Question 5 of 8: Ventilation, infiltration and attic ventilation

Question 5: Ventilation, infiltration and attic ventilation (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.

Part (a) — ventilation and infiltration heat loss (10 points)

Given. A fast-food cafeteria in Ottawa, Ontario, 30 × 100 × 9 ft, with windows and doors on the east and north elevations only and a humidified HVAC system. Winter design conditions are 72 °F and 30% RH inside and −7 °F outside.

Find. The winter design heat loss attributable to ventilation and/or infiltration.

Approach. Size the mechanical outdoor air from ASHRAE Standard 62.1 for a dining space, evaluate its sensible and latent load, then check the infiltration that would occur if the building were not pressurised and take the governing figure.

1. Floor area, volume and occupancy. $$A = 30 \times 100 = 3000\ \text{ft}^{2}, \qquad V = 3000 \times 9 = 27{,}000\ \text{ft}^{3}$$ASHRAE Standard 62.1 gives a default occupant density of 70 people per 1000 ft² for restaurant dining rooms, so the design population is $0.070 \times 3000 = 210$ people.
2. Outdoor-air requirement. Standard 62.1 combines a per-person and a per-area term: $$\dot V_{OA} = R_p\,P_z + R_a\,A_z = 7.5(210) + 0.18(3000) = 1575 + 540 = \boxed{2115\ \text{cfm}}$$This is the dominant outdoor-air stream and it is delivered deliberately, through the air handler.
3. Air states. Indoors at 72 °F and 30% RH the humidity ratio is $W_i = 0.004976$ lb/lb. Outdoor air at −7 °F is taken as saturated, which at that temperature means almost no moisture at all, $W_o = 0.000540$ lb/lb, so the humidifier must add $$\Delta W = 0.004976 - 0.000540 = 0.004436\ \text{lb/lb}$$
4. Ventilation load. With $\Delta t = 72 - (-7) = 79\ ^\circ\text{F}$, $$q_S = 1.10\,\dot V\,\Delta t = 1.10(2115)(79) = 1.84\times10^{5}\ \text{Btu/h}$$$$q_L = 4840\,\dot V\,\Delta W = 4840(2115)(0.004436) = 4.54\times10^{4}\ \text{Btu/h}$$$$q_{vent} = \boxed{2.29\times10^{5}\ \text{Btu/h}}$$One fifth of the load is latent — the price of the humidifier the brief requires.
5. Infiltration cross-check. The building has openings on two of four elevations only, so by the crack method only the east and north walls admit wind-driven air, and ASHRAE practice is to charge the larger of the windward crack or half the total crack. For a building of this age and tightness the air-change method with 0.5 air changes per hour is representative: $$\dot V_{inf} = \frac{0.5 \times 27{,}000}{60} = 225\ \text{cfm}$$$$q_{inf} = 1.10(225)(79) + 4840(225)(0.004436) = 19{,}600 + 4{,}830 = 2.44\times10^{4}\ \text{Btu/h}$$
6. Which governs. The mechanical ventilation air is 9.4 times the estimated infiltration. Because that air is supplied by the air handler, the building runs at positive pressure and the infiltration is largely suppressed — outdoor air enters through the coil rather than through the cracks. The two are therefore alternatives, not additions, and the design figure is the ventilation load of 2.29 × 10⁵ Btu/h. A door vestibule on the east and north entrances is strongly advisable in Ottawa to keep the door-traffic component of that figure from growing.

Part (b) — ventilated attic (10 points)

Given. An attic ventilated at 59 l/s with outdoor air at −12 °C. Roof area 244 m² with $U_{roof} = 2.7$ W/m²·K; ceiling area 203 m² with $U_{ceiling} = 0.30$ W/m²·K; inside design temperature 22 °C.

Find. The ceiling heat loss, and the comparison with the unventilated case.

Approach. Treat the attic as one isothermal node and write a steady-state energy balance on it; the ceiling loss then follows from the attic temperature.

1. Conductances. $$U_cA_c = 0.30(203) = 60.9\ \text{W/K}, \qquad U_rA_r = 2.7(244) = 658.8\ \text{W/K}$$The ventilation air is a third conductance. At −12 °C the density is $\rho = 101{,}325/(287 \times 261.15) = 1.352$ kg/m³, so $$\dot m = 0.059(1.352) = 0.0797\ \text{kg/s}, \qquad \dot m c_p = 0.0797(1005) = 80.1\ \text{W/K}$$
2. Attic energy balance. What enters through the ceiling leaves through the roof and with the air: $$U_cA_c\,(t_i - t_a) = U_rA_r\,(t_a - t_o) + \dot m c_p\,(t_a - t_o)$$Solving for the attic temperature, $$t_a = \frac{U_cA_c\,t_i + (U_rA_r + \dot m c_p)\,t_o}{U_cA_c + U_rA_r + \dot m c_p} = \frac{60.9(22) + 738.9(-12)}{799.8} = \boxed{-9.4\ ^\circ\text{C}}$$
3. Ceiling heat loss. $$q_{ceiling} = U_cA_c\,(t_i - t_a) = 60.9\,[22 - (-9.4)] = \boxed{1913\ \text{W}}$$
4. The unventilated comparison. Deleting the ventilation term, $$t_a = \frac{60.9(22) + 658.8(-12)}{719.7} = -9.1\ ^\circ\text{C}, \qquad q_{ceiling} = 60.9\,(31.1) = 1895\ \text{W}$$so ventilating the attic increases the ceiling heat loss by only $$\Delta q = 1913 - 1895 = \boxed{17.6\ \text{W}\ (+0.9\ \%)}$$

The result is the point of the question. The roof is thermally almost transparent — a U-value of 2.7 W/m²·K is little better than bare sheathing — so the attic already floats within about three degrees of outdoor temperature before any air is introduced. Adding 59 l/s of ventilation, a conductance of 80 W/K against the roof's 659 W/K, moves the attic temperature by only three tenths of a degree. The moisture benefit is bought almost free of thermal penalty, which is precisely why code-required attic ventilation (Part 9 of the National Building Code calls for a vent area of about 1/300 of the insulated ceiling area) is not a significant energy issue. It would be a different matter if the roof were well insulated and the ceiling poorly so.