Particulate technology
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Transcript of Particulate technology
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7/28/2019 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.