Question

# A cylindrical fluidized bed of diameter 0.5 m is used in flour mill for flour production....

A cylindrical fluidized bed of diameter 0.5 m is used in flour mill for flour production. Air of density 1.2 kg/m3 and viscosity 1.84 x 10-5 Pa.s is fluidizing the flour powder of mean sieve size 60 µm and particle density 1800 kg/m3 . 240 kg of flour powder is charged to the bed and air flowrate to the bed is 140 m3 /hr. The average bed voidage at incipient fluidization is 0.45. Compute the following problems:

(a) the minimum fluidization velocity

(b) the bed height at incipient fluidization

(c) the superficial air velocity in unit meter per second

(d) the minimum fluidization velocity using Wen & Yu correlation, compare the answer with Q1(a).

Cylindrical bed diameter = 0.5 m

Air density = 1.2 Kg/m3

viscosity = 1.84×10-5 Pas

Diameter of particle(Dp) = 60 ×10-6 m

Density of particle () = 1800 Kg/m3

= 0.45

At incipient fluidization pressure drop is maximum and is given by

∆P/L = 9705.371 Pas/m

Sphericity of particle = 1

Ergun equation is given by

In above equation sphericity term = 1 (so not mentioned)

9705.371 = 2545038.866 (v) + 211248.285 v2

Solving this we get

v = 0.0038122 m/s

Minimum fluidization velocity (v)

= 0.0038122 m/s = 3.8122 mm/s

B)

Mass of flour fed = 240 Kg

Density of particle =1800 Kg/m3

Volume of solid particles in bed = m/(density)

= 240/(1800) = 0.1333 m3

= (volume of bed - volume of particles) /(volume of bed)

0.45 = (V-0.1333) /(V)

V = 0.24236 m3

Diameter of bed = 0.5 m

Crossectional area(A) = π(d2) /4 = (π) (0.5) 2/4 = 0.19634 m2

V = A× L

0.24236 = 0.19634×L

L = 1.2343 m

L- total height of column

Volume of particles = 0.1333 m3

0.1333 = A× L'

0.1333= 0.19634×L'

L' = 0.67892 m

Height of packed bed (L') at incipient fluidization = 0.67892 m

C)

Superficial velocity (v) = Q/A

Q = 140 m3/h

Cross sectional area (A) = 0.19634 m2

v = Q/A = 140/(0.19634) = 713.048 m/h

v = 0.198069 m/s

D)

Wen and yu correlation is

Ref is Reynolds number at minimum fluidization

Ar - archimedes number

Ar = 13.5098

13.5098 = 6456.521(v) + 375.14(v2)

Solving we get

v = 0.0020921 m/s

v = 2.0921 mm/s

Minimum fluidization velocity according to correlation = 2.0921 mm/s

From

1)

v = 3.8221 mm/s

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