For Laminar flow through a packed bed, the pressure drop is proportional to (Vs is the superficial liquid velocity and Dp is the particle diameter)
A. $$\frac{{{{\text{V}}_{\text{s}}}}}{{{{\text{D}}_{\text{p}}}^2}}$$
B. $$\frac{{{{\text{V}}_{\text{s}}}^2}}{{{{\text{D}}_{\text{p}}}^2}}$$
C. $$\frac{{{{\text{V}}_{\text{s}}}^2}}{{{{\text{D}}_{\text{p}}}^3}}$$
D. $$\frac{{{{\text{V}}_{\text{s}}}}}{{{{\text{D}}_{\text{p}}}^3}}$$
Answer: Option A
Solution (By Examveda Team)
For laminar flow in a packed bed (imagine a pipe filled with small balls), the pressure drop (how much the pressure decreases as the fluid flows) depends on how fast the fluid is moving and the size of the balls.Ergun Equation describes this relationship. For laminar flow, a simplified version of the Ergun equation is used.
This simplified version shows that the pressure drop is directly proportional to the superficial velocity (Vs) and inversely proportional to the square of the particle diameter (Dp).
Superficial velocity (Vs) means the velocity of the fluid if the bed were empty.
Particle diameter (Dp) is the size of the balls in the packed bed.
Therefore, if the superficial velocity increases, the pressure drop increases. If the particle diameter increases, the pressure drop decreases (because bigger particles offer less resistance to flow).
Looking at the options, only option A shows this direct proportionality to Vs and inverse proportionality to the square of Dp (Dp2).
Option A: Vs/Dp2 correctly represents this relationship for laminar flow in a packed bed.
This Question Belongs to Chemical Engineering >> Fluid Mechanics
The drag co-efficient for a bacterium moving in water at 1 mm/s, will be of the following order of magnitude (assume size of the bacterium to be 1 micron and kinematic viscosity of water to be 10-6m2/s).
24000
24
0.24
0.44