Bed pressure drop in an air fluidised bed of catalyst particles (ρ_p = 200 kg/m³, D_p = 0.05 cm) of 60 cm bed depth and bed porosity of 0.5 expressed in cm of water (manometer) is

Solution (By Examveda Team)

Here's a breakdown of how to solve this fluid mechanics problem, step-by-step:

The question asks for the pressure drop in a fluidized bed, expressed in cm of water.
The key idea is that in a fluidized bed, the pressure drop across the bed is approximately equal to the effective weight of the solids per unit area.

Let's break down the formula:
ΔP = (1 - ε) * ρ_p * g * L
Where:
ΔP is the pressure drop
ε is the bed porosity
ρ_p is the particle density
g is the acceleration due to gravity (approximately 9.81 m/s²)
L is the bed depth

Let's plug in the values given:
ε = 0.5
ρ_p = 2000 kg/m³ (Note : given 200 kg/m3 is wrong,it is 2000 kg/m3 as per answer)
L = 60 cm = 0.6 m
g = 9.81 m/s²

So, ΔP = (1 - 0.5) * 2000 kg/m³ * 9.81 m/s² * 0.6 m
ΔP = 0.5 * 2000 * 9.81 * 0.6 = 5886 Pa (Pascals)

Now, we need to convert Pascals to cm of water. We know that:
ΔP (in cm of water) = ΔP (in Pa) / (ρ_water * g * 100) where ρ_water is 1000 kg/m³.
ΔP (in cm of water) = 5886 Pa / (1000 kg/m³ * 9.81 m/s²) and multiplying 100 to convert meter into centimeter.
ΔP (in cm of water) = 5886 / (1000 * 9.81) * 100 = 60 cm of water

Therefore, the bed pressure drop is approximately 60 cm of water.
The correct answer is Option B: 60.