Question 7 of 8: Design heat loss from a Toronto conference room (20 marks)

Question 7: Design heat loss from a Toronto conference room (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.

Given. A single conditioned room with two exposed walls, no floor, ceiling or party-wall loss, and infiltration driven by a stated pressure difference.

Find. (a) the indoor and outdoor design conditions, with justification, and (b) the design heat loss from the room.

Approach. Choose the design conditions from Canadian climatic design data, build the wall U-value from the layer resistances, take the glazing from published fenestration data, compute infiltration by the crack method on the stated pressure difference, and sum the three.

1. Part (a) — outdoor design condition. The heating design temperature for Toronto (Pearson) is 0 °F (−18 °C), taken as the ASHRAE 99.6 % annual heating dry bulb, which for that station is −18.2 °C (−0.7 °F); the National Building Code of Canada 2020 January 2.5 % design temperature for Toronto is the same −18 °C. The percentile matters: a heating plant is sized so that it just meets the load for all but roughly 35 hours a year, and picking a colder record temperature would oversize the plant and force it to cycle at part load for the whole season. Design wind is 15 mph, which is the condition the surface films and the crack-flow coefficients below assume; the outdoor air is taken as saturated, giving $W_{o} = 0.00078\ \text{lb/lb}$.
2. Part (a) — indoor design condition, and a humidity warning. Take 72 °F (22 °C) dry bulb, the middle of the ASHRAE Standard 55 winter operative-temperature range for sedentary occupants at 1.0 clo, which is what a seated conference audience wears indoors in winter. Humidity has to be chosen against the glazing, not against comfort: with $U_{g} = 0.98$ and an inside film coefficient of $1/0.68 = 1.47\ \text{Btu/h}\cdot\text{ft}^{2}\cdot^\circ\text{F}$, the inner glass surface sits at $$t_{si} = t_{i} - \frac{U_{g}}{h_{i}}(t_{i} - t_{o}) = 72 - \frac{0.98}{1.47}(72) = 24.0\ ^\circ\text{F}$$ so any indoor relative humidity above about 16 % will frost the single glazing. The practical answer is therefore 72 °F at 20 °C–equivalent comfort with the humidity held near the 20 to 25 % that ASHRAE 55 tolerates, accepting some condensation at the glass edge, and to note in the design report that the single glazing — not the humidifier — is the constraint.
3. Part (b) — areas. Only the two exposed walls lose heat: $$A_{\text{gross}} = 2 \times 120 \times 20 = 4{,}800\ \text{ft}^{2},\qquad A_{\text{win}} = 30 \times 3 \times 5 = 450\ \text{ft}^{2}$$ $$A_{\text{wall}} = 4{,}800 - 450 = 4{,}350\ \text{ft}^{2}$$ The floor, ceiling and the two 60 ft end walls face conditioned space and carry no load by the question's own instruction.
4. Part (b) — the wall U-value. Summing the layer resistances in $\text{h}\cdot\text{ft}^{2}\cdot^\circ\text{F}/\text{Btu}$:

   Wall thermal resistance

   Layer	R
   outside surface film, 15 mph winter	0.17
   face brick, 4 in.	0.43
   concrete block, sand and aggregate, 12 in.	1.28
   fibreglass board insulation, 4 in. at R 4.0 per inch	16.00
   vapour barrier (plastic film)	0.00
   plaster board, ½ in.	0.45
   inside surface film, still air	0.68
   Total	19.01

   $$U_{\text{wall}} = \frac{1}{19.01} = 0.0526\ \frac{\text{Btu}}{\text{h}\cdot\text{ft}^{2}\cdot{}^\circ\text{F}}$$ The vapour barrier contributes nothing thermally — it is there for moisture control, and the fibreglass board alone carries 84 % of the resistance.
5. Part (b) — transmission losses. With $\Delta t = 72 - 0 = 72\ ^\circ\text{F}$, and taking $U_{\text{win}} = 0.98$ for a single-glazed, wood-framed, operable window: $$\dot{Q}_{\text{wall}} = U A \Delta t = 0.0526 \times 4{,}350 \times 72 = 16{,}476\ \text{Btu/h}$$ $$\dot{Q}_{\text{win}} = 0.98 \times 450 \times 72 = 31{,}752\ \text{Btu/h}$$ The windows occupy 9 % of the exposed wall area and lose almost twice as much as the other 91 % — a ratio of nineteen to one in loss per square foot.
6. Part (b) — infiltration by the crack method. A double-hung sash leaks along its perimeter and along the meeting rail, so the crack length per window is $$L_{c} = 2H + 3W = 2(5) + 3(3) = 19\ \text{ft}$$ Only the windward wall infiltrates — the leeward wall exfiltrates — so 15 windows are counted, giving 285 ft of crack. (The alternative convention of taking half the total crack length gives $\tfrac{1}{2}(30)(19) = 285\ \text{ft}$ as well, so the two rules agree here.) The crack flow follows a power law in the pressure difference, $$\frac{\dot{V}}{L_{c}} = K\,(\Delta p)^{n} = 3.4\,(0.2)^{0.65} = 1.194\ \frac{\text{cfm}}{\text{ft}}$$ with $K = 3.4$ and $n = 0.65$ for an average-fit, non-weather-stripped double-hung wood sash; the caulked frame-to-wall joint means no separate frame leakage need be added. Hence $$\dot{V}_{\text{inf}} = 1.194 \times 285 = \boxed{340\ \text{cfm}}$$ which is 0.14 air changes per hour on the 144,000 ft³ room — a believable figure for leaky sash windows, and a useful sanity check that the coefficient has been applied correctly.
7. Part (b) — the infiltration load and the total. $$\dot{Q}_{\text{inf}} = 1.08\,\dot{V}\,\Delta t = 1.08 \times 340 \times 72 = 26{,}470\ \text{Btu/h}$$ $$\dot{Q}_{\text{total}} = 16{,}476 + 31{,}752 + 26{,}470 = \boxed{74{,}700\ \text{Btu/h}} \equiv 21.9\ \text{kW}$$ Expressed against the floor area that is 10.4 Btu/h per square foot, which is a high but entirely plausible intensity for a 1960s-vintage single-glazed perimeter space.
8. Interpret the split, and note what has been left out. Windows take 42.5 % of the loss, infiltration 35.4 % and the insulated wall only 22.1 %. Replacing the single glazing with a sealed double unit at $U = 0.49$ would remove 15,900 Btu/h and weather-stripping the sashes would remove roughly half the infiltration — together a 34 % cut, for far less money than any change to the already well-insulated wall. Two deliberate omissions should be declared: the humidification load, which at 20 % indoor relative humidity is a further $4{,}840 \times 340 \times (0.00331 - 0.00078) = 4{,}160\ \text{Btu/h}$ of latent duty and is normally carried by the central plant rather than by the room; and any internal gain from the occupants, which a design heat-loss calculation properly ignores because the plant must be able to bring the room up from cold before they arrive.