Office heat loss, and moisture flow through wall structures: 22-Mec-B2 Environmental Control in Buildings, December 2014, Q5

Question 5: Office heat loss, and moisture flow through wall structures (20 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. A single office cell bounded by one external wall with a window, one wall to a corridor, and party surfaces to identical offices above, below and on both sides.

Given data — Problem 5(a)
Quantity	Symbol	Value
External wall length × height	—	5 m × 3 m
Window area	$A_g$	4 m²
Room depth (window to corridor wall)	—	4 m
Room / corridor / outdoor temperature	$t_r$ / $t_c$ / $t_o$	$20^\circ\text{C}$ / $16^\circ\text{C}$ / $-1^\circ\text{C}$
$U$ external wall / window / internal wall	$U_w$ / $U_g$ / $U_i$	1.0 / 5.6 / 2.7 W/m²K
Ventilation rate	$n$	1 air change per hour

Find. (a) The steady-state heat input the room requires at the design condition; (b) a discussion of moisture transport through wall assemblies and of vapour-barrier installation.

Problem 5(a) — plan of the office cell. Only the external wall, the window, the corridor wall and the ventilation air lose heat; the surfaces to the offices above, below and on either side see no temperature difference and carry no heat.

Approach. Take each bounding surface in turn with its own area, $U$-value and temperature difference — noting that the party surfaces are adiabatic because the neighbouring offices are at the same temperature — then add the ventilation load computed from the room volume and the air change rate.

Establish the areas. The external wall is 5 m long and 3 m high, so its gross area is 15 m², of which the window occupies 4 m²; the net opaque wall is therefore $A_w = 15-4 = 11\ \text{m}^{2}$. The corridor wall is the opposite face, also $5 \times 3 = 15\ \text{m}^{2}$. The room volume is $5 \times 4 \times 3 = 60\ \text{m}^{3}$.
Identify which surfaces carry heat. The offices above, below and on either side are held at the same 20°C, so those five surfaces see zero temperature difference and are treated as adiabatic. Only three fabric elements and the ventilation air are active — recognising this is most of the marks in part (a).
Fabric losses. Applying $q = U A \Delta t$ to each element, with $\Delta t = 21\ \text{K}$ to outdoors and $4\ \text{K}$ to the corridor: $$q_w = 1.0(11)(21) = 231\ \text{W},\qquad q_g = 5.6(4)(21) = 470\ \text{W},\qquad q_i = 2.7(15)(4) = 162\ \text{W}$$ The window alone accounts for more loss than the wall and corridor combined, even though it is a quarter of the external wall area — the price of a 5.6 W/m²K glazing, which is roughly single glazing with a thermally broken frame.
Ventilation loss. One air change per hour of 60 m³ is $60/3600 = 0.0167\ \text{m}^{3}/\text{s}$, and $$q_v = \rho\,c_p\,\dot V\,\Delta t = 1.2(1{,}005)(0.0167)(21) = 422\ \text{W}$$ using $\rho c_p = 1{,}206\ \text{J}/\text{m}^{3}\text{K}$ for air at room conditions. Note that the ventilation air must be heated from the outdoor temperature, not from the corridor temperature.
Total heat input. Summing the fabric and ventilation terms, $$q = 231+470+162+422 = \boxed{1{,}286\ \text{W}}$$ or about 1.29 kW, which is 64 W per square metre of floor — a high but not unreasonable figure for a perimeter office with poor glazing at −1°C.

(b) Moisture flow through wall structures and vapour barriers. Water vapour crosses a wall assembly by two quite different mechanisms, and confusing them is the root of most moisture failures. The first is vapour diffusion, driven by the difference in vapour pressure between the warm humid interior and the cold dry exterior and resisted by the permeance of each layer; it obeys a Fick's-law relation exactly analogous to conduction, with permeance in place of conductance. The second is air leakage, in which humid indoor air is carried bodily through gaps in the assembly by the same stack and wind pressures that drive infiltration. Air leakage typically moves one to two orders of magnitude more moisture than diffusion through the same wall, which is why a continuous air barrier matters more than the vapour barrier in practice, and why the two functions should be thought about separately even when one membrane performs both.

Damage occurs where vapour reaches a surface colder than its dew point and condenses — typically on the back face of the sheathing in a Canadian heating climate. The governing rule for installation follows directly: in a heating-dominated climate the vapour barrier belongs on the warm (interior) side of the insulation, so that vapour never reaches a cold surface at high concentration. The practical requirements are that it be continuous — sealed at laps, at the floor and ceiling junctions, and around every penetration for outlets, ducts and services, since a small hole passes far more moisture than the surrounding membrane does; that it have a permeance low enough to control diffusion (the National Building Code of Canada requires no more than 60 ng/Pa·s·m², about 1 US perm, for a Type I or II barrier); and that the assembly be able to dry outward, which means the exterior layers should be more permeable than the interior ones. The classic errors are installing a second, exterior vapour barrier — an impermeable sheathing membrane or vinyl wall covering — which traps moisture between the two and prevents any drying, and puncturing the interior barrier during electrical rough-in without sealing the repair. Where a wall must be vapour-tight on both faces, or in a cooling-dominated or swimming-pool application where the vapour drive reverses seasonally, the correct response is a hygrothermal analysis of the assembly rather than a rule of thumb.

Problem 5(a) — heat loss by component
Component	Calculation	Heat loss (W)
External wall	$1.0 \times 11 \times 21$	231
Window	$5.6 \times 4 \times 21$	470
Corridor wall	$2.7 \times 15 \times 4$	162
Fabric subtotal	—	863
Ventilation (1 ach of 60 m³)	$1{,}206 \times 0.0167 \times 21$	422
Total heat input	—	1,286 (1.29 kW)

Topic. Room-by-room design heat loss with adiabatic party surfaces, and moisture transport through building envelopes.

Key relations. $q = UA\,\Delta t$ per element; ventilation $q_v = \rho c_p (nV/3600)\,\Delta t$ with $\rho c_p \approx 1{,}206\ \text{J}/\text{m}^{3}\text{K}$; vapour diffusion $g = M\,A\,\Delta p_v$ with $M$ the assembly permeance; condensation when the local temperature falls below the local dew point.

Why this works. A room-by-room heat loss is an accounting exercise over the room's boundary. Each surface contributes in proportion to its area, its conductance and the temperature difference across it, and a surface with no temperature difference contributes nothing — which is why identical offices above, below and beside this one remove five of the eight bounding surfaces from the calculation. Ventilation is handled separately because it is a mass flow rather than a conductance, but it scales with the same temperature difference and is here a third of the total. The moisture discussion is the same physics applied to a second potential: vapour moves down its own pressure gradient, and the designer's job is to keep the point where it would condense either warm or unreachable.

Common pitfall. Using the room-to-outdoor temperature difference on the corridor wall (which would give 850 W instead of 162 W), or forgetting that the ventilation air enters at the outdoor temperature. On part (b), the standard error is placing the vapour barrier on the cold side, which in a Canadian climate converts the membrane from a control layer into a condensing surface.

Source. W. P. Jones, Air Conditioning Engineering, Ch. 5 (heat loss calculations); ASHRAE Handbook — Fundamentals Ch. 25 (Heat, Air and Moisture Control in Building Assemblies) and Ch. 27; National Building Code of Canada, Div. B §9.25 (air barriers and vapour barriers).

Question 5 of 8: Office heat loss, and moisture flow through wall structures

Question 5: Office heat loss, and moisture flow through wall structures (20 points)