Question 2 of 6

Question 2 (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) — air-flow models for a wall cavity. Two complementary models are used to analyze air movement through a wall's air space, depending on what drives the flow. The ORIFICE-FLOW (power-law) model treats each leakage path (crack, gap, joint) as a discrete opening and relates volumetric flow to the pressure difference across it by $Q=C(\Delta P)^n$, with the flow exponent n ranging from 0.5 (fully turbulent, sharp-edged orifice) to 1.0 (fully laminar, e.g. flow through a porous batt); this is the standard model for AIR-BARRIER-DRIVEN leakage testing (ASTM E283/E779) and for predicting infiltration through known crack geometries. The POROUS-MEDIUM (Darcy) model instead treats the cavity's fill (insulation batt, loose fibre) as a continuum with an effective air permeability, appropriate when the "leakage path" is distributed through a porous material rather than concentrated at discrete cracks; flow is proportional to $\Delta P$ directly (n=1), consistent with laminar Darcy flow. In parallel, a THERMALLY-DRIVEN convection loop can develop within an open (unfilled or lightly filled) cavity even with zero net pressure difference across the wall, driven purely by the temperature difference between the cavity's warm and cold faces setting up a local buoyancy-driven circulation cell that degrades the nominal R-value of an air space; this natural-convection behaviour is analyzed separately from the pressure-driven orifice/Darcy models using a cavity Rayleigh number.

Part (A)(ii) — what cavity air-flow resistance depends on. Resistance to air flow through a wall cavity is governed by: (1) the cavity's GEOMETRY — its width (gap dimension), continuity, and compartmentalization by vertical/horizontal blocking (fire-stopping also interrupts convective loops); (2) the PERMEABILITY of any fill material occupying the cavity — a lightly-filled or fibrous batt cavity is far more resistant to bulk flow than a truly open air space, but also conducts more convective heat if the fill is loose enough to allow circulation; (3) the ROUGHNESS and TORTUOSITY of the leakage path where flow is concentrated at discrete cracks (orifice model), which sets the flow exponent n and the flow coefficient C; and (4) the PRESSURE DIFFERENTIAL itself and its DRIVER (wind gusting vs. a steady stack-effect gradient), since a turbulent, gusting driver interacts differently with a given cavity geometry than a slow steady one. A well-designed drainage/ventilation cavity behind a rainscreen deliberately balances these: enough resistance to avoid becoming a bulk air-leakage bypass path, but enough continuity to drain and dry (Question 6).

Part (B).

Given.

Find. (i) The total heat loss through the 300 m² wall; (ii) the extra fibreglass thickness needed to cut that heat loss by 20%.

Approach. Sum the four conduction/surface-film resistances in series to get the base total resistance and heat loss, then use the fact that heat loss is inversely proportional to total resistance (area and ΔT fixed) to find how much EXTRA fibreglass resistance — and hence thickness — a 20% cut requires.

1. Surface and layer resistances. Standard ASHRAE winter-design surface films (still air inside, h_i=8.29 W/m²K; 24 km/h wind outside, h_o=34 W/m²K) and layer conduction R=L/k: $$R_{si}=\frac{1}{8.29}=0.1206,\ \ R_{brick}=\frac{0.100}{0.9}=0.1111,\ \ R_{fg}=\frac{0.100}{0.04}=2.500,\ \ R_{gyp}=\frac{0.010}{0.16}=0.0625,\ \ R_{so}=\frac{1}{34}=0.0294\ \ (\text{m}^2\text{K/W})$$
2. Total resistance and heat flux. $$R_{total}=0.1206+0.1111+2.500+0.0625+0.0294=2.824\ \text{m}^2\text{K/W}$$ $$q''=\frac{\Delta T}{R_{total}}=\frac{20-(-15)}{2.824}=12.40\ \text{W/m}^2$$
3. (i) Total heat loss. $$\boxed{Q = q''\times A = 12.40\times300 = 3719\ \text{W}\approx 3.72\ \text{kW}}$$
4. (ii) Set up the 20%-reduction condition. With A and ΔT unchanged, $Q\propto 1/R_{total}$, so cutting Q by 20% (to 0.8Q) requires $$R'_{total}=\frac{R_{total}}{0.8}=\frac{2.824}{0.8}=3.530\ \text{m}^2\text{K/W}$$ i.e. an ADDED resistance of $\Delta R = 3.530-2.824=0.706\ \text{m}^2\text{K/W}$, supplied entirely by extra fibreglass thickness (the only layer being changed).
5. Convert the added resistance to thickness using $\Delta R = \Delta L_{fg}/k_{fg}$: $$\Delta L_{fg} = \Delta R\times k_{fg} = 0.706\times0.04 = 0.0282\ \text{m} = 28.2\ \text{mm}$$ $$\boxed{L_{fg,new} = 100+28.2 = 128.2\ \text{mm (i.e. the fibreglass blanket must roughly, but not quite, double)}}$$