Particulate technology

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    Problem

    Table salt is being fed to a vibrating screen at therate of 3000 kg/hr. The desired product is the 48/65mesh fraction. A -48 + 65-mesh screen aretherefore used, the feed being introduced on the 48mesh screen, the product being discharged from the65-mesh screen. During the operation it wasobserved that the average proportion of the

    oversize: product: undersize was 2:1 :1. Calculatethe effectiveness of the screens? What conclusionscan you draw from the values computed versus thegiven proportion?

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    Screen Analysis

    Screen Mesh Feed Oversize Product Undersize

    -10+14 0.0003 0.0008

    -14+20 0.0037 0.0082 0.0005

    -20+28 0.089 0.0189 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 0.020 0.299

    -150+200 0.025 0.0011 0.002 0.337

    1 1 1 1

    F1

    Q3F3 P3R3

    R1P1 Q1

    F2 R2 P2 Q2

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    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)

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    SolutionF= 3000 kg/hr

    Q (undersized

    particles)

    Q1= 0.00105Q

    Q2=0.0136QQ3=0.98535Q

    P ( desired

    product)

    P1=0.377PP2=0.526P

    P3=0.097P

    R (oversized particles)

    R1= 0.7939R

    R2=0.176RR3=0.0301R

    S

    S1

    S2

    S3

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    Solution

    From table:

    F1= 0.537(3000)= 1611 kg/hr

    F2= 0.285(3000)= 855 kg/hr

    F3= 0.178(3000)=534 kg/hr

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    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/hrP=1113.6192 kg/hr

    Q=386.4945 kg/hr

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    Solution

    R (oversized particles)

    R1= 0.7939R= 1190.7595kg/hr

    R2=0.176R= 263.98 kg/hrR3=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

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    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

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    COMPUTING FOR EFFICIENCY

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    Given the proportion

    R : P : Q

    2:1.5:1proportion of oversize: product : undersize

    Therefore OMB can be 3000= P + R +Q

    3000 = 1.5Q + 2Q +Q

    Solving Q= 666.67 kg/hrR= 2 (666.67) = 1333.33 kg/hr

    P= 1.5( 666.67)= 1000 kg/hr

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    Solution

    R (oversized particles)

    R1= 0.7939R= 1058.53kg/hr

    R2=0.176R= 234.666kg/hr

    R3=0.0301R= 40.133 kg/hr

    P (desired product)

    P1=0.377P= 377kg/hrP2=0.526P= 526 kg/hr

    P3=0.097P= 97 kg/hr

    Q (undersized particles)

    Q1= 0.00105Q=0.7

    Q2=0.0136Q= 9.07

    Q3=0.98535Q= 656.903

    S1= 552.47

    S2= 620.334S3= 493.867

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    Efficiency

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    CONCLUSION

    The efficiency as to when the proportions were

    given is greater than the computed value.

    If the efficiency of separation (E) is definedasthe mass ratio of solids (of all sizes) in the

    underflow to that in the feed, then a clarified

    overflow corresponds to an efficiency of 100percent.