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)(i) — criteria of a proper air barrier system. A code-recognized air barrier SYSTEM (not just a material) must be judged against four independent criteria, all of which must hold simultaneously for the whole enclosure, not just typical field-of-wall areas. IMPERMEABLE: the material itself must have a low air permeance, typically ≤0.02 L/(s·m²) at 75 Pa per ASTM E2178. CONTINUOUS: every seam, lap, penetration, and transition (wall-to-roof, wall-to-window, wall-to-foundation) must be detailed and sealed so the barrier forms one unbroken plane around the entire enclosure — a perfect sheet material with unsealed penetrations is not a system. STRUCTURALLY ADEQUATE: the assembly must resist the full design wind, stack, and fan-pressurization loads over the building's service life without rupture, delamination or excessive deflection (rated per ASTM E330/E2357). DURABLE: it must maintain its air-tightness and structural adequacy for the design service life of the enclosure, resisting UV exposure during construction, thermal cycling, and loss of adhesion at seams. A system missing any one of these four (e.g., an impermeable but discontinuous membrane, or a continuous but structurally inadequate one) fails to control air leakage in practice, however good its rated material permeance.

Part (A)(ii) — air barriers vs. vapour barriers. The two layers control physically distinct transport mechanisms and are frequently, but not necessarily, the same material. An AIR BARRIER stops bulk, PRESSURE-DRIVEN air flow (and everything the moving air carries: heat, water vapour, dust, pollutants) through the assembly; its performance metric is air PERMEANCE (L/(s·m²) at a reference pressure) and it must be perfectly continuous to work at all, since even a small breach lets disproportionate airflow through. A VAPOUR BARRIER (more precisely, a vapour RETARDER, since a true zero-permeance barrier is rarely used) slows DIFFUSION of water vapour molecules through the material driven purely by a vapour-pressure (concentration) gradient, independent of any air pressure difference; its metric is vapour PERMEANCE (ng/(s·m²·Pa), or US perms) and, unlike an air barrier, a vapour retarder can still function usefully even with minor gaps, since diffusion (unlike bulk flow) is not dominated by the largest hole. Critically, vapour diffusion moves roughly two orders of magnitude LESS moisture than air leakage through an equivalent unsealed gap, so a wall with an excellent vapour retarder and a poor air barrier will still suffer moisture damage — air barrier continuity is the higher-leverage control in most climates. A single material (e.g., polyethylene sheet, or a sealed OSB sheathing) can serve BOTH functions if it is both sufficiently air-impermeable and made continuous AND has an appropriate vapour permeance for the climate and assembly, but treating "vapour barrier" and "air barrier" as synonyms is a common and consequential design error, since a material with excellent vapour performance (very low permeance) but unsealed joints provides essentially no air-barrier function.

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): $$P_{sat}(22^\circ\text{C})=2639\ \text{Pa} \Rightarrow P_{v,room}=0.55\times2639=1451\ \text{Pa}$$ $$P_{sat}(-15^\circ\text{C})=165\ \text{Pa} \Rightarrow P_{v,out}=0.80\times165=132\ \text{Pa}$$
2. Humidity ratio, W = 0.622·Pₜ/(Pₔₜ₞−Pₜ), Pₔₜ₞=101.325 kPa: $$W_{room}=\frac{0.622\times1451}{101\,325-1451}=9.04\ \text{g/kg (dry air)},\quad W_{out}=\frac{0.622\times132}{101\,325-132}=0.81\ \text{g/kg}$$
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 $$\boxed{\dot m_{moisture}=\dot m_{da}(W_{room}-W_{out})=0.1113\times(0.00904-0.00081)=9.16\times10^{-4}\ \text{kg/s} = 3.30\ \text{kg/hr}}$$ Since Wₛₔₔ₁ > W₀₦ₜ, this is moisture LOST by the room.
5. Sensible heat, using humid specific heat cₖ ≈ 1.02 kJ/(kg·K): $$\boxed{Q_{sens}=\dot m_{da}\,c_p\,(T_{room}-T_{out})=0.1113\times1.02\times37=4.20\ \text{kW}}$$ Warm room air leaving is replaced by much colder outside air, so this is sensible heat LOST by the room.
6. Latent heat, using hᵤᵤ ≈ 2501 kJ/kg: $$\boxed{Q_{lat}=\dot m_{da}\,h_{fg}\,(W_{room}-W_{out})=0.1113\times2501\times0.00823=2.29\ \text{kW}}$$ Moist room air leaving carries its latent heat with it, so this is also latent heat LOST by the room.
7. Total. $$Q_{total}=Q_{sens}+Q_{lat}=4.20+2.29=6.49\ \text{kW, lost by the room.}$$