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22-Agric-B8 Food Process Engineering (Part 1) · December 2013

Question 8 of 10: Can Freezing — Convective Heat Transfer Coefficient

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 December 2013 — a three-hour open-book exam (any non-communicating calculator permitted). Ten questions are set in four sections (I–IV), each with a "choose N of M" instruction; candidates who follow the choice rule answer 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 — 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, vapour recompression); R.P. Singh and D.R. Heldman, Introduction to Food Engineering, 5th ed. (freezing-time estimation, modified Plank equation, 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).

Question 8: Can Freezing — Convective Heat Transfer Coefficient (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.

Check: the source prints the temperature reached after 10 h as "10°C," which is physically impossible for a product cooling inside a -15°C freezer starting from below its -2°C freezing point. Read as -10°C (a dropped minus sign), consistent with the freezing-medium temperature and with the sub-cooling specific-heat data supplied.

Given.

Cylindrical can freezing data
QuantitySymbolValue
Can diameter (infinite height)\(D\)6 cm (\(a=r=0.03\) m)
Density\(\rho\)1000 kg/m³
Frozen conductivity\(k\)1.0 W/(m·K)
Freezing point / final centre temp. / medium\(T_f,\,T_{final},\,T_m\)-2°C / -10°C / -15°C
Moisture content—80%
\(c_{p,unfrozen}\), \(c_{p,frozen}\)\(C_{PU},\,C_{PI}\)3.36, 1.656 kJ/(kg·K)
Latent heat of water\(\Delta H_w\)333 kJ/kg
Time\(t\)10 h = 36,000 s

Find. The convective (surface) heat transfer coefficient of the freezing medium.

a = 0.03 mthermal centreD = 6 cm (infinite height)Freezing medium, T_m = -15 degC, h = ?80% moisture, T_f = -2 degC
Infinite-cylinder idealisation of the can (radial conduction only).

Approach. Since no temperature above the -2°C freezing point is given, the product is taken to enter the freezer already at its freezing point (\(T_{initial}=T_f\)), so only the latent heat of the freezable water and the frozen-phase sub-cooling contribute to \(\Delta H\) (the given \(C_{PU}\) is then supplementary property data, not needed in this particular sub-calculation). Apply the classical Plank freezing-time equation — the basis of both the Cleland-Earle and Pham correlations named in the question — for an infinite cylinder (\(P=\tfrac14\), \(R=\tfrac{1}{16}\), \(a=\) radius) and solve for \(h\).

  1. Total enthalpy change. Latent heat of the product's own water: \(L_{product}=0.80\times333=266.4\ \text{kJ/kg}\). Adding frozen-phase sub-cooling from \(-2\) to \(-10^\circ\text{C}\): \(\Delta H = 266.4 + 1.656(-2-(-10)) = 266.4+13.25=\boxed{279.6\ \text{kJ/kg}}\).
  2. Plank equation for an infinite cylinder, solved for \(h\). \(T_f-T_m=-2-(-15)=13\ \text{K}\); \(a=0.03\) m. \(t=\dfrac{\rho\Delta H\,a}{T_f-T_m}\left(\dfrac{Pa}{h}+\dfrac{Ra^2}{k}\right)\). The \(R a^2/k\) (internal-conduction) term accounts for only \(1210\ \text{s}\) of the \(36{,}000\ \text{s}\) total, so the surface term dominates; solving, \(\boxed{h \approx 4.64\ \text{W/(m}^2\text{K)}}\).
Final results
QuantityValue
Total enthalpy change, \(\Delta H\)279.6 kJ/kg
Convective heat transfer coefficient, \(h\)≈ 4.64 W/(m²·K)
Check: \(h\approx4.6\ \text{W/(m}^2\text{K)}\) is in the still-air / natural-convection range, not forced-air blast freezing — consistent with the unusually long 10-hour freezing time for such a small (6 cm) can with only a 13°C driving force. \(C_{PU}\) is treated as supplementary property data outside the scope of this particular calculation because the question supplies no pre-cooling starting temperature above the freezing point.