Centrifugal filtration without cake formation: for a liquid layer of thickness δ at radius R in a rotor spinning at angular speed ω, what is the initial pressure drop across the layer (liquid density ρ_L)?
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A0.5 * ω^2 * R^2 * ρ_L
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B0.5 * ω^2 * δ^2 * ρ_L
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C0.5 * ω^2 * δ * ρ_L * (2R + δ)
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D0.5 * ω^2 * R * ρ_L * (R + 2δ)
Answer
Correct Answer: 0.5 * ω^2 * δ * ρ_L * (2R + δ)
Explanation
Introduction / Context:In centrifugal filtration, the driving force is the radial pressure gradient generated by rotation. With no cake yet formed, the initial pressure drop arises from the centrifugal head across the liquid film.
Concept / Approach:For a rotating fluid, pressure varies with radius as dP/dr = ρ_L * ω^2 * r. Integrating from r = R to r = R + δ gives ΔP = ∫RR+δ ρ_L * ω^2 * r dr = 0.5 * ρ_L * ω^2 * [(R + δ)^2 − R^2] = 0.5 * ρ_L * ω^2 * δ * (2R + δ).
Step-by-Step Solution:Start from dP/dr relation.Integrate over the liquid thickness δ.Obtain ΔP expression shown above.
Why Other Options Are Wrong:Forms using only R^2 or δ^2 ignore the cross term and do not result from the correct integral.
Final Answer:0.5 * ω^2 * δ * ρ_L * (2R + δ)