22-Agric-B8 Food Process Engineering (Part 1) · May 2017
Question 1 of 10: Chilling honeydew melons before shipping
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-Agric-B8, Food Process Engineering (Part 1) — National Exams, May 2017. 3 hours duration, closed book (one aid sheet permitted). Ten questions are printed, grouped into four sections (I–IV); a complete exam paper requires six. All ten are solved below as a complete study set.
Reference texts: Toledo, R.T., Fundamentals of Food Process Engineering (this exam's own cited source for its Stumbo g-table and steam-table appendix); Geankoplis, C.J., Transport Processes and Separation Process Principles (evaporator design and steam economy); Incropera, F.P. & DeWitt, D.P., Fundamentals of Heat and Mass Transfer (unsteady-state Heisler-chart conduction, composite cylindrical walls); Singh, R.P. & Heldman, D.R., Introduction to Food Engineering, and Cleland, A.C., Food Refrigeration Processes (freezing-time prediction, modified Plank equation).
Question 1: Chilling honeydew melons before shipping (15 marks)
Given. A honeydew melon is modelled as a sphere at uniform initial temperature, plunged into a 5°C cold room and cooled by convection at its surface.
Given data
Quantity
Symbol
Value
Melon radius
r₀
0.13 m
Thermal conductivity
k
0.58 W/(m·K)
Density
ρ
1000 kg/m³
Specific heat
C
4200 J/(kg·K)
Surface coefficient
h
6 W/(m²·K)
Initial temperature
Tᵢ
30°C
Cold-room air temperature
T∞
5°C
Available cooling time (6 PM Jul 2 → 10 AM Jul 3)
t
16 h = 57 600 s
Find. The melon centre temperature at t = 16 h, and whether it is ≤ 10°C.
Approach. This is unsteady-state conduction into a sphere with surface convection — compute the Biot and Fourier numbers, read (or reproduce) the one-term Heisler/Gröber sphere-centre solution for the dimensionless temperature ratio Y, and convert back to a centre temperature.
Biot and Fourier numbers. Bi = hr₀/k = (6)(0.13)/0.58 = 1.345. Thermal diffusivity α = k/(ρC) = 0.58/(1000×4200) = 1.381×10⁻&sup7; m²/s. Fo = αt/r₀² = (1.381×10⁻&sup7;)(57 600)/(0.13²) = 0.471. Since Fo > 0.2, the one-term series (equivalently, the Heisler-chart curve for k/(hr₀) = 1/Bi = 0.744) is valid on its own, without needing the extra terms.
Sphere-centre dimensionless temperature. Chart 1 gives the centre ratio Y₀ = (T₀−T∞)/(Tᵢ−T∞) for the parameter k/(hr₀) = 0.744 at Fo = 0.471. The first root λ₁ of 1−λcotλ = Bi (Bi = 1.345) is λ₁ = 1.764 rad, with series coefficient A₁ = 4(sinλ₁−λ₁cosλ₁)/(2λ₁−sin2λ₁) = 1.352.
$$Y_0 = A_1 e^{-\lambda_1^2 Fo} = 1.352\,e^{-(1.764)^2(0.471)} = 1.352 \times 0.231 = 0.313$$
Recover the centre temperature. $$T_0 = T_\infty + Y_0(T_i-T_\infty) = 5 + 0.313(30-5) = 5+7.8 = \boxed{12.8\ ^\circ\text{C}}$$
Compare to the target. The target centre temperature for a premium-price shipment is 10°C. The computed centre temperature after the full 16-hour cold-room soak is 12.8°C — ABOVE the target, so the melons will not reach 10°C or below by 10 AM on July 3. Only Y₀ = 0.2 (i.e. 10°C) is achieved at a later Fo ≈ 0.62, which back-calculates to about 21.1 hours of cooling — roughly 5 more hours than the distributor has available.
