Question 5 of 8: Stack and wind pressures, and infiltration, in a 10-storey building

Problem 5: Stack and wind pressures, and infiltration, in a 10-storey building (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 130 ft, ten-storey rectangular tower (13 ft per storey), sealed glazing on curtain walls, vestibule entrances on the two long facades, at sea level.

Find. (a) the stack and wind pressure difference on each of the four walls at floors 1, 5 and 10; (b) the infiltration those pressures drive on the same floors.

Approach. Compute the stack gradient from the density difference and place the neutral pressure level at mid-height (which is what “balanced for neutral pressure” means), add the wind pressure wall by wall through surface pressure coefficients, then push the resulting net pressures through a curtain-wall leakage law and, at grade, through a door-traffic model.

1. Part (a) — the stack gradient. With $\rho = p/RT = 39.667/T$ lb/ft³ at 14.696 psia and $1\ \text{in. wg} = 5.2023\ \text{lb}_f/\text{ft}^2$, $$\Delta p_s = C_d\,\frac{39.667}{5.2023}\left(\frac{1}{T_o}-\frac{1}{T_i}\right)(H_{NPL}-H) = 0.65(7.625)(1.9679\times10^{-4})(H_{NPL}-H)$$ $$\Delta p_s = 9.75\times10^{-4}\,(H_{NPL}-H)\ \ \text{in. wg},\qquad H_{NPL}=65\ \text{ft}$$ The 0.65 draft coefficient is what accounts for the floors resisting vertical flow; without it the tower would behave as one open chimney and the gradient would be 50 % higher.
2. Stack pressure at the three floors. Taking each floor at its mid-height (6.5, 58.5 and 123.5 ft): $$\Delta p_{s,1}=9.75\times10^{-4}(58.5)=+0.0571,\quad \Delta p_{s,5}=+0.0063,\quad \Delta p_{s,10}=-0.0571\ \text{in. wg}$$ Positive means outdoor pressure exceeds indoor, i.e. air is pushed in. The tenth floor is the mirror image of the first, as it must be with the neutral level exactly at mid-height.
3. Wind velocity pressure. At 20 °F, $\rho_o = 39.667/479.7 = 0.0827$ lb/ft³, so $$p_v=\frac{\rho_o V^2}{2g_c}=\frac{0.0827(22.0)^2}{2(32.174)}=0.622\ \frac{\text{lb}_f}{\text{ft}^2} =0.1196\ \text{in. wg}$$
4. Wall pressures from the surface pressure coefficients. The wind runs parallel to the short facades, so the two 100 ft walls are windward and leeward and the two 60 ft walls are sidewalls. Using conventional surface-averaged coefficients for a tall rectangular block with the wind normal to a face — $C_p = +0.60$ windward, $-0.30$ leeward, $-0.50$ on the sides: $$\Delta p_w=C_p p_v:\qquad +0.0717\ \text{(windward)},\quad -0.0359\ \text{(leeward)}, \quad -0.0598\ \text{(sides)}\ \ \text{in. wg}$$
5. Add them: the answer to part (a). The net pressure on each wall is the sum, and the result is the table below. Three features are worth reading off it: on floor 1 every wall except the two sides is pushed inward; on floor 5 the stack term has nearly vanished and the wind alone decides, so only the windward wall infiltrates; and on floor 10 the stack has reversed so strongly that even the windward wall is only just positive while the leeward wall exfiltrates at −0.093 in. wg.
6. Part (b) — curtain-wall leakage law. The glazing is airtight, so only the opaque half of each wall leaks. Taking the ASHRAE “average” curtain-wall rating of 0.30 cfm/ft² at 0.30 in. wg with the usual $n = 0.65$ exponent, $$q'' = K\,\Delta p^{0.65},\qquad K=\frac{0.30}{0.30^{0.65}}=0.656\ \frac{\text{cfm}}{\text{ft}^2(\text{in. wg})^{0.65}}$$ The leaking area per floor is $0.5 \times L \times 13$: 650 ft² on a 100 ft wall, 390 ft² on a 60 ft wall. Only walls with a positive net pressure infiltrate; the rest exfiltrate and contribute nothing to the load.
7. Curtain-wall infiltration, floor by floor. $$\text{Floor 1}:\ 0.656(650)(0.1288)^{0.65}+0.656(650)(0.0212)^{0.65}=112.6+34.8=\boxed{147\ \text{cfm}}$$ $$\text{Floor 5}:\ 0.656(650)(0.0781)^{0.65}=\boxed{81\ \text{cfm}}\qquad \text{Floor 10}:\ 0.656(650)(0.0147)^{0.65}=\boxed{27\ \text{cfm}}$$ The trend is the whole point of part (b): the wall leakage falls by a factor of five from grade to roof, because stack and wind reinforce each other at the bottom and oppose each other at the top.
8. Door traffic at grade. Gross floor area is $100\times60\times10=60{,}000$ ft², so at one occupant per 150 ft² there are 400 occupants; at four passages each per 10 hours the traffic is $$n = \frac{400(4)}{10}=160\ \text{passages per hour}$$ shared over the four doors, i.e. 80 per hour through the windward pair and 80 through the leeward pair.
9. Volume admitted per door passage. No door-flow chart is attached, so the passage volume is derived from an orifice model and stated as an assumption. Half a 3 ft × 7 ft leaf is open on average, $A = 10.5\ \text{ft}^2$, for $t = 2.0$ s per passage with a discharge coefficient of 0.60; a vestibule is two door banks in series, so each leaf sees half the facade pressure: $$V_p = C_D A\, t \sqrt{\frac{2\Delta p_{leaf}}{\rho_o}}$$ On the windward facade $\Delta p_{leaf} = 0.0644$ in. wg $= 0.335\ \text{lb}_f/\text{ft}^2$, giving $v = 16.1$ ft/s and $V_p = 203\ \text{ft}^3$; on the leeward facade $\Delta p_{leaf} = 0.0106$ in. wg, giving $v = 6.5$ ft/s and $V_p = 83\ \text{ft}^3$.
10. Door infiltration and the floor-1 total. $$q_{\text{door}}=\frac{80(203)+80(83)}{60}=271+110=\boxed{381\ \text{cfm}}$$ $$q_{\text{floor 1}}=147+381=\boxed{529\ \text{cfm}}$$ Doors carry 2.6 times the whole curtain-wall leakage of the same floor. That ranking is robust: halving either the open time or the discharge coefficient still leaves the doors dominant, which is the design message — a revolving door or a properly interlocked vestibule buys far more at grade than any tightening of the wall.
11. Sanity check on the whole building. Repeating the wall calculation on all ten floors gives roughly 700 cfm of curtain-wall infiltration; adding the doors, about 1,080 cfm total, or 0.083 air changes per hour on the 780,000 ft³ volume. That is a credible number for a sealed curtain-wall tower and confirms nothing has gone astray by an order of magnitude.

All values in inches of water gauge; positive means outdoor pressure exceeds indoor, i.e. infiltration.