Particulate technology

Problem Table salt is being fed to a vibrating screen at the rate of 3000 kg/hr. The desired desired product is the 48/65 mesh fraction. A -48 + 65-mesh 65-mesh screen are therefore used, the feed being introduced on the 48 mesh screen, the product being discharged from the 65-mesh screen. During the operation operation it was observed that the average proportion of the oversize: product: undersize was 2:1 ½:1. Calculate the effectiveness of the screens? s creens? What conclusions can you draw from the values computed versus the given proportion?

Screen Analysis Screen Mesh

Feed

Oversize

-10+14

0.0003

0.0008

-14+20

F1

R1

0.0037

0.0082

-20+28

0.089

0.0189

-28+35

0.186

0.389

-35+48

F2

-48+65

0.258

R2

0.285

F3

0.377

Product

P1

0.025

0.0155

Q1

0.039

P2

0.176

R3

0.0005

Undersize

0.322

0.00012

Q2

0.526

P3

0.075

0.00003 0.0009 0.0136

Q3

-65+100

0.091

0.34935

-100+150

0.062

0.004

0.020

0.299

-150+200

0.025

0.0011

0.002

0.337

1

1

1

1

Solution Given: F=3000 kg/hr  Let: 1: subscript of particles larger than 48 mesh(+48) 2: subscript of particles smaller than 48 but l arger than 65 mesh (-48+65) 3: subscript of particles smaller than 65 mesh (-65)

Solution F= 3000 kg/hr 

R (oversized particles) R1= 0.7939R R2=0.176R R3=0.0301R

S S1 S2 S3

Q (undersized particles) Q1= 0.00105Q Q2=0.0136Q

P ( desired product) P1=0.377P P2=0.526P P3=0.097P

Solution From table: F1= 0.537(3000)= 1611 kg/hr  F2= 0.285(3000)= 855 kg/hr  F3= 0.178(3000)=534 kg/hr 

Solution OMB: 3000= R + P + Q (1-MB) : 1611= 0.7939R+0.377P+0.00105Q (2-MB) : 855= 0.176R+0.526P+0.0136Q (3-MB) : 534= 0.0301R+0.097P+0.98535Q Solving for R P and Q: R= 1499.886 kg/hr  P=1113.6192 kg/hr  Q=386.4945 kg/hr 

Solution R (oversized particles) R1= 0.7939R= 1190.7595 kg/hr  R2=0.176R= 263.98 kg/hr  R3=0.0301R= 45.1466 kg/hr  P (desired product) P1=0.377P= 419.83 kg/hr  P2=0.526P= 585.764 kg/hr  P3=0.097P= 108.021 kg/hr 

Q (undersized particles) Q1= 0.00105Q=0.4058 Q2=0.0136Q= 5.2563 Q3=0.98535Q= 380.832

Solution To solve for S ( particles leaving mesh 48 and entering mesh 65) Sn= Fn- Rn (where n is the subscript for particles) S1= 1611-1190.7595=420.2405 S2= 855-263.98= 591.02 S3= 534-45.1466= 488.8534

COMPUTING FOR EFFICIENCY

Given the proportion R:P:Q

2:1.5:1—proportion of oversize: product : undersize Therefore OMB can be 3000= P + R +Q 3000 = 1.5Q + 2Q +Q Solving Q= 666.67 kg/hr  R= 2 (666.67) = 1333.33 kg/hr  P= 1.5( 666.67)= 1000 kg/hr 

Solution R (oversized particles) Q (undersized particles) R1= 0.7939R= 1058.53kg/hr  Q1= 0.00105Q=0.7 R2=0.176R= 234.666kg/hr  Q2=0.0136Q= 9.07 R3=0.0301R= 40.133 kg/hr  Q3=0.98535Q= 656.903 P (desired product) P1=0.377P= 377kg/hr  P2=0.526P= 526 kg/hr  P3=0.097P= 97 kg/hr 

S1= 552.47 S2= 620.334 S3= 493.867

Efficiency

CONCLUSION The efficiency as to when the proportions were given is greater than the computed value. If the efficiency of separation (E) is defined as the mass ratio of solids (of all sizes) in the underflow to that in the feed, then a clarified overflow corresponds to an efficiency of 100 percent.