A single strength fruit juice is concentrated from 6 % total solids (TS) to 24 % by ultrafiltration. The feed stream has a flow rate of 12 kg min⁻¹. The ultrafiltration membrane tube has an inside diameter of 80 mm and the pressure difference applied across the membrane is 2 MPa. If the permeability constant is 4×10⁻⁵ kg water m⁻² kPa⁻¹ s⁻¹, the length of the membrane tube is _________ m.
Official Answer Key:
7.0 to 7.6
Step-by-Step Solution & Explanation
Solids balance. Solids in feed = 12 × 0.06 = 0.72 kg min⁻¹. All solids are retained.
Concentrate flow = 0.72/0.24 = 3.0 kg min⁻¹.
Permeate (water removed) = 12 − 3.0 = 9.0 kg min⁻¹ = 0.15 kg s⁻¹.
Flux. J = K ΔP = 4×10⁻⁵ × 2000 kPa = 0.08 kg water m⁻² s⁻¹.
Membrane area required A = 0.15/0.08 = 1.875 m².
Length. A = πDL ⇒ L = 1.875/(π × 0.08) = 1.875/0.25133 = 7.46 m — within the official range 7.0–7.6.
Section 7 — Dairy and Food Engineering
6 questions · 9 marks
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