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 satisfy four independent criteria simultaneously, over the whole enclosure rather than 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 building enclosure — a perfect sheet material with unsealed penetrations does not function as 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 both its air-tightness and its structural adequacy for the enclosure's full design service life, resisting UV exposure during construction, thermal cycling, and loss of adhesion at seams and laps. A system missing any single one of these four criteria (e.g., an impermeable but discontinuous membrane) fails to control air leakage in practice however good its rated material-level permeance.

Part (A)(ii) — stack effect in a tall building. Stack effect arises because warm interior air is LESS DENSE than cold exterior air, creating a buoyancy-driven pressure difference that grows with building height. In winter, this pressure difference PRESSURIZES the upper portion of the building relative to outside (driving exfiltration of warm, moisture-laden air out through the upper enclosure and roof) and DEPRESSURIZES the lower portion (driving infiltration of cold outside air in through the lower enclosure and entrance doors); somewhere between the two lies a NEUTRAL PRESSURE PLANE (NPP) where the interior and exterior pressures are equal. The magnitude of the stack pressure at any height is approximately $$\Delta P_{stack} \propto H\left(\frac{1}{T_{out}}-\frac{1}{T_{in}}\right)$$ so it scales directly with the building's height H above (or below) the NPP and with the indoor–outdoor temperature difference — which is exactly why stack effect is a comparatively minor nuisance in a two-storey house but a dominant, sometimes structurally significant driver of air leakage, door-opening forces, and elevator/stairwell smoke movement in a high-rise. The effect REVERSES in summer whenever the interior is cooler than the exterior (a much weaker reversal in practice, since the summer indoor–outdoor ΔT is normally far smaller than the winter one). Tall-building enclosure design must therefore provide a continuous air barrier capable of resisting a pressure difference that varies systematically with FLOOR LEVEL, not a single uniform design pressure, and compartmentalize the vertical shafts (elevator, stair, service risers) that would otherwise short-circuit the stack flow path from bottom to top of the building.

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 own 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): $$P_{sat}(22^\circ\text{C})=2639\ \text{Pa} \Rightarrow P_{v,room}=0.55\times2639=1451\ \text{Pa}$$ $$P_{sat}(-10^\circ\text{C})=260\ \text{Pa} \Rightarrow P_{v,out}=0.80\times260=208\ \text{Pa}$$
2. Humidity ratio, W = 0.622·P_v/(P_atm−P_v), P_atm=101.325 kPa: $$W_{room}=\frac{0.622\times1451}{101\,325-1451}=9.04\ \text{g/kg (dry air)},\quad W_{out}=\frac{0.622\times208}{101\,325-208}=1.28\ \text{g/kg}$$
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 $$\boxed{\dot m_{moisture}=\dot m_{da}(W_{room}-W_{out})=0.1113\times(0.00904-0.00128)=8.64\times10^{-4}\ \text{kg/s} = 3.11\ \text{kg/hr}}$$ Since W_room > W_out, this is moisture LOST by the room.
5. Sensible heat, using humid specific heat c_p ≈ 1.02 kJ/(kg·K): $$\boxed{Q_{sens}=\dot m_{da}\,c_p\,(T_{room}-T_{out})=0.1113\times1.02\times32=3.63\ \text{kW}}$$ Warm room air leaving is replaced by much colder outside air, so this is sensible heat LOST by the room.
6. Latent heat, using h_fg ≈ 2501 kJ/kg: $$\boxed{Q_{lat}=\dot m_{da}\,h_{fg}\,(W_{room}-W_{out})=0.1113\times2501\times0.00776=2.16\ \text{kW}}$$ Moist room air leaving carries its latent heat with it, so this is also latent heat LOST by the room.
7. Total. $$Q_{total}=Q_{sens}+Q_{lat}=3.63+2.16=5.79\ \text{kW, lost by the room.}$$