ION EXCHANGEDESIGN PROCEDURES Design Procedures • The breakthrough curves for an ion exchange column and an adsorption column are similar. • The contacting techniques are almost identical. • Therefore, the same procedures used for the design of adsorption columns may be used for ion exchange columns. o the scale-up approach o the kinetic approach A laboratory- or pilot-scale breakthrough curve is required for both procedures. The breakthrough curve for a column shows the solute or ion concentration in the effluent on the y-axis versus the effluent throughput volume on the x-axis. The area above the breakthrough curve represents the amount of solute or ions taken up by the column and is C0 C dV from V = 0 to V = the allowable throughput volume under consideration. At the allowable breakthrough volume, VB, the area above the breakthrough curve is equal to the amount of ions removed by the column. .At complete exhaustion. the ion concentration in the influent is equal to the ion concentration in the effluent. C = Co and the area above the breakthrough curve is equal to the maximum amount of ions removed by the column. the entire exchange column is in equilibrium with the influent and effluent flows. At complete exhaustion. Also. The ratio of the amount of ions removed per unit mass of exchanger is computed. . VB.Another design approach is to determine the meqs or equivalent weights of the ions removed by a test column using the breakthrough curve. The throughput volume under consideration should be the allowable breakthrough volume. at the allowable breakthrough concentration. Ca. Then. the flowrate used for the test column in terms of bed volumes per hour must be similar to the flowrate of the design column. For this method to be valid. using this ratio. the mass of exchanger required is calculated from the allowable breakthrough volume for the design column and the concentration of the polyvalent metallic ions to be removed from the liquid flow. . g.Co V Q = effluent solute concentration = influent solute concentration = rate constant = maximum solid phase concentration of the sorbed solute e.C Co C Co k1 qo M V Q 1 1+e k1 q o M . mg/g = mass of the adsorbent = volume treated = flow rate . Co V Q kt q o M . of both sides yield the design equation. y = b + mx .1+e e kt q o M .Co V Q Co = C Co = -1 C k1q o M k1Co V Co ln -1 = Q Q C taking natural log. The breakthrough curves for an ion exchange column and an adsorption column are similar. C0 C dV from V = 0 to V = the allowable throughput volume. The area above the breakthrough curve represents the amount of solute or ions taken up by the columns. . A breakthrough curve shows the solute or ion concentration in the effluent on the y – axis versus the effluent throughout volume on the x – axis. A laboratory or pilot scale breakthrough curve is required. moisture = 44%. is 5% Co. mass of resin = 41. 2. The design column flowrate will be 378. the allowable breakthrough time is 7 days of flow. . The allowable effluent concentration.1 SI Ion Exchange in Waste Treatment An industrial wastewater with 107 mg/L of Cu2+ (3.500 L/d. Ca. determine : 1.3 cm.0428 L/d. shown in Figure 13.6. Data concerning the column are as follows : inside diameter = 1.50 g on a moist basis (23.Example 13.24 gm on a dry basis). The diameter and depth. A breakthrough curve. and the resin depth is approximately twice the column diameter. The height of the sorption zone. Using the kinetic approach to column design. The kilograms of resin required.37 meq/L) is to be treated by an exchange column. bulk density of resin = 716.5 kg/m3 on a moist basis. length = 45. 3.7 cm. has been obtained from an experimental laboratory column on the sodium cycle. and liquid flowrate = 1. 7 cm Massresin = 41.37 meq/L) Ca = 5% Co d = 1.5 kg/m3 on a moist basis Q = 1.3 cm L = 45.500 L/d breakthrough time = 7 days resin depth = 2 x column diameter .0428 L/d Qdesign = 378.Solution Co = 107 mg/L (3.50 g on a moist basis = 23.24 g on a dry basis moisture = 44% bulk densityresin = 716. 1. Mass of resin required k1q o M k1Co V Co ln -1 = Q Q C . 16 0.54 0.0 49.0 86.045 23.235 1.000 .644 1.863 2.962 1.374 2.15 0.092 10.673 0.00 3.4 98.040 0.553 0.138 -1.03 2.159 0.879 1.93 3.41 2.238 0.9 94.41 24.882 1.090 0.138 0.0 102.34 1.45 0.701 -0.588 1.235 5.89 2.43 22.040 -3.51 20.90 1.673 1.808 1.85 0.2 68.9 9.V (L) C(mg/L) C(meq/L) C/Co Co/C Co/C-1 ln(Co/C-1) 15.0 107.14 0.045 3.160 6.258 3.66 19.55 0.23 26.000 0.5 40.06 19.29 18.1 27.03 1.042 24.1 17.56 1.22 0.882 2.56 0.9 4.14 16.463 2.97 0.31 0.15 20.98 23.59 22.863 9.70 0.25 0.94 0.86 0.7 62.36 21.07 0.238 -1.917 1.159 1.553 -0.701 0.090 -2.000 1.17 3. 120 100 C (mg/L) 80 60 40 20 0 0 5 10 15 20 Volume treated (L) 25 30 . 5 0 0 5 10 15 20 Volume treated (L) 25 30 .4 3.5 1 0.5 2 1.5 C (meq/L) 3 2. 00 -4.0 25.0 5.00 ln(Co/C .00 -3.0 V (L) 20.341 0.00 1.0 10.0 15.00 -1.1) 2.00 3.0 30.00 -2.0 .00 0.4.00 y = -0.7603x + 15. Solution Co = 107 mg/L (3.3 cm L = 45.24 g on a dry basis moisture = 44% bulk densityresin = 716.0428 L/d Qdesign = 387.50 g on a moist basis = 23.7 cm Massresin = 41.5 kg/m3 on a moist basis Q = 1.500 L/d breakthrough time = 7 days resin depth = 2 x column diameter .37 meq/L) Ca = 5% Co d = 1. 37 meq k1 = 236 L/d.7603 L = Q -1 k1 = 0.0428 L/d .k1Co 0.341 = Q 15.eq x qo x 23.7603 L -1 L 1 L x 1.24 g 1.0428 x d 3.eq k1q o M 15.341 = 236 L/d. qo = 1.341 236 L/d.92 g kg -3 Compute the mass of resin required for the design column from .eq x 23. k1q o M k1Co V Co ln -1 = Q Q C .24 g eq eq = 2.0428 L/d x 15.92x10 = 2. 5.56 g dry wt.2.82 x 10-3 M .94 = 1. Diameter and depth : or 1 g wet wt. because 44% moisture . 0.57 M = 4676 kg 2. D = 1.95 m Depth = 2 x 1.90 m .95 = 3. 3. The height of the sorption zone the length of the column in which adsorption occurs Sorption zone. VT . is related to the column height. VB volume of exhaustion. Z breakthrough volume. Zs. 500 x 7 d = 2.95 Co L VB = 378.65x106 L d .The exhaustion is considered to occur at C = 0. 51 .1 x 10-6 VT 2.8x106 L 2.8x10 .65x106 L = 2.8x106 Zs = 4.45 x 106 L VZ = 5.2.-2.5 x 2.1x10-6 VT = 11.45x10 . 2.79 m 6 6 5.45 VT = 5.04m x = 2.94 = 8.0.45x106 L . e.1 m3 of ion – exchange resin with an exchange capacity of 57 kg/m3 (i. The occupants use 2000 L of water per day. Assumption : All (100%) hardness in the water which passes through the ion exchange column is removed. How much water should be bypassed? 2. What is the time between regeneration cycle “Breakthrough time” ? . The water contains 280 mg/L of hardness as CaCO3 and it is desired to soften it to 85 mg/L as CaCO3. 57 kg of hardness as CaCO3 per m3 of resin volume). 1.Example A home water softener has 0.. Q Mass balance equations Accumulation = Input – Output ± Reactions Q. Cin with no accumulation and no reaction (Q – Qb). C : concentration Q : flowrate Loading rate = C. Cin Qb Cin (Q – Qb).Solution : 1. Ce ~ 0 Q. . Cp Input = Outputs. If Ce = 0 Qb =0.3 x 2000 L/d = 600 L/d .3Q=0. 2.1 m Total capacity of resin = = 14. Breakthrough time = 57 kg/m x 0.5 d Mass of ions removed/time 392000 mg/d x 10 kg/mg 3 3 -6 . Use the design criteria a)The maximum loading rate = 5 gpm/ft3 of resin. . The resin has an exchange capacity of 20 kilograins of CaCO3 per ft3 when regenerated at the rate of 15 Ib of salt per ft3.1 mg/L). b)The bed depth = 30-72 inches. The raw water has total hardness of 100 mg/L as CaCO3. Assume this process can achieve 95% hardness removal efficiency. A synthetic zeolite resin will be packed in shells with diameter of 5 ft.Example : An ion exchange process is to be used to soften water at the rate of 500 gpm. (1 grain/gal = 17. 3. Calculate total hardness (mg/L as CaCO3). Calculate hardness to be removed (grains/gal). Calculate total resin required (ft3). Calculate the bed depth and shell diameter for the ion exchange equipment (considering two units). 4. 2. Calculate hardness to be removed (kilograins/d). 6. . 5.1. Calculate the amount of salt required for regeneration.