Screening and Size Reduction Group Problem Set Analysis

problem set on screening and size reduction group n.w
1 / 31
Embed
Share

Analyzing a problem set on screening and size reduction involving the use of vibrating screens for table salt separation, determining screen effectiveness, and drawing conclusions from computed values. The solution involves simultaneous equations and calculations for undersize, product, and oversize quantities.

  • Problem Set
  • Screening
  • Size Reduction
  • Vibrating Screens
  • Analysis
rayaanacharya
bo_131 Follow

Uploaded on | 0 Views

Download Presentation

Please find below an Image/Link to download the presentation.

The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.

You are allowed to download the files provided on this website for personal or commercial use, subject to the condition that they are used lawfully. All files are the property of their respective owners.

Download Presentation

The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.

E N D

Presentation Transcript

  1. Problem Set on Screening and Size Reduction Group 5 5ChE A Bigso, Jhullie Ann Magalong, Marinela Kris Micaller, Ian Kenneth Miranda, Kenneth Paulo Tamayo, Benedict Problems Assigned: No. 5 Challenge Problem No. 13 No. 18 Problem from Lecture 2A Slide No. 45

  2. Problem No. 5 Benedict Tamayo

  3. 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=pwgRPZMGrjE

  4. F R P Q Screen Mesh -10+14 -14+20 -20+28 -28+35 -35+48 -48+65 -65+100 -100+150 -150+200 Feed Oversize Product Undersize 0.0003 0.0037 0.089 0.186 0.258 0.285 0.091 0.062 0.025 1 0.0008 0.0082 0.0189 0.389 0.377 0.176 0.025 0.004 0.0011 1 0.0005 0.0155 0.039 0.322 0.526 0.075 0.02 0.002 1 0.00003 0.00012 0.0009 0.0136 0.34935 0.299 0.337 1

  5. +28 -28+100 -100 TOTAL Comput ed, Kg Streams x Kg x Kg x Kg 0.0000 3 0.016 0.3639 7 0.962 0.636 Q P 0.022 R 0.0279 0.967 0.0051 F 0.093 0.82 0.087

  6. Given: F= 3000 kg +28 -28+100 -100 TOTAL Computed , Kg Streams x Kg x Kg x Kg Q 0.00003 0.36397 0.636 P 0.016 0.962 0.022 F= Q + P + R 3000= Q + P + R - eqn 1 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

  7. F1= 279 F2= 2460 F3= 261 +28 -28+100 -100 TOTAL Streams x Kg x Kg x Kg Computed, Kg Q (undersize) 0.00003 0.026537 0.36397 321.9538 0.636 562.5811 884.5614859 P (product) 0.016 -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 F (feed) 0.093 279 0.82 2460 0.087 261 3000 R1= 574.7082 R2= 19919.1 R3= 105.0541 3000.000006 S1= 574.735 S2= 20241.1 S3= 667.635 +28 Q1=0.026537 Q2= 321.9538 Q3= 562.5811 100 P1= -295.735 P2= -17781.1 P3= -406.635

  8. For +28: ? ?= 0.093 0.0279 0.0279 0.00003= 2.335844995 ? ?= 0.093 0.00003 0.0279 0.00003= 3.368478261 ?? ?? ? =? ? ? 1 ?? 1 ?? 0.00003 0.093 1 0.0279 1 0.093 = 2.3358 3.3685 = 0.0027203 ? For +100: ? ?=0.087 0.0051 ? ?= ? =? ? ?? 0.636 0.0051= 0.1298145506 0.636 0.087 0.636 0.0051= 0.8701854494 ?? ? ? 1 ?? 1 ?? 0.0051 0.087 1 0.636 1 0.087 = 0.1298 0.8702 = 0.0026398

  9. 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.

  10. Challenge Problem Marinela Kris Magalong

  11. 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=IzPWEL6Oj8s

  12. 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 B C D 0.087 0.208 0.417 0.243 0.045 0.039 0.01 0.01 0.521 0.156 0.182 0.052 0.029 0.075 0.017 0.061 0.238 0.395 0.17 0.041 0.003 0.088 0.059 0.588 0.206 0.059

  13. Basis: 100g of entering Feed OMB: 100 = A + B + C + D MB: +10: 29.8 = 0.955A + 0.059B -10+35: 40.8 = 0.045A + 0.911B + 0.153C -35+150: 26.4 = 0.029B + 0.844C + 0.147D -150: 3 = 0.003C + 0.853D; D = 3000/853 3/853C eq. 6 in eq. 1 100 = A + B + C + (3000/853 3/853C) 82300/853 = A + B + 850/853C eq. 1 eq. 2 eq. 3 eq. 4 eq. 5 eq. 6 eq. 7

  14. Solving eq. 2, eq.3, and eq. 7 simultaneously, A = 28.8295g C = 29.3187g B = 38.4379g D : 3000/853 3/853(29.3187) = 3.4139

  15. F1 = 29.8 F2 = 4038 F3 = 26.4 F4 =3 A1 = 0.955(28.8295) = 27.5332 A2 = 0.045(28.8295) = 1.2973 A3 = 0 A4 = 0 E1 = 2.2678 E2 = 39.5027 E3 = 26.4 E4 =3 10 B1 = 0.059(38.4379) = 2.2678 B2 = 0.911(38.4379) = 35.0169 B3 = 0.029(38.4379) = 1.1147 B4 = 0 G1 = 0 G2 = 4.4858 G3 = 25.2853 G4 =3 35 C1 = 0 C2 = 0.153(29.3187) = 4.4858 C3 = 0.844(29.3187) = 24.745 C4 = 0.003(29.3187) = 0.08796) D1 =0 D2 = 0 D3 = 0.147(3.4139) = 0.5018 D4 = 0.853(3.4139) = 2.912 150

  16. Efficiency: Mesh 10: = 45.03% Mesh 35: = 79.76% Mesh 150: = 95.31%

  17. Problem No. 13 Ian Kenneth Micaller

  18. Problem No. 13 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 Calculate the surface per unit volume in square centimeters per cubic centimeters (cm2/cm3) 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=1Wf1q5FHdk0

  19. Problem No. 13 Cum Analysis 0 0.12 0.21 0.22 0.28 0.42 0.58 0.65 0.87 0.88 0.94 Dpi mean 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 Mesh xi Sample Calculation for -65+100 Mesh -3+4 -4+6 -6+8 -8+10 -10+14 -14+20 -20+28 -28+35 -35+48 -48+65 -65+100 0 = x CSA CSA = 0.12 0.09 0.01 0.06 0.14 0.16 0.07 0.22 0.01 0.06 + i n 1 n = x . 0 94 . 0 88 . 0 06 i 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 + Dp of mesh a Dp of mesh b = i i Dp i 2 + . 0 208 . 0 147 = = Dp 1775 . 0 mm i 2 0.1255 0.1594 -100+150 0.96 0.02 Excluding mesh -200 0.089 0.3371 -150+200 0.99 0.03 -200 1 0.01 Total 0.074 19.177 2.2406 0.1351 Source: Tyler Standard Sieve Scale

  20. Problem No. 13 Cum Analysis 0.94 Dpi mean 0.1775 xi/Dpi mean 0.3380 Mesh xi From Table 28.2 of MSH book, 7th ed: Mica flakes: s = 0.28 -65+100 0.06 0.1255 0.1594 -100+150 0.96 0.02 6 V 0.089 0.3371 -150+200 0.99 0.03 p = s D S -200 1 0.01 0.074 Total 0.9696 0.1351 p p S 3 6 D s 6 10 p = = x 2 . 0 ( 28 )( 19 . 177 ) 10 V p p S 1 1 2 cm p = = = 11 . 1741 D s 3 cm x n V 9696 . 0 = i p D pi i 1 0313 . 1 = D mm (for smaller th mesh) 65 an s

  21. Problem No. 18 Kenneth Paulo Miranda

  22. Problem No. 18 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 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=LHygXfdsPqU

  23. 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

  24. Plotting Cum Analysis vs Dp 1.2 1 0.8 Cum Analysis 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 Dp Approximately = 1.49

  25. 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 kW

  26. Problem from Lecture 2A Slide No. 45 Jhullie Ann Bigso

  27. Problem from Lecture 2A Slide No. 45 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 Fine Dp where 80% of the particles of given size analysis in the table, passes through.

  28. Mesh 3 4 6 8 10 14 20 28 35 48 65 100 150 200 pan Dp Cum Analysis Cum Analysis 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 -- 0 1 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 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

  29. Dp = 2.5mm

  30. 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

  31. Thank You!

Related


No related presentations.

More Related Content