Question 7 of 8: Dilution ventilation for an indoor NO x source; stack and wind pressures on a tall building

Question 7: Dilution ventilation for an indoor NO_x source; stack and wind pressures on a tall building (20 marks: a — 5, b — 15)

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) — outdoor airflow needed to dilute the NO_x source (5 marks)

Given. Indoor generation rate S = 115 µg/s; outdoor (supply) concentration C_o = 55 µg/m³; the room is perfectly mixed and at steady state.

Find. The outdoor airflow rate that holds the indoor concentration at the recommended guideline value.

1. Write the steady-state mass balance on the well-mixed space. Contaminant in with the ventilation air plus contaminant generated equals contaminant out with the exhaust: $$\dot V C_o + S = \dot V C_i \quad\Longrightarrow\quad \dot V = \frac{S}{C_i - C_o}$$ The result is independent of the room volume, which is why none is given: at steady state the volume sets only how quickly the concentration gets there, not where it settles.
2. Select the guideline concentration. Because these are Canadian examinations, the governing value is Health Canada's Residential Indoor Air Quality Guideline for nitrogen dioxide, 170 µg/m³ as a one-hour exposure (the long-term value of 20 µg/m³ is below the stated outdoor concentration and therefore unreachable by dilution — a point worth making explicitly rather than silently). Substituting, $$\dot V = \frac{115}{170 - 55} = \frac{115}{115} = \boxed{1.00\ \text{m}^3\!/\text{s}}$$ that is 1 000 L/s, or about 2 120 cfm.
3. Report the alternative standard, because the answer is sensitive to it. ASHRAE 62.1 Table 4-1 adopts the US national ambient standard for NO₂ of 100 µg/m³ as an annual mean, which gives 115/(100 − 55) = 2.56 m³/s (5 415 cfm), and the World Health Organization's one-hour value of 200 µg/m³ gives 0.79 m³/s. The three answers span a factor of three, so the standard must be named with the number.
4. Note what the arithmetic reveals about the design. The outdoor air is already at one-third of the guideline, so only 115 µg/m³ of headroom is available for dilution. Halving the outdoor concentration to 27 µg/m³ would cut the required airflow to 0.80 m³/s, whereas raising it to 100 µg/m³ would demand 1.64 m³/s — and at 170 µg/m³ outdoors no finite airflow would work at all. That hyperbolic sensitivity is the practical argument for source control before dilution ventilation: capturing the NO_x at source with local exhaust, or replacing a combustion appliance with an electric one, removes the problem instead of diluting it, and costs far less to run than conditioning 1 m³/s of outdoor air through a Canadian winter.

Part (b) — stack and wind pressure differences on the first and tenth floors (15 marks)

Given.

Find. The indoor-minus-outdoor pressure difference on the first and tenth floors, separating the stack and wind contributions.

Approach. Compute the two mechanisms separately — the stack pressure from the indoor-to-outdoor air density difference acting over the height above or below the neutral pressure level, and the wind pressure from the free-stream velocity pressure scaled by a surface pressure coefficient — then add them algebraically for each face and floor.

1. Fix the two air densities. Taking 70 °F indoors and 0 °F outdoors to give the stated 70 °F difference, and treating the air as dry at sea level, $$\rho = \frac{p}{R_a T}: \qquad \rho_i = \frac{2\,116.2}{53.35 \times 529.67} = 0.07489\ \text{lb/ft}^3, \qquad \rho_o = \frac{2\,116.2}{53.35 \times 459.67} = 0.08629\ \text{lb/ft}^3$$ $$\Delta\rho = \rho_o - \rho_i = 0.01140\ \text{lb/ft}^3$$ The cold outdoor air is 15 % heavier, and that density difference is the entire stack effect.
2. Locate the neutral pressure level. For a building with leakage distributed uniformly over its height and no imbalance between mechanical supply and exhaust, the plane at which indoor and outdoor pressures are equal sits at mid-height: h_NPL = 50 ft. This is an assumption, and a consequential one — the grade-level vestibule doors named in the question are a large low-level opening that in practice pushes the neutral level upward, worsening ground-floor infiltration. The sensitivity is given in the callout below.
3. Compute the stack pressure at each floor. With the sign convention that positive means the outdoor pressure exceeds the indoor (so air is driven in), $$\Delta p_s = \Delta\rho\,g\,(h_{NPL} - h)$$ Evaluating at the mid-height of each storey — 5 ft for the first floor and 95 ft for the tenth — both are 45 ft from the neutral level, so the magnitudes are equal and the signs opposite: $$\Delta p_{s,1} = 0.01140 \times 45 = 0.513\ \text{lbf/ft}^2 = \boxed{+0.099\ \text{in.\ wg}}$$ $$\Delta p_{s,10} = \boxed{-0.099\ \text{in.\ wg}}$$ using 1 in. wg = 5.202 lbf/ft². Air is driven into the first floor and out of the tenth — the classic winter chimney behaviour of a tall building.
4. Compute the free-stream velocity pressure. At 15 mph = 22.0 ft/s in the cold outdoor air, $$p_v = \frac{\rho_o U^2}{2g_c} = \frac{0.08629 \times 22.0^2}{2 \times 32.174} = 0.6486\ \text{lbf/ft}^2 = 0.125\ \text{in.\ wg}$$ The tabulated shortcut p_v = 0.000482 U² gives 0.108 in. wg because it is written for standard 0.075 lb/ft³ air; the 15 % denser winter air raises it by the same 15 %.
5. Apply the surface pressure coefficients. The wall pressure is Δp_w = C_pp_v, so for the standard average coefficients on a rectangular building, $$\Delta p_{w} = \begin{cases} +0.6 \times 0.125 = +0.075\ \text{in.\ wg} & \text{windward (the 150 ft face)}\\ -0.3 \times 0.125 = -0.037\ \text{in.\ wg} & \text{leeward}\\ -0.5 \times 0.125 = -0.062\ \text{in.\ wg} & \text{sidewalls (the 50 ft faces)} \end{cases}$$ Note that only the windward face is pushed inward; the other three are in the wake and are suctioned, so wind assists exfiltration over three-quarters of the building's perimeter.
6. Add the two mechanisms for each floor and face. Since both are pressures acting on the same envelope, they superpose:

   Floor	Face	Stack (in. wg)	Wind (in. wg)	Net (in. wg)	Direction
   1st (5 ft)	windward	+0.099	+0.075	+0.174	infiltration
   	leeward	+0.099	−0.037	+0.061	infiltration
   	sidewall	+0.099	−0.062	+0.036	infiltration
   10th (95 ft)	windward	−0.099	+0.075	−0.024	slight exfiltration
   	leeward	−0.099	−0.037	−0.136	exfiltration
   	sidewall	−0.099	−0.062	−0.161	exfiltration

   The two extremes are the ones a designer needs: +0.174 in. wg driving air into the windward ground floor and −0.161 in. wg driving it out of the tenth-floor sidewall. The whole first floor infiltrates on every face because stack dominates there, while on the tenth floor the windward wall is nearly balanced — wind almost exactly cancels stack, and a small change in either would reverse the flow through those windows.
7. Interpret the numbers as a design engineer would. A pressure difference of 0.17 in. wg across a swinging door is about 5 lbf of opening force on a 3 ft × 7 ft leaf, near the limit at which doors become difficult and start to be held open — which is precisely why the question specifies double vestibule-type doors: two door banks in series each see half the total difference. On the tenth floor the same magnitude of difference acting outward is what drives moisture-laden indoor air into the curtain-wall cavity, where it meets a cold surface; that, not heat loss, is the usual reason top-floor curtain walls fail. The standard remedies are exactly the ones the question hints at: vestibules and revolving doors at grade, compartmentation of shafts and stairwells to break the vertical connection, and a slight net mechanical pressurisation of the building to shift the neutral level below the ground floor.