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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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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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