()α 4-43 1/2 Eq. (4.4-33) is the Fenske equation (4.4-22) where α = αave = ( α1αB) x1− x log D B 1− x x N = D B (4.4-22) m ()α log ave b. Minimum reflux ratio . As the reflux ratio is reduced, the distance between the operating line and the equilibrium curve becomes smaller. The minimum reflux ratio Rm is the limiting reflux where the operating line either touches the equilibrium curve or intersects the equilibrium curve at the q-line. The minimum reflux ratio will require an infinite number of trays to attain the specified separation of xD and xB. Figure 4.4-10 shows an equilibrium plate n with streams Ln-1 and Vn+1 entering and streams Ln and Vn leave the plate. If the two steams Ln-1 and Vn+1 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-1 Ln-1 Vn n V Ln n+1 n+1 Figure 4.4-10 Equilibrium plate n with vapor and liquid streams. Given q, xD and xF, the feed line is fixed and the upper operating line depends on the reflux ratio R. At total reflux, the operating line coincides with the 45 o line. As R is decreased, the slope of the enriching operating line R/( R + 1) is decreased. The operating line will rotate clockwise around the point ( x = x D, y = x D ) 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 reflux is then defined. R 1 y = m x + x n+1 + n + D Rm 1 Rm 1 The minimum reflux Rm can be obtained from either the intercept of the slope of the enriching operating line. The operating flux ratio is between the minimum Rm and total reflux. Usual value is between 1.2 Rm to 1.5 Rm. Figure 4.4-11 shows the pinch points for case 1 where the operating line intersects the equilibrium curve and case 2 where the operating line touches the equilibrium curve. 4-44 Figure 4.4-11 The pinch points for minimum reflux. Example 4.4-4 ---------------------------------------------------------------------------------- A distillation column receives a feed that is 40 mole % n-pentane and 60 mole % n-hexane. Feed is saturated liquid with a flow rate of 2,500 lbmol/hr. The column is at 1 atm. A distillate of 90 mole % n-pentane is desired. A total condenser is used. Reflux is a saturated liquid. Bottoms from the reboiler is 98 mole % n-hexane. Determine the minimum number of equilibrium trays and the minimum reflux ratio. Data: Vapor pressure, Psat , data: ln Psat = A − B/( T + C), where Psat is in kPa and T is in K. Compound A B C n-pentane (1) 13.9778 2554.6 − 36.2529 n-hexane (2) 14.0568 2825.42 − 42.7089 o Heat of evaporation for n-pentane, λC5 = 11,369 Btu/lbmol, CpL,C5 = 39.7 Btu/lbmol ⋅ F o Heat of evaporation for n-hexane, λC6 = 13,572 Btu/lbmol, CpL,C6 = 51.7 Btu/lbmol ⋅ F Solution ------------------------------------------------------------------------------------------ (a) Minimum number of equilibrium trays The equilibrium data for n-pentane and n-hexane at 1 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-hexane system at 1 atm. x = mole fraction of n-pentane in the liquid y = mole fraction of n-pentane in the vapor x = 0.00000 , y = 0.00000, T(K) = 342.06 x = 0.05000 , y = 0.12705, T(K) = 339.40 x = 0.10000 , y = 0.23699, T(K) = 336.91 x = 0.15000 , y = 0.33263, T(K) = 334.58 x = 0.20000 , y = 0.41626, T(K) = 332.39 x = 0.25000 , y = 0.48975, T(K) = 330.32 x = 0.30000 , y = 0.55462, T(K) = 328.38 x = 0.35000 , y = 0.61214, T(K) = 326.53 x = 0.40000 , y = 0.66335, T(K) = 324.79 x = 0.45000 , y = 0.70911, T(K) = 323.14 x = 0.50000 , y = 0.75016, T(K) = 321.56 x = 0.55000 , y = 0.78711, T(K) = 320.07 x = 0.60000 , y = 0.82048, T(K) = 318.64 x = 0.65000 , y = 0.85070, T(K) = 317.28 x = 0.70000 , y = 0.87816, T(K) = 315.97 x = 0.75000 , y = 0.90317, T(K) = 314.72 x = 0.80000 , y = 0.92601, T(K) = 313.53 x = 0.85000 , y = 0.94692, T(K) = 312.38 x = 0.90000 , y = 0.96610, T(K) = 311.28 x = 0.95000 , y = 0.98374, T(K) = 310.22 x = 1.00000 , y = 1.00000, 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 xB = 0.1 to xD = 0.9.
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