Infiltration

Infiltration is from combined effect of buoyancy and wind. Infiltration rate: O

Where Q = infiltration rate (L/s)

L = effective leakage area (cm2)

Cs = stack coefficient (L2 s’2 cm’4 K’1)

AT = average indoor-outdoor temperature difference (K)

Cw = wind coefficient (L2 s’4 cm’4 K’1 m’2)

V = average wind speed measured at local weather station (m/s)

Table 7 Local Shielding Classes

Description

Table 5 Example of Calculation of Building Effective Air Leakage Area Based on Component Leakage Areas

Class

Size or x Al Description Number per unit

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 heights, thick hedge, solid fence, or one neighboring house

4 Heavy shielding; obstructions around most of perimeter, buildings or trees within 10 m in most directions; typical suburban shielding

5 Very heavy shielding; large obstructions surrounding perimeter within two house heights; typical downtown shielding

2

Cm

Component

4.0 cm2/m 0.5 cm2 ea

4.0 cm2/m2

1.7 crrr/m2

7.7 cm2/m2

1.7 cm2/m2 350 cm2 ea

6.0 cm2 ea 144 cm2 ea

173

10

75

54

350

42

144

Uncaulked

Sliding

Single

Without damper Pipes

Ducts untaped.

43.2 m 20

13.1m2

13.1m2

5.7 m2

5.7 m2 1

7

1

Sills

Electrical outlets Windows Framing Exterior doors Framing Fireplace Penetrations Heating ducts

Calculated total building air leakage area Ac = 848 cm2

Class

One

Two

Three

1

0.000 319

0.000 420

0.000494

Table 6 Stack Coefficient Cs

2

0.000 246

0.000 325

0.000 382

House Height (Stories)

3

0.000 174

0.000 231

0.000 271

One Two Three

4

0.000 104

0.000 137

0.000 161

Stack coefficient 0.000 145 0.000 290 0.000 435

5

0.000 032

0.000 042

0.000049

Table 8 Wind Coefficient C*,

,. … House Height (Stories)

Infiltration

Example 6.2

Estimate the infiltration at design conditions for a two-story house in State College. The house has an effective leakage area of 500 cm2, a volume of 340 m3, and is surrounded by a thick hedge (shielding class 3).

Solution:

Under winter design conditions: TD = -13 °C, V = 8 m/s, T; = 22 °C.

From A Tables 25.6 and 25.8 of the handouts, Cs = 0.000290, Cw = 0.000231

 

O

Air change rate = 284 / 340 = 0.84 ACH.

0.1 ACH (air change rate per hour) 0.5 ACH

1 ACH

2 ACH

3 ACH

подпись: 0.1 ach (air change rate per hour) 0.5 ach
1 ach
2 ach
3 ach
Air exchange method:

Extremely low low normal high

Extremely high Crack method:

Q = A C Apn

Where A = effective leakage area of cracks (m2)

C = flow coefficient

Ap = pressure difference between outdoor and indoor (Pa)

Posted in Fundamentals of Heating. Ventilating, and Air-Conditioning


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