Question 4 of 8: Tall-building infiltration — stack effect, wind and door traffic

Question 4: Tall-building infiltration — stack effect, wind and door traffic (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.

Find. For floors 1, 10 and 22: the stack-effect and wind pressure differences acting on each of the four walls, and the resulting total infiltration rate through the curtain wall and (on the ground floor) the entrance doors.

Approach. Because the ventilation system is balanced, the neutral pressure level sits at mid-height, 125 ft. Compute the stack pressure at each floor from the density difference, add the wind pressure wall-by-wall using surface-averaged pressure coefficients, and apply the net inward pressure to the leakage characteristic of the opaque curtain wall. Ground-floor doors are handled separately by the traffic-rate method.

1. Air densities and the neutral pressure level. With $p = 14.696$ psia and $R = 53.35\ \text{ft}\cdot\text{lbf/(lbm}\cdot\text{R)}$, $$\rho_o = \frac{144 \times 14.696}{53.35 \times 479.67} = 0.0827\ \text{lbm/ft}^3,\qquad \rho_i = \frac{144 \times 14.696}{53.35 \times 534.67} = 0.0742\ \text{lbm/ft}^3$$ Because the mechanical ventilation is balanced for neutral pressure and the leakage is uniformly distributed over the height, the neutral pressure level is at mid-height, $H_{\text{NPL}} = 125\ \text{ft}$.
2. Stack pressure at each floor. For a building with resistance to vertical flow the theoretical stack pressure is reduced by the draft coefficient: $$\Delta p_s = C_d\,(\rho_o - \rho_i)\,(H_{\text{NPL}} - H) = 0.65\,(0.0085)(125 - H)\ \text{lbf/ft}^2$$ Dividing by 5.202 lbf/ft² per inch of water and evaluating at the mid-height of each floor ($H$ = 5.7, 108.0 and 244.3 ft): $$\Delta p_{s,1} = +0.127,\qquad \Delta p_{s,10} = +0.018,\qquad \Delta p_{s,22} = -0.127\ \text{in. wg}$$ Positive means the outdoor pressure exceeds the indoor pressure, i.e. air is driven in. Floors 1 and 10 lie below the NPL; floor 22 lies above it and therefore exfiltrates on stack effect alone.
3. Wind pressure on each wall. The wind blows parallel to the 150-ft facades, so it is normal to the 200-ft facades: those become the windward and leeward walls and the 150-ft facades are side walls. At 15 mph = 22.0 ft/s the free-stream dynamic pressure is $$p_v = \frac{\rho_o U^2}{2g_c} = \frac{0.0827 \times 22.0^2}{2 \times 32.174} = 0.622\ \text{lbf/ft}^2 = 0.120\ \text{in. wg}$$ Applying surface-averaged wall pressure coefficients for a tall building with the wind normal to a face ($C_p = +0.80$ windward, $-0.40$ leeward, $-0.70$ on the side walls): $$\Delta p_{w} = C_p\,p_v \;\Rightarrow\; +0.096\ \text{(windward)},\quad -0.048\ \text{(leeward)},\quad -0.084\ \text{in. wg (sides)}$$ These are independent of floor level to the accuracy of this calculation (see the note below on the wind-speed profile).
4. Net pressure difference on each wall. Stack and wind superpose, so $\Delta p_{\text{net}} = \Delta p_s + \Delta p_w$, and only walls with $\Delta p_{\text{net}} > 0$ admit air. The table that follows part (a) collects the nine wall/floor combinations. The pattern is the classic winter one: at grade every wall infiltrates; near the NPL only the windward wall does; at the top the whole envelope exfiltrates.
5. Curtain-wall leakage characteristic. Taking an average curtain wall, ASHRAE gives 0.15 cfm per ft² of wall at a reference 0.30 in. wg, with the usual flow exponent $n = 0.65$: $$\dot Q = 0.15\,A_{\text{leak}} \left(\frac{\Delta p}{0.30}\right)^{0.65}$$ Because the windows are fixed and airtight and the window-wall ratio is 0.5, only half of each wall leaks: $$A_{\text{leak, 200-ft wall}} = 200 \times 11.36 \times 0.5 = 1136\ \text{ft}^2,\qquad A_{\text{leak, 150-ft wall}} = 852\ \text{ft}^2$$
6. Curtain-wall infiltration, floor by floor (part b). Substituting each positive net pressure: $$\text{Floor 1: } 140 + 72 + 2(36) = \boxed{284\ \text{cfm}}\qquad \text{Floor 10: } \boxed{91\ \text{cfm}}\ \text{(windward only)}$$ $$\text{Floor 22: } \boxed{0\ \text{cfm}}\ \text{(every wall exfiltrates)}$$
7. Door infiltration at grade. The building holds $N_{\text{occ}} = 22 \times 150 \times 200 / 150 = 4400$ occupants, each making 4 passages per 10 h, so the traffic rate is $$\dot n = \frac{4400 \times 4}{10} = 1760\ \text{persons/h} = 440\ \text{per door}$$ Modelling each passage as an effective opening of half the 3 ft × 7 ft leaf area, held open for an equivalent 2.0 s, with a discharge coefficient of 0.60, and noting that a vestibule puts two door banks in series so each leaf sees half the net pressure: $$V_{\text{passage}} = C_dA_{\text{eff}}\sqrt{\tfrac{2\,\Delta p_{\text{leaf}}}{\rho_o}}\;t \;\Rightarrow\; 267\ \text{ft}^3 \text{ (windward)},\quad 159\ \text{ft}^3 \text{ (leeward)}$$ $$\dot Q_{\text{doors}} = 2(1961) + 2(1168) = \boxed{6260\ \text{cfm}}$$
8. Total infiltration for the three floors (part b). Adding the door flow, which occurs only at grade: $$\dot Q_1 = 284 + 6260 = \boxed{6540\ \text{cfm}},\qquad \dot Q_{10} = \boxed{91\ \text{cfm}},\qquad \dot Q_{22} = \boxed{0\ \text{cfm}}$$ The dominant result is that the entrance doors admit more than twenty times the curtain-wall leakage of the whole ground floor. That is the engineering message of the problem: in a busy tower the lobby, not the envelope, is the infiltration load, which is why revolving doors, longer vestibules, entrance heating and lobby pressurisation earn their cost.

Collecting the pressures asked for in part (a):