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).

Homework Answers

Answer #1

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

Please upvote if helpful

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT