Question 7 of 8: Stack and wind pressures and infiltration in a 20-storey tower

Question 7: Stack and wind pressures and infiltration in a 20-storey tower (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 sealed-curtain-wall office tower with vestibule entrances at grade, a balanced ventilation system, and a winter design condition with wind.

Find. (a) The stack and wind pressure differences on each wall at floors 1, 10 and 20; (b) the total infiltration rate at each of those floors.

Approach. Compute the indoor and outdoor air densities, place the neutral pressure level at mid-height because the ventilation system is balanced, evaluate the stack pressure at each floor's mid-height and the wind pressure on each wall from the dynamic pressure and its pressure coefficient, add the two to get the net driving pressure, and convert that pressure into flow through the published curtain-wall leakage characteristic and through an orifice model of the doors.

1. Air densities and the storey height. From the perfect-gas law at standard barometric pressure, $\rho_o = 0.0827$ and $\rho_i = 0.0749\ \text{lbm}/\text{ft}^{3}$ at 20°F and 70°F respectively, a difference of $0.00781\ \text{lbm}/\text{ft}^{3}$. The storey height is $220/20 = 11\ \text{ft}$, so the mid-heights of floors 1, 10 and 20 are 5.5, 104.5 and 214.5 ft above grade.
2. Locate the neutral pressure level. The ventilation system is balanced for neutral pressure, meaning supply and exhaust are equal and the building neither pressurises nor depressurises itself. The neutral pressure level therefore sits at mid-height, $H_{\mathrm{NPL}} = 110\ \text{ft}$, and the stack effect drives air inward below it and outward above it.
3. (a) Stack pressure difference. With heights in feet and densities in lbm/ft³, and the constant 5.202 converting lbf/ft² to inches of water, $$\Delta p_s = C_d\,\frac{(\rho_o-\rho_i)(H_{\mathrm{NPL}}-H)}{5.202}$$ which gives $+0.1019$ in. wg at floor 1, $+0.0054$ at floor 10 and $-0.1019$ at floor 20. The draft coefficient of 0.65 accounts for the flow resistance of the floors and shafts, which prevents the building from behaving as a single open chimney.
4. (a) Wind pressure difference. The wind speed is $15\ \text{mph} = 22.0\ \text{ft}/\text{s}$, so the free-stream dynamic pressure is $$p_v = \frac{\rho_o U^{2}}{2 g_c \times 5.202} = \frac{0.0827(22.0)^{2}}{2(32.174)(5.202)} = 0.1196\ \text{in. wg}$$ Because the wind runs parallel to the 60-ft facades, the two 120-ft facades are the windward and leeward walls; the 60-ft facades are sidewalls. With the standard pressure coefficients $C_p = +0.60$, $-0.30$ and $-0.65$, the wind contributions are $+0.0717$, $-0.0359$ and $-0.0777$ in. wg, independent of height for this purpose.
5. (a) Net pressure across each wall. Adding stack and wind gives the table below. The pattern is the one the question is testing: at grade everything is positive and air comes in on all four sides; at mid-height only the windward wall still admits air; at the top every wall is exfiltrating. The sidewalls are the most sensitive, because their large negative wind coefficient cancels most of the stack effect even at floor 1.
6. (b) Curtain-wall leakage. Fixed, airtight glazing means the leakage is through the opaque curtain wall, taken at the ASHRAE tight-curtain-wall value of 0.15 cfm/ft² at a reference 0.30 in. wg with a flow exponent $n = 0.65$: $$Q = 0.15\,A\left(\frac{\Delta p}{0.30}\right)^{0.65}\ \text{cfm, for }\Delta p \gt 0$$ The leakage area per floor is the wall area less the glazed half: $120(11)(0.5) = 660\ \text{ft}^{2}$ on each long facade and $60(11)(0.5) = 330\ \text{ft}^{2}$ on each short one. Walls under negative net pressure exfiltrate and contribute nothing to infiltration.
7. (b) Door infiltration at grade. The building holds $20(60)(120)/150 = 960$ occupants, each making 4 passages per 10 hours, so the traffic is 384 passages per hour shared over the four doors. A vestibule is two door banks in series, so each leaf sees half the net wall pressure. Treating an opening leaf as an orifice of effective area $21\ \text{ft}^{2} \times 0.5$ with $C_d = 0.60$ and an open time of 2.0 s per passage, $$V_{\mathrm{passage}} = C_d\,A_{\mathrm{eff}}\sqrt{\frac{2\,\Delta p_{\mathrm{leaf}}}{\rho_o}}\;t_{\mathrm{open}}$$ which is 236 ft³ per windward passage and 146 ft³ per leeward passage, giving 756 and 466 cfm respectively.
8. (b) Total infiltration by floor. Adding the wall and door contributions, $$Q_1 = 126 + 1{,}222 = \boxed{1{,}348\ \text{cfm}},\qquad Q_{10} = \boxed{41\ \text{cfm}},\qquad Q_{20} = \boxed{0}$$ Floor 20 lies entirely above the neutral pressure level with every wall under net outward pressure, so it exfiltrates rather than infiltrates.

The numbers carry a clear design message. On the ground floor the doors admit roughly ten times as much cold air as the entire curtain wall of that floor, which is why entrance design — vestibules, revolving doors, air curtains and properly sized entrance heating — dominates the comfort and the heating load of a tower lobby in a Canadian winter. Higher up, infiltration collapses to a few tens of cfm and then reverses, so the top floors of the building are exfiltrating warm humid air into the curtain wall — a moisture problem rather than a heating one, and the reason condensation and frost damage in tall buildings is usually found near the top. The neutral pressure level is the hinge, and because it is set by the ventilation system's balance, it is one of the few things in this calculation the designer can actually move.