Tiara Dwi Aisha – 2006578614 Pengolahan Mineral – 03
Jawab:
a. Buatlah tabulasi neraca material (material balance) shift 1 Diketahui:
M(F) = 210 ton A(F) = 2.5%
A(C) = 45%
A(T) = 0.2%
Maka,
• Mencari nilai Massa Konsentrat (M(C)) 𝑀(𝐶)
𝑀(𝐹) =𝐴(𝐹)− 𝐴(𝑇) 𝐴(𝐶)− 𝐴(𝑇) 𝑀(𝐶)
210 𝑡𝑜𝑛= 2.5% − 0.2%
45% − 0.2%
𝑀(𝐶) = 2.3
44.8× 210 𝑡𝑜𝑛 𝑴(𝑪) = 𝟏𝟎. 𝟕𝟖𝟏𝟐𝟓 𝒕𝒐𝒏
• Mencari nilai Massa Tailing (M(T)) M(F) = M(C) + M(T)
M(T) = M(F) – M(C)
M(T) = 210 ton – 10.78125 ton M(T) = 199.21875 ton
Stream Mass of Stream (M), ton
Metal Assay (A), %
Mass of Metal, (MxA), ton
Distribution ((MxA)/∑)x100%
Feed 210 2.5 5.25 100%
Concentrate 10.78125 45 4.8515625 (4.8515625/5.25)x100%
= 92.411%
Tailings 199.21875 0.2 0.3984375 (0.3984375/5.25)x100%
= 7.589%
b. Buatlah tabulasi neraca material (material balance) shift 2 Diketahui:
M(F) = 300 ton A(F) = 2%
A(C) = 40%
A(T) = 0.15%
Maka,
• Mencari nilai Massa Konsentrat (M(C)) 𝑀(𝐶)
𝑀(𝐹) =𝐴(𝐹)− 𝐴(𝑇) 𝐴(𝐶)− 𝐴(𝑇) 𝑀(𝐶)
300 𝑡𝑜𝑛= 2% − 0.15%
40% − 0.15%
𝑀(𝐶) = 1.85
39.85× 300 𝑡𝑜𝑛 𝑴(𝑪) = 𝟏𝟑. 𝟗𝟐𝟕 𝒕𝒐𝒏
• Mencari nilai Massa Tailing (M(T)) M(F) = M(C) + M(T)
M(T) = M(F) – M(C)
M(T) = 300 ton – 13.927 ton M(T) = 286.073 ton
Stream Mass of Stream (M),
ton
Metal Assay (A),
%
Mass of Metal, (MxA), ton
Distribution ((MxA)/∑)x100%
Feed 300 2 6 100%
Concentrate 13.927 40 5.5708 (5.5708/6)x100%
= 92.847%
Tailings 286.073 0.15 0.4291 (0.4291/6)x100%
= 7.15167%
c. Buat tabulasi neraca material gabungan M(F) = 210 ton + 300 ton = 510 ton 𝐴(𝐹) = (2.5% × 210) + (2% × 300)
510 = 2.20588%
𝐴(𝐶) = (45% × 210) + (40% × 300)
510 = 42.0588%
𝐴(𝑇) =(0.2% × 210) + (0.15% × 300)
510 = 0.170588%
• Mencari nilai Massa Konsentrat
𝑀(𝐶)
𝑀(𝐹) =2.20588% − 0.170588%
42.0588% − 0.170588%
𝑀(𝐶)
510 𝑡𝑜𝑛= 2.20588% − 0.170588%
42.0588% − 0.170588%
𝑀(𝐶) = 2.035292
41.888212× 510 𝑡𝑜𝑛 𝑴(𝑪) = 𝟐𝟒. 𝟕𝟖𝟎𝟐𝟏𝟓𝟓 𝒕𝒐𝒏
• Mencari nilai Massa Tailing (M(T)) M(F) = M(C) + M(T)
M(T) = M(F) – M(C)
M(T) = 510 ton – 24.7802155 ton M(T) = 485.2197845 ton
Stream Mass of Stream (M),
ton
Metal Assay (A), %
Mass of Metal, (MxA),
ton
Distribution ((MxA)/∑)x100%
Feed 510 2.20588 11.24998 8
100%
Concentrat e
24.7802155 42.0588 10.42226 (10.42226/11.249988)x100
% = 92.6424%
Tailings 485.219784 5
0.17058 8
0.827726 7
(0.8277267/11.249988)x100
% = 7.3576%
d. Tentukan: (i) recovery, (ii) rasio pengayaan, (iii) rasio konsentrasi untuk neraca material shift 1, shift 2 dan gabungan.
• Shift 1
- 𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑦 = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑀𝑒𝑡𝑎𝑙(𝐶)
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑀𝑒𝑡𝑎𝑙(𝐹)× 100% =4.8515625
5.25 × 100%
𝑹𝒆𝒄𝒐𝒗𝒆𝒓𝒚 = 𝟗𝟐. 𝟒𝟏𝟏%
- 𝑅𝑎𝑠𝑖𝑜 𝑝𝑒𝑛𝑔𝑎𝑦𝑎𝑎𝑛 =%𝑀𝑒𝑡𝑎𝑙 𝐴𝑠𝑠𝑎𝑦(𝐶)
%𝑀𝑒𝑡𝑎𝑙 𝐴𝑠𝑠𝑎𝑦(𝐹)= 45
2.5
𝑹𝒂𝒔𝒊𝒐 𝒑𝒆𝒏𝒈𝒂𝒚𝒂𝒂𝒏 = 𝟏𝟖 - 𝑅𝑎𝑠𝑖𝑜 𝑘𝑜𝑛𝑠𝑒𝑛𝑡𝑟𝑎𝑠𝑖 =𝑀(𝐹)
𝑀(𝐶)= 210
10.78125
𝑹𝒂𝒔𝒊𝒐 𝒌𝒐𝒏𝒔𝒆𝒏𝒕𝒓𝒂𝒔𝒊 = 𝟏𝟗. 𝟒𝟕𝟖𝟐𝟔
• Shift 2
- 𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑦 = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑀𝑒𝑡𝑎𝑙(𝐶)
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑀𝑒𝑡𝑎𝑙(𝐹)× 100% =5.5708
6 × 100%
𝑹𝒆𝒄𝒐𝒗𝒆𝒓𝒚 = 𝟗𝟐. 𝟖𝟒𝟕%
- 𝑅𝑎𝑠𝑖𝑜 𝑝𝑒𝑛𝑔𝑎𝑦𝑎𝑎𝑛 =%𝑀𝑒𝑡𝑎𝑙 𝐴𝑠𝑠𝑎𝑦(𝐶)
%𝑀𝑒𝑡𝑎𝑙 𝐴𝑠𝑠𝑎𝑦(𝐹)= 40
2
𝑹𝒂𝒔𝒊𝒐 𝒑𝒆𝒏𝒈𝒂𝒚𝒂𝒂𝒏 = 𝟐𝟎 - 𝑅𝑎𝑠𝑖𝑜 𝑘𝑜𝑛𝑠𝑒𝑛𝑡𝑟𝑎𝑠𝑖 =𝑀(𝐹)
𝑀(𝐶)= 300
13.927
𝑹𝒂𝒔𝒊𝒐 𝒌𝒐𝒏𝒔𝒆𝒏𝒕𝒓𝒂𝒔𝒊 = 𝟐𝟏. 𝟓𝟒𝟏
• Gabungan
- 𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑦 = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑀𝑒𝑡𝑎𝑙(𝐶)
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑀𝑒𝑡𝑎𝑙(𝐹)× 100% = 10.42226
11.249988× 100%
𝑹𝒆𝒄𝒐𝒗𝒆𝒓𝒚 = 𝟗𝟐. 𝟔𝟒𝟐𝟒%
- 𝑅𝑎𝑠𝑖𝑜 𝑝𝑒𝑛𝑔𝑎𝑦𝑎𝑎𝑛 =%𝑀𝑒𝑡𝑎𝑙 𝐴𝑠𝑠𝑎𝑦(𝐶)
%𝑀𝑒𝑡𝑎𝑙 𝐴𝑠𝑠𝑎𝑦(𝐹)= 42.0588
2.20588
𝑹𝒂𝒔𝒊𝒐 𝒑𝒆𝒏𝒈𝒂𝒚𝒂𝒂𝒏 = 𝟏𝟗. 𝟎𝟔𝟔𝟔𝟕𝟔 - 𝑅𝑎𝑠𝑖𝑜 𝑘𝑜𝑛𝑠𝑒𝑛𝑡𝑟𝑎𝑠𝑖 =𝑀(𝐹)
𝑀(𝐶)= 510
24.7802155
𝑹𝒂𝒔𝒊𝒐 𝒌𝒐𝒏𝒔𝒆𝒏𝒕𝒓𝒂𝒔𝒊 = 𝟐𝟎. 𝟓𝟖𝟏
