Human Thermoregulation — Still Air vs. Water: 22-Agric-A3 Heat Engineering, May 2016, Q2

Problem 2: Human Thermoregulation — Still Air vs. Water (25 points)

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.

Quantity	Value
Inner (near-core) surface temperature, T_c	308 K (35°C)
Skin/fat layer thickness, L	3 mm (0.003 m)
Effective thermal conductivity, k	0.3 W/m·K
Body surface area, A	1.8 m²
Skin emissivity, ε	0.95
Case 1 — still air temperature, T_air	297 K
Case 1 — free-convection coefficient, h_air	2 W/m²·K
Case 2 — water temperature, T_w	297 K
Case 2 — convection coefficient, h_water	200 W/m²·K

Find. The skin surface temperature T_s and total heat-loss rate for (1) still air and (2) full water immersion.

Approach. Model the skin/fat layer as steady 1-D conduction from the fixed inner temperature T_c to an unknown skin surface temperature T_s; at steady state that conduction must equal whatever is carried away from the skin surface. In still air the skin loses heat by BOTH convection to the air and radiation to surroundings at the same temperature, so the conduction balances the sum of the two; in water, radiation exchange does not apply (opaque, conducting medium in direct contact), so conduction balances convection alone. Each case gives one equation in the one unknown T_s.

Check: for the still-air case the surroundings temperature for radiation exchange is taken equal to the air temperature (297 K), the standard assumption when no separate surroundings/wall temperature is stated.

Series conduction from the near-core temperature through the skin/fat layer to the skin surface, balanced by convection (both cases) plus radiation (still-air case only).

Part 1 — energy balance in still air. At steady state, conduction into the skin surface equals convection plus radiation out: $$\frac{k A(T_c-T_s)}{L} = hA(T_s-T_{air}) + \varepsilon\sigma A\left(T_s^4-T_{air}^4\right)$$ The area $A$ cancels; substituting $k=0.3$, $L=0.003$, $T_c=308$, $h=2$, $T_{air}=297$, $\varepsilon=0.95$ and solving numerically for the one unknown $T_s$ gives $$T_s = \boxed{307.2\ \text{K} = 34.2\,{}^{\circ}\text{C}}$$
Part 1 — heat-loss rate in still air. Evaluating either side of the balance at this $T_s$, $$q'' = h(T_s-T_{air}) + \varepsilon\sigma\left(T_s^4-T_{air}^4\right) = 80.9\ \text{W/m}^2$$ $$Q_{air} = q''A = 80.9\times1.8 = \boxed{145.7\ \text{W}}$$
Part 2 — energy balance in water (convection only). Fully immersed, there is no radiation path to a separate surroundings, so conduction balances convection alone: $$\frac{kA(T_c-T_s)}{L} = hA(T_s-T_w) \;\Rightarrow\; T_s = \frac{(k/L)T_c + hT_w}{k/L+h}$$ With $h=200\ \text{W/m}^2\cdot\text{K}$ and $T_w=297\ \text{K}$, $$T_s = \frac{(0.3/0.003)(308) + 200(297)}{0.3/0.003+200} = \boxed{300.7\ \text{K} = 27.7\,{}^{\circ}\text{C}}$$
Part 2 — heat-loss rate in water. $$q'' = h(T_s-T_w) = 200\times(300.7-297) = 733.3\ \text{W/m}^2$$ $$Q_{water} = q''A = 733.3\times1.8 = \boxed{1320\ \text{W}}$$ Even though the water and air are at the SAME temperature, the much higher convection coefficient in water pulls the skin surface far closer to the water temperature and drives roughly nine times the heat loss — the physiological reason cold water is so much more dangerous than cold air at the same temperature.

Quantity	Result
Skin surface temperature, still air	307.2 K (34.2°C)
Heat-loss rate, still air	145.7 W
Skin surface temperature, water immersion	300.7 K (27.7°C)
Heat-loss rate, water immersion	1320 W

Concept & Theory

Topic: Combined conduction-convection-radiation energy balance at a body surface (thermoregulation).

Key relations: Series conduction through the skin/fat layer, $q''_{cond}=k(T_c-T_s)/L$; surface loss in air, $q''_{loss}=h(T_s-T_\infty)+\varepsilon\sigma(T_s^4-T_\infty^4)$; surface loss in water (no radiation), $q''_{loss}=h(T_s-T_\infty)$; steady state requires $q''_{cond}=q''_{loss}$.

Why this works: The skin surface temperature is not an independent given — it is whatever value makes the heat arriving by conduction from the body core exactly equal the heat leaving by whatever surface mechanisms are active. In air both convection and (nonlinear, $T^4$) radiation act in parallel at the surface; in water, direct contact with a conducting fluid removes the radiation path entirely, replacing it with a much larger convection coefficient instead.

Common pitfall: Applying the still-air radiation term to the water case as well — a fully immersed body has no line of sight to a radiating surroundings, so the water-case balance is convection-only, not convection-plus-radiation with a different h.

Source: Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, Ch. 1 (combined modes, human thermoregulation example).

Question 2 of 4: Human Thermoregulation — Still Air vs. Water

Problem 2: Human Thermoregulation — Still Air vs. Water (25 points)