Separation Process Principles- 2n - Seader & Henley - Solutions Manual

5 lbmol of adsorbate we only need 250 lb of silica

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Unformatted text preview: ium curve in Fig. 4.30a, it is seen that when x = 0.25, y = 0.47, which is far from the desired value of 0.90. Therefore, cannot achieve the desired separation. So, calculate what can be achieved. To solve, assume W, plot the equation on Fig. 4.30a and find intersection with the equilibrium curve to find y and x. Read the mmole adsorbate/g adsorbent = lbmoles adsorbate/1000 lb adsorbent from Fig. 4.30b. However, we see in Fig. 4.30b that the smallest adsorbate amount is 1.57 lbmol adsorbate/1000 lb adsorbent. But we only have a total of 1.0 lbmol of feed gas. Therefore, we can use a lot less adsorbent. Assume a value of W = 0.5 lbmole (50% of the feed gas and equal to the amount of A in the feed gas). Then Eq. (1) becomes: y = 1 - x. From Fig. 4.30a, the intersection with the equilibrium curve is y = 0.635 and x = 0.365. From Fig. 4.30b, for x = 0.365, we have 2.0 lbmoles adsorbate/1000 lb of silica gel. Since we only have 0.5 lbmol of adsorbate, we only need 250 lb of silica gel. The separation achieved is far less than desired. Multiple stages might achieve the desired separation. Other values of W will give other separations. Exercise 4.69 Subject: Crystallization of MgSO4 from an aqueous by evaporation. Given: 1,000 lb of MgSO4 dissolved in 4,000 lb of water at 160oF. 90% of the water is evaporated at 160oF, causing the monohydrate to crystallize. At 160oF, MgSO4 has a solubililty of 36 wt%. Assumptions: Equilibrium. Find: (a) Pounds of water evaporated. (b) Pounds of monohydrate, MgSO 4 ⋅ H 2 O , crystals produced. (c) Crystallizer pressure. Analysis: (a) Evaporate 0.9(4,000) = 3,600 lb of water Therefore, water left = 4,000 - 3,600 = 400 lb (b) MW of water = 18 and MW of MgSO4 = 120.4 Let W = lb of MgSO4 remaining in solution. Then MgSO4 in the crystals = 1,000 - W. From crystal stoichiometry, water of hydration in the crystals = 18 (1) (1,000 − W ) = 0.15(1,000 − W ) 120.4 Therefore, the water in solution = 400 - 0.15(1,000 - W) = 250 + 0.15W The total solution (water + dissolved sulfate) = 250 + 0.15W + W = 250 + 1.15W From solubility of the sulfate, W 0.36 = (2) 250 + 1.15W Solving Eq. (2), W = 153.6 lb MgSO4 dissolved. MgSO4 in the crystals = 1,000 - 153.6 = 846.4 lb Water of crystallization = 18 (846.4) = 126.5 lb 120.4 Total crystals = 846.4 + 126.5 = 972.9 lb of MgSO 4 ⋅ H 2 O (c) Since the crystallizer pressure depends only on temperature and the composition of the mother liquor in the crystallizer, it is the same as in Example 4.17 or 4.38 psia. Exercise 4.70 Subject: Crystallization of Na2SO4 from an aqueous solution by evaporation. Given: 1,000 kg/h of Na2SO4 dissolved in 4,000 kg/h of water. Water is evaporated at 60oC to crystallization 80% of the Na2SO4. Solubility data in Fig. 4.24. Assumptions: Equilibrium. Raoult's law for the mother liquor. Find: (a) The kg/h of water that must be evaporated. (b) The crystallizer pressure in torr. Analysis: (a) From Fig. 4.24, at 60oC, the stable crystals are Na2SO4 with no water of crystallization. The solubility of Na2SO4 at 60oC is 0....
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