Continuous Screening

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Problem Set
Problem # 1 Continuous Screening

Given:
a.

Effective dimension of 35 mesh screen
(6ft x 20ft)

b.

Undersize flow rate = 50 tons/hr

Mesh Size

Feed Size

Oversize

Undersize

6/8

0.075

0.08

0.02

8/10

0.125

0.145

0.055

10/20

0.1

0.17

0.09

20/28

0.125

0.15

0.085

28/35

0.125

0.28

0.1

35/48

0.175

0.175

0.15

48/65

0.225

0.15

65/100

0.05

0.25

100/150

0.1

Required:

a.

Effectiveness of screening operation

b.

Capacity of each 35-mesh screen

Solution:Effectiveness
Let A= desired material (undersize)

Pxp
effectiveness =
Fxf

1 − xp P
1−
1 − xf F



xf = 0.225 + 0.050 + 0.175 = 0.45



xp = 0.150 + 0.250 + 0.100 + 0.150 = 0.65



xr = 0.175

Overall mass balance:
F=R+P
F = R + 50

(equation 1)

Component mass balance:

Fxf = Rxr + Pxp
0.45F = 0.175R + 32.5 (equation 2)

Solving equations 1 and 2 simultaneously,
F=86.3636 tons/hr
R=36.3636 tons/hr
Pxp
effectiveness =
Fxf

1 − xp P
1−
1 − xf F

(50 tons/hr)(0.65)
effectiveness =
(86.3636 tons/hr)(0.45)
1 − 0.65 50 tons/hr
1−
1 − 0.45 86.3636 tons/hr
effectiveness = 0.53

Capacity
Using the throughflow method of Matthews,

A = 0.4Ct /Cu Foa Fs
Where

A = screen area, ft2
Ct=throughflow rate
Cu=unit capacity
Foa=open-area factor
Fs=slotted-area factor

Assuming that the screen has square
openings,
 Fs=1
 Foa=(a/(a+d))2
where

a=clear opening dimension
d=diameter of wire

From Table 19-6 of Perry’s Chemical Engineers’ Handbook,
7th edition
d=0.0114in and a=0.165in (mesh 35 screen)

Foa=(a/(a+d))2
=(0.0165/(0.0165+0.0114))2
Foa=0.3497514163

𝐀 = 𝟎. 𝟒𝑪𝒕 /𝑪𝒖 𝑭𝒐𝒂 𝑭𝒔
0.4(50 𝑡𝑜𝑛𝑠/ℎ𝑟)
6𝑓𝑡 20𝑓𝑡 =
𝐶𝑢 (0.3497514163)(1)
Cu=0.4765289256 tons/hr-ft2
=11.44tons/24hr-ft2 =22880
tons/24hr-ft2

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