Question 6 of 8: Summer cooling load method; sheathing thickness and vapour control

Question 6: Summer cooling load method; sheathing thickness and vapour control (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) — calculating the summer cooling load (10 points)

The summer load calculation differs from the winter one in a way that governs the whole method: in winter the loss is essentially steady and can be treated as a simple conductance problem, whereas in summer the driving influences — solar radiation, outdoor temperature swing and occupancy — all vary through the day, and the massive elements of the building store and release that energy with a delay of several hours. The distinction that must be made first is therefore between heat gain, the rate at which energy enters the space at a given instant, and cooling load, the rate at which energy must be removed to hold the room temperature constant. Radiant heat gain does not become cooling load until it has been absorbed by the floor, walls and furnishings and released again by convection, so the cooling load peak is smaller and later than the heat gain peak. Only convective gains — infiltration, ventilation air delivered to the space, and the convective fraction of people, lights and equipment — convert immediately.

The calculation proceeds in five stages. First the design conditions are fixed: the outdoor dry-bulb and coincident wet-bulb temperatures for the chosen risk level (ASHRAE gives 0.4%, 1% and 2% annual percentiles by station, and Ottawa or Toronto data would be taken from Chapter 14 of the Handbook or from the supplementary climatic tables of the National Building Code), the daily range, the latitude, and the indoor conditions from ASHRAE Standard 55 — typically 24–26 °C at 50% relative humidity.

Second, the envelope gains are computed hour by hour. For opaque walls and roofs the classical approach uses the cooling load temperature difference, $q = U\,A\,\mathrm{CLTD}_{corrected}$, in which the tabulated CLTD already contains the combined effect of solar absorption, outdoor air temperature and the thermal lag of the construction, and is corrected for latitude, month, colour, indoor and outdoor design temperatures. Modern practice replaces this with the radiant time series method, which convolves the hourly conduction heat gain with a set of conduction time factors and then converts the radiant part to load through radiant time factors; the heat balance method solves the surface energy balances directly and is what building simulation software uses. Fenestration is treated separately and usually dominates: the transmitted solar gain is the product of the incident radiation, the glass area, the solar heat gain coefficient and the shading factor, and it is converted to load through cooling load factors that account for the mass of the room.

Third, the internal gains are assembled. People contribute both sensible and latent heat at rates that depend on activity, roughly 70 W sensible and 45 W latent each for seated restaurant occupancy. Lighting gain is the input wattage multiplied by a use factor and a special allowance factor for ballasts, with a cooling load factor applied because much of the output is radiant. Appliances and equipment are taken at their measured or nameplate load with a diversity factor. Fourth, infiltration and ventilation are added, the ventilation air being a coil load rather than a room load if it is treated in the air handler. Fifth, the hourly totals for each orientation are summed and the calculation is repeated for several hours and months, because the peak room load, the peak zone load and the peak plant load generally occur at different times; a west-facing zone peaks in late afternoon while the building as a whole may peak at mid-afternoon.

Finally, the sensible and latent components are kept separate throughout, since together they define the room sensible heat factor that fixes the supply air condition, and a safety allowance is added for fan heat, duct gain and duct leakage. The result is a set of loads at the peak hour from which the airflow, the coil and the refrigeration plant are sized.

Part (b) — sheathing thickness to prevent freezing (10 points)

Given. A wall of 4 in. face brick, pressed fibre board sheathing of conductivity $k = 0.44$ Btu·in./ft²·h·°F, a 3.5 in. air space containing water pipes, and 0.5 in. lightweight gypsum plaster on 0.5 in. plaster board. Inside air is at 70 °F and outside air at −15 °F.

Find. The sheathing thickness that keeps the pipes above freezing.

Approach. The pipes sit in the air space, so the governing temperature is that of the cold face of the cavity, i.e. the inner surface of the sheathing. Temperature falls through a series assembly in proportion to resistance, so require the fraction of the total resistance lying outboard of that plane to correspond to a drop of no more than $70 - 32 = 38\ ^\circ\text{F}$ below the inside air.

1. Set up the proportionality. With $R_{out} = 0.17 + 0.44 = 0.61$ outboard of the sheathing, $R_s = x/0.44$ for the sheathing itself and $R_{in} = 1.01 + 0.32 + 0.45 + 0.68 = 2.46$ inboard of the cavity face, $$\frac{t_{cavity} - t_o}{t_i - t_o} = \frac{R_{out} + R_s}{R_{out} + R_s + R_{in}}$$The requirement is $t_{cavity} \ge 32\ ^\circ\text{F}$, i.e. the left-hand side must reach $$\frac{32 - (-15)}{70 - (-15)} = \frac{47}{85} = 0.5529$$
2. Solve for the sheathing resistance. $$0.61 + R_s = 0.5529\,(0.61 + R_s + 2.46) \;\Rightarrow\; 0.4471\,R_s = 1.087 \;\Rightarrow\; R_s = 2.43\ \text{ft}^{2}\!\cdot\!\text{h}\!\cdot\!{}^\circ\text{F/Btu}$$
3. Convert to a thickness. $$x = R_s\,k = 2.43\,(0.44) = \boxed{1.07\ \text{in.}}$$so a nominal 1⅝ in. board is specified. The completed wall then has $$R_T = 5.50\ \text{ft}^{2}\!\cdot\!\text{h}\!\cdot\!{}^\circ\text{F/Btu}, \qquad U = 0.182\ \text{Btu/ft}^{2}\!\cdot\!\text{h}\!\cdot\!{}^\circ\text{F}$$and a design heat flux of $q = 0.182(85) = 15.4$ Btu/h·ft².
4. Check the result. Working the temperature back through the assembly with 1.07 in. of sheathing gives a cavity cold-face temperature of exactly 32 °F, confirming the algebra. Because the pipes exchange heat with both faces of the cavity they will in practice run a little warmer than this, so taking the cold face is the conservative choice.

Moisture flow and vapour barriers. Water vapour crosses a wall by two quite different mechanisms. Diffusion is driven by the vapour pressure difference and is resisted by the permeance of each layer, in exact analogy with heat conduction; air leakage carries vapour bodily through gaps and in Canadian heating climates typically moves an order of magnitude more moisture than diffusion does. Both drive vapour outward in winter, from the warm humid interior toward the cold dry exterior. Condensation occurs wherever the vapour reaches a plane colder than its dew point, and the analysis is done by plotting the temperature profile computed above against the dew-point profile implied by the vapour-pressure gradient: wherever the two cross, moisture accumulates. In the wall above, an indoor condition of 70 °F at 35% relative humidity has a dew point near 41 °F, and the plaster/air-space interface is already below that, so an unprotected assembly would wet the cavity and the sheathing.

The remedy is a vapour barrier of low permeance placed on the warm side of the insulation — in Canada, the interior face — so that vapour is stopped before it reaches any surface below its dew point. The National Building Code requires a vapour barrier with a permeance not exceeding 60 ng/Pa·s·m² on the warm side of insulated assemblies, and a separate air barrier system that is continuous, structurally supported and sealed at every penetration; polyethylene sheet commonly serves both functions in residential construction. Three rules follow from the physics. Never install a second low-permeance layer on the cold side, or moisture that does get in cannot dry out — the sheathing and cladding should be progressively more permeable outward. Seal the barrier continuously, since a small unsealed gap leaks far more vapour by air transport than the whole intact area passes by diffusion. And ventilate the cavity behind an absorptive cladding such as face brick with a drained rain-screen space, so that solar-driven inward vapour and any incidental water can escape.