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Mass Transfer Operations – Robert TreybalSample Problems 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...

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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 E 1 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

𝐹𝑥𝐹 8000(0.30) = 0.0857 𝑥𝑀 = = 8000 + 20 000 𝐹 +FB. 𝐵 Line 𝑅𝑁 𝑀 extended provides E1 at y1=0.10. Line and point M is located as shown on line 𝐹 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 0.10 − 0.02 = 𝑦 −𝑥 𝐹 𝑅𝑁𝐹1 = 𝑀𝑁− 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 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟 1 − 0.01

0.001

𝑘𝑔 𝑘𝑔𝑛𝑖𝑐𝑜𝑡𝑖𝑛𝑒 𝑤𝑎𝑡𝑒𝑟 = 0.001 001 1 − .001 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, 𝑥′𝑁𝐹 =

𝑥𝑁𝐹 = 0.001

𝐴

𝐵𝑚

=

0.0093−0

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

14.30

17.25

20.75

27.8

40.0

50.0

50.0

1 𝑥−𝑥∗

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

isopropyl ether is very small, so that r can be taken as 𝑁𝑇𝑈 = 8.40 + ln 1

2

1−0.02

1−0.30

+ ln 1

2

0.02(0.3−1)+1 0.30(0.3−1)+1

18

60

= 0.30. Eq.(10.110):

= 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 + xwb Lb = 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) 1.261 0.755(0.761)

Acetone = H 2O =

= 0.418

1.261 0.21(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 We have:

𝑀𝐹 = 1.022 𝑀𝑆

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 B:

𝐵𝑣𝑜𝑙𝑢𝑚𝑒 =

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡

𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑝𝑖𝑐𝑟𝑖𝑐 𝑎𝑐𝑖𝑑 𝐶𝑆1

= 0.0155 = 𝟓. 𝟏𝟔𝒍 solvent required per liter feed

Cross-current operation:

0.08

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 =

For stage 1:

1

= 0.02

𝑚𝑜𝑙

𝑙𝑖𝑡𝑒𝑟

𝐶𝐴3 = 0.02𝑚𝑜𝑙/𝑙𝑖𝑡𝑒𝑟𝐶𝑆3 = 0.0155

− For stage 3:

0.02

𝑚𝑜𝑙 (𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑣𝑎𝑙𝑢𝑒) 𝑙𝑖𝑡𝑒𝑟

𝐴 𝐶𝑆1 − 𝐶𝑠0 = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑙𝑖𝑛𝑒 = 𝐶𝐴1 − 𝐶𝐹 𝐵 −𝐵 = 𝐴

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=−

𝐵=

Benzene required per stage= 1.3L Total benzene...