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22-Agric-B8 Food Process Engineering (Part 1) · May 2014

Question 3 of 10: Grapefruit Freezing Time — Modified Plank (Levy)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 04-Agric-B8 Food Process Engineering (Part 1), National Exams May 2014 — a three-hour open-book exam (any non-communicating calculator permitted). Ten questions are set in four sections (I–IV), each with a "do one/any N of M" instruction; a candidate following the choice rules answers six questions for a 100-mark paper. All ten are worked here so the set is a complete study resource.

Reference texts. R.T. Toledo, Fundamentals of Food Process Engineering, 3rd ed. (thermal-process lethality, D and z values, Ball/Stumbo process calculation, aseptic holding-tube residence time, evaporator design — this is the exam's own appendix source); C.J. Geankoplis, Transport Processes and Separation Process Principles, 4th ed. (evaporator heat and mass balances, multiple-effect steam economy); R.P. Singh and D.R. Heldman, Introduction to Food Engineering, 5th ed. (freezing-time estimation, modified Plank and Cleland-Earle equations, unsteady-state heat transfer in canned foods); A.C. Cleland, Food Refrigeration Processes: Analysis, Design and Simulation (Plank/Cleland-Earle freezing-time correlations); F.P. Incropera and D.P. DeWitt, Fundamentals of Heat and Mass Transfer (transient conduction, Heisler charts, composite-wall resistance).

Check: this paper's four roman-numeral section headers ("I. Heat transfer", "II. Food freezing and freeze concentration", "III. Thermal processing", "IV. Several assumptions (retort come-up correction factor, reference temperature for spore D-values, evaporator steam temperature reused for Question 9) are flagged inline where the source leaves a value implicit.

Question 3: Grapefruit Freezing Time — Modified Plank (Levy) (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Grapefruit freezing data
QuantitySymbolValue
Diameter (treated as infinite cylinder)\(D\)4 cm → \(a=2\) cm
Surface coefficient\(h\)20 W/(m²·K)
Initial temperature\(T_i\)2°C
Air (medium) temperature\(T_a\)-20°C
Initial freezing point\(T_{fi}\)-2°C
Target centre temperature\(T_{f}\)-10°C
Latent heat of fusion of water\(\Delta H_{fus}\)333 kJ/kg
Moisture content—90%
Frozen-phase conductivity\(k_I\)1.108 W/(m·K)
Frozen-phase specific heat\(C_{PI}\)2.05 kJ/(kg·K)
Unfrozen-phase specific heat\(C_{PU}\)4.22 kJ/(kg·K)
Density\(\rho\)1000 kg/m³

Find. The freezing time to reach a centre temperature of -10°C, using the Levy-modified Plank equation.

Approach. The stated \(\Delta H\) is the water-fusion latent heat, not a true sublimation term, so it is used to build the total volumetric enthalpy change the product must give up: sensible heat above the freezing point (precooling to \(T_{fi}\)), the phase-change latent heat scaled by the moisture fraction, and sensible heat below the freezing point (subcooling the frozen phase down to the -10°C target). That combined \(\Delta H_{tot}\) replaces the pure latent heat in Plank's original cylinder equation.

  1. Latent heat of freezing per kg of product. Only the water fraction freezes: \(\Delta H_f = 0.90\times333 = \boxed{299.7\ \text{kJ/kg}}\).
  2. Total enthalpy change (precool + latent + subcool). \(\Delta H_{tot} = C_{PU}(T_i-T_{fi}) + \Delta H_f + C_{PI}(T_{fi}-T_f)\) \(= 4.22(2-(-2)) + 299.7 + 2.05((-2)-(-10)) = 16.88+299.7+16.4 = \boxed{333.0\ \text{kJ/kg}}\).
  3. Plank geometry factors (infinite cylinder). \(P=1/4=0.25\), \(R=1/16=0.0625\), radius \(a=0.02\ \text{m}\).
  4. Freezing time. \(t=\dfrac{\rho\,\Delta H_{tot}}{T_{fi}-T_a}\left[\dfrac{Pa}{h}+\dfrac{Ra^2}{k_I}\right]\) \(=\dfrac{1000\times332{,}980}{-2-(-20)}\left[\dfrac{0.25\times0.02}{20}+\dfrac{0.0625\times0.02^2}{1.108}\right]\) \(= \boxed{5042\ \text{s} = 84.0\ \text{min} \approx 1.40\ \text{h}}\).
Final results
QuantityValue
\(\Delta H_f\) (latent, per kg product)299.7 kJ/kg
\(\Delta H_{tot}\) (precool+latent+subcool)333.0 kJ/kg
Freezing time \(t\)84.0 min (1.40 h)
Check: the source labels \(\Delta H=333\ \text{kJ/kg}\) as an "enthalpy for sublimation of water," but 333 kJ/kg is the well-known heat of FUSION of ice, not the (much larger, ≈2834 kJ/kg) heat of sublimation — treated here as a labelling slip and used as the fusion latent heat, consistent with the freezing-time physics the question actually calls for.