In a juice filtration process, solid concentration per m³ of filtrate is 0.2 kg. During filtration of 12.49 m³ of juice, 0.02 m thick cake (porosity of 0.32) is deposited. If 2.5 kg of solid is collected in 180 s, the pressure drop across the cake in kPa is [Absolute viscosity of juice is 2.12×10⁻³ kg m⁻¹s⁻¹, and specific cake resistance is 1.2×10⁸ m kg⁻¹]
Options:
(A) 0.18
(B) 1.81
(C) 18.06
(D) 180.60
Official Answer Key:
C
Step-by-Step Solution & Explanation
The governing cake-filtration equation is ΔP = μαc V²/(A²t), where μ = viscosity, α = specific cake resistance, c = solid concentration per unit filtrate volume, V = total filtrate volume, A = filter area, t = filtration time. A mass-balance check confirms the given data are self-consistent: total solids = c×V = 0.2×12.49 = 2.498 kg ≈ the stated 2.5 kg collected in t=180 s.
Verification note: the filter area A is not stated explicitly in the reproduced problem, and cannot be uniquely back-calculated from cake thickness, porosity and total solids without the particle (solid) density, which is also not given — so we could not independently re-derive A from first principles to fully cross-check the numeric value. Applying the standard cake-filtration formula with the officially expected filter area (consistent with option C) gives ΔP ≈ 18.06 kPa, which we adopt as the answer, while noting that a complete independent derivation of A was not possible from the data as transcribed. Answer: (C) 18.06.
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