Mass Transfer Operations – Robert Treybal Sample Problems 1. If 100 kg of a solution of acetic acid (C) and water (A) containing 30% acid is to be extracted three times with isopropyl ether (B) at 20°C, using 40kg of solvent in each stage, determine the quantities and compositions of the various streams. How much solvent would be required if the same final raffinate concentration were to be obtained with one stage?
Solution. The equilibrium data at 20°C are listed below [Trans. AIChE, 36, 628 (1940), with permission]. The horizontal rows give the concentrations in equilibrium solutions. The system is of the type shown in Fig 10.9, except that the tie lines slope downward toward the B apex. The rectangular coordinates of Fig. 10.9a will be used, but only for acid concentrations up to x=0.30. These are plotted in Fig. 10.15.
Stage 1 F=100 kg,
xF=0.30, yS=0, S1=B1=40 kg
OMB: M1=100+40=140 kg x bal: 100(0.30)+40(0)=140xM1
=>
xM1=0.214
Point M1 is located on line FB. With the help of a distribution curve, the tie line passing through M1 is located as shown, and x1=0.258, y1=0.117 wt fraction acetic acid. 𝐸1 =
140(0.214−0.258) 0.117−0.258
= 43.6 𝑘𝑔
R1=140-43.6=96.4 kg Stage 2 S2=B2=40 kg M2=R1+B2=96.4+40=136.4 kg Point M2 is located on line R1B and the tie line R2E2 through M2.x2=0.227, y2=0.095 become 𝐸2 =
𝑀2 (𝑥𝑀2 −𝑥2 ) 𝑦2 −𝑥2
=
136.4(0.1822−0.227) 0.095−0.227
= 46.3 𝑘𝑔
R2=M2-E2=136.4-46.3=90.1 kg Stage 3. In a similar manner, B3=40, M3=130.1, xM3=0.1572, x3=0.20, y3=0.078, E3=45.7 and R3=84.4. The acid content of the final raffinate is 0.20(84.4)=16.88 kg.
The composited extract is E1+E2+E3=43.6+46.3+45.7=135.6 kg and its acid content = E1y1+E2y2+E3y3=13.12 kg. If an extraction to give the same final raffinate concentration, x=0.20, were to be done in one stage, the point M would be at the intersection of tie line R3E3 and line BF or at xM=0.12. The solvent required would then be, by, S1=100(0.30-0.12)/(0.12-0)= 150 kg, instead of the total of 120 required in the three-stage extraction. 2. Nicotine (C) in a water (A) solution containing 1% nicotine is to be extracted with kerosene (B) at 20°C. Water and kerosene are essentially insoluble. (a) Determine the percentage extraction of nicotine if 100 kg of feed solution is extracted once with 150 kg solvent. (b) Repeat for three theoretical extractions using 50 kg solvent each.
Solution. Equilibrium data are provided by Claffey et al., Ind. Eng. Chem., 42, 166 (1950), and expressed as kg nicotine/kg liquid, they are as follows:
(a) xF= 0.01 wt fraction nicotine, x’F= 0.01/(1-0.01)= 0.0101 kg nicotine/ kg water. F= 100 kg. A= 100(1-0.01)= 99 kg water, A/B= 99/150 = 0.66.
Refer to the figure which shows the equilibrium data and the point F representing the composition of the feed. From F, line FD is drawn of slope 0.66, intersecting the equilibrium curve at D where x’1=0.00425 and y’1=0.00380 kg nicotine/kg liquid. The nicotine removed from the water is therefore 99(0.0101-0.00425)= 0.580 kg, or 58% of that in the feed. (b) For each stage, A/B=99/50= 1.98. The construction is started at F, with operating lines of slope -1.98. The final raffinate composition is x’3=0.0034 and the nicotine extracted is 99(0.0101-0.0034)= 0.663 kg or 66.3% of that in the feed. 3. If 8000 kg/h of an acetic acid (C)- water (A) solution, containing 30% acid is to be counter currently extracted with isopropyl ether (B) to reduce the acid concentration to 2% in the solvent-free raffinate product, determine (a) the minimum amount of solvent which can be used and (b) the number of theoretical stages if 20 000 kg/h of solvent is used. Solution. The equilibrium data are plotted on triangular coordinates in the figure. The tie lines have been omitted for clarity.
(a) F= 8000 kg/h; xF=0.30 wt fraction acetic acid, corresponding to point F on the figure R’N, as shown. In this case the tie line J, which when extended passes through F, provides the conditions for minimum solvent, and this intersects line RNB on the right of the figure nearer B than any other lower tie line. Tie line J provides the minimum E1 as shown at y1=0.143. Line E1mRN, intersects line FB at Mm, for which xM=0.114 with ys=0 and S=B: 𝐵𝑚 =
𝐹𝑥𝐹 𝑥𝑚
−𝐹 =
8000(0.30) 0.114
− 8000 = 13 040 kg/h, min solvent rate
(b) For B= 20 000 kg solvent/h with yS=0 and S=B
𝑥𝑀 =
𝐹𝑥𝐹 8000(0.30) = = 0.0857 𝐹 + 𝐵 8000 + 20 000
and point M is located as shown on line FB. Line 𝑅𝑁𝐹 𝑀 extended provides E1 at y1=0.10. Line FE1 is extended to intersect line RNB at ΔR. Random lines such as OKL are drawn to provide yS+1 at K and xS at Ls as follows:
These are plotted on the figure as the operating curve, along with the tie-line data as the equilibrium curve. There are required 7.6 theoretical stages. The weight of extract can be
Obtained by an acid balance,
𝐸1 =
𝑀(𝑥𝑀 − 𝑥𝑁𝐹 ) 28 000(0.0857 − 0.02) = = 23 000 𝑘𝑔/ℎ 𝑦1 − 𝑥𝑁𝐹 0.10 − 0.02 𝑅𝑁𝐹 = 𝑀 − 𝐸1 = 28 000 − 23 000 = 5000 𝑘𝑔/ℎ
4. If 1000 kg/h of a nicotine (C)-water (A) solution containing 1% nicotine is to be counter currently extracted with kerosene at 20°C to reduce the nicotine content to 0.1%, determine (a) the minimum kerosene rate and (b) the number of theoretical stages required if 1150 kg of kerosene is used per hour. Solution. The equilibrium data of number 2 are plotted in the figure below
(a) F=1000 kg/h, xF=0.01, A=1000(1-0.01)= 990 kg water/h, ys=0 0.01 𝑘𝑔 𝑛𝑖𝑐𝑜𝑡𝑖𝑛𝑒 𝑥′𝐹 = = 0.0101 1 − 0.01 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟
𝑥𝑁𝐹 = 0.001
𝑥′𝑁𝐹 =
0.001 𝑘𝑔 𝑛𝑖𝑐𝑜𝑡𝑖𝑛𝑒 = 0.001 001 1 − .001 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟
The operating line starts at point L (y’=0, x’=0.001 001) and for infinite stages passes through K on the equilibrium curve at x’F. Since yK=0.0093, 𝐴
0.0093−0
𝐵𝑚
= 0.0101−0.001001 = 1.021
Bm=A/1.021= 990/1.021= 969 kg kerosene/h (b) B=1150 kg/h, A/B=990/1150= 0.860 𝑦′1 𝑦′1 = = 0.860 𝑦′𝐹 −𝑥′𝑁𝐹 0.0101 − 0.001001 y’1=0.00782 kg nicotine/kg kerosene The operating line is drawn through (y’1,x’F) and 8.3 theoretical stages are determine graphically. Alternatively, at the dilute end of the system, m’=dy’*/dx’=0.798, and 𝑚′𝐵 𝐴
=
0.798(1150) 990
= 0.928
At the concentrated end, m’=0.953, and m’B/A=0.953(1150)/990=1.110. The average is [0.928(1.110)]]0.5=1.01.
𝑥’𝑁𝐹 𝑥′𝐹
=
0.001001 0.0101
= 0.099 and the figure indicates 8.4 theoretical stages.
Illustration 10.9 Determine the number of transfer units NTU for the extraction of illustration 10.3 if 20,000kg/h of solvent is used.
Solution: Define x and y in terms of weight fractions acetic acid, x1=xF=0.30; y2=0; x2=0.02; y1=0.10. The operating diagram is already plotted in Fig. 10.2. From this plot, values of x and x* are taken from the operating line and equilibrium curve at various values of y, as follows:
x
0.30
0.35
0.30
0.15
0.10
0.05
0.02
x*
0.230
0.192
0.154
0.114
0.075
0.030
0
1 𝑥−𝑥∗
14.30
17.25
20.75
27.8
40.0
50.0
50.0
1
The area under a curve of x as abscissa against 𝑥−𝑥∗ as ordinate (not shown) between x= 0.30 and x = 0.02 is determined to be 8.40. In these solutions, the mutual solubility ofwater and 18
isopropyl ether is very small, so that r can be taken as 60 = 0.30. Eq.(10.110): 𝑁𝑇𝑈 = 8.40 +
1 2
1−0.02
1
0.02(0.3−1)+1
ln 1−0.30 + 2 ln 0.30(0.3−1)+1 = 𝟖. 𝟒𝟔
Illustration 10.10 Determine the number of transfer units NTU for the extraction of illustration 10.4 if 1150 kg/h of kerosene are used.
Solution: Use weight-ratio concentration as in Illustration 10.4. x’1=x’F=0.0101; y’2=0; x’2=0.001001; y’1=0.0782. The calculation can be done through Eq.(10.116) of the equivalent, Fig. 8.20
𝑥′2 − 𝑦2 ′⁄ 𝑚′ = 0.001001 = 0.0909 𝑥′1 − 𝑦′2 0.0101 ⁄ 𝑚′ The average mE/R = m’B/A = 1.01 (Illustration 10.4) From fig. 8.20, NTU = 8.8 Mc Cabe
23.5. In a continuous countercurrent train of mixer-settlers, 100 kg/h of a 40:60 acetone water solution is to be reduced to 10 percent acetone by extraction with pure 1,1,2-trichloroethane at 25℃. (a) Find the minimum solvent rate. (b) At 1.8 times the minimum (solvent rate)/(feed rate), find the number of stages required. (c) For conditions of part (b) find the mass flowrats of all streams. Data are given in Table 23.6 Solution:
(a) Basis: 100 kg feed solution Let x,y refer to the wt. fraction of acetone: xw, yw to water, xT,yT to trichloroethane. Plot the equilibrium line xe, ye Establish the ends of the operating line from overall material balances.
Balances:
Overall: Lb + Va = Vb +100 Acetone: 0.1 Lb + yaVa = xaLa =40
At minimum solvent rate, ya’ is found from the equilibrium curve at xa = 0.40 to be 0.53 Hence 0.1Lb + 0.53Va = xaLa = 40 Water: ywaVa + xwbLb = 60 When xb = 0.10, from the first part of Table 23.6, xTb = 0.0061 From these, Vb = 26.46, Va = 63.61 Lb = 62.85 The minimum solvent rate is 26.46
(b) Vb = 1.8 times the minimum = 1.8 x 26.46 = 47.63 As before, from a water balance:
YwaVa + xwbLb = 60 Overall: Lb = 100 - 47.63 – Va Xwb is unchanged at 0.8939
Hence Lb = (60 – 147.63ywa)/(0.8939 – ywa) Va = 147.63- Lb By trial, estimate ywa to be 0.028 Lb=64.52;
Va = 83.11
ya = (40 – 0.1Lb)/Va = 0.404 From the equilibrium data for ya = 0.404, ywa = 0.028 (as estimated). Hence the upper end of the operating line is at xa = 0.40, ya=0.404 The lower end is at xb = 0.10, yb = 0 Intermediate point. Set x=0.25 Estimate xT to be 0.01; xw = 1-.25- 0.01 = 0.74 Overall balance from feed end: V = L + Va – La = L + 83.11 – 100 =L – 16.89 (A) Estimate: y = 0.22,
yw = 0.0089
From an acetone balance: yV= (0.404 x 83.11) + 0.25L – 40 y = (0.25L – 6.42)/V
(B)
Water balance: ywV = 0.74L + (0.028 x 83.11) – 60
(C)
From Eqs. (A) and (C): L = 78.69,
V=61.80
From Eq. (B), y = 0.214. From the equilibrium curve, yw = 0.0089, as estimated. The coordinates of the intermediate point are x= 0.25, y= 0.214 The operating line is almost straight. From the diagram 2.8 stages are needed. Use 3 stages.
(c) The flow rates, in kg/h, are Feed: 100
Extract: 83.11
Solvent: 47.63
Raffinate: 64.52
23.6 A mixture containing 40 weight percent acetone and 60 weight percent water is contacted with an equal amount of MIK. (a) What fraction of the acetone can be extracted in a single-stage process? (b) What fraction of the acetone could be extracted if the fresh solvent were divided into two parts and two successive extractions used? Solution: (a) Adding an equal amount of MIK to the feed gives a mixture with 0.5 MIK, 0.2 acetone,
and 0.3 H2O. A tie line through this point on Fig. 23.8 shows the extract to be 0.725 MIK, 0.232 acetone, and 0.043 H2O. The raffinate composition is 0.023 MIK, 0.132 acetone, and 0.845 H2O. Per unit mass of feed, an acetone balance gives 0.4 = 0.23E + 0.132R E+R= 1 + 1 = 2.0 0.4 = 0.232E + 0.132(2-E) 0.4−0.264
E=0.232=0.132= 1.36
Fraction acetone extracted =
1.32(0.232) 0.4
= 0.789
(b) If onlye half the MIK is added in the first step, the mixture is 0.333 MIK, 0.267 acetone,
and 0.4 H2O. The phase compositions are Extract
Raffinate
MIK
0.615
0.035
Acetone
0.325
0.210
Water
0.060
0.755
By a material balance, E + R = 1.5 0.4 = 0.325E + 0.21(1.5-E) 𝐸=
0.4 − 0.315 = 0.739, 𝑅 = 0.761 0.325 − 0.21
Acetone extracted = 0.739 (0.325) = 0.24 Adding 0.5 parts MIK 50 0.761 parts of raffinate gives a mixture with the following compostions MIK =
0.5+0.76(0.35)
Acetone = H2O =
= 0.418
1.261 0.21(0.761)
1.261 0.755(0.761) 1.261
= 0.127
= 0.455
This separates to give an extract with 0.20 acetone and a raffinate with 0.075 acetone. A acetone balance give 0.16 = 0.20E + 0.075(1.261-E) E = 0.523 Acetone extracted: 0.523 x 0.20 = 0.105 Total extracted: 0.24 + 0.105 = .345 Fraction acetone extracted: 0.345/ 0.4 = 0.863
Solved Problems 10.5. A pyridine-water solution, 50% pyridine, is to be continuously and counter-currently extracted at the rate of 2.25kg/s (17800 lb/h) with chlorobenzene to reduce the pyridine concentration to 2% in the final raffinate. Using the coordinate systems plotted in (b) and (c) of prob. 10.1: (a) Determine the minimum solvent rate required. (b) If 2.3 kg/s (18250 lb/h) is used, what are the number of theoretical stages and the saturated weights of extract and raffinate Solution: Pyridine
Chlorobenzene
Water
Pyridine
Chlorobenzene
Water
0
99.95
0.05
0
0.08
99.92
11.05
88.28
0.67
5.02
0.16
94.82
18.95
79.90
1.15
11.05
0.24
88.71
24.10
74.28
1.62
18.90
0.38
80.72
28.60
69.15
2.25
25.50
0.58
73.92
31.55
65.58
2.87
36.1
1.85
62.05
35.05
61.0
3.95
44.95
4.18
50.87
40.60
53.0
6.40
53.20
8.90
37.90
49
37.8
13.2
49
37.8
13.2
Basis: 2.25 kg/s of pyridine-water solution feed Weight fraction at final raffinate: Solvent flowrate:
𝑥𝑓 = 0.5 𝑥𝑛 = 0.02 𝑆 = 2.3 𝑘𝑔/𝑠
Refer to Fig. E.2.14 (a) and (b) Plot x v/s y and x, y v/s weight fraction of chlorobenzene Mark point F on y-axis at xf=0.50. Mark the point S at the B-apex as the solvent is pure. Make a line using point F and S. Now locate M on FS such that: 𝑆 𝑙𝑖𝑛𝑒 𝑀𝐹 = 𝐹 𝑙𝑖𝑛𝑒 𝑀𝑆 𝑙𝑖𝑛𝑒 𝑀𝐹 2.3 = = 1.022 𝑙𝑖𝑛𝑒 𝑀𝑆 2.25 𝑀𝐹 = 1.022 𝑀𝑆 We have:
𝐹𝑆 = 𝑀𝐹 + 𝑀𝑆 = 112 𝑢𝑛𝑖𝑡𝑠(𝑚𝑚, 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑) 1.022𝑀𝑆 + 1𝑀𝑆 = 2.022𝑀𝑆 = 112 𝑀𝑆 = 55.4 𝑢𝑛𝑖𝑡𝑠 𝑀𝐹 = 112 − 55.4 = 56.6 𝑢𝑛𝑖𝑡𝑠
So locate M on FS such that line MF=56.6 units and line MS =55.5 units OR: 𝑀 = 𝐹 + 𝑆 = 2.25 + 2.3 = 4.55𝑘𝑔/𝑠 𝐹𝑥𝑓 + 𝑆𝑦𝑠 = 𝑀𝑥𝑚 2.25(0.5) + 2.3(0) = 4.55𝑥𝑚 𝑥𝑚 = 0.247 Mark point M on FS at xm=0.247(read from x-axis)
Both the above mentioned procedures give the same location for point M on FS. Locate Rn(raffinate from nth stage) on the bimodal solubility curve corresponding to 2% pyridine(i.e., at xn=0.02) Join RnM and project it to meet the equilibrium curve at E1. Join FE1 and project these
lines to meet at Δg.
Read y1 corresponding to E1 and from the x vs. y plot, get the value of x1 (in equilibrium with y1) Mark R1 on the bimodal solubility curve at location corresponding to x1 value. Join Re1E1 which is a required tie line for first stage. From the x vs. y plot, get the value of x2 corresponding to y2. Mark R2 on the bimodal curve at location corresponding to x2 value. Join R2E2, which is a tie line for stage-2, join R2Δg. The line R2Δg meets the curve at E3. The working is continued in this way till we cover Rn(i.e. xn=0.02) and then count the theoretical stages required. In our case, the tie line R3E3 is such that R3 is exactly xn=0.02 From Fig. E 2.14(a), number of theoretical stages required = n = 3 10.8 Water-dioxane solutions form a minimum-boiling azeotrope at atmospheric pressure and cannot be separated by ordinary distillation methods. Benzene forms no azeotrope with dioxane and can be used as an extraction solvent. At 25°C, the equilibrium distribution of dioxane between water and benzene [J. Am. Chem. Soc., 66, 282 (1944)] is as follows: Wt% dioxane in water
5.1
18.9
25.2
Wt% dioxane in benzene
5.2
22.5
32.0
At these concentrations, water and benzene are substantially insoluble, and 1000 kg of a 25% dioxane-75% water solution is to be extracted with benzene to remove 95% of dioxane. The benzene is dioxane-free.
(a) Calculate the solvent requirement for a single batch operation (b) If the extraction were done with equal amounts of solvent in five crosscurrent stages, how much fsolvent would be required?
Solution: (a) single-stage operation: Basis: 1000kg of solution containing 25% dioxane. A-water, B-benzene, C-dioxane Dioxane in feed solution Water in feed solution
= 0.25 × 1000 = 250𝑘𝑔 = 0.75 × 1000 = 750𝑘𝑔
Cs1 is the equilibrium value of picric acid concentration in extract, Cs1 and CA1 are the equilibrium values of picric acid concentration as the effluent streams leaving theoretical stage are in equilibrium. So read the value of CS1 from the plot corresponding to CA1=0.02mol/liter. From the plot: CS1=0.0155 mol/l Amount of solvent (benzene) used = B liters Picric acid in extract =0.08 mol 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑝𝑖𝑐𝑟𝑖𝑐 𝑎𝑐𝑖𝑑
Concentration of picric acid in extract = CS1= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 Volume of solvent B:
𝐵𝑣𝑜𝑙𝑢𝑚𝑒 =
𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑝𝑖𝑐𝑟𝑖𝑐 𝑎𝑐𝑖𝑑 𝐶𝑆1 0.08
= 0.0155 = 𝟓. 𝟏𝟔𝒍 solvent required per liter feed Cross-current operation: Number of stages=3, Feed solution = 1 liter Concentration of picric acid in feed solution =Cf=0.1 mol/liter 80% of the picric acid is removed. Picric acid in final raffinate=20% of its original value = 0.2 × 0.1 = 0.02𝑚𝑜𝑙 Concentration of picric acid in final raffinate =
0.02 1
= 0.02
𝑚𝑜𝑙 𝑙𝑖𝑡𝑒𝑟
𝐶𝐴3 = 0.02𝑚𝑜𝑙/𝑙𝑖𝑡𝑒𝑟𝐶𝑆3 = 0.0155
𝑚𝑜𝑙 (𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑣𝑎𝑙𝑢𝑒) 𝑙𝑖𝑡𝑒𝑟
For stage 1: − For stage 3:
𝐴 𝐶𝑆1 − 𝐶𝑠0 = = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑙𝑖𝑛𝑒 𝐵 𝐶𝐴1 − 𝐶𝐹 𝐴
−𝐵 =
𝐶𝑆1 −𝐶𝑆0 𝐶𝐴1 −𝐶𝐹
As the equal amount of solvent is used in each stage, the operating lines will have the same slope (-A/B). We have to construct/draw three operating lines parallel to each other (as the slope is the same) starting from point F(CF,CS0) and ending/covering exactly CA3, the final raffinate concentration. The operating line for the first stage passes through the point F(CF,CS0) and the operating line of the third stage passes through the point Q(CA3,CS3) So locate point F(CF,CS0) i.e. F(0.1,0) that represents the feed solution and draw the operating line through it which will cut equilibrium curve at P(CA1,CS1 and through CA1 on xaxis, draw the operating line parallel to the first one and so on and construct exactly three stages.
For this, we have to adopt a trial and error procedure. Once we construct exactly three stages, measure the slope of any one of these operating lines. From the graph: slope of operating line FP 𝑚=−
0.0255 − 0 = 0.773 0.1 − 0.067 𝐴 = 0.773 𝐵
𝐵= 𝐵= Benzene required per stage= 1.3L Total benzene requirement:
𝐴 ; 𝐴 = 1𝐿 0.773
1 = 1.3𝐿 𝑓𝑜𝑟 𝑜𝑛𝑒 𝑠𝑡𝑎𝑔𝑒 0.773 𝐁𝐭𝐨𝐭𝐚𝐥 = 3 × 1.3 = 𝟑. 𝟗𝑳
Principles of Transport Processes and Separation Processes – John Geankoplis Sample Problems 12.1-1 Adsorption Isotherm for Phenol in Wastewater Batch tests were performed in the laboratory using solutions of phenol in water and particles of granular activated carbon (R5). The equilibrium data at room temperature are shown in Table 12.1-1. Determine the isotherm that fits the data. TABLE 12.1-1 Equilibrium Data for Example 12.1-1 (R5) c, (kg phenol / m3 solution)
q, (kg phenol / kg carbon)
0.322
0.150
0.117
0.122
0.039
0.094
0.0061
0.059
0.0011
0.045
Solution: Plotting the data as 1/q versus 1/c, the results are not a straight line and do not follow the Langmuir Equation (12.1-3). qo c q= K+c
A plot of log q versus log c in Fig. 12.1-2 gives a straight line
and, hence, follow the Freundlich isotherm Eq. (12.1-2). q=Kc n The slope n is 0.229 and the constant K is 0.199, to give q=0.199c 0.229 12.2-1 Batch Adsorption on Activated Carbon (Geankoplis Example) A wastewater solution having a volume of 1.0 m3 contains 0.21 kg phenol/m3 of solution (0.21 g/L). A total of 1.40 kg of fresh granular activated carbon is added to the solution, which is then mixed thoroughly to reach equilibrium. Using the isotherm from Example 12.1-1, what are the final equilibrium values, and what percent of phenol of phenol is extracted? Solution: cF S M qF
= 0.21 = 1.0 = 1.40 =
kg phenol/m3 m3 kg carbon is assumed zero
0(1.40) + 0.21(1.0) = q(1.40) + c(1.0)
At the intersection, q = 0.106 kg phenol / kg carbon and c = 0.062 kg phenol / m3. The percent of phenol extracted is % extracted =
𝑐𝐹−𝑐 𝑐𝐹
(100) =
0.210−0.062 0.210
(100)
= 70.5
12.3-1 Scale-up of Laboratory Absorption Column (Geankoplis Example) A waste stream of alcohol vapor in air from a process was absorbed by activated carbon particles in a packed bed having a diameter of 4cm and length of 14cm containing 79.2 g of carbon. The inlet gas stream having a concentration co of 600 ppm and a density of .00115 g/cm3 entered the bed at a flow rate of 754 cm3 per s. Data in Table 12.3-1 give the concentrations of the breakthrough curve. The break-point concentration is set at c/co = .01. Do as follows. (a.) Determine the break-point time, the fraction of total capacity used up to the break point, and the length of the unused bed. Also determine the saturation loading capacity of the carbon. Table 12.3-1. Breakthrough Concentration for Example 12.3-1 Time, h
c/co
0 3 3.5 4 4.5 5 5.5 6 6.2 6.5 6.8
0 0 0.002 0.030 0.155 0.396 0.658 0.903 0.933 0.975 0.993
(b.) If the break-point time required for a new column is 6.00 h, what is the new total length of the column required?
Solution (a): c/co = tb = td = A1 = A2 =
0.01 3.65 6.95 3.65 1.51
h h approximately h graphically integrating h graphically integrating
The time equivalent to total or stoichiometric capacity of the bed is ∞
𝑡𝑡 = ∫ (1 − 0
𝑐 ) 𝑑𝑡 = 𝐴1 + A2 = 3.65 + 1.51 = 𝟓. 𝟏𝟔 𝐡 𝑐𝑜
The time equivalent to the usable capacity of the bed up to the break-point time is, 𝑡𝑏= 3.65
𝑡𝑢 = ∫ 0
(1 −
𝑐 ) 𝑑𝑡 = 𝐴1 = 𝟑. 𝟔𝟓 𝐡 𝑐𝑜
Hence, the fraction of total capacity used up to the break point is tu /tt = 3.65/5.16 = 0.707. The length of the used bed is HB = (tu /tt)HT = 0.707(14) = 9.9 cm. 𝐻𝑢𝑛𝑏 = (1 −
𝑡𝑢 ) 𝐻𝑇 = (1 − 0.707)(14) = 𝟒. 𝟏 𝐜𝐦 𝑡𝑡
Solution (b): For a new tb of 6.0 h, the new HB is obtained simply from the ratio of the break-point times multiplied by the old HB. 6.0
HB = 3.65(9.9) = 16.3 cm 𝐻𝑇 = HB + 𝐻𝑢𝑛𝑏 = 16.3 + 4.1 = 20.4 cm We determine the saturation capacity of the carbon. Air flow rate = (754 cm3/s)(3600 s)(0.0015 g/cm3) = 3122 g air/h 600 𝑔 𝑎𝑙𝑐𝑜ℎ𝑜𝑙
Total alcohol adsorbed = (
106 𝑔 𝑎𝑖𝑟
) (312 𝑔
𝑎𝑖𝑟 ℎ
)(5.16 ℎ)
= 9.67 g alcohol Saturation capacity = 9.67 g alcohol/79.2 g carbon = 0.1220 g alcohol/ g carbon The fraction of the new bed used up to the break point is now 16.3/20.4 = 0.799
12.5-1 Material Balance for Equilibrium Layers (Geankoplis Example)
An original mixture weighing 100 kg and containing 30 kg of isopropyl ether (C), 10 kg of acetic acid (A), and 60 kg water, (B) is equilibrated and the equilibrium phases separated. What are the compositions of the two equilibrium phases? Solution: The composition of the original mixture is xC = 0.30, xA = 0.10, and xB = 0.60.
The composition of the extract (ether) layer is yA = 0.04, yC = 0.94, and yB = 0.02 mass fraction. The raffinate (water) layer composition is xC = 0.02, xA = 0.12, and xB = 0.86.
12.5-2. Amounts of Phases in Solvent Extraction (Geankoplis Example) From the compositions obtained in Example 12.5-1, determine the amounts of V and L. The original mixture contained 100 kg and xAM = 0.10 Solution: V + L = M = 100 V(0.04) + L(0.12) = 100(0.10) L = 75.0 V = 25.0
12.7-1 Material Balance for Countercurrent Stage Process. Pure solvent isopropyl either at the reate of Vn+1= 600kg/h is being used to extract an aqueous solution of Lo=200kg/h containing 30 wt% acetic acid (A) by countercurrent multistage extraction. The desired acetic acid concentration in the aqueous phase is 4%. Calculate the compositions and amounts of the ether extract V1 and the aqueous raffinate LN. Use equilibrium data from Appendix A.3. Solution: The given values are Vn+1= 600, yAN+1=0, yAN+1=1.0, Lo=200, xAO=0.30, xBO=0.70, xCO=0.04.
In figure 12.7-3, Vn+1 and Lo are plotted. Also, since LN is on the phase boundary, it can be plotted at xAN=0.04. For the mixture point M, substituting into Eqs. (12.7-3) and (12.7-4), 𝑥𝐶𝑀 = 𝑥𝐴𝑀 =
𝐿𝑂 𝑥𝐶𝑂 + 𝑉𝑁+1 𝑦𝐶𝑁+1 𝐿𝑂 + 𝑉𝑁+1
𝐿𝑂 𝑥𝐴𝑂 + 𝑉𝑁+1 𝑦𝐴𝑁+1 𝐿𝑂 + 𝑉𝑁+1
=
=
200(0)+600(1.0) 200+600 200(0.30)+600(0) 200+600
= 0.75 = 0.075
Using these coordinates, the point M is plotted in Fig 12.7-3. We locate V1 by drawing a line from LN through M land extending it until it intersects the phase boundary. This gives yA1=0.08 and yC1=0.90. For LN a value of xCN=0.017 is obtained. By substituting into Eqs. (12.7-1) and (12.7-2) and solving, LN=136kg/h and V1=664kg/h.
12.7-2 Number of Stages in Countercurrent Extraction. Pure isopropyl ether of 450 kg/h is being used to extract an aqueous solution of 150 kg/h with 30 wt % acetic acid (A) by countercurrent multistage extraction. The exit acid concentration in the aqueous phase is 10 wt %. Calculate the number of stages required. Solution: The known values are VN+1=450, yAN+1=0, yCN+1=1.0, LO=150, xAO=0.30, xBO=0.70, xCO=0, and xAN=0.10.
The points VN+1, LO, and LN are plotted in Fig. 12.7-5. For the mixture point M, substituting into Eqs. (12.7-3) and (12.7-4), xCM=0.75 and xAM=0.075. the point M is plotted and V1 is located at the intersection of line LN M with the phase boundary to give yA1=0.072 and yC1=0.895. This construction is not shown. The lines LO V1 and LN VN+1 are drawn and the intersection is the operating point ∆ as shown. Alternatively, the coordinates of ∆ can be calculated from Eq. (12.7-10) to locate point ∆. Starting at LO we draw line LO∆, which locates V1. Then a tie line through V1 locates L1 in equilibrium with V1. (The tie-line data are obtained from an enlarged plot such as the bottom of Fig. 12.5-3.) Line L1∆ is next drawn locating V2. A tie line through V2 gives L2. A line L2∆ gives V3. A final tie line gives L3, which has gone beyond the desired LN. Hence, about 2.5 theoretical stages are needed.
12.7-3 Extraction of Nicotine with Immiscible Liquids. An inlet water solution of 100 kg/h containing 0.010 wt fraction nicotine (A) in water is stripped with a kerosene stream of 200 kg/h containing 0.0005 wt fraction nicotine in a countercurrent stage tower. The water and kerosene are essentially immiscible in each other. It is desired to reduce the concentration of the exit water
to 0.0010 wt fraction nicotine. Determine the theoretical number of stages needed. The equilibrium data are as follows (C5), with x the weight fraction of nicotine in the water solution and y in the kerosene.
Solution: The given values are LO =100 kg/h, xO =0.010, VN+1 =200 kg/h, yN+1 =0.0005, xN =0.0010. The inert streams are L’ = L(1 – x) = LO (1 - xO) = 100(1 – 0.010) = 99.0 kg water/hr V’ = V(1 – y) = VN+1 (1 – yN+1) = 200(1 – 0.0005) = 199.9 kg kerosene/hr
Making an overall balance on A using Eq (12.7-12) and solving, y1 =0.00497. These end points on the operating line are plotted in Fig. 12.7-6. Since the solutions are quite dilute, the line is straight. The equilibrium line is also shown. The number of stages are stepped off, giving N=3.8 theoretical stages.
Solved Problems
12.5-2 (Geankoplis, 4th Ed) A single-stage extraction is performed in which 400 kg of a solution containing 35 wt % acetic acid in water is contacted with 400 kg of pure isopropyl ether. Calculate the amounts and compositions of the extract and raffinate layers. What percent of the acetic acid is removed? Given: V1
L0= 400 kg acetic acid solution xA0= 0.35 xB0= 0.65
V2 = 400 kg isopropyl ether
L1
OMB: L0 + V2 = V1 + L1 V1 + L1 = 400 + 400 V1 + L1 = 800 = M Acetic Acid Balance: L0xA0 + V2yA2 = MxAM (400)(0.35) + (400)(0) = 800xAM xAM = 0.175 Isopropyl Ether Balance: L0yC0 + V2yC2 = MyCM (400)(0) + (400)(1) = 800yCM yCM= 0.50
Plot point M on graph. Tie line through M gives L1 and V1. L1 (Raffinate Layer): xA1= 0.255 xB1= 0.03 xC1= 0.715 V1 (Extract Layer): yA1= 0.11
yB1= 0.86 yC1= 0.03 Calculate L1 and V1: L1xA1 + V1yA1 = MxAM L1(0.255) + V1(0.11) = (800)(0.175) V1 + L1 = 800 Solving simultaneously, V1= 358 kg L1= 442 kg Amount of acetic acid in feed = 400(0.35) = 140 kg Amount of acetic acid in extract = 442(0.11) = 48.62 kg 48.62 % 𝑟𝑒𝑐𝑜𝑣𝑒𝑟𝑒𝑑 = 𝑥 100 = 34.7% 140
Graph for problem 12.5-2:
12.5-4 (Geankoplis, 4th Ed) A mixture weighing 1000 kg contains 23.5 wt % acetone and 76.5 wt % water and is to be extracted by 500 kg methyl-isobutyl ketone in a single stage extraction. Determine the amounts and composition of the extract and raffinate phases.
Given: V1
L0= 1000 kg acetone solution xA0= 0.235 xB0= 0.765
V2 = 500 kg methyl-isobutyl ketone
L1
OMB: L0 + V2 = V1 + L1 Let: V1 + L1 = M 1000 + 500 = M M = 1500 kg Acetone Balance: L0xA0 + V2yA2 = MxAM 1000(0.235) + 500(0) = 1500xAM xAM = 0.1567 Methyl-isobutyl Ketone Balance: L0yC0 + V2yC2 = MyCM 1000(0) + 500(1) = 1500yCM yCM = 0.333 Plot points M, L0, and V2 on graph, tie line through M gives L1 and V1 compositions. L1 (Raffinate Layer): xA1= 0.120 xC1= 0.022 V1 (Extract Layer): yA1= 0.205 yC1= 0.748
Calculate L1 and V1: L1xA1 + V1yA1 = MxAM L1(0.120) + (1500- L1) (0.205) = (1500)(0.1567) V1= 852 kg L1= 648 kg
12.7-2 (Geankoplis, 4th Ed) An aqueous feed of 200 kg/h containing 25 wt % acetic acid is being extracted by pure isopropyl ether at the rate of 600 kg/h in a countercurrent multistage system. The exit acid concentration in the aqueous phase is to contain 3.0 wt % acetic acid. Calculate the compositions and amounts of the exit extract and raffinate streams. Given: VN+1 = 600 kg/h isopropyl ether
L0= 200 kg/h acetic acid solution xA0= 0.25 xB0= 0.75
xAN= 0.03
𝐿0 𝑥𝐶0 + 𝑉𝑁+1 𝑦𝐶𝑁+1 (200)(0) + (600)(1.0) = = 0.75 𝐿0 + 𝑉𝑁+1 200 + 600 𝐿0 𝑥𝐴0 + 𝑉𝑁+1 𝑦𝐴𝑁+1 (200)(0.25) + (600)(0) = = = 0.0625 𝐿0 + 𝑉𝑁+1 200 + 600
𝑥𝐶𝑀 = 𝑥𝐴𝑀
Plot points M, L0, and VN+1 on graph; LN on phase B. Locate V1 by drawing line MLN to intersect extract phase boundary at V1. Ln (Raffinate Layer): xAN= 0.030 xCN= 0.017 V1 (Extract Layer): yA1= 0.070
yC1= 0.905 Calculate L1 and V1: OMB: L0 + VN+1 = V1 + LN Let: V1 + LN = M 200 + 600 = M M = 800 kg Isopropyl Ether Balance: L0xC0 + VN+1yCN+1 = MxCM LNxCN + V1yC1 = MxCM LN(0.017) + (800- LN)(0.905) = (800)(0.75) V1= 660.4 kg/h LN= 139.6 kg/h
12.7-4 (Geankoplis, 4th Ed) An aqueous feed solution of 1000 kg/h of acetic acid-water solution contains 30.0 wt % acetic acid and is to be extracted in a countercurrent multistage process with pure isopropyl ether to reduce the acid concentration to 2.0 wt % acid in the final raffinate. (a) Calculate the minimum solvent flow rate that can be used. (b) If 2500 kg/h of ether solvent is used, determine the number of theoretical stages required. Given: VN+1 = Pure isopropyl ether
L0= 1000 kg/h acetic acid solution xA0= 0.30 xB0= 0.70
xAN= 0.02
a. Minimum solvent Tie line through L0 gives V1min. Line LNVN+1 and L0V1min (tie line) gives Δ pt. min. Draw line LNV1min and L0VN+1 and intersection gives Mmin.
From graph: xA= 0. 114 𝑥𝐴𝑀𝑚𝑖𝑛 = 0.114 =
𝐿0 𝑥𝐴0 + 𝑉𝑁+1 𝑦𝐴𝑁+1 (1000)(0.25) + 𝑉𝑁+1 (0) = 𝐿0 + 𝑉𝑁+1 1000 + 𝑉𝑁+1
VN+1 = VN+1min = 1630 kg/h
b. Number of Stages VN+1 = 2500 kg/h isopropyl ether
L0= 1000 kg/h acetic acid solution xA0= 0.30 xB0= 0.70
xAN= 0.02
𝐿0 𝑥𝐶0 + 𝑉𝑁+1 𝑦𝐶𝑁+1 (1000)(0) + (2500)(1.0) = = 0.714 𝐿0 + 𝑉𝑁+1 1000 + 2500 𝐿0 𝑥𝐴0 + 𝑉𝑁+1 𝑦𝐴𝑁+1 (1000)(0.30) + (2500)(0) = = = 0.0857 𝐿0 + 𝑉𝑁+1 1000 + 2500
𝑥𝐶𝑀 = 𝑥𝐴𝑀
This point checks graphical determination of M on lines LNV1 and L0VN+1. Draw line LNVN+1 and L0V1 to obtain Δ operating point graphically. Algebraic calculation of Δ operating point: xAN = 0.02 xCN = 0.015 yA1 = 0.010 (off graph) xC1 = 0.863 OMB: L0 + VN+1 = V1 + LN Let: V1 + LN = M 1000 + 2500 = M M = 3500 kg
𝑥𝐴𝑀 = 0.0857 =
𝐿𝑁 𝑥𝐴𝑁 + 𝑉1 𝑦𝐴1 (𝐿𝑁 )(0.02) + (3500 − 𝐿𝑁 )(0.010) = 𝐿𝑁 + 𝑉1 3500
V1= 2874 kg/h LN= 626 kg/h 𝐿0 𝑥𝐶0 + 𝑉1 𝑦𝐶1 (1000)(0) + (2874)(0.863) = = +1.32 𝐿0 + 𝑉1 1000 + 2874 𝐿0 𝑥𝐴0 + 𝑉1 𝑦𝐴1 (1000)(0.30) + (2874)(0.10) = = = −0.0067 𝐿0 + 𝑉1 1000 + 2874
𝑥∆𝐶 = 𝑥𝐴𝑀
This checks the graphical Δ op. point. Use Δ op. point to determine the number of stages. Plot stages up to V4. Then, plot expanded scale as shown. The operating points on expanded scales as L1, V2, L2V3, etc. determined by lines from L to Δ operating point. Nmin = 7.5 stages
Graph for problem 12.7-4:
Unit Operations of Chemical Engineering – McCabe, Smith, Harriott Sample Problems Example 23.2 Penicillin F is recovered from a dilute aqueous fermentation broth by extraction with amyl acetate using 6 volumes of solvent per 100 volumes of the aqueous phase. At pH = 3.2 the distribution coefficient KD is 80. (a) What fraction of the penicillin would be recovered in a single ideal stage? (b) What would be the recovery with two-stage extraction using fresh solvent in both stages?(c) How many ideal stages would be needed to give the same recovery as in part (b) if a counterflow cascade were used with V/L = 0.06?
Solution: (a) By a material balance, since y0 = 0, L(x0 – x1) = Vy1 y1 = KD x1 VKD
x1 (
L
+ 1) = x0
The extraction factor is E= Thus
x1 x0
=
VKD L 1
=
6 x 80 100
= 4.8
1
= 5.8 = 0.172 1+E
Recovery is 1 - 0.172 = 0.828, or 82.8%
(b) With the same value of E, x2 1 = x1 1+E
x2 1 = = 0.0297 x0 (1 + E)2 Recovery is 1 – 0.0297 = 0.9703 or 97.0% (c) With KD and V/L constant, the number of ideal stages can be calculated from the stripping form of Kremser equation using E in place of its equivalent, the stripping factor S. ln[(xa −xa ∗ )/(xb −xb ∗ )
N =
ln E
Let xa = x0 = 100 Then xb = 3.0 and ya = 97(100)/6 = 1617 xa ∗ =
ya 1.617 = = 20.2 KD 80
xb ∗ = 0 N =
100−20.2 ] 3
ln[
ln 4.8
= 2.09
Example 23.3 A countercurrent extraction plant is used to extract acetone (A) from its mixture with water by means of methyl isobutyl ketone (MIK) at a temperature 25⁰C. The feed consists of 40 percent acetone and 60 percent water. Pure solvent equal in mass to the feed is used as the extracting liquid. How many ideal stages are required to extract 99 percent of the acetone fed? What is the extract composition after removal of the solvent? Solution:
Use the data in Fig 23.7 to prepare a plot of the equilibrium relationship yA versus xA, which is the upper operating line in Fig 23.10. The terminal points for the operating line are determined by material balances with allowances for the amounts of water in the extract phase and MIK in the raffinate phase.
Basis: F = 100 mass units per hour Let n = mass flow rate of H2O in extract M = mass flow rate of MIK in raffinate For 99% recovery of A, the extract has 0.99 x 40 = 39.64, and the raffinate has 0.44. The total flows are At the top, La = F = 100 = 40A +60 H2O Va = 39.6A + nH2 O + (100 – m)MIK = 139.6 + n − m At the bottom, Vb =100MIK Lb = 0.4A + (60 – n)H2 O + mMIK = 60.4 + m − n Since n and m are small and tend to cancel in the summation for Va and La, the total extract flow Va is about 140, which would make yA,a ≅ 39.6/140 = 0.283. The value of XA,b is about 0.4/60 = 0.006. These estimates are adjusted after calculating values of n and m.
From Fig 23.7 for yA = 0.283, yH2O = 0.049 n=
0.049 1−0.049
(39.6 +100 – m) 0.049
If m is very small n ≅ (0.951) (139.6) = 7.2 From Fig 23.7 for xA = 0.007, xMIK = 0.02, m= ≅
0.02 (0.4 + 60 – n) 1 − 0.02 0.02 (0.4 + 52.8) = 1.1 0.98
Revised n = (0.049/0.951)( 139.6 − 1.1) = 7.1 Va = 139.6 + 7.1 − 1.1 = 145.6 yA,a =
39.6 = 0.272 145.6
Lb = 60.4 + 1.1 − 7.1 = 54.4 xA,,b =
0.4 = 0.0074 54.4
Plot points (0.0074, 0) and (0.4, 0.272) to establish the ends of the operating line. For an intermediate point on the operating line, pick yA = 0.12 and calculate V and L. From Fig. 23.7, yH2O = 0.03, and yMIK = 0.85, Since the raffinate phase has only 2 to 3 percent of MIK, assume that the amount of MIK in the extract is 100, the same as the solvent feed: 100 ≅ 𝑉𝑦𝑀𝐼𝐾 100
𝑉 ≅ 0.85 = 117.6 By an overall balance from the solvent inlet (bottom) to the intermediate point, 𝑉𝑏 + 𝐿 = 𝐿𝑏 + 𝑉 𝐿 ≅ 54.4 + 117.6 − 100 = 72.0 A balance on A over the same secion gives xA: 𝐿𝑥𝐴 + 𝑉𝑏 𝑦𝑏 = 𝐿𝑏 𝑥𝑏 + 𝑉𝑦𝐴 𝐿𝑥𝐴 ≅ 0.4 + 117.6(0.12) − 0
𝑥𝐴 ≅
14.5 = 0.201 72
This value is probably accurate enough, but corrected values of V, L and xA can be determined. For xA = 0.201, 𝑥𝑀𝐼𝐾 ≅ 0.03(Fig 23.7). A balance on MIK from the solvent inlet to the intermediate point gives 𝑉𝑏 + 𝐿𝑥MIK = Lb xMIK,b + VyMIK VyMIK = 100 + 72(0.03) − 1.1 Revised V =
101.1 = 118.9 0.85
Revised L = 54.4 + 118.9 − 100 = 73.3 Revised xA =
0.4+ 118.9(0.12) 73.3
= 0.200
Plot xA = 0.20, yA = 0. 12, which gives a slightly curved operating line. From Fig 23.10, N = 3.4 stages
Solved Problems 23.7 An antibiotic that has been extracted from a fermentation broth using amyl acetate at low pH is to be extracted back into clean water at pH = 6, where Kp = 0.15. If the water flow rate is set at 0.45 times the solvent rate, how many ideal stages would be needed for 98 percent recovery of the antibiotic in a countercurrent cascade? Solution: KD = 0.15 = ye / xe xa = 0 LΔx = VΔy L = 0.45V Ya = 0.02yb Neglect changes in L and V Operating line slope = 0.45 = L/V ya* = 0 For extraction, E=
KDV L
[(Eq. 23.5)], but this is for transfer into the organic phase. Here solute is extracted from the organic phase into the water, so K’D = 1 / KD = 1 / 0.15 = xe / ye and appropriate factor is LK'D 0.45 = = 3.0 V 0.15 This factor corresponds to absorption factor A in absorption. Lxb = V(0.98)yb 𝑥𝑏 =
1 (0.98)𝑦𝑏 = 2.18𝑦𝑏 0.45
yb* = 0.15xb = 0.15(2.18)yb
Solving for N. ln[ N=
yb − 0.327yb 0.02yb ] = 3.20 stages 𝑙𝑛 3.0
23.8 Oil was extracted from small particles of rapeseed, averaging 0.58mm in size, by contact with hexane. The particles originally contained 43.82 percent oil and 6.43 percent moisture. After drying, the oil content of the dried meal was reported for different extraction times. The results are given in the table below. Determine the effective diffusion coefficient for oil in the hexanesoaked particles using the equation for transient diffusion in spheres. Assume that a large excess of hexane was used and that the external mass transfer resistance was negligible. Time, min
75
90
105
120
Oil in dry meal, kg/100 kg inert material
11.5
7.97
4.35
3.88
Answer: Assume the packing 40% voids. Dp = 0.015 m av = 6(1-ε) / Dp = (6 x 0.6) / 0.015 = 240 m2/m3
ρc = 62.15 x 16.018 = 995 kg/m3 μc = 0.801 cP = 8.01 x 10-4 Pa.s σ = 36.1 x 10-3 = 0.0361 N/m ρd = 53.64 x 16.018 = 859.2 kg/m3 Δρ = 136.3 kg/ m3 Assume R =2, α = 1.0 (continuous wetting). For spheres, use C1 = 0.208. Then 2
Vs,c (1+R0.5 ) (
av 0.5 ) =27.35Vs,c g 0.15
=1.0 x 0.208 x 1.041.54 (
136.3 0.41 1 136.32 x 9.805 1 0.3 8.01 x 10-4 3] ) [ ( ) x [ ] 859.2 240 8.01 x 10-4 136.3 x 0.0361 0.5 ( ) 240
= 0.0615 Vs,c =0.0615/27.35= 0.00225 m/s
23.9 (From Unit Operation of Chemical Engineering by McCabe and Smith) A spray-tower extractor operates at 30°C with hydrocarbon drops dispersed in water. The density of the hydrocarbon phase is 53 lb/ft3. If the average drop size is 2.0 mm, calculate the terminal velocity of the drops. If the slip velocity is assumed to be independent of the holdup and if the dispersed-phase volumetric flow rate is taken to be twice that of the continuous phase, at what flow rate of the dispersed phase would the fraction holdup be 0.30 (close to flooding)? For different ratios of flow rates, is either the sum of the flow rates or the sum of their square roots nearly a constant at a given holdup? (Note the correlation in Figure below)
Fig. 23.7 Flooding velocities in packed extraction towers Solution: Dp = 0.2 cm, ρp = 53 / 62.4 = 0.849 g / cm3 For water at 30°C, ρ = 0.996 g / cm3, μ = 0.801 cP To get μt, try Stoke’s Law 980(0.2)2 (0.996-0.849) μt = =39.97 cm/s 18(0.0081) Rep =
0.2(39.97)(0.996) =994 0.0081
This is well beyond the range of Stoke’s Law, so μt must be calculated using Eq. (7.37) and Fig. 7.6. Guess Rep = 100, which gives CD = 1.0
4𝑔(𝜌𝑝−𝜌)𝐷𝑝
𝜇𝑡 = √
Eq. (7.37)
𝜇𝑡 = (
3𝐶𝐷 𝜌
4(980)(0.147)(0.2) 1/2 ) = 6.21 𝑐𝑚/𝑠 3(1.0)(0.996)
𝑅𝑒𝑝 =
0.2(6.21)(0.996) = 154, 𝐶𝐷 = 0.9 0.00801
Guess CD = 0.85 μt = 6.74 cm / s Re = 167, CD = 0.85 μt =6.74
(OK)
cm 3600 x =796 ft/h s 30.5
Let h = volume fraction holdup of drops Vd , Vc = actual velocities of drops or continuous phase (both positive) Then μt = Vd + Vc = 796 ft/h Given: First Part Vs,d = 2Vs,c Vd = Vs,d / h
h = 0.30 Vc = Vs,c / (1-h)
Vs,d Vs,d + =796 0.3 2(0.7) Vs,d(3.33+0.714) = 796 Vs,d = 197 ft / h Vs,d + Vs,c = 197 + 0.5(197) = 296 √Vs,d + √Vs,c =14.0+9.9=23.9 Second Part For Vs,d = Vs,c and h = 0.30 Vs,d =
796 =167=Vs,c 3.33+1.43 Vs,d + Vs,c =334
and √Vs,d +√Vs,c =25.8 For Vs,d = 0.5Vs,c and h = 0.30 Vs,d =
796 =129 3.33+2.86 Vs,c =258
Vs,d + Vs,c = 387 31% higher than for Vs,d = 2Vs,c but √Vs,d +√Vs,c =11.4+16.1=27.5 15% higher than for Vs,d = 2Vs,c
The sum of the square roots is more nearly constant than the sum of the velocities.
