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

# Total balance around stripper qst q1 qe qs 158 qs 1

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Unformatted text preview: g/h of combined HNO3 and H2O. For 5.1 N HNO3 = 5.1 mol/L of HNO3, have 5.1(63.01) = 321.4 g/h HNO3 in the feed. Therefore, the H2O in the feed = 1,166 - 321.4 = 844.6 g/h. Since the feed includes 127 g/h of combined Hf and Zr oxides, the total feed flow rate = 1,166 + 127 = 1,293 g/h. Since the feed contains 22,000 parts of Hf per 1,000,000 parts of Zr, we have: (22,000/1,000,000 = 0.022 g Hf/g Zr. This corresponds to 0.022(210.6/178.6)/(123.22/91.22) = 0.0192 g HfO2/g ZrO2. Thus, of the 127 g/h of oxides, we have 127(0.0192/1.0192) = 2.39 g/h of HfO2 and 127 - 2.39 = 124.61 g/h of ZrO2. In summary, the feed contains: Component g/h g/L mol/L HfO2 2.39 2.39 0.0194 ZrO2 124.61 124.61 0.5917 HNO3 321.4 321.4 5.1 H2 O 884.6 884.6 49.1 Total: 1,293.0 1,293.0 Exercise 8.21 (continued) Analysis: (continued) HfO2 balance: From the given Stagewise Analyses table, the aqueous phase leaving Stage 14 is the final raffinate and contains 3.54 g total oxides /L and g Hf/g Zr = 0.72. Therefore, g HfO2/g ZrO2 = 0.72(210.6/178.6)/(123.22/91.22) = 0.6285. Then, g HfO2/L = 3.54(0.6285/1.6285) = 1.366 g HfO2/L and g ZrO2/L = 3.54 - 1.366 = 2.174. From the Stagewise Analyses table, it appears that the concentration of Hf in the extract leaving Stage 1 is negligible. Therefore, all of the HfO2 entering in the feed exits in the raffinate. Therefore, with a feed of , QF, of 1 L/h with 2.39 g/L of HfO2 and a raffinate with 1.366 g/L HfO2, the volumetric flow rate of raffinate = 1.0(2.39/1.366) = 1.75 L/h = QR . Total oxide balance: All of the oxides in the feed leave in either the raffinate from Stage 14 or the aqueous product (stripped extract, E) from the Stripper. Using the Stagewise Analyses table, a total oxide material balance gives: 127QF = 3.54QR + 76.4QE , where QF = 1.0 L/h and QR = 1.75 L/h. Therefore, solving, QE = [127 - 3.54(1.75)]/76.4 = 1.58 L/h of stripped extract, which contains 76.4 g ZrO2/L. Total balance: Assume that the total volume flows in = total volume flows out. Let QST = volumetric flow rate of stripping water, and QSC = volumetric flow rate of scrubbing water. Then a volumetric flow balance gives: QST + QSC = QE + QR - QF = 1.58 + 1.75 - 1.0 = 2.33 L/h. Nitric acid balance: The Stagewise Analyses data around Stage 1 indicate that the scrubbing water contains HNO3. Let x = mol/L of HNO3 in the scrubbing water, but assume no HNO3 in the stripping water. An overall nitric acid balance gives: 5.1(1) +xQSC = 2.56QR + 3.96QE . Therefore, xQSC = [2.56(1.75) + 3.96(1.58) - 5.1] = 5.64 mol/h = HNO3 entering in scrub water. Total balance around stripper: QST + Q1 = QE + QS = 1.58 + QS (1) HNO3 balance around stripper: Using the Stagewise Analyses table, 1.95Q1 + 0 = 3.96(1.58) + 0.65Qs = 6.26 + 0.65Qs (2) Oxide balance around stripper: Using the Stagewise Analyses table, 22.2Q1 = 76.4(1.58) = 120.7 g/h of ZrO2 Therefore, solving, Q1 = 120.7/22.2 = 5.44 L/h. From Eq. (2), QS = [1.95(5.44) - 6.26]/0.65 = 6.69 L/h From Eq. (1), QST = 1.58 + 6.69 - 5.44 = 2.83 L/h From above, QST + QSC = 2.33. Therefore, th...
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