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

View Answer

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, ε_{mf} is the porosity and Φ_{S} is the shape factor.

a) U_{mf}^{2} = \(\frac {Dp (ρP-ρF)g}{1.75ρF}\) × ε_{mf}^{3} × Φ_{S}

b) U_{mf}^{2} = \(\frac {Dp (ρP-ρF)g}{1.75ρP}\) × ε_{mf}^{3} × Φ_{S}

c) U_{mf}^{2} = \(\frac {Dp (ρP-ρF)g}{1.75ρP}\) × ε_{mf}^{2} × Φ_{S}

d) U_{mf}^{2} = \(\frac {Dp (ρP-ρF)g}{1.75ρF}\) × ε_{mf}^{2} × Φ_{S}

View Answer

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/m^{3}. 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/s^{2}.

a) 0.07

b) 0.004

c) 0.08

d) 0.04

View Answer

Explanation: Given,

Since particle are assumed to be spherical, Φ

_{S}= 1

Dp = 0.002 m

ρP = 2500 kg/m

^{3}

Since the fluid is water, ρF = 1000 kg/m

^{3}

Viscosity of water = 8.11 × 10

^{-4}kg/ms

g = 9.8 m/s

^{2}

ε

_{mf}= 0.45

When the particles have Reynolds number less than 20, the minimum fluidization velocity is given as

U

_{mf}= \(\frac {(ρP – ρF)g}{150 µf (1 – εmf)}\) × Dp

^{2}× ε

_{mf}

^{3}× Φ

_{S}

^{2}

Substituting the values we get U

_{mf}= 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

View Answer

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, U

_{t}= \(\frac {g(ρP – ρF)}{18 µf}\) × Dp

^{2}and U

_{mf}= \(\frac {(ρP – ρF)g}{150 µf (1 – εmf)}\) × Dp

^{2}× ε

_{mf}

^{3}× Φ

_{S}

^{2}

where Dp is the diameter of the particle, ρP is the density of the particle, ρF is the density of fluid medium, ε

_{mf}is the porosity, Φ

_{S}is the shape factor and µf is the viscosity of the fluid medium.

U

_{t}÷ U

_{mf}= \(\frac {150 (1 – εmf)}{18}\) ε

_{mf}

^{-2}× Φ

_{S}

^{-2}

Since the particles are spherical Φ

_{S}= 1

ε

_{mf}= 0.45

Substituting the above values, we get

U

_{t}= 50 × U

_{mf}

For U

_{mf}= 1 cm/s U

_{t}= 50 cm/s

Therefore, the operating velocity should be less than 50 cm/s.

**Sanfoundry Global Education & Learning Series – Mechanical Operations.**

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