A commercial water treatment system has a packed section that is 4 meters long w

Question

A commercial water treatment system has a packed section that is 4 meters long with a 30 cm diameter and is packed with a void fraction of 0.4. The resin in the bed obeys a Langmuir isotherm for calcium which has the form: q = q_max Ks/Ks + 1, K = 50 liters/equivalent, q_max = 2 equivalents/liters, and mass transfer parameter: K_OC a_I = 0.002 s^-1 where c, s and q have units of equivalents per liter. The untreated tap water has a calcium concentration of 1 mM and, at a flow rate of 100 liters/hr, the resin has the overall mass transfer rate given above. Remember that calcium is divalent and assume that, at equilibrium, the calcium concentration is equal in the mobile phase and in the pores, i.e., m = 1. a. Assuming ideal operation, determine how many liters of tap water this packed bed can treat before it needs to be regenerated. b. Determine what fraction of q_max is used at the feed concentration c. Determine the velocity of the adsorption front and an approximate cycle time at this feed flow rate, d. Determine the length of the mass transfer zone in the bed under these operating conditions, and e. Determine what fraction of the bed's equilibrium capacity is used if the feed is turned off when the mass transfer zone begins to break through.

Explanation / Answer

Part A:

Concentration of Ca2+ is 10-3 moles/L = 2*10-3 equivalents/L (as Calcium is divalent, the factor of 2 comes into play).

Total volume of packed resin = 4*{pi*(0.6/2)2} m3 = 4*{pi*(0.6/2)2}*103 L = 1131 L [because radius = 0.6/2 metres; length = 4 metres]

Therefore, total void = 0.4*1131 L = 452.39 L (=Vvoid)

Here s = 2*10-3 equivalents/L [because given m=1]

Now, q = qmaxKs/(1+Ks) = 2*50*2*10-3/(1+50*2*10-3) = 0.182 equivalents/L

Total equivalent of Ca2+ that can be treated = q*Vvoid = 82.253 equivalents

SInce 1 litre water contains 2*10-3 equivalents of water, therefore total water that can be treated before regeneration is = (82.253equv)/(2*10-3equv/L) = 41126 L (approx.)

Part B:

Fraction of qmax is given by q/qmax that can be evaluated from the results in part A and that is 0.182/2 = 0.091.

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