# Mechanical Operations Questions and Answers – Minimum Fluidisation Velocity

This set of Mechanical Operations Multiple Choice Questions & Answers (MCQs) focuses on “Minimum Fluidisation Velocity”.

1. Which of the following is the best condition for minimum fluidization?
a) Drag force imparted by the down flowing gas equals the weight of the particles, and void space in the bed increases slightly
b) Drag force imparted by the upward flowing gas equals the weight of the particles, and void space in the bed increases slightly
c) Drag force imparted by the upward flowing gas equals the weight of the particles, and void space in the bed decreases slightly
d) Drag force imparted by the down flowing gas equals the weight of the particles, and void space in the bed decreases slightly

Explanation: Minimum fluidization takes place when the superficial velocity becomes minimum fluidization velocity. This happens when the drag force due to the upward moving gas, becomes equal to the weight of the bed particles and voids in the bed increase a little.

2. Which of the following is the correct equation for minimum fluidization velocity for large particles? Here, Dp is the diameter of the particle, ρP and ρF are the density of particle and fluid medium respectively, εmf is the porosity and ΦS is the shape factor.
a) Umf2 = $$\frac {Dp (ρP-ρF)g}{1.75ρF}$$ × εmf3 × ΦS
b) Umf2 = $$\frac {Dp (ρP-ρF)g}{1.75ρP}$$ × εmf3 × ΦS
c) Umf2 = $$\frac {Dp (ρP-ρF)g}{1.75ρP}$$ × εmf2 × ΦS
d) Umf2 = $$\frac {Dp (ρP-ρF)g}{1.75ρF}$$ × εmf2 × ΦS

Explanation: Particles are said to be large when the Reynolds number is greater than 1000. The minimum fluidization velocity for large particles is directly proportional to the bed porosity at the point of minimum fluidization and square root of the shape factor of the bed particles.

3. A fluidized bed has small particles with Reynolds number 18 and diameter 2 mm. The fluid medium is water. The density of the particle is 2500 kg/m3. Consider viscosity of water to be 8.11 × 10-4 kg/ms. Assume the particles to be completely spherical. Porosity of bed is 0.45. Assume acceleration due to gravity as 9.8 m/s2.
a) 0.07
b) 0.004
c) 0.08
d) 0.04

Explanation: Given,
Since particle are assumed to be spherical, ΦS = 1
Dp = 0.002 m
ρP = 2500 kg/m3
Since the fluid is water, ρF = 1000 kg/m3
Viscosity of water = 8.11 × 10-4 kg/ms
g = 9.8 m/s2
εmf = 0.45
When the particles have Reynolds number less than 20, the minimum fluidization velocity is given as
Umf = $$\frac {(ρP – ρF)g}{150 µf (1 – εmf)}$$ × Dp2 × εmf3 × ΦS2
Substituting the values we get Umf = 0.08 m/s

4. For a bed of small spherical particles that fluidizes at the rate of 1cm/s, the operating velocity should be less than which of the following? Consider porosity of the bed to 0.45.
a) 25 cm/s
b) 50 cm/s
c) 40 cm/s
d) 48 cm/s

Explanation: To avoid the carryover of particles in a fluidized bed, we need to select a velocity greater than the minimum fluidization velocity and lesser than the terminal velocity.
For small particles, Ut = $$\frac {g(ρP – ρF)}{18 µf}$$ × Dp2 and Umf = $$\frac {(ρP – ρF)g}{150 µf (1 – εmf)}$$ × Dp2 × εmf3 × ΦS2
where Dp is the diameter of the particle, ρP is the density of the particle, ρF is the density of fluid medium, εmf is the porosity, ΦS is the shape factor and µf is the viscosity of the fluid medium.
Ut ÷ Umf = $$\frac {150 (1 – εmf)}{18}$$ εmf-2 × ΦS-2
Since the particles are spherical ΦS = 1
εmf = 0.45
Substituting the above values, we get
Ut = 50 × Umf
For Umf = 1 cm/s Ut = 50 cm/s
Therefore, the operating velocity should be less than 50 cm/s.

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