Search
Home / Leslie Marie Red - Bs Che3A - Che 126 Pe 1: 1 1 A A B B [PDF]

Leslie Marie Red - Bs Che3A - Che 126 Pe 1: 1 1 A A B B [PDF]

  • 0 0 0
  • Suka dengan makalah ini dan mengunduhnya? Anda bisa menerbitkan file PDF Anda sendiri secara online secara gratis dalam beberapa menit saja! Sign Up

Leslie Marie Red - Bs Che3A - Che 126 Pe 1: 1 1 A A B B [PDF]

LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 12.10-3. Countercurrent Multistage Washing of Ore. A treated ore containing i

5 0 691 KB

Report DMCA / Copyright

DOWNLOAD FILE

Text 1 (A & B)
Text 1 (A & B)

0 0 53 KB Read more

B A B Ii
B A B Ii

0 0 96 KB Read more

Resume Solas Chapter 1 A B
Resume Solas Chapter 1 A B

0 0 367 KB Read more

Tugas Kelompok Ke-1 Week 3: A B A B A
Tugas Kelompok Ke-1 Week 3: A B A B A

0 0 115 KB Read more

A 1
A 1

0 0 4 MB Read more

Proposal Nadia 1 B-1
Proposal Nadia 1 B-1

0 0 194 KB Read more

Materi A Dan B
Materi A Dan B

2 0 81 KB Read more

A Pengantar - B
A Pengantar - B

0 0 55 KB Read more

Form A Dan B
Form A Dan B

2 0 184 KB Read more

B Indo Tugas 1
B Indo Tugas 1

1 0 99 KB Read more

File loading please wait...
Citation preview

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: 𝐿𝑏 =



𝐡𝑏 π‘˜π‘” π‘Žπ‘žπ‘’π‘’π‘œπ‘’π‘  π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› ; 𝑠𝑖𝑛𝑐𝑒 𝐡 π‘Žπ‘›π‘‘ 𝑁 𝑖𝑠 π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ ∴ 𝐿𝑏 = (10,000 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘) ( ) 𝑁𝑏 0.5 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘ 𝑳𝒃 = 𝟐𝟎, 𝟎𝟎𝟎 π’Œπ’ˆ



CuSO4 recovered: πΏπ‘Žπ‘‹π‘Ž(0.95) = 1,200 π‘˜π‘” 𝐢𝑒𝑆𝑂4 (0.95) = 1,140 π‘˜π‘” 𝐢𝑒𝑆𝑂4 CuSO4 retained: πΏπ‘Žπ‘‹π‘Ž(0.05) = 1,200 π‘˜π‘” 𝐢𝑒𝑆𝑂4 (0.05) = 60 π‘˜π‘” 𝐢𝑒𝑆𝑂4



We can now compute for Xb: 𝑋𝑏 =



60 π‘˜π‘” 𝐢𝑒𝑆𝑂4 20,000 π‘˜π‘” π‘Žπ‘žπ‘’π‘’π‘œπ‘’π‘  π‘ π‘œπ‘™β€²π‘› 𝑿𝒃 = 𝟎. πŸŽπŸŽπŸ‘



We can find the water in exit flow from multiplying the water-solute ratio to the amount of recovered solute. 92 π‘˜π‘” 𝐻2 𝑂 ) 8 π‘˜π‘” 𝐢𝑒𝑆𝑂4



H2O in exit flow: 1,140 π‘˜π‘” 𝐢𝑒𝑆𝑂4 (



We can now compute for Va and Ya:



= 13,110 π‘˜π‘” 𝐻2 𝑂



LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 π‘‰π‘Ž = 1,140 𝐾𝑔 𝐢𝑒𝑆𝑂4 + 13,110 π‘˜π‘” 𝐻2 𝑂 𝑽𝒂 = πŸπŸ’, πŸπŸ“πŸŽ π’Œπ’ˆ π‘Œπ‘Ž =



1,140 π‘˜π‘” 𝐢𝑒𝑆𝑂4 14,250 π‘˜π‘” π‘Žπ‘žπ‘’π‘’π‘œπ‘’π‘  π‘ π‘œπ‘™β€²π‘› 𝒀𝒂 = 𝟎. πŸŽπŸ–



OMB: πΏπ‘Ž + 𝑉𝑏 = 𝐿𝑏 + π‘‰π‘Ž (1,600 π‘˜π‘”) + 𝑉𝑏 = (20,000 π‘˜π‘”) + (14,250 π‘˜π‘”) 𝑽𝒃 = πŸ‘πŸ, πŸ”πŸ“πŸŽ π’Œπ’ˆ



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. 𝑉1 π‘Œ1 + πΏπ‘Ž π‘‹π‘Ž = 𝐿1 𝑋1 + π‘‰π‘Ž π‘Œπ‘Ž (32,650 π‘˜π‘” 𝐻2 𝑂)π‘Œ1 + (1,600 π‘˜π‘”)(0.75) = (20,000π‘˜π‘”)(0.08) + (14,250 π‘˜π‘”)(0.08)



LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 0.0472 π‘˜π‘” 𝐢𝑒𝑆𝑂4 (32,650 π‘˜π‘” 𝐻2 𝑂) π‘˜π‘” 𝐻2 𝑂 π‘Œ1 = 0.0472 π‘˜π‘” 𝐢𝑒𝑆𝑂4 32,650 π‘˜π‘” 𝐻2 𝑂 + (32,650 π‘˜π‘” 𝐻2 𝑂) π‘˜π‘” 𝐻2 𝑂 π’€πŸ = 𝟎. πŸŽπŸ’πŸ“πŸ π‘½πŸ = πŸ‘πŸ’, πŸπŸ—πŸ. πŸŽπŸ–



Oil Composition (Oil kg/kg sol’n) Xa



0.75



Ya



0.08



π‘βˆ’1=



π‘Œ βˆ’π‘‹ ln (π‘Œπ‘Ž βˆ’ π‘‹π‘Ž ) 𝑏



𝑏



𝑋 βˆ’ π‘‹π‘Ž ln ( 𝑏 ) π‘Œπ‘ βˆ’ π‘Œπ‘Ž



=



π‘Œ βˆ’π‘‹ ln (π‘Œ1 βˆ’ 𝑋1 ) 𝑏



𝑏



𝑋 βˆ’ 𝑋1 ln ( 𝑏 ) π‘Œπ‘ βˆ’ π‘Œ1



We use points in N-1 stages: Xb, Yb, X1, Y1 X1



0.08



Y1



0.0451



Xb



0.003



Yb



0



0.0451 βˆ’ 0.08 ln ( 0 βˆ’ 0.003 ) 𝑁= +1 0.003 βˆ’ 0.08 ln ( ) 0 βˆ’ 0.0451 𝑡 = πŸ“. πŸ“πŸ— β‰ˆ πŸ” π’”π’•π’‚π’ˆπ’†π’”



We can make a graph for the stages using Ponchon Savarit Method. πΏπ‘Ž = 1,600 π‘˜π‘” ; 𝐿𝑏 = 20,000 π‘˜π‘”; 𝐡1 = 𝐡𝑏 = 10,000 π‘˜π‘” π‘π‘Ž =



π΅π‘Ž 10,000 π‘˜π‘” = = 6.25 πΏπ‘Ž 1,600 π‘˜π‘”



𝑁𝑏 =



𝐡𝑏 10,000 π‘˜π‘” = = 0.5 𝐿𝑏 20,000 π‘˜π‘”



Coordinates for βˆ† π‘βˆ† = π‘‹π‘Žβˆ† =



N 6.25 0.5 -0.79 0.5 (constant)



𝐡 10,000 π‘˜π‘” = = βˆ’0.79 πΏπ‘Ž βˆ’ π‘‰π‘Ž 1,600 π‘˜π‘” βˆ’ 14,250 π‘˜π‘”



πΏπ‘Ž π‘‹π‘Ž βˆ’ π‘‰π‘Ž π‘Œπ‘Ž 1,600 π‘˜π‘”(0.75) βˆ’ 14,250(0.08) π‘˜π‘” = = βˆ’0.0047 πΏπ‘Ž βˆ’ π‘‰π‘Ž 1,600 π‘˜π‘” βˆ’ 14,250 π‘˜π‘” COORDINATES FOR GRAPH X/Y 0.75 0.003 -0.11



LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 Ponchon Savarit Diagram



𝑡 = πŸ“. πŸ— β‰ˆ πŸ” π’”π’•π’‚π’ˆπ’†π’”



Basing N-1, 𝐿1 = 𝐿𝑏 = 169.97 π‘˜π‘”; 𝐡1 = 𝐡𝑏 = 743 π‘˜π‘” 𝑁1 =



𝐡1 10,000 π‘˜π‘” = = 0.5 𝐿1 20,000 π‘˜π‘”



𝑁𝑏 =



𝐡𝑏 10,000 π‘˜π‘” = = 0.5 𝐿𝑏 20,000 π‘˜π‘”



Coordinates for βˆ† π‘βˆ† = π‘‹π‘Žβˆ† =



𝐡 10,000 π‘˜π‘” = = βˆ’0.7 𝐿1 βˆ’ 𝑉1 20,000 π‘˜π‘” βˆ’ 34,191.08 π‘˜π‘”



𝐿1 𝑋1 βˆ’ 𝑉1 π‘Œ1 20,000 π‘˜π‘” (0.08) βˆ’ 34,191.08 π‘˜π‘”(0.0451) = = βˆ’0.004 𝐿1 βˆ’ 𝑉1 20,000 π‘˜π‘” βˆ’ 34,191.08 π‘˜π‘”



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. 𝑁 βˆ’ 1 = 4.6 π‘ π‘‘π‘Žπ‘”π‘’π‘  𝑡 = πŸ“. πŸ” β‰ˆ πŸ” π’”π’•π’‚π’ˆπ’†π’”



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: πΏπ‘Žπ‘‹π‘Ž(0.95) = 257 π‘˜π‘” π‘œπ‘–π‘™ (0.95) = 244.15 π‘˜π‘” π‘œπ‘–π‘™ Oil retained: πΏπ‘Žπ‘‹π‘Ž(0.05) = 257 π‘˜π‘” π‘œπ‘–π‘™ (0.05) = 12.85 π‘˜π‘” π‘œπ‘–π‘™



We can now compute for Va: π‘‰π‘Ž = 244.15 π‘˜π‘” π‘œπ‘–π‘™ (



1 π‘˜π‘” π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› ) 0.7 π‘˜π‘” π‘œπ‘–π‘™



𝑽𝒂 = πŸ‘πŸ’πŸ–. πŸ•πŸ— π’Œπ’ˆ Oil Balance 𝑉𝑏 π‘Œπ‘ + πΏπ‘Ž π‘‹π‘Ž = 𝐿𝑏 𝑋𝑏 + π‘‰π‘Ž π‘Œπ‘Ž 𝑉𝑏 (0) + 257 π‘˜π‘” = 𝐿𝑏 𝑋𝑏 + 244.15 π‘˜π‘”



LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1



𝑿𝒃 =



𝟏𝟐. πŸ–πŸ“ π’Œπ’ˆ 𝑳𝒃



𝟏 𝑳𝒃 = πŸ•πŸ’πŸ‘ π’Œπ’ˆ π’Šπ’π’†π’“π’• π’”π’π’π’Šπ’… ( ) 𝑡 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: 𝑋𝑏 = 0.2, 𝑁 = 3.5 𝐿𝑏 = 743 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘ ( 𝑋𝑏 =



1 ) = 212.29 π‘˜π‘” 3.5



12.85 π‘˜π‘” = 0.06 212.29 π‘˜π‘”



Trial 2: 𝑋𝑏 = 0.4, 𝑁 = 2.47 𝐿𝑏 = 743 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘ ( 𝑋𝑏 =



1 ) = 300.8 π‘˜π‘” 2.47



12.85 π‘˜π‘” = 0.043 ; π‘‘β„Žπ‘’ π‘’π‘Ÿπ‘Ÿπ‘œπ‘Ÿ π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’π‘‘ 300.8 π‘˜π‘”



From trial 2 we can assume that 0 > 𝑋𝑏 > 0.2 Trial 3: 𝑋𝑏 = 0.1, 𝑁 = 4.19 𝐿𝑏 = 743 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘ ( 𝑋𝑏 =



1 ) = 177.33 π‘˜π‘” 4.19



12.85 π‘˜π‘” = 0.07 177.33 π‘˜π‘”



LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1



Trial 4: 𝑋𝑏 = 0.05, 𝑁 = 4.535 𝐿𝑏 = 743 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘ ( 𝑋𝑏 =



1 ) = 163.84 π‘˜π‘” 4.535



12.85 π‘˜π‘” = 0.078 163.84 π‘˜π‘”



Trial 5: 𝑋𝑏 = 0.03, 𝑁 = 4.673 𝐿𝑏 = 743 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘ ( 𝑋𝑏 =



1 ) = 101.22 π‘˜π‘” 4.673



12.85 π‘˜π‘” = 0.12 ; 𝑋𝑏 π‘šπ‘’π‘ π‘‘ 𝑏𝑒 π‘”π‘Ÿπ‘’π‘Žπ‘‘π‘’π‘Ÿ π‘‘β„Žπ‘Žπ‘› 0.05 101.22 π‘˜π‘”



Trial 6: 𝑋𝑏 = 0.08, 𝑁 = 4.328 𝐿𝑏 = 743 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘ ( 𝑋𝑏 =



1 ) = 171.67 π‘˜π‘” 4.328



12.85 π‘˜π‘” = 0.075 171.67 π‘˜π‘”



Trial 7: 𝑋𝑏 = 0.078, 𝑁 = 4.357 𝐿𝑏 = 743 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘ ( 𝑋𝑏 =



1 ) = 170.53 π‘˜π‘” 4.357



12.85 π‘˜π‘” = 0.075 170.53 π‘˜π‘”



Trial 8: 𝑋𝑏 = 0.076, 𝑁 = 4.369 𝐿𝑏 = 743 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘ ( 𝑋𝑏 =



1 ) = 170.06 π‘˜π‘” 4.369



12.85 π‘˜π‘” = 0.0756 170.06 π‘˜π‘”



Trial 8: 𝑋𝑏 = 0.0758, 𝑁 = 4.3702 𝐿𝑏 = 743 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘ ( 𝑋𝑏 =



1 ) = 170.02 π‘˜π‘” 4.3702



12.85 π‘˜π‘” = 0.0756 170.06 π‘˜π‘”



Trial 10: 𝑋𝑏 = 0.0756, 𝑁 = 4.3714 𝐿𝑏 = 743 π‘˜π‘” π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘ (



1 ) = 169.97 π‘˜π‘” 4.3714



LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1



𝑋𝑏 =



12.85 π‘˜π‘” = 0.0756 169.99 π‘˜π‘”



𝑳𝒃 = πŸπŸ”πŸ—. πŸ—πŸ• π’Œπ’ˆ 𝑿𝒃 = 𝟎. πŸŽπŸ•πŸ“πŸ” OMB: 𝑉𝑏 + πΏπ‘Ž = π‘‰π‘Ž + 𝐿𝑏 𝑉𝑏 + 257 π‘˜π‘” = 348.79 π‘˜π‘” + 169.97 π‘˜π‘” 𝑽𝒃 = πŸπŸ”πŸ. πŸ•πŸ” π’Œπ’ˆ 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. 𝑉1 π‘Œ1 + πΏπ‘Ž π‘‹π‘Ž = π‘‰π‘Ž π‘Œπ‘Ž + 𝐿1 𝑋1 (261.76 π‘˜π‘” π‘’π‘‘β„Žπ‘¦π‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ)π‘Œ1 + 257π‘˜π‘” π‘œπ‘–π‘™ = 244.15 π‘˜π‘” π‘œπ‘–π‘™ + 118.98 π‘˜π‘” π‘œπ‘–π‘™ 0.405 π‘˜π‘” π‘œπ‘–π‘™ (261.76 π‘˜π‘” π‘’π‘‘β„Žπ‘¦π‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ) π‘˜π‘” π‘’π‘‘β„Žπ‘¦π‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ π‘Œ1 = 0.405 π‘˜π‘” π‘œπ‘–π‘™ 261.76 π‘˜π‘” π‘’π‘‘β„Žπ‘¦π‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ + (261.76 π‘˜π‘” π‘’π‘‘β„Žπ‘¦π‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ) π‘˜π‘” π‘’π‘‘β„Žπ‘¦π‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ π’€πŸ = 𝟎. πŸπŸ–πŸ– π‘½πŸ = πŸ‘πŸ”πŸ•. πŸ•πŸ• π’Œπ’ˆ Oil Composition (Oil kg/kg sol’n)



π‘βˆ’1=



π‘Œ βˆ’π‘‹ ln (π‘Œπ‘Ž βˆ’ π‘‹π‘Ž ) 𝑏



𝑏



=



π‘Œ βˆ’π‘‹ ln (π‘Œ1 βˆ’ 𝑋1 ) 𝑏



𝑏



Xa



1



𝑋 βˆ’π‘‹ ln ( π‘Œπ‘ βˆ’ π‘Œπ‘Ž ) 𝑏 π‘Ž



𝑋 βˆ’π‘‹ ln ( π‘Œπ‘ βˆ’ π‘Œ1 ) 𝑏 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 βˆ’ 0.7 ln ( ) 0 βˆ’ 0.0756 + 1 𝑁= 0.0756 βˆ’ 0.7 ln ( 0 βˆ’ 0.288 )



Yb



0



𝑡 = πŸ‘. πŸπŸ— β‰ˆ πŸ‘. 𝟐 π’”π’•π’‚π’ˆπ’†π’”



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



πΏπ‘Ž = 257 π‘˜π‘”; 𝐿𝑏 = 169.97 π‘˜π‘”; 𝐡1 = 𝐡𝑏 = 743 π‘˜π‘” π‘π‘Ž = 𝑁𝑏 =



π΅π‘Ž 743 π‘˜π‘” = = 2.89 πΏπ‘Ž 257 π‘˜π‘”



𝐡𝑏 743 π‘˜π‘” = = 4.37 𝐿𝑏 169.97 π‘˜π‘”



Coordinates for βˆ† π‘βˆ† = π‘‹π‘Žβˆ† =



𝐡 743 π‘˜π‘” = = βˆ’8.09 πΏπ‘Ž βˆ’ π‘‰π‘Ž 257 π‘˜π‘” βˆ’ 348.79π‘˜π‘”



πΏπ‘Ž π‘‹π‘Ž βˆ’ π‘‰π‘Ž π‘Œπ‘Ž 257 π‘˜π‘”(1) βˆ’ 348.79 π‘˜π‘”(0.7) = = βˆ’0.14 πΏπ‘Ž βˆ’ π‘‰π‘Ž 257 π‘˜π‘” βˆ’ 348.79π‘˜π‘”



𝑡 = πŸ”. 𝟐 π’”π’•π’‚π’ˆπ’†π’”



LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 For N-1: 𝐿1 = 𝐿𝑏 = 169.97 π‘˜π‘”; 𝐡1 = 𝐡𝑏 = 743 π‘˜π‘” 𝑁1 =



𝐡1 743 π‘˜π‘” = = 4.37 𝐿1 169.97 π‘˜π‘”



𝑁𝑏 =



𝐡𝑏 743 π‘˜π‘” = = 4.37 𝐿𝑏 169.97 π‘˜π‘”



Coordinates for βˆ† π‘βˆ† = π‘‹π‘Žβˆ† =



𝐡 743 π‘˜π‘” = = βˆ’3.75 𝐿1 βˆ’ 𝑉1 169.97 π‘˜π‘” βˆ’ 367.77π‘˜π‘”



𝐿1 𝑋1 βˆ’ 𝑉1 π‘Œ1 169.97 π‘˜π‘”(0.7) βˆ’ 367.77π‘˜π‘”(0.288) = = βˆ’0.066 𝐿1 βˆ’ 𝑉1 169.97 π‘˜π‘” βˆ’ 367.77π‘˜π‘”



L1



PS diagram for N-1 𝑁 βˆ’ 1 = 2.3 π‘ π‘‘π‘Žπ‘”π‘’π‘  𝑡 = πŸ‘. πŸ‘ ; 𝒄𝒍𝒐𝒔𝒆 𝒕𝒐 𝒂𝒑𝒑𝒓𝒐𝒙



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: πΏπ‘Žπ‘‹π‘Ž(0.95) = 4.3 π‘”π‘Žπ‘™ π‘œπ‘–π‘™ (0.95) = 4.085 π‘”π‘Žπ‘™ π‘œπ‘–π‘™ Oil retained: πΏπ‘Žπ‘‹π‘Ž(0.05) = 257 π‘˜π‘” π‘œπ‘–π‘™ (0.05) = 0.215 π‘”π‘Žπ‘™ π‘œπ‘–π‘™ We can now compute for Va: π‘‰π‘Ž = 4.085 π‘˜π‘” π‘œπ‘–π‘™ (



1 π‘”π‘Žπ‘™ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› ) 0.65 π‘”π‘Žπ‘™ π‘œπ‘–π‘™



𝑽𝒂 = πŸ”. πŸπŸ–πŸ“ π’ˆπ’‚π’



LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1 Oil Balance 𝑉𝑏 π‘Œπ‘ + πΏπ‘Ž π‘‹π‘Ž = 𝐿𝑏 𝑋𝑏 + π‘‰π‘Ž π‘Œπ‘Ž 𝑉𝑏 (0) + 4.3 π‘”π‘Žπ‘™ = 𝐿𝑏 𝑋𝑏 + 4.085π‘”π‘Žπ‘™ 𝑿𝒃 =



𝟎. πŸπŸπŸ“ π’ˆπ’‚π’ 𝑳𝒃



𝑳𝒃 = 𝟏𝟎𝟎 𝒍𝒃 π’Šπ’π’†π’“π’• π’”π’π’π’Šπ’…(𝑯) 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: 𝑋𝑏 = 0.1, 𝐻 = 0.042 𝐿𝑏 = 100 𝑙𝑏 π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘(0.042) = 4.2 π‘”π‘Žπ‘™ 𝑋𝑏 =



0.215 π‘”π‘Žπ‘™ = 0.051 4.2 π‘”π‘Žπ‘™



Trial 2: 𝑋𝑏 = 0.2, 𝐻 = 0.05 𝐿𝑏 = 100 𝑙𝑏 π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘(0.05) = 5 π‘”π‘Žπ‘™ 𝑋𝑏 =



0.215 π‘”π‘Žπ‘™ = 0.043; π‘‘β„Žπ‘’ π‘’π‘Ÿπ‘Ÿπ‘œπ‘Ÿ π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’π‘‘ 5 π‘”π‘Žπ‘™



From trial 2 we can assume that 0 > 𝑋𝑏 > 0.1 Trial 3: 𝑋𝑏 = 0.05, 𝑁 = 0.0385 𝐿𝑏 = 100 𝑙𝑏 π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘(0.0385) = 3.85 π‘”π‘Žπ‘™ 𝑋𝑏 =



0.215 π‘”π‘Žπ‘™ = 0.056 3.85 π‘”π‘Žπ‘™



Trial 4: 𝑋𝑏 = 0.055, 𝑁 = 0.03885 𝐿𝑏 = 100 𝑙𝑏 π‘–π‘›π‘’π‘Ÿπ‘‘ π‘ π‘œπ‘™π‘–π‘‘(0.03885) = 3.885 π‘”π‘Žπ‘™



LESLIE MARIE RED | BS CHE3A | CHE 126 PE 1



𝑋𝑏 =



0.215 π‘”π‘Žπ‘™ = 0.055 3.885 π‘”π‘Žπ‘™



𝑳𝒃 = πŸ‘. πŸ–πŸ–πŸ“ π’ˆπ’‚π’ 𝑿𝒃 = 𝟎. πŸŽπŸ“πŸ“



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: 𝑉𝑏 + πΏπ‘Ž = π‘‰π‘Ž + 𝐿𝑏 𝑉𝑏 + 4.3 π‘”π‘Žπ‘™ = 6.615 π‘”π‘Žπ‘™ + 3.885 π‘”π‘Žπ‘™ 𝑽𝒃 = πŸ”. 𝟐 𝒐𝒇 𝒆𝒕𝒉𝒆𝒓 Solving for Y1 we make an oil balance on the first stage. 𝑉1 π‘Œ1 + πΏπ‘Ž π‘‹π‘Ž = π‘‰π‘Ž π‘Œπ‘Ž + 𝐿1 𝑋1 (6.2 π‘”π‘Žπ‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ)π‘Œ1 + 4.3 π‘”π‘Žπ‘™ π‘œπ‘–π‘™ = 4.085 π‘”π‘Žπ‘™ π‘œπ‘–π‘™ + 2.53 π‘”π‘Žπ‘™ π‘œπ‘–π‘™ 0.37 π‘”π‘Žπ‘™ π‘œπ‘–π‘™ (6.2 π‘”π‘Žπ‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ) π‘”π‘Žπ‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ π‘Œ1 = 0.37 π‘”π‘Žπ‘™ π‘œπ‘–π‘™ 6.2 π‘”π‘Žπ‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ + (6.2 π‘”π‘Žπ‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ) π‘”π‘Žπ‘™ π‘’π‘‘β„Žπ‘’π‘Ÿ π’€πŸ = 𝟎. πŸπŸ•πŸ π‘½πŸ = πŸ–. πŸ’πŸ—πŸ’ π’ˆπ’‚π’ Oil Composition (Oil kg/kg sol’n)



π‘βˆ’1=



π‘Œ βˆ’π‘‹ ln (π‘Œπ‘Ž βˆ’ π‘‹π‘Ž ) 𝑏



𝑏



𝑋 βˆ’π‘‹ ln ( π‘Œπ‘ βˆ’ π‘Œπ‘Ž ) 𝑏 π‘Ž



=



π‘Œ βˆ’π‘‹ ln (π‘Œ1 βˆ’ 𝑋1 ) 𝑏



𝑏



𝑋 βˆ’π‘‹ ln ( π‘Œπ‘ βˆ’ π‘Œ1 ) 𝑏 1



Xa



1



Ya



0.65



X1



0.65



Y1



0.277



Xb



0.055



0.272 βˆ’ 0.65 ln ( ) 0 βˆ’ 0.055 + 1 𝑁= 0.055 βˆ’ 0.65 ln ( 0 βˆ’ 0.272 )



Yb



0



𝑡 = πŸ‘. πŸ’πŸ” π’”π’•π’‚π’ˆπ’†π’”



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



πΏπ‘Ž = 4.3 π‘”π‘Žπ‘™; 𝐿𝑏 = 3.885 π‘”π‘Žπ‘™; π΅π‘Ž = 𝐡𝑏 = 100 𝑙𝑏 π‘π‘Ž = 𝑁𝑏 =



π΅π‘Ž 100𝑙𝑏 = = 23.26 πΏπ‘Ž 4.3 π‘”π‘Žπ‘™



𝐡𝑏 100𝑙𝑏 = = 25.74 𝐿𝑏 3.885 π‘”π‘Žπ‘™



Coordinates for βˆ† π‘βˆ† = π‘‹π‘Žβˆ† =



𝐡 100 𝑙𝑏 = = βˆ’50.38 πΏπ‘Ž βˆ’ π‘‰π‘Ž 4.3 π‘”π‘Žπ‘™ βˆ’ 6.285 π‘”π‘Žπ‘™



πΏπ‘Ž π‘‹π‘Ž βˆ’ π‘‰π‘Ž π‘Œπ‘Ž 4.3 π‘”π‘Žπ‘™(1) βˆ’ 6.285 π‘”π‘Žπ‘™(0.65) = = βˆ’0.11 πΏπ‘Ž βˆ’ π‘‰π‘Ž 4.3 π‘”π‘Žπ‘™ βˆ’ 6.285 π‘”π‘Žπ‘™



𝑡 = πŸ”. πŸ•πŸ“ π’”π’•π’‚π’ˆπ’†π’”



____________________________________________________________________________ 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.