Question 5 of 10: Scaling freezing time with product size, air velocity, and medium temperature

Question 5: Scaling freezing time with product size, air velocity, and medium temperature (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. A sphere of diameter 18 cm freezes in 5 h in the same freezer whose air is at −30°C; Plank's equation splits the freezing time into a surface-controlled (linear in diameter) term and a conduction-controlled (quadratic in diameter) term.

Find. (a) freezing time at D=15 cm; (b) whether raising air velocity is worthwhile; (c) the effect of a colder (−40°C) medium.

Approach. Plank's equation for a sphere, t = (ρΔH₁/ΔT)[Pa/h + Ra²/k] with a=D, has exactly two terms: one ∝ D/h (surface-film controlled) and one ∝ D²/k (conduction controlled). Neither h, k, ρ nor ΔH₁ is given, so the single data point (D=18 cm → t=5 h) cannot by itself separate the two terms — the question's own "assume any unknown" licenses a representative split, taken here as Biot number Bi=hD/k≈4 (a mid-range value typical of blast-freezing spheres of this size), which puts roughly half of the 5-hour total in each term.

1. Split the 5-hour total (sphere: R/P = 1/4, so conduction/surface = Bi/4). With Bi=4, conduction and surface terms are equal: tᵣ⁽ᵥᵤᵚᵌᵩᵤ=tᵣⁿ₀ᵢᵏᵎᵏᵤ=2.5 h at D₁=18 cm. Writing t=K₁D+K₂D²: $$K_1=\frac{2.5}{18}=0.1389\ \text{h/cm},\qquad K_2=\frac{2.5}{18^2}=0.007716\ \text{h/cm}^2$$
2. Part (a) — predict t at D=15 cm (same freezer, same K₁,K₂). $$t_2=K_1D_2+K_2D_2^2=(0.1389)(15)+(0.007716)(15^2)=2.083+1.736=\boxed{3.82\ \text{h}}$$ The smaller product freezes faster, as expected, but not in simple proportion to D² alone — the linear surface term keeps the ratio from being as favourable as pure conduction scaling would suggest (3.82/5=0.764, vs (15/18)²=0.694 if conduction alone controlled).
3. Part (b) — is a faster air velocity worthwhile? Only the surface term (∝1/h) responds to velocity; h∝v⁰·⁸, so doubling velocity scales h (and shrinks the surface term) by 2⁰·⁸=1.741. At D=18 cm: $$t_{surf,new}=\frac{2.5}{1.741}=1.436\ \text{h},\qquad t_{new}=1.436+2.5=3.94\ \text{h}$$ That is a 21% reduction in total freezing time for a doubled fan speed. Since the surface term is a genuinely large share (about half) of the total time at this Biot number, increasing air velocity IS worthwhile — it would not be, only if Bi were very large (conduction-dominated, thick/low-k product) so that the surface term were already negligible.
4. Part (c) — effect of colder air (−40°C instead of −30°C). Both terms of Plank's equation scale as 1/ΔT. Taking a nominal product reference temperature of 0°C (so ΔT₁=30°C, ΔT₂=40°C): $$t_{new}=t_1\frac{\Delta T_1}{\Delta T_2}=5\times\frac{30}{40}=\boxed{3.75\ \text{h}}$$ a 25% reduction — somewhat larger than the 21% gained from doubling the air velocity, so of the two options a colder freezing medium is the more effective lever here, though both help because neither resistance is negligible.