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LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 12.10-3. Countercurrent Multistage Washing of Ore. A treated ore containing inert solid gangue and copper sulfate is to be leached in a countercurrent multistage extractor using pure water to leach CuSO4. The solid charge rate per hour consists of 10,000 kg of inert gangue (B), 1,200 kg of CuSO4 (solute A), and 400 kg of water (solute C). The exit wash solution is to contain 92 wt% water and 8 wt% CuSO4. A total of 95% of the CuSO4 in the inlet ore is to be recovered. The underflow is constant at N=0.5 kg inert gangue/kg aqueous solution. Calculate the number of stages required. Given: V1 Y1
Vb Yb =0 (pure H2O)
Va Ya
92 wt% H2O 8 wt% CuSO4
1
N-1 L1 X1
Lb Xb
La = 1,600 kg Xa = 0.75 B=10,000 kg inert solid 1,200 kg CuSO4 400 kg H2O
N = 0.5 kg inert solid/kg aqueous solution 95% recovery of oil Required: N Solution: πΏπ =
π΅π ππ πππ’πππ’π π πππ’π‘πππ ; π ππππ π΅ πππ π ππ ππππ π‘πππ‘ β΄ πΏπ = (10,000 ππ πππππ‘ π ππππ) ( ) ππ 0.5 ππ πππππ‘ π ππππ π³π = ππ, πππ ππ
CuSO4 recovered: πΏπππ(0.95) = 1,200 ππ πΆπ’ππ4 (0.95) = 1,140 ππ πΆπ’ππ4 CuSO4 retained: πΏπππ(0.05) = 1,200 ππ πΆπ’ππ4 (0.05) = 60 ππ πΆπ’ππ4
We can now compute for Xb: ππ =
60 ππ πΆπ’ππ4 20,000 ππ πππ’πππ’π π ππβ²π πΏπ = π. πππ
We can find the water in exit flow from multiplying the water-solute ratio to the amount of recovered solute. 92 ππ π»2 π ) 8 ππ πΆπ’ππ4
H2O in exit flow: 1,140 ππ πΆπ’ππ4 (
We can now compute for Va and Ya:
= 13,110 ππ π»2 π
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 ππ = 1,140 πΎπ πΆπ’ππ4 + 13,110 ππ π»2 π π½π = ππ, πππ ππ ππ =
1,140 ππ πΆπ’ππ4 14,250 ππ πππ’πππ’π π ππβ²π ππ = π. ππ
OMB: πΏπ + ππ = πΏπ + ππ (1,600 ππ) + ππ = (20,000 ππ) + (14,250 ππ) π½π = ππ, πππ ππ
V1 Y1
Vb = 32,650 kg Yb =0 (pure H2O)
Va = 14,250 kg 92 wt% H2O 8 wt% CuSO4 Ya = 0.08 1
N-1
La = 1,600 kg Xa = 0.75
L1 X1
Lb = 20,000 kg Xb = 0.003
1,140 kg CuSO4 13,110 kg H2O
B=10,000 kg inert solid 1,200 kg CuSO4 400 kg H2O
60 kg CuSO4 19,940 kg H2O
V1= 32,650 kg H2O Y1
Va = 14,250 kg Ya = 0.08
92 wt% H2O 8 wt% CuSO4 1,140 kg CuSO4 13,110 kg H2O
1
L1=20,000 X1=0.08
La = 1,600 kg Xa = 0.75
60 kg CuSO4 19,940 kg H2O
B=10,000 kg inert solid 1,200 kg CuSO4 400 kg H2O
We now make an oil balance on the first stage to find Y1. π1 π1 + πΏπ ππ = πΏ1 π1 + ππ ππ (32,650 ππ π»2 π)π1 + (1,600 ππ)(0.75) = (20,000ππ)(0.08) + (14,250 ππ)(0.08)
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 0.0472 ππ πΆπ’ππ4 (32,650 ππ π»2 π) ππ π»2 π π1 = 0.0472 ππ πΆπ’ππ4 32,650 ππ π»2 π + (32,650 ππ π»2 π) ππ π»2 π ππ = π. ππππ π½π = ππ, πππ. ππ
Oil Composition (Oil kg/kg solβn) Xa
0.75
Ya
0.08
πβ1=
π βπ ln (ππ β ππ ) π
π
π β ππ ln ( π ) ππ β ππ
=
π βπ ln (π1 β π1 ) π
π
π β π1 ln ( π ) ππ β π1
We use points in N-1 stages: Xb, Yb, X1, Y1 X1
0.08
Y1
0.0451
Xb
0.003
Yb
0
0.0451 β 0.08 ln ( 0 β 0.003 ) π= +1 0.003 β 0.08 ln ( ) 0 β 0.0451 π΅ = π. ππ β π ππππππ
We can make a graph for the stages using Ponchon Savarit Method. πΏπ = 1,600 ππ ; πΏπ = 20,000 ππ; π΅1 = π΅π = 10,000 ππ ππ =
π΅π 10,000 ππ = = 6.25 πΏπ 1,600 ππ
ππ =
π΅π 10,000 ππ = = 0.5 πΏπ 20,000 ππ
Coordinates for β πβ = ππβ =
N 6.25 0.5 -0.79 0.5 (constant)
π΅ 10,000 ππ = = β0.79 πΏπ β ππ 1,600 ππ β 14,250 ππ
πΏπ ππ β ππ ππ 1,600 ππ(0.75) β 14,250(0.08) ππ = = β0.0047 πΏπ β ππ 1,600 ππ β 14,250 ππ COORDINATES FOR GRAPH X/Y 0.75 0.003 -0.11
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 Ponchon Savarit Diagram
π΅ = π. π β π ππππππ
Basing N-1, πΏ1 = πΏπ = 169.97 ππ; π΅1 = π΅π = 743 ππ π1 =
π΅1 10,000 ππ = = 0.5 πΏ1 20,000 ππ
ππ =
π΅π 10,000 ππ = = 0.5 πΏπ 20,000 ππ
Coordinates for β πβ = ππβ =
π΅ 10,000 ππ = = β0.7 πΏ1 β π1 20,000 ππ β 34,191.08 ππ
πΏ1 π1 β π1 π1 20,000 ππ (0.08) β 34,191.08 ππ(0.0451) = = β0.004 πΏ1 β π1 20,000 ππ β 34,191.08 ππ
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1
We can also make a PS diagram for N-1, to have a better view of L1, and Lb and to support the Kremser equation. π β 1 = 4.6 π π‘ππππ π΅ = π. π β π ππππππ
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 McCabe Diagram X 0.08 0.003 0.75
COORDINATES FOR GRAPH Y 0.0451 0 0.08
From the graph we can see that N is approximately 6 stages.
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 12.10-4. Countercurrent Multistage Leaching of Halibut Livers. Fresh halibut livers containing 25.7 wt% oil is to be extracted with pure ethyl ether to remove 95% of the oil in a countercurrent multistage leaching process. The feed rate is 1000 kg of fresh livers per hour. The final exit overflow solution is to contain 70 wt% oil. The retention of solution by the inert solids (oil-free liver) of the liver varies as follows (C1), where N is the kg inert solid/kg retained and yA is kg oil/kg solution. N 4.88 3.50 2.47 1.67 1.39
yA 0 0.2 0.4 0.6 0.81
Calculate the amounts and compositions of the exit streams and the total number of theoretical stages. Given: Vb Yb =0 (pure ethyl ether)
V1 Y1
70 wt% oil Va Ya=0.7
1
N-1
Lb Xb
La = 257 kg Xa = 1
L1 X1
1,000 kg solid+oil 25.7 wt% oil 257 kg oil 743 kg solid
95% recovery of oil
Required: N and composition Solution: Oil recovered: πΏπππ(0.95) = 257 ππ πππ (0.95) = 244.15 ππ πππ Oil retained: πΏπππ(0.05) = 257 ππ πππ (0.05) = 12.85 ππ πππ
We can now compute for Va: ππ = 244.15 ππ πππ (
1 ππ π πππ’π‘πππ ) 0.7 ππ πππ
π½π = πππ. ππ ππ Oil Balance ππ ππ + πΏπ ππ = πΏπ ππ + ππ ππ ππ (0) + 257 ππ = πΏπ ππ + 244.15 ππ
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1
πΏπ =
ππ. ππ ππ π³π
π π³π = πππ ππ πππππ πππππ
( ) π΅ From here, we will use trial and error to find Lb and Xb using the values given in the table. The values were interpolated to give the estimation. N 4.88 4.752 4.673 4.535 4.381 4.3714 4.3702 4.369 4.357 4.328 4.314 4.19 3.50 2.47 1.67 1.39
yA 0 0.02 0.03 0.05 0.074 0.0756 0.0758 0.076 0.078 0.08 0.085 0.1 0.2 0.4 0.6 0.81
Trial 1: ππ = 0.2, π = 3.5 πΏπ = 743 ππ πππππ‘ π ππππ ( ππ =
1 ) = 212.29 ππ 3.5
12.85 ππ = 0.06 212.29 ππ
Trial 2: ππ = 0.4, π = 2.47 πΏπ = 743 ππ πππππ‘ π ππππ ( ππ =
1 ) = 300.8 ππ 2.47
12.85 ππ = 0.043 ; π‘βπ πππππ πππππππ ππ 300.8 ππ
From trial 2 we can assume that 0 > ππ > 0.2 Trial 3: ππ = 0.1, π = 4.19 πΏπ = 743 ππ πππππ‘ π ππππ ( ππ =
1 ) = 177.33 ππ 4.19
12.85 ππ = 0.07 177.33 ππ
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1
Trial 4: ππ = 0.05, π = 4.535 πΏπ = 743 ππ πππππ‘ π ππππ ( ππ =
1 ) = 163.84 ππ 4.535
12.85 ππ = 0.078 163.84 ππ
Trial 5: ππ = 0.03, π = 4.673 πΏπ = 743 ππ πππππ‘ π ππππ ( ππ =
1 ) = 101.22 ππ 4.673
12.85 ππ = 0.12 ; ππ ππ’π π‘ ππ πππππ‘ππ π‘βππ 0.05 101.22 ππ
Trial 6: ππ = 0.08, π = 4.328 πΏπ = 743 ππ πππππ‘ π ππππ ( ππ =
1 ) = 171.67 ππ 4.328
12.85 ππ = 0.075 171.67 ππ
Trial 7: ππ = 0.078, π = 4.357 πΏπ = 743 ππ πππππ‘ π ππππ ( ππ =
1 ) = 170.53 ππ 4.357
12.85 ππ = 0.075 170.53 ππ
Trial 8: ππ = 0.076, π = 4.369 πΏπ = 743 ππ πππππ‘ π ππππ ( ππ =
1 ) = 170.06 ππ 4.369
12.85 ππ = 0.0756 170.06 ππ
Trial 8: ππ = 0.0758, π = 4.3702 πΏπ = 743 ππ πππππ‘ π ππππ ( ππ =
1 ) = 170.02 ππ 4.3702
12.85 ππ = 0.0756 170.06 ππ
Trial 10: ππ = 0.0756, π = 4.3714 πΏπ = 743 ππ πππππ‘ π ππππ (
1 ) = 169.97 ππ 4.3714
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1
ππ =
12.85 ππ = 0.0756 169.99 ππ
π³π = πππ. ππ ππ πΏπ = π. ππππ OMB: ππ + πΏπ = ππ + πΏπ ππ + 257 ππ = 348.79 ππ + 169.97 ππ π½π = πππ. ππ ππ V1= 261.76 kg ethyl ether Y1
Vb= 261.76 kg Yb =0 (pure ethyl ether)
Va=348.79 kg 70 wt% oil Ya=0.7 1
N-1
Lb=169.97 kg Xb=0.0756
La Xa
L1=169.97 kg X1=0.7
1,000 kg solid+oil 25.7 wt% oil 257 kg oil 743 kg solid
Solving for Y1 we make an oil balance on the first stage. π1 π1 + πΏπ ππ = ππ ππ + πΏ1 π1 (261.76 ππ ππ‘βπ¦π ππ‘βππ)π1 + 257ππ πππ = 244.15 ππ πππ + 118.98 ππ πππ 0.405 ππ πππ (261.76 ππ ππ‘βπ¦π ππ‘βππ) ππ ππ‘βπ¦π ππ‘βππ π1 = 0.405 ππ πππ 261.76 ππ ππ‘βπ¦π ππ‘βππ + (261.76 ππ ππ‘βπ¦π ππ‘βππ) ππ ππ‘βπ¦π ππ‘βππ ππ = π. πππ π½π = πππ. ππ ππ Oil Composition (Oil kg/kg solβn)
πβ1=
π βπ ln (ππ β ππ ) π
π
=
π βπ ln (π1 β π1 ) π
π
Xa
1
π βπ ln ( ππ β ππ ) π π
π βπ ln ( ππ β π1 ) π 1
Ya
0.7
We use points in N-1 stages: Xb, Yb, X1, Y1
X1
0.7
Y1
0.288
Xb
0.0756
0.288 β 0.7 ln ( ) 0 β 0.0756 + 1 π= 0.0756 β 0.7 ln ( 0 β 0.288 )
Yb
0
π΅ = π. ππ β π. π ππππππ
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 We can make a graph for the stages using Ponchon Savarit Method. COORDINATES FOR GRAPH N X/Y 2.89 1 4.37 0.0756 -8.09 -0.14 4.88 4.752 4.673 4.535 4.19 3.50 2.47 1.67 1.39
0 0.02 0.03 0.05 0.1 0.2 0.4 0.6 0.81
πΏπ = 257 ππ; πΏπ = 169.97 ππ; π΅1 = π΅π = 743 ππ ππ = ππ =
π΅π 743 ππ = = 2.89 πΏπ 257 ππ
π΅π 743 ππ = = 4.37 πΏπ 169.97 ππ
Coordinates for β πβ = ππβ =
π΅ 743 ππ = = β8.09 πΏπ β ππ 257 ππ β 348.79ππ
πΏπ ππ β ππ ππ 257 ππ(1) β 348.79 ππ(0.7) = = β0.14 πΏπ β ππ 257 ππ β 348.79ππ
π΅ = π. π ππππππ
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 For N-1: πΏ1 = πΏπ = 169.97 ππ; π΅1 = π΅π = 743 ππ π1 =
π΅1 743 ππ = = 4.37 πΏ1 169.97 ππ
ππ =
π΅π 743 ππ = = 4.37 πΏπ 169.97 ππ
Coordinates for β πβ = ππβ =
π΅ 743 ππ = = β3.75 πΏ1 β π1 169.97 ππ β 367.77ππ
πΏ1 π1 β π1 π1 169.97 ππ(0.7) β 367.77ππ(0.288) = = β0.066 πΏ1 β π1 169.97 ππ β 367.77ππ
L1
PS diagram for N-1 π β 1 = 2.3 π π‘ππππ π΅ = π. π ; πππππ ππ ππππππ
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 23.4 Oil is to be extracted from halibut livers by means of ether in a countercurrent extraction battery. The entrainment of solution by the granulated liver mass was found by experiment to be as shown in Table 20.5. In the extraction battery, the charge per cell is to be 100 lb, based on completely exhausted livers. The unextracted livers contain 0.043 gal of oil per pound of exhausted material. A 95 percent recovery of oil is desired. The final extract is to contain 0.65 gal of oil per gallon of extract. Solution retained by Solution 1 lb exhausted livers, concentration, gal gal oil/gal solution 0.035 0 0.042 0.1 0.050 0.2 0.058 0.3 0.068 0.4 0.081 0.5 0.099 0.6 0.120 0.68 a. How many gallons of ether are needed per charge of livers? b. How many extractors (stages) are needed?
Given: V1 Y1
Vb Yb =0 (pure ether)
Va Ya=0.65 1
N-1
Lb Xb
0.65 gal oil/ gal of soln
La Xa = 1
L1 X1
0.043 gal oil/lb liver Basis:100 lb liver 4.3 gal oil
95% recovery of oil
Required: N and composition Solution: Oil recovered: πΏπππ(0.95) = 4.3 πππ πππ (0.95) = 4.085 πππ πππ Oil retained: πΏπππ(0.05) = 257 ππ πππ (0.05) = 0.215 πππ πππ We can now compute for Va: ππ = 4.085 ππ πππ (
1 πππ π πππ’π‘πππ ) 0.65 πππ πππ
π½π = π. πππ πππ
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 Oil Balance ππ ππ + πΏπ ππ = πΏπ ππ + ππ ππ ππ (0) + 4.3 πππ = πΏπ ππ + 4.085πππ πΏπ =
π. πππ πππ π³π
π³π = πππ ππ πππππ πππππ
(π―) Solution retained by 1 lb exhausted livers, gal (1/N or H) 0.035 0.0385 0.03885 0.042 0.050 0.058 0.068 0.081 0.099 0.120
Solution concentration, gal oil/gal solution 0 0.05 0.055 0.1 0.2 0.3 0.4 0.5 0.6 0.68
Trial 1: ππ = 0.1, π» = 0.042 πΏπ = 100 ππ πππππ‘ π ππππ(0.042) = 4.2 πππ ππ =
0.215 πππ = 0.051 4.2 πππ
Trial 2: ππ = 0.2, π» = 0.05 πΏπ = 100 ππ πππππ‘ π ππππ(0.05) = 5 πππ ππ =
0.215 πππ = 0.043; π‘βπ πππππ πππππππ ππ 5 πππ
From trial 2 we can assume that 0 > ππ > 0.1 Trial 3: ππ = 0.05, π = 0.0385 πΏπ = 100 ππ πππππ‘ π ππππ(0.0385) = 3.85 πππ ππ =
0.215 πππ = 0.056 3.85 πππ
Trial 4: ππ = 0.055, π = 0.03885 πΏπ = 100 ππ πππππ‘ π ππππ(0.03885) = 3.885 πππ
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1
ππ =
0.215 πππ = 0.055 3.885 πππ
π³π = π. πππ πππ πΏπ = π. πππ
V1=6.2 gal Y1
Vb=6.2 gal Yb =0 (pure ether)
Va=6.285 gal Ya=0.65 1
N-1
Lb=3.885 gal Xb=0.055
0.65 gal oil/ gal of soln
La =4.3 gal Xa = 1
L1=3.885 gal X1=0.65
0.043 gal oil/lb liver Basis:100 lb liver 4.3 gal oil
OMB: ππ + πΏπ = ππ + πΏπ ππ + 4.3 πππ = 6.615 πππ + 3.885 πππ π½π = π. π ππ πππππ Solving for Y1 we make an oil balance on the first stage. π1 π1 + πΏπ ππ = ππ ππ + πΏ1 π1 (6.2 πππ ππ‘βππ)π1 + 4.3 πππ πππ = 4.085 πππ πππ + 2.53 πππ πππ 0.37 πππ πππ (6.2 πππ ππ‘βππ) πππ ππ‘βππ π1 = 0.37 πππ πππ 6.2 πππ ππ‘βππ + (6.2 πππ ππ‘βππ) πππ ππ‘βππ ππ = π. πππ π½π = π. πππ πππ Oil Composition (Oil kg/kg solβn)
πβ1=
π βπ ln (ππ β ππ ) π
π
π βπ ln ( ππ β ππ ) π π
=
π βπ ln (π1 β π1 ) π
π
π βπ ln ( ππ β π1 ) π 1
Xa
1
Ya
0.65
X1
0.65
Y1
0.277
Xb
0.055
0.272 β 0.65 ln ( ) 0 β 0.055 + 1 π= 0.055 β 0.65 ln ( 0 β 0.272 )
Yb
0
π΅ = π. ππ ππππππ
We use points in N-1 stages: Xb, Yb, X1, Y1
LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1
We can make a graph for the stages using Ponchon Savarit Method. COORDINATES FOR GRAPH N X/Y 23.26 1 25.74 0.055 -50.38 -0.11 28.57 0 20 0.2 17.24 0.3 14.70 0.4 12.36 0.5 10.1 0.6 8.33 0.68
πΏπ = 4.3 πππ; πΏπ = 3.885 πππ; π΅π = π΅π = 100 ππ ππ = ππ =
π΅π 100ππ = = 23.26 πΏπ 4.3 πππ
π΅π 100ππ = = 25.74 πΏπ 3.885 πππ
Coordinates for β πβ = ππβ =
π΅ 100 ππ = = β50.38 πΏπ β ππ 4.3 πππ β 6.285 πππ
πΏπ ππ β ππ ππ 4.3 πππ(1) β 6.285 πππ(0.65) = = β0.11 πΏπ β ππ 4.3 πππ β 6.285 πππ
π΅ = π. ππ ππππππ
____________________________________________________________________________ Observations: For problems dealing with variable countercurrent multistage leaching, the P-S graph yields twice the stages calculated using kremser equation. (both cases occurred in 12.10-4 and 23.4). But if we make a P-S diagram for N-1 stage, the approximation matches. In constant countercurrent multistage leaching, both the kremser equation and P-S diagrams yields the same approximation of N. The McCabe-thiele diagram also showed the same results.