Problem Set on Screening and Size Reduction Group
Problem Set on Screening and Size Reduction Group 5 Problems Assigned: 5 Ch. E – A • No. 5 Bigso, Jhullie Ann • Challenge Problem Magalong, Marinela Kris • No. 13 Micaller, Ian Kenneth • No. 18 Miranda, Kenneth Paulo • Problem from Lecture 2 A Slide No. 45 Tamayo, Benedict
Problem No. 5 Benedict Tamayo
Problem No. 5 Table salt is being fed to a vibrating screen at the rate of 3000 kg/hr. The desired product is the 28/100 mesh fraction. A -28 + 100 -mesh screens are therefore used, the feed being introduced on the 28 mesh screen, the product being discharged from the 100 -mesh screen. Calculate the effectiveness of the screens. What conclusions can you draw from the values computed versus the given proportion? Video source: https: //www. youtube. com/watch? v=pwg. RPZMGrj. E
Screen Mesh F Feed R Oversize P Product Q Undersize -10+14 0. 0003 0. 0008 -14+20 0. 0037 0. 0082 0. 0005 -20+28 0. 089 0. 0155 0. 00003 -28+35 0. 186 0. 389 0. 039 0. 00012 -35+48 0. 258 0. 377 0. 322 0. 0009 -48+65 0. 285 0. 176 0. 526 0. 0136 -65+100 0. 091 0. 025 0. 075 0. 34935 -100+150 0. 062 0. 004 0. 02 0. 299 -150+200 0. 025 0. 0011 0. 002 0. 337 1 1
+28 Streams x -28+100 Kg x Kg -100 x Q 0. 0000 3 0. 3639 7 0. 636 P 0. 016 0. 962 0. 022 R 0. 0279 0. 967 0. 0051 F 0. 093 0. 82 0. 087 TOTAL Kg Compute d, Kg
Given: F= 3000 kg F= Q + P + R 3000= Q + P + R - eqn 1 +28 Streams x -28+100 Kg x -100 Kg x Q 0. 00003 0. 36397 0. 636 P 0. 016 0. 962 0. 022 R 0. 0279 0. 967 0. 0051 F 0. 093 0. 82 0. 087 3000(0. 093)= Q(0. 00003) + P(0. 016) + R(0. 0279) -eqn 2 3000(0. 82)= Q(0. 36397) + P(0. 962) + R(0. 967) -eqn 3 3000(0. 087)= Q(0. 636) + P(0. 022) + R(0. 0051) -eqn 4 Solving simultaneously: Q(undersize) = 884. 56149 kg P(product) = -18, 483. 42257 kg R(oversize) = 20, 598. 86109 kg TOTAL Kg Computed, Kg
F 1= 279 F 2= 2460 F 3= 261 Streams Q (undersize) 0. 00003 0. 026537 0. 36397 321. 9538 0. 636 562. 5811 884. 5614859 P (product) Kg -295. 735 0. 962 -17781. 1 0. 022 -406. 635 -18483. 42257 R (oversize) 0. 0279 574. 7082 0. 967 19919. 1 0. 0051 105. 0542 20598. 86109 0. 82 2460 0. 087 261 3000. 000006 +28 Q 1=0. 026537 Q 2= 321. 9538 Q 3= 562. 5811 P 1= -295. 735 P 2= -17781. 1 P 3= -406. 635 x TOTAL Computed, Kg 0. 016 Kg -100 Kg R 1= 574. 7082 R 2= 19919. 1 R 3= 105. 0541 100 x -28+100 x F (feed) S 1= 574. 735 S 2= 20241. 1 S 3= 667. 635 +28 0. 093 279
Conclusion: With a computed value of efficiencies (very small) and obtained value of the product, Q (negative), the given proportions do not satisfy the type of sample screened. There is wrong with the stoichiometry of the given values.
Challenge Problem Marinela Kris Magalong
Challenge Problem It is desired to separate a mixture of crushed stone clinker in a rotary trommel to obtain 3 products D, C, and B passing through 150, 35, and 10 mesh screens respectively. Find he effectiveness of each screen using the screen analysis below. Video source: https: //www. youtube. com/watch? v=Iz. PWEL 6 Oj 8 s
MESH -3+4 -4+6 -6+8 -8+10 -10+14 -14+20 -20+28 -28+35 -35+48 -48+65 -65+100 -100+150 -150+170 -170+200 -200+270 FEED 2. 5 7. 5 12. 4 7. 4 21. 3 8. 2 7. 5 3. 8 8. 1 11. 6 5. 3 1. 4 2. 1 0. 7 0. 2 A 0. 087 0. 208 0. 417 0. 243 0. 045 B 0. 039 0. 01 0. 521 0. 156 0. 182 0. 052 0. 029 C 0. 075 0. 017 0. 061 0. 238 0. 395 0. 17 0. 041 0. 003 D 0. 088 0. 059 0. 588 0. 206 0. 059
Basis: 100 g of entering Feed OMB: 100 = A + B + C + D eq. 1 MB: +10: 29. 8 = 0. 955 A + 0. 059 B eq. 2 -10+35: 40. 8 = 0. 045 A + 0. 911 B + 0. 153 C -35+150: 26. 4 = 0. 029 B + 0. 844 C + 0. 147 D eq. 5 -150: 3 = 0. 003 C + 0. 853 D; D = 3000/853 – 3/853 C eq. 6 in eq. 1 100 = A + B + C + (3000/853 – 3/853 C) 82300/853 = A + B + 850/853 C eq. 7 eq. 3 eq. 4
Solving eq. 2, eq. 3, and eq. 7 simultaneously, A = 28. 8295 g C = 29. 3187 g B = 38. 4379 g D : 3000/853 – 3/853(29. 3187) = 3. 4139
F 1 = 29. 8 F 2 = 4038 F 3 = 26. 4 F 4 =3 E 1 = 2. 2678 E 2 = 39. 5027 E 3 = 26. 4 E 4 =3 G 1 = 0 G 2 = 4. 4858 G 3 = 25. 2853 G 4 =3 A 1 = 0. 955(28. 8295) = 27. 5332 A 2 = 0. 045(28. 8295) = 1. 2973 A 3 = 0 A 4 = 0 10 B 1 = 0. 059(38. 4379) = 2. 2678 B 2 = 0. 911(38. 4379) = 35. 0169 B 3 = 0. 029(38. 4379) = 1. 1147 B 4 = 0 35 C 1 = 0 C 2 = 0. 153(29. 3187) = 4. 4858 C 3 = 0. 844(29. 3187) = 24. 745 C 4 = 0. 003(29. 3187) = 0. 08796) D 1 =0 150 D 2 = 0 D 3 = 0. 147(3. 4139) = 0. 5018 D 4 = 0. 853(3. 4139) = 2. 912
Efficiency: Mesh 10: Mesh 35: Mesh 150: = 45. 03% = 79. 76% = 95. 31%
Problem No. 13 Ian Kenneth Micaller
Problem No. 13 Calculate the surface per unit volume in square centimeters per cubic centimeters (cm 2/cm 3) of shale having the screen analysis shown in the table. What is the volume surface mean diameter of the particles smaller than 65 mesh? Assume the shale to be of the same shape as mica flakes. Video source: https: //www. youtube. com/watch? v=1 Wf 1 q 5 FHdk 0 Mesh -3+4 -4+6 -6+8 -8+10 -10+14 -14+20 -20+28 -28+35 -35+48 -48+65 -65+100 -100+150 -150+200 -200 Cum Analysis 0 0. 12 0. 21 0. 22 0. 28 0. 42 0. 58 0. 65 0. 87 0. 88 0. 94 0. 96 0. 99 1
Problem No. 13 xi Dpi mean -3+4 -4+6 -6+8 -8+10 -10+14 -14+20 -20+28 -28+35 -35+48 -48+65 -65+100 Cum Analysis 0 0. 12 0. 21 0. 22 0. 28 0. 42 0. 58 0. 65 0. 87 0. 88 0. 94 0 0. 12 0. 09 0. 01 0. 06 0. 14 0. 16 0. 07 0. 22 0. 01 0. 06 5. 6895 4. 013 2. 8445 2. 0065 1. 4095 1. 0005 0. 711 0. 503 0. 356 0. 2515 0. 1775 xi/Dpi mean 0. 0000 0. 0299 0. 0316 0. 0050 0. 0426 0. 1399 0. 2250 0. 1392 0. 6180 0. 0398 0. 3380 -100+150 0. 96 0. 02 0. 1255 0. 1594 -150+200 0. 99 0. 03 0. 089 0. 3371 -200 1 0. 01 Total 0. 074 0. 1351 19. 177 2. 2406 Mesh Sample Calculation for -65+100 Mesh Excluding mesh -200 Source: Tyler Standard Sieve Scale Mesh 3 4 6 8 10 14 20 28 35 48 65 100 150 200 Dpi 6. 68 4. 699 3. 327 2. 362 1. 651 1. 168 0. 833 0. 589 0. 417 0. 295 0. 208 0. 147 0. 104 0. 074
Problem No. 13 From Table 28. 2 of MSH book, 7 th ed: Mica flakes: ɸs = 0. 28 Cum Analysis -65+100 0. 94 xi Dpi mean 0. 06 0. 1775 xi/Dpi mean 0. 3380 -100+150 0. 96 0. 02 0. 1255 0. 1594 -150+200 0. 99 0. 03 0. 089 0. 3371 -200 1 0. 074 0. 1351 Total 0. 9696 Mesh
Problem No. 18 Kenneth Paulo Miranda
Problem No. 18 Find the power requirement to dry grind 200 MT/day of phosphate rock from 10 mm particles to a crushed product with analysis as given in Problem 14. Use Bond’s Law. Video source: https: //www. youtube. com/watch? v=LHyg. Xfds. Pq. U Mesh -3+4 -4+6 -6+8 -8+10 -10+14 -14+20 -20+28 -28+35 -35+48 -48+65 -65+100 -100+150 -150+200 % 100 97. 3 94. 4 88. 5 62. 4 56. 1 44. 2 38. 8 21. 1 11. 7 5. 8 3. 2 0
Relevant Data for Computing Power: Mesh 3 4 6 8 10 14 20 28 35 48 65 100 150 200 Dpi 6. 68 4. 699 3. 327 2. 362 1. 651 1. 168 0. 833 0. 589 0. 417 0. 295 0. 208 0. 147 0. 104 0. 074 Cum Analysis 1 1 0. 973 0. 944 0. 885 0. 624 0. 561 0. 442 0. 388 0. 211 0. 117 0. 058 0. 032 0
Plotting Cum Analysis vs Dp 1. 2 1 Cum Analysis 0. 8 0. 6 0. 4 0. 2 0 0 1 2 Approximately = 1. 49 3 4 Dp 5 6 7 8
Solving: Dsa = 10 mm Dsb = 1. 49 mm (from plot) Wi = 9. 92 (from table 28. 2) m = 200 MT/hr = 8. 33333 MT/hr KB = 0. 3162 * 9. 92 = 3. 136704 P = 13. 1481 k. W
Problem from Lecture 2 A Slide No. 45 Jhullie Ann Bigso
Problem from Lecture 2 A Slide No. 45 Fine Dp where 80% of the particles of given size analysis in the table, passes through. Mesh -3+4 -4+6 -6+8 -8+10 -10+14 -14+20 -20+28 -28+35 -35+48 -48+65 -65+100 -100+150 -150+200 -200 pan Cum Analysis 0 0. 121 0. 214 0. 221 0. 280 0. 422 0. 585 0. 652 0. 871 0. 881 0. 924 0. 951 0. 982 0. 987 1. 000
Mesh 3 4 6 8 10 14 20 28 35 48 65 100 150 200 pan Dp 6. 680 4. 699 3. 327 2. 362 1. 651 1. 168 0. 833 0. 589 0. 417 0. 295 0. 208 0. 147 0. 104 0. 074 -- Cum Analysis 0 0. 121 0. 214 0. 221 0. 280 0. 422 0. 585 0. 652 0. 871 0. 881 0. 924 0. 951 0. 982 0. 987 1 Cum Analysis 1 0. 879 0. 786 0. 779 0. 720 0. 578 0. 415 0. 348 0. 129 0. 119 0. 076 0. 049 0. 018 0. 013 0
Dp = 2. 5 mm
Team Honor Code: We certify that each of us whose name appears below has contributed to the solution of this Problem Set and that no part of the solution was copied from other teams or other sources nor did we allow other teams to copy from us. Jhullie Ann Bigso Marinela Kris Magalong Ian Kenneth Micaller Kenneth Paulo Miranda Benedict Tamayo
Thank You!
- Slides: 31