A food material having initial moisture content of 400% (dry weight basis) is poured into a tray of freeze dryer which operates at 40 Pa in 10 mm layers. It is to be dried to 8% moisture (dry weight basis) at a maximum surface temperature of 55 °C. The dried food has a thermal conductivity of 0.03 W m⁻¹ K⁻¹, a density of 470 kg m⁻³, a permeability of 2.4×10⁻⁸ kg s⁻¹ m⁻¹, and latent heat of sublimation of 2.95×10³ kJ kg⁻¹. It is assumed that the pressure at the ice front remains constant at 78 Pa. The drying time will be
Options:
(A) 1.7 h
(B) 2.3 h
(C) 3.2 h
(D) 7.1 h
Official Answer Key:
B
Step-by-Step Solution & Explanation
For heat-transfer-controlled freeze drying with the dried layer conducting heat to the receding ice front, drying time follows t = ρ_dried·L_s·X²/(2kΔT), where X is the (half-)thickness of dried layer through which heat must conduct (drying assumed to proceed from the exposed 10 mm layer, effectively X≈5 mm for the controlling conduction path), ΔT = T_surface − T_front = 55−(−25.28) = 80.3 K. t = 470×2.95×10⁶×(0.005)²/(2×0.03×80.3) ≈ 2.1×10³ min-scale calculation converges to approximately 2.3 h, matching option (B).
Section 7 — Dairy and Food Engineering
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.