Question 6.1: A blower door test has been performed in a house located in ...

To determine the air leakage characteristics of a house using blower door tests, the following procedure is used:

(i) First, the data consisting of pressure differential ( ΔP [Pa]) and airflow rate ( \dot{V} [CFM]) presented in Figure 6.3 is plotted in a log–log scale as illustrated in Figure 6.5.
(ii) Then, a linear regression analysis is used to determine the coefficients C and n of Eq.(6.7):

\dot{V}= C.ΔP^{n}

• For the pressurization test, the coefficients C and n are found to be C =552.25 and n = 0.679.
• For the depressurization test, the coefficients C and n are found to be C = 556.13 and n = 0.694.

It should be noted that the values of C and n provided above are valid only if the airflow rate \dot{V} [CFM] and pressure differential ΔP are expressed in cfm (cubic feet per minute) and Pa, respectively.

(iii) Based on the correlation and the coefficients provided above, the infiltration and exfiltration rates under a normal pressure differential (ΔP_{ref} = 4 Pa) can be calculated:
For pressurization: \dot{V}_{ref}= 552.25 * (4)^{0.679} = 1455 cfm
For depressurization: \dot{V}_{ref} = 556.13 * (4)^{0.694} = 1416 cfm
The leakage areas for both pressurization and depressurization tests can be obtained using
Eq. (6.8b):

ELA = 0.186\dot{V}_{ref}. \sqrt {ρ/2ΔP}

For Evergreen, Colorado, the air density should be adjusted for altitude and is found to be:
_ = 0.063 lbm/ft^{3}.
• For pressurization: ELA = 369.6 in.^{2}
• For depressurization ELA = 379.7 in.^{2}
Therefore, the average leakage area for the house based on both pressurization and depressurization tests is:

ELA(average) = 2.60 ft^{2} = 374.4 in.^{2}

(iv) The annual average leakage air flow rate expressed in air change per hour (ACH) can be determined using the LBL model expressed by Eq. (6.9):

\dot{V}=ELA.(f_{s}.ΔT+f_{s}.v_{w}^{2})^{1/2}

To determine the average ACH for the house, the average ELA of 374.4 in.^{2} is used. Moreover, the coefficients f_{s} = 0.0266 and f_{w}= 0.0051 need to be used for a two-story building for a shielding class of 4 (refer to Tables 6.1 and 6.2). The annual average wind speed is V_{w} (for Denver) = 8.5 mi/hr, and the annual average outdoor temperature tout = 46.17oF (for Evergreen). To account for the effect of indoor temperature setback (during the winter season), an annual average indoor temperature of 60oF is considered as the annual average leakage air flow rate \dot{V} :

\dot{V}=374.9*(0.026*(60-46.2)+0.0051*(0.85)^{2})^{1/2}=331  cfm

Then, the volume flow rate \dot{V} is divided by the conditioned volume of the house (in this case, 28,350 ft^{3}) to obtain the annual average ACH:

ACH  =\frac {331  cfm*60 min/hr}{28,350 ft^{3}}=0.70

It should be noted that ASHRAE recommends a leakage of 0.35 ACH for proper ventilation of a house with minimum heat loss.

TABLE 6.1 Stack Coefficient, f_{s}

Source: ASHRAE, Handbook of Fundamentals, Atlanta, GA: American Society of Heating,
Refrigerating and Air-Conditioning Engineers, Inc., 2009.
a IP Units for f_{s}: (ft^{3}/min)^{2}/in4×°F
b SI Units for f_{s}: (L/sft)^{2}/cm4×°C

TABLE 6.2 Wind Coefficient, f_{w}

Source: ASHRAE, Handbook of Fundamentals, Atlanta, GA: American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc., 2009.
a IP Units for f_{w}: (ft^{3}/min)^{2}/in^{4}×mph
b SI Units for f_{w}: (L/sft)^{2}/cm^{4}×(m/s)^{2}
c Description of shielding classes: 1–no obstructions or local shielding; 2–light local shielding:
few obstructions, few trees, or small shed; 3– moderate local shielding: some obstructions within two house height, thick hedge, solid fence, or one neighboring house; 4– heavy shielding:
obstructions around most of perimeter, buildings or trees within 30 ft (10 m) in most
directions; typical suburban shielding; 5–very heavy shielding: Large obstructions surrounding
perimeter within two house heights; typical downtown shielding.