Question 5 of 6

Question 5 (20 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.

Part (A) — temperature vs. emitted wavelength. Every surface above absolute zero emits thermal (electromagnetic) radiation across a continuous spectrum, and Wien's displacement law fixes where that spectrum peaks: λ₀ₕₓ = b/T, with b = 2898 µm·K and T in kelvin. The relationship is inverse — a hotter surface radiates more total energy (per the Stefan–Boltzmann law, ∝T⁴) AND shifts its emission to shorter wavelengths, while a cooler surface radiates less energy and shifts to longer wavelengths. At "normal" building surface temperatures (roughly 0–30 °C, i.e. 273–303 K), this gives $$\lambda_{max} = \frac{2898}{293} \approx 9.9\ \mu\text{m}$$ — the AVERAGE wavelength of thermal radiation exchanged between building surfaces sits around 10 µm, deep in the long-wave (far) infrared. This is the physical reason building-science treats "solar" radiation (peaked near 0.5 µm, from the sun's ~5800 K surface) and "long-wave" or "thermal" radiation (peaked near 10 µm, exchanged between building surfaces near room temperature) as two distinct bands with different glazing and coating properties — a low-e coating, for instance, is engineered to be reflective in the 8–13 µm long-wave band while staying transparent in the visible/near-IR solar band.

Part (B).

Given.

Find. The mass flow rate of moisture exchanged, the sensible and latent heat exchanged, and whether the room loses or gains each.

Approach. Use the psychrometric relations to get the humidity ratio of the room and outside air, convert the volumetric flow leaving the room to a dry-air mass flow using the room's specific volume, then form the moisture-flow, sensible-heat and latent-heat balances for air LEAVING the room (necessarily replaced one-for-one by incoming outside air at the outside condition).

1. Saturation and actual vapour pressure, each condition (Magnus-Tetens, ice curve below 0°C): $$P_{sat}(22^\circ\text{C})=2639\ \text{Pa} \Rightarrow P_{v,room}=0.55\times2639=1451\ \text{Pa}$$ $$P_{sat}(-5^\circ\text{C})=402\ \text{Pa} \Rightarrow P_{v,out}=0.80\times402=321\ \text{Pa}$$
2. Humidity ratio, W = 0.622·Pₜ/(Pₔₜ₞−Pₜ), Pₔₜ₞=101.325 kPa: $$W_{room}=\frac{0.622\times1451}{101\,325-1451}=9.04\ \text{g/kg (dry air)},\quad W_{out}=\frac{0.622\times321}{101\,325-321}=1.98\ \text{g/kg}$$
3. Dry-air mass flow leaving the room, from the room's moist-air specific volume v = 0.287T(1+1.6078W)/P (T in K, P in kPa): $$v_{room}=\frac{0.287\times295.15\times(1+1.6078\times0.00904)}{101.325}=0.848\ \text{m}^3\text{/kg}$$ $$\dot m_{da}=\frac{\dot V}{v_{room}}=\frac{0.0944}{0.848}=0.1113\ \text{kg/s}$$
4. Moisture exchanged. Air leaving the room carries the room's (higher) humidity ratio out, and is replaced by outside air at the (lower) outside humidity ratio, so the room's net moisture change is $$\boxed{\dot m_{moisture}=\dot m_{da}(W_{room}-W_{out})=0.1113\times(0.00904-0.00198)=7.86\times10^{-4}\ \text{kg/s} = 2.83\ \text{kg/hr}}$$ Since Wₛₔₔ₁ > W₀₦ₜ, this is moisture LOST by the room.
5. Sensible heat, using humid specific heat cₖ ≈ 1.02 kJ/(kg·K): $$\boxed{Q_{sens}=\dot m_{da}\,c_p\,(T_{room}-T_{out})=0.1113\times1.02\times27=3.07\ \text{kW}}$$ Warm room air leaving is replaced by cold outside air, so this is sensible heat LOST by the room.
6. Latent heat, using hᵤᵤ ≈ 2501 kJ/kg: $$\boxed{Q_{lat}=\dot m_{da}\,h_{fg}\,(W_{room}-W_{out})=0.1113\times2501\times0.00706=1.97\ \text{kW}}$$ Moist room air leaving carries its latent heat with it, so this is also latent heat LOST by the room.
7. Total. $$Q_{total}=Q_{sens}+Q_{lat}=3.07+1.97=5.03\ \text{kW, lost by the room.}$$