Frost threshold on a grapefruit (radiation + convection balance): 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2018, Q7

Question 7 — Frost threshold on a grapefruit (radiation + convection balance) (Part B, equal value)

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 grapefruit at steady state exchanges radiation with a black night sky at $-45\ ^\circ\text{C}$ and convection with the surrounding air; frost forms when its surface reaches $0\ ^\circ\text{C}$.

Quantity	Value
Frost (surface) temperature, $T_s$	0 °C = 273.15 K
Sky (black body) temperature, $T_\text{sky}$	−45 °C = 228.15 K
Emissivity, $\varepsilon$	0.93
Convection coefficient, $h$	17 W/m²·°C

Find. the lowest air temperature $T_\text{air}$ at which the surface just reaches 0 °C (below this, frost forms).

Figure 7 — At the frost threshold the surface sits at 0 °C. Radiation to the cold sky drains heat; convection from the warmer air replaces it. The lowest air temperature is the one that exactly balances the radiative loss.

Approach. Impose a steady surface energy balance at $T_s=0\ ^\circ\text{C}$: convective gain from the air equals radiative loss to the sky. Solve for the air temperature (use absolute temperatures in the radiation term).

Radiative loss at the threshold. With $T_s=273.15$ K and $T_\text{sky}=228.15$ K: $$q_\text{rad}=\varepsilon\sigma\!\left(T_s^{4}-T_\text{sky}^{4}\right)=0.93(5.67\times10^{-8})\!\left(273.15^{4}-228.15^{4}\right)=150.7\ \text{W/m}^2$$
Surface energy balance. At steady state the convective supply from the air equals this loss: $$h\,(T_\text{air}-T_s)=q_\text{rad}$$
Solve for the lowest air temperature. $$T_\text{air}=T_s+\frac{q_\text{rad}}{h}=0+\frac{150.7}{17}=8.9\ ^\circ\text{C}$$ $T_\text{air,min}\approx 8.9\ ^\circ\text{C}$ If the air drops below about 8.9 °C, convection can no longer make up the radiative loss, the surface falls below 0 °C, and frost forms — even though the air itself is well above freezing.

Quantity	Result
Radiative loss at threshold	151 W/m²
Convective driving $\Delta T$ needed	8.9 °C
Lowest safe air temperature	≈ 8.9 °C

Concept & Theory

Topic. Radiative frost (radiation cooling below air temperature) via a steady surface energy balance.

Key relations. Surface balance $h(T_\text{air}-T_s)=\varepsilon\sigma(T_s^4-T_\text{sky}^4)$; Stefan–Boltzmann constant $\sigma=5.67\times10^{-8}\ \text{W/m}^2\text{K}^4$; radiation temperatures must be absolute (K).

Why this works. Because the grapefruit exchanges long-wave radiation with a very cold sky, its surface can chill below the surrounding air — the familiar reason frost appears on clear, calm nights even when the thermometer reads several degrees above zero. Setting the surface at exactly 0 °C and balancing the radiative drain against the convective gain gives the warmest air temperature that still lets frost just form; any warmer air keeps the surface above freezing.

Common pitfall. Using Celsius inside the $T^4$ terms (the fourth-power law requires kelvin), or assuming frost needs sub-zero air — the whole point is that the surface, not the air, is what must reach 0 °C.

Source. Çengel & Ghajar, Heat and Mass Transfer §1 (combined convection and radiation at a surface) and §12 (radiation exchange with surroundings).

Question 7 of 8: Frost threshold on a grapefruit (radiation + convection balance)

Question 7 — Frost threshold on a grapefruit (radiation + convection balance) (Part B, equal value)