Transcript Infiltration

Chapter 6: INFILTRATION
Agami Reddy (July 2016)
1.
2.
3.
4.
5.
Definitions
Energy implications
Infiltration rates across building stock
Typical air leakage locations and background leakage
Scientific background for estimating infiltration
- Flow thru large and small orifices
- Pressure difference due to wind effect and stack effect
6. Empirical methods for air leakage
- Air change method (for residences and small commercial)
7. Basic LBNL model for one-zone buildings
8. Engineering model (residential-type doors and windows, closed
swinging doors, open doors, curtain wall)
9. Multi-zone network models
10. Measuring air infiltration
HCB-3: Chap 6 Infiltration
1
Definitions
Three mechanisms that contribute to the total air exchange:
• Infiltration is the uncontrolled air flow rate through all the unintentional
openings such as little cracks and gaps between different components (such
as ill-fitting windows or doors). It is balanced by an equal mass flow called
exfiltration, since mass must be conserved
• Natural ventilation is the air flow rate induced by deliberate opening of
windows or doors. It is a variable quantity depending on prevailing outdoor
conditions and one cannot control it properly.
• Mechanical or forced ventilation is the air flow rate intentionally drawn-in by
mechanical ventilation such as fans. It can be controlled and varied as
necessary.
The term passive ventilation is also used, and applies to both infiltration and
natural ventilation
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Example 6.1: Energy implications of air infiltration
Air leaks into a residential building is 1 ACH. If the outdoor air temperature is −10°C and
the interior temperature is 22°C, how much heat must be provided by the heating system to
heat the infiltrating air? The building has 150 m2 of floor area and is two stories high with a
net interior height of 7 m.
Given: Ti = 22°C, To = −10°C, V = 150 m2 × 7 m = 1050 m3, V = 1 volume/h,
Lookup values: Specific heat of air cp = 1.01 kJ/(kg. K). Density of air at 22°C and sealevel barometric pressure is 1.19 kg/m3 (it can be found from the ideal gas law).
SOLUTION
The solution uses the rate form of the open-system first law of thermodynamics. The kinetic
and potential energy terms are identical on both sides of the equation if the small openings
through which air infiltrates and leaves (exfiltrates) the building are uniformly distributed.
From the first law of thermodynamics

Q inf  mc p To  Ti 
The mass flow rate can be found from the density and volumetric flow rate:
(1.19 kg/m ) 1050 m
3
m  V 
3
/ h
 0.347 kg/s
3600 s/h
Finally, the heating to be provided by the heating system is:
(6.1b)
(6.2)

Q inf  (0.347 kg/s)[1.01 kJ /(kg · K)][( 10)  22] C  11.2 kW (heating load)
HCB-3: Chap 6 Infiltration
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Infiltration Rates across Building Stock
Number of air changes per hour (ACH): 1 ACH is equal
to a flow rate equal to one volume of the house per hour
Newer
Older
Older typical US homes: 0.5 – 2 ACH
(seasonal averages)
Newer homes: 0.3 – 0.7 ACH
HCB-3: Chap 6 Infiltration
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Moderately tight buildings
Tight buildings
In Swedish homes, it is
standard practice to limit
infiltration to about 0.2 ACH
HCB-3: Chap 6 Infiltration
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Mechanisms
The cause for infiltration is basically a pressure difference due to
wind, stack effect and mechanical HVAC system imbalances:
p  pwind  pstack  pvent
In the U.S., residential buildings typically rely on infiltration to
meet ventilation needs (though this is changing)
In commercial and institutional buildings, infiltration may not
be desirable from the view point of energy conservation and
comfort. Hence, efforts are made to reduce it. However, it may
be significant especially in tall buildings
HCB-3: Chap 6 Infiltration
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Air leakage locations and background leakage
Three sources of air leakage:
a) Component perforations –
easy to identify
(vents, stacks, chimneys)
b) Openings – easy to identify
(windows, doors,…)
c) Background or fabric leakage
- depends on construction
impossible to identify all cracks,..
- generally assumed to be
uniformly distributed over
surface area of the building
Represented in terms of
“leakage/unit area of envelope”
HCB-3: Chap 6 Infiltration
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Scientific Background
Given in textbook: Flow thru opening in building envelope treated as ORIFICE flow
2
V  CD Aop [ p ]1/ 2

where
(6.3)
V = air flow rate, m3/s (ft3 /s)
Cd  0.61 for sharp edge orifices
CD = discharge coefficient of the opening
2
2
Aop = area of opening, m (ft )
Δp = po−pi = pressure difference between outside and inside, Pa (inches WG)
 =air density, kg/m3 (lb/ft3).
On the other hand, in extremely narrow openings or fine hair-line cracks with relatively long
flow paths (such as in mortar-joints and tight-fitting components), the flow is laminar or viscous.
In reality, flow through a building crack is neither laminar nor fully turbulent, but some
combination
both.thru
Hence,
the crack
flow isequation
is often used
(ASHRAE
In reality,offlow
building
cracks
a combination
of laminar
andFundamentals,
turbulent
2013):
and corrections have been proposed based
on experimental evaluations
n
V  C p
(6.5)
where C = flow coefficient which is function of crack geometry and flow path,
n
HCB-3: Chap 6m/(s.Pa
Infiltration
8
often determined experimentally,
) [ft/(min.inWGn)]
n = flow exponent which depends on the nature of flow through the crack
Scientific backgroundWind EffectVery complex
HCB-3: Chap 6 Infiltration
From Liddament, 1986
9
For Time Averaged Wind Pressure
When wind strikes a building, it creates a static pressure distribution on the exterior of the
building’s envelope. This wind pressure or velocity pressure is given by Bernoulli's equation:
pwind 

2
(v 2  v 2f )
(6.7SI)
Pa
where
v = wind speed (undisturbed by building), m/s
vf = final speed of air at building boundary, m/s
ρ = air density, kg/m3
Since the final speed vf is difficult to determine, a convenient shortcut is to use Eq. (6.7)
with vf = 0, multiplying it instead by a pressure coefficient Cp:
pwind  C p

2
v2
HCB-3: Chap 6 Infiltration
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where
v = wind speed (undisturbed by building), m/s
vf = final speed of air at building boundary, m/s
ρ = air density, kg/m3
For Time Averaged Wind Pressure
Since the final speed vf is difficult to determine, a convenient shortcut is to use Eq. (6.7)
with vf = 0, multiplying it instead by a pressure coefficient Cp:
pwind  C p

2
v2
Plots to determine Cp are shown in next 2 slides
Actually we are interested in the pressure difference between the interior and exterior of a
building. Since that average is approximately −0.2 for a low rise building1, the local pressure
difference (po−pi) averaged across the wall is, in that case,
pwind  C p

v2
(6.10)
2
with ΔCp = Cp−(−0.2) being the difference between the local pressure coefficient and the average
over all orientations of the building. This approach presumes, obviously, that air leakage is
uniformly distributed across all four walls.
HCB-3: Chap 6 Infiltration
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Wind
Pressure
Coefficient
a) For roofs of tall
buildings for three
different aspect ratios of
length L and width W
b) For roofs of low rise buildings
inclined at less than 20o, C p 0.5
c) For walls of tall buildings for
three different aspect ratios of
length L and width W
HCB-3: Chap 6 Infiltration
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d) Walls of low
rise buildings
6.9
HCB-3: Chap 6 Infiltration
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Effective Wind Speed
This value should correspond to the eaves height for a low-rise building and the building
height for a high-rise building. Since meteorological data for wind velocity are usually collected
at 10 m height, and since wind speed is strongly modified by terrain and obstacles, being
significantly higher far above the ground, a conversion equation based on boundary layer theory
is used to correct for this height differential (Spitler, 2010):

h
v  v met ( met ) amet ( ) a
hmet

(6.11)
v and v met = effective and reference wind speeds, m/s (ft/min)
where
 and  met = boundary layer thickness for the local terrain and for the
meteorological station, m (ft)
h and hmet = average height above local obstacles and height at which wind speed
is measured, m (ft)
a and amet = exponent for the local terrain and for the meteorological station.
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Terrain
Description
Category
1
Large city centers, in which at least 50% of buildings are
higher than 21.3 m (70 ft), over a distance of at least 0.8
km (0.5 mi) or 10 times the height of the structure
upwind, whichever is greater
2
Urban and suburban areas, wooden areas, or other
terrain with numerous closely spaced obstructions
having the size of single-family dwellings or larger, over
a distance of at least 460 m (1,500 ft) or 10 times the
height of the structure upwind, whichever is greater
3
Open terrain with scattered obstructions having heights
generally less than 9.1 m (30 ft), including flat open
country typical of meteorological station surroundings
4
Flat, unobstructed areas exposed to wind flowing over
water for at least 1.6 km (1 mi), over a distance of 460 m
(1,500 ft) or 10 times the height of the structure inland,
whichever is greater
Exponent a
0.33
Layer thickness
 , m (ft)
460 (1,509)
0.22
370 (1,214)
0.14
270 (886)
0.10
210 (689)
Since meteorological stations are usually located in flat open terrain, they often correspond
to terrain category 3 in the table so that amet = 0.14 and  met =270 m . It is recommended that Eq.
(6.11) not be used when h < 0, i.e., when average obstacle height is greater than the building
height.
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Example 6.3: Wind pressure on four sides of a tall building
Find the pressure difference across all four sides on the top floor of a 20 storey building
located in a large city center. The reference wind velocity is 10 m/s. The building is square and
one of the sides is normal to the wind direction. Each floor is 3 m high and the average height of
local obstacles around the building is 23.5 meters.
v met = 10 m/s, building (L/W) = 1, building is uniformly leaky on all sides
h = mid-height of 20th floor minus obstacles height = 3 x 20- 1.5 – 23.5= 35 m,
Meteorological station is located 10 m above ground and is at terrain category 3
(Table 6.1) with amet = 0.14 and  met =270 m
Building corresponds to terrain category 1 with a = 0.33 and  =460 m
Given:
Lookup values:   1.20 kg/m3 (   0.075 lb m /ft 3 )
pwind on windward, leeward and the two sides.
Find:
SOLUTION
We first use Eq. (6.11) to calculate the effective wind velocity on the 20th floor
270 0.14 35 0.33
v  10.(
) (
)  6.8 m/s
10
460
This velocity is also assumed to hold for the roof (strictly speaking, the height should be
35+1.5=36.5 m not 35 m; but this is a small correction). The following table illustrates the
intermediate and final calculations:
Face
Face
of of
Building
Building
Windward
Windward
Leeward
Leeward
Faces (2 sides)
Faces (2 sides)
Roof
Roof
PressureCoeff.
Coeff.Diff.
Diff.
Wind
Pressure
Pressure
Wind
Pressure
Coefficient
ΔC=
Angle Coefficient
p)
p = Cp−(Average)
(C(C
ΔC
Cp−(Average)
Angle
p)
p
o
0
(Fig.
6.9)
0.62
0.62-(-0.308)=
0.928
00 o (Fig. 6.9) 0.62
0.62-(-0.308)= 0.928
180
-0.30
-0.30-(-0.308)= 0.008
1800 o
-0.30 -0.49-(-0.308)= - 0.182
90
-0.58
-0.58-(-0.308)= -0.272
900 o
-0.58
-0.50-(-0.308)= -0.192
0
(Fig.
6.10)
-0.70
-0.70-(-0.308)= -0.392
00
(Fig. 6.10) -0.70 -0.70-(-0.308)= -0.392
Average= - 0.308
Average= - 0.308
HCB-3: Chap 6 Infiltration
Pressure Diff.
Diff.
Pressure
(Eq. 6.10)
6.10)
wind (Eq.
ppwind
25.7 Pa
25.7 Pa
0.22 Pa
-5.0 Pa
-7.55 Pa
-5.3 Pa
-10.88 Pa
-10.9 Pa
16
Scientific Background- Stack Effect
Caused by temperature differences (and hence air densities) on the inside and outside
Winter stack effect in tall buildings
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The outside-inside pressure difference at a specific location on the wall during winter (when
outside air is colder and, hence, denser) is given by
pstack  ( o  i ) g h
where
6.12
(6.12a)
ρi = density of air inside building,
ρo = density of air outdoors,
Δh = vertical distance of specific location minus that of neutral pressure height,
g = acceleration due to gravity.
Or alternatively
pstack
o
Ti  To
  g h i (  1)   i g h (
)
i
To
6.14
h- height from neutral pressure line
T is in absolute units
HCB-3: Chap 6 Infiltration
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Variation of pressure
difference due to stack
effect with vertical distance
from neutral pressure line
pstack
 0.04 Pa / (m · K)
Cd h T
pstack
 0.00014 lbf / (ft 2 · ft ·  R)
Cd h T
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For real buildings, the airflow resistance of the exterior walls relative to that between floors due
to doors and stairways is modeled by introducing a thermal draft coefficient Cd,
pstack  Cd i g h
Ti  To
To
6.15
(5.13SI)
where
ρi = density of air in building = 1.20 kg/m3 (0.075 lbm/ft3)
Δh = vertical distance from neutral pressure level, up being positive, (m) (ft)
g = 9.81 m/s2 (32.17 ft/s2) = acceleration due to gravity [gc = 32.17 (lbm • ft)/(lbf • s2)]
Ti and To = indoor and outdoor absolute temperatures, K (°R)
Cd = draft coefficient, a dimensionless number from about 0.63 (tight) to 0.82 (loose)
for typical modern office buildings to 1.0 if there is no resistance at all.
HCB-3: Chap 6 Infiltration
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Example 6.4: Stack pressure on top floor of a tall building
Consider the same high-rise 20 storey building as in Example 6.3. Calculate the pressure
difference due to the stack effect on the 20th floor when the outdoor temperature is -20C and the
indoor at 220 C. Assume a draft coefficient of 0.65 and that the neutral pressure line is 26.5 m
from ground level.
Given:
Δh = mid-height of 20th floor minus height of neutral pressure line
= (3 x 20- 1.5) – 26.5 = 32 m,
T0 = -2o C= 271 K and Ti = 220 C = 295 K, Cd = 0.65

 1.20 kg/m3 (   0.075 lb m /ft 3 )
Lookup values:
, g = 9.81 m/s2
Find:
pstack (which is assumed to be the same for all four walls as well as for the roof)
SOLUTION
Equation
(6.13SI) is used directly to yield
Eq. 6.12
pstack  0.65(1.2)(9.81)(32) (
295  271
)  19.9 Pa
295
HCB-3: Chap 6 Infiltration
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Combining wind, stack and mechanical
ventilation effects
The law of conservation of mass implies that the net airflow provided by the ventilation

system equals the net leakage V across the envelope by summing over all leakage sites j of the
building envelope:
n
V   Aj K j p j j
p j  pwind, j  pstack, j  pvent
If a building is pressurized to Δpvent by mechanical ventilation and if wind
and stack pressures are smaller than Δpvent, then it is indeed a fair
approximation to neglect them altogether
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Superimposition of stack and wind pressures along height of a building
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Example 6.5 Combining wind and stack effects
Calculate the combined effect of the wind and stack pressure on the top floor of the 20 storey
building considered in Examples 6.3 and 6.4.
Since the pressures are additive, the calculation is simple to perform using the pressure
differences computed previously. A table better illustrates the procedure:
Face of
Building
Windward
Leeward
Faces (2 sides)
Roof
Wind Pressure
Diff. pwind
(Example 6.3)
25.75
0.22
-7.55
-10.88
Stack Pressure
Diff. pstack
(Example 6.4)
-19.92
-19.92
-19.92
-19.92
Combined Pressure
Diff. p
(Pa)
5.83
-19.7
-27.47
-30.8
We note that the roof and all walls (except the windward side) experience higher pressures
indoors than outdoors resulting in substantial exfiltration. The stack effect has enhanced the wind
pressure difference on these walls, while reducing the outdoor-indoor pressure difference on the
windward side. Despite the stack effect, outdoor air will infiltrate from the windward wall. The
exfiltration rate will be much higher than the infiltration rate for this floor; the difference is being
supplied from the floors underneath.
HCB-3: Chap 6 Infiltration
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Empirical Models
Air change method: for residences and small commercial
assumes that a portion of the air in the building is replaced with
outdoor air which must be heated/cooled.
Number of air changes per hour (ACH): 1 ACH is equal to a flow rate
equal to one volume of the house per hour
Range of ACH: 0.5 1.5
tight
loose
Air infiltration volume = (ACH) x (room volume) / 60 min/hr
Tables in next slide allow determination of
ACH for summer and winter conditions
HCB-3: Chap 6 Infiltration
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Construction Type
Winter* Outdoor Design Temperature oC (oF)
Tight
10
4
-1
-7
-12 -18
(50) (40) (30) (20) (10) (0)
0.41 0.43 0.45 0.47 0.49 0.51
-23
(-10)
0.53
-29
(-20)
0.55
-34
-40
(-30) (-40)
0.57 0.59
Medium
0.69 0.73 0.77 0.81 0.85 0.89
0.93
0.97
1.00
1.05
Loose
1.11 1.15 1.20 1.23 1.27 1.30
1.35
1.40
1.43
1.47
Summer** Outdoor Design Temperature oC (oF)
Tight
29
32
35
38
41
43
(85) (90) (95) (100) (105) (110)
0.33 0.34 0.35 0.36 0.37 0.38
Medium
0.46 0.48 0.50 0.52
0.54
0.56
Loose
0.68 0.70 0.72 0.74
0.76
0.78
Method
used by some
professionals
because of its
simplicity
*Design values for winter are for 6.7 m/s or 24 km/h (15 mph) and indoor temperature of 20o C (68o F)
** Design values for summer are for 3.4 m/s or 12 km/h (7.5 mph) and indoor temperature of 24o C (75o F)
©ASHRAE, www.ashrae.org. (2001) ASHRAE Handbook - Fundamentals.
HCB-3: Chap 6 Infiltration
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Basic LBNL Model for Air leakage
• Applicable for 1-zone small buildings WITHOUT mechanical
ventilation:
V  Aleak as T  aw v 2 L/s
(6.23a)
6.25
where
Aleak = effective leakage area of building, cm2
as = stack coefficient of Table 6.3, (L/s)2/(cm4. K) [(ft3/min)2/(in4. oF)]
T  Ti  To , K
aw = wind coefficient of Table 6.4, (L/s)2/[cm4. (m/s)2] [(ft3/min)2/(in.4 (mi/h)2)]
v = wind speed, m/s.
Effective leakage area (ELA) is the equivalent amount of free open area of an
orifice that allows the same volume of air by infiltration as the actual building
(Eqn can be used for specific days as well as seasonal averages depending
on how the temperatures and wind velocity values are selected)
HCB-3: Chap 6 Infiltration
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Table 6.3 Stack Coefficient as
Number of Stories
One
Two
Three
Stack coefficient as,
(ft3/min)2/(in4 • °F)
0.0150
0.0299
0.0449
Stack coefficient as,
(L/s)2/(cm4 • K)
0.000145
0.000290
0.000435
TABLE 6.4 Wind Coefficient aw
Shielding
Class
Description
Wind Coefficient aw,
2
4
2
(L/s) /[cm • (m/s) ]
Wind Coefficient aw,
3
2
4
2
(ft /min) /[in • (mi/h) ]
Number of Stories
Number of Stories
One
Two
Three
One
Two
Three
1
No obstructions or local
shielding
0.000319
0.000420
0.000494
0.0119
0.0157
0.0184
2
Light local shielding; few
obstructions, a few trees or
small shed
0.000246
0.000325
0.000382
0.0092
0.0121
0.0143
3
Moderate local shielding; some
obstructions within two house
heights, thick hedge, solid
fence, or one neighboring
house
0.000174
0.000231
0.000271
0.0065
0.0086
0.0101
4
Heavy shielding; obstructions
around most of perimeter,
buildings, or trees within 10 m
in most directions; typical
suburban shielding
0.000104
0.000137
0.000161
0.0039
0.0051
0.0060
5
Very heavy shielding; large
obstructions surrounding
perimeter within two house
heights; typical downtown
shielding
0.000032
0.000042
0.000049
0.0012
0.0016
0.0018
HCB-3: Chap 6 Infiltration
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Table 6.2
Data for Effective Leakage Area Coefficient of Building Components at 4 Pa (0.016 inWG)
Component
Best
Estimate
Maximum
Minimum
0.04
0.19
0.06
0.19
0.02
0.05
0.07
0.12
0.02
0.011
0.023
0.017
0.034
0.006
0.011
0.011
0.023
0.017
0.034
0.006
0.011
0.032
0.063
0.042
0.083
0.026
0.052
0.043
0.086
0.063
0.126
0.023
0.046
0.026
0.052
0.039
0.077
0.013
0.026
0.037
0.074
0.054
0.110
0.02
0.04
0.114
0.157
0.215
0.243
0.043
0.086
0.114
0.215
0.043
Sill foundation—Wall
Caulked, in2/ft of perimeter
Not caulked, in2/ft of perimeter
Joints between ceiling and walls
Joints, in2/ft of wall (only if not taped or plastered and no vapor
barrier)
Windows
- Casement
Weatherstripped, in2/ft2 of window
Not weatherstripped, in2/ft2 of window
- Awning
Weatherstripped, in2/ft2 of window
Not weatherstripped, in2/ft2 of window
- Single-hung
Weatherstripped, in2/ft2 of window
Not weatherstripped, in2/ft2 of window
- Double-hung
Weatherstripped, in2/ft2 of window
Not weatherstripped, in2/ft2 of window
- Single-slider
Weatherstripped, in2/ft2 of window
Not weatherstripped, in2/ft2 of window
- Double-slider
Weatherstripped, in2/ft2 of window
Not weatherstripped, in2/ft2 of window
Doors
- Single door
Weatherstripped, in2/ft2 of door
Not weatherstripped, in2/ft2 of door
- Double door
Weatherstripped, in2/ft2 of door
HCB-3: Chap 6 Infiltration
29
Example 6.8: Effective leakage area of a building
Estimate the effective leakage area for a simple wood-frame building. It is built as a rectangular box 12
m×12 m×2.5 m (39.4 ft×39.4 ft×8.2 ft) with a flat roof, and the walls at the sill are uncaulked and the
walls at the roof are not taped or plastered. The windows, which are double-hung and not weatherstripped, have no caulking and cover 20 percent of the sides.
HCB-3: Chap 6 Infiltration
30
Component
Walls at sill (sill
uncaulked)
Walls at roof not taped or
plastered, no vapor
barrier
Windows: double-hung, not
weather-stripped
Window frames: no
caulking
Area, m2, or
Perimeter, m
(given)
48 m
Leakage Area per
Area or Perimeter
(from Table 6.2)
0.19 in2/ft = 4 cm2/m
48 m
0.07 in2/ft =
1.5 cm2/m
72
24 m2
0.086 in2/ft2 =
6 cm2/m2
0.023 in2/ft2 =
1.7 cm2/m2
144
24 m2
Total
Leakage
Area, cm2
192
41
449
HCB-3: Chap 6 Infiltration
31
Example 6.9: Air changes of a building
Find the air exchange rate for the building of Example 6.8 when ΔT = Ti−To = 20 K (36° F)
and effective design wind speed v = 6.7 m/s (15 mi/h).
Given: Building of Example 6.8 in suburban neighborhood
ΔT = Ti−To = 20 K, v = 6.7 m/s, Aleak = 449 cm2
Intermediate quantities: Stack coefficient as = 0.000145 (L/s)2/(cm4.K) from Table 6.3.
Wind coefficient aw = 0.000104 (L/s)2/[cm4.(m/s)2] from Table 6.4.
SOLUTION
6.25 one obtains the infiltration rate
From Eq. (6.23)
V  449 cm2  0.000145(L/s) 2 (cm 4· K )  20 K  0.000104(L/s) 2 / [cm4·(m/s) 2 ]  (6.7 m/s) 2
 39.1 L/s  140.6 m3 /h
Since the volume V = 360 m3, the air change rate is 0.39 ACH.
HCB-3: Chap 6 Infiltration
32
Engineering Models
1) Leakage Thru identifiable Components: Based on FIELD
tests on Actual Buildings:
• Residential windows and doors
• Commercial swinging doors when closed
• Opening of commercial swinging doors
2) Background leakage
• Curtain walls for commercial buildings
HCB-3: Chap 6 Infiltration
Use
Table 6.2
33
Background Leakage
For background leakage, it is better to assume these leaks to be uniformly distributed about
the surface area of the building and to specify leakage in terms of surface area of the building
envelope element. For the other types of identifiable openings, it is convenient to characterize the
infiltration in terms of a single equivalent amount of free open area of an orifice that allows the
same volume of air flow as the actual component. Such a normalized area has been referred to as
the overall leakage coefficient per unit area of the element. This allows Eq. (5.5) to be recast as:
6.20
V  AK p n
(5.18)
where A = area of building element such as a wall, m2 (ft2); note that sometime a leakage
length is used instead for elements such as window sills;
K = overall leakage area coefficient of building element, m/(s • Pan) [ft/(min •
inWGn)]
n = flow exponent, between 0.4 and 1.0 and usually around 0.65 for buildings
To apply Eq. (5.18), one needs data for leakage area coefficients and flow coefficients of all the
components of a building. Much research has been done to obtain such data, both for components and
for complete buildings, e.g., by pressurizing the component or even the whole building. Data for leakage
area coefficients K can be found in Table 5.2 for a wide variety of building elements that are part of the
envelope “background” leakage. Note that
someChap
elements
have very large leakage areas. A door may
HCB-3:
6 Infiltration
34
2
have as much leakage as that through a 100 m exterior wall. Air leakage through curtain walls, operable
IDENTIFIABLE Components
Residential Windows and doors
Figure 6.15
V
 k ( p)n
lp
where lp is the perimeter
and
n=0.65 doors and
residential
windows : n=0.65
k=1 is tight gap width
k=2 is average
k= 6 is loose
HCB-3: Chap 6 Infiltration
35
Infiltration for commercial-type swinging doors
when CLOSED
V
 k ( p)n
lp
where lp is the perimeter
and n=0.65n=0.5
because gaps
are larger
Figure 6.16
HCB-3: Chap 6 Infiltration
36
Infiltration through opening of doors due to traffic
V  C ( p)0.5
Figure 6.17
Similar plots are available for revolving doors and automatic doors
HCB-3: Chap 6 Infiltration
37
Figure 6.18
Infiltration per area of curtain wall for one room or one floor.
Coefficient (K)
SI
Description
Construction
IP
0.031
0.22
Tight
0.093
0.66
Average
0.183
1.30
Loose
Close supervision of workmanship; joints are redone
When they appear inadequate
Conventional
Poor quality control, or older building where joints
have not been redone
Curtain wall
V
 k ( p)0.65
A
Figure 6.18
HCB-3: Chap 6 Infiltration
38
Example 6.7: Air leakage from single story building
Estimate how much air leaks out of a single-story office building of average construction
when it is pressurized to Δp = 50 Pa (0.2 inWG) (assume that wind and stack pressures can
be neglected).
Given: One-story building 30 m×30 m×2.5 m (98.4 ft×98.4 ft×8.2 ft); curtain wall,
alternating with floor-to-ceiling windows 3 m×2.5 m each (9.84 ft×8.2 ft); windows
cover 50 percent of the entire wall area; four swinging doors 2 m×1 m (6.56 ft×3.28
1
ft), single-bank type, crack width 3.2 mm ( in); floor area 15 m2 (161 ft2) per
8
occupant.
Estimate the average traffic over 10 h by assuming occupants make two entries and
two exits per day. Assume that 500 visitors enter and leave the building per day.
HCB-3: Chap 6 Infiltration
39
SOLUTION
Total area of all four walls = 4×30 m×2.5 m = 300 m2
Curtain wall area = 300 x 0.5 = 150 m2 assuming K = 0.66
Total number of windows = 150 / (3 x 2.5) = 20
Perimeter of windows = 20 × [2  (3 + 2.5) = 220 m assuming k = 2.0 (“average” value)
Perimeter of doors = 4 x [2  (2 + 1)] =24 m assuming k = 40
900 m 2
Number of occupants 
 60
2
15 m
(60  4)  (500x2) visitors
Traffic per door 
 31 passages/h
4 door  10 h
At Δp = 50 Pa, we read from Fig. 6.15,
V
 1.1 L / (s·m) from perimeter of windows
lp
Hence,
Vwindows  1.1 (L/s) / m  220 m  242 L/s
From Fig. 6.16,
V
 28 L / (s·m) from perimeter of doors when closed
lp
Hence, Vdoors  28 (L/(s.m)  24 m  672 L/s
HCB-3: Chap 6 Infiltration
40
From Fig. 6.17b, coefficient C is around 40 in SI units, and from Eq. (6.20) or Fig. 6.17(a)
the extra flow through the doors due to traffic, per door, is:
V  40  (50)0.5  282.8 L/s/door
Thus for 4 doors, V doors  4  282.8=1,131.4 L/s .
From Fig. 6.18, V / A = 1.2 L/(s .m2) from curtain wall; hence
V curtain wall = 1.2 (L/s)/m2×150 m2 = 180 L/s.
Finally, the total flow is
V  242  672  1,131.4  180  2,225.4 L/s
 2.225m3 /s  3600 s/h  8,011m3 /h (4,582 ft 3 / min)
Since the building volume is V = 2250 m3, this corresponds to (8,011 m3/h/2,250 m3) = 3.56
ACH (air changes per hour).
Note that the dominant contribution comes from the doors, 672 L/s when they are closed and
1100 L/s from traffic. In the interest of energy conservation, tighter doors should be
considered and replacing them by vestibule-type doors.
COMMENTS
There are 60 permanent occupants and the total floor area of the building is 900 m2.
According to ASHRAE Standard 62.1-2013 (ASHRAE Fundamentals, 2013) discussed in
sec. 3.5.3, one should use the multipliers of 2.5 L/s per person and 0.3 L/s per m2 for an
office building. Thus, according to the standard, the ventilation flow required is 1.77 m 3 /s.
We note that the calculated value (1.94 m3 /s) is about 10% higher than this value, but is
within the design safety factor.
HCB-3: Chap 6 Infiltration
41
Multizone Models
Set up and solve
a set of simultaneous
non-linear equations
HCB-3: Chap 6 Infiltration
42
Multizone Models
-For two leakage paths in series (Fig. 6.21a):

pt  p1 +p2 and V  C1p  C2 p
n1
1
n2
2

=>pt  (V/ C1 )

 (V/ C2 )1/ n2
(6.24)
=> V t  C1p1n1  C2p2n2
(6.25)
1/ n1
-For two leakage paths in parallel (Fig. 6.21b):





Vt  V1  V 2 and V1  C1p and V 2  C2 p
n1
1
n2
2
HCB-3: Chap 6 Infiltration

43
Opening of Windows and Doors
Figure 7.2 Measured ventilation rates, as
a function of wind speed, in a two-story
house with windows open on lower floor
HCB-3: Chap 6 Infiltration
44
Natural Ventilation Air Flow through Large Openings
Due to wind:

V wind  Cv Aop v
Eq. 6.28
where,
Aop = area of opening, either inlet or outlet (assumed equal) (m2, ft2)
v = local wind speed (m/s, mi/h),
Cv = dimensionless opening effectiveness, assumed to be in the range of 0.5-0.6 for
perpendicular winds and 0.25-0.35 for diagonal winds.
If the openings are
not of the same size,
the correction curve
should be used:
Figure 6.24 Increase in flow caused
by excess area of one opening over
the other
HCB-3: Chap 6 Infiltration
45
Lab Testing for Airtightness of a
Component
Controlled tests are done in the lab without
the influence of climatic parameters.
A static pressure difference (about 200 Pa)
is created across the test specimen
from which the ELA can be deduced.
Often results are presented in terms of
- Flow per hour per area
or
- Flow per hour per unit crack length
HCB-3: Chap 6 Infiltration
46
Blower Door Tests
Blower door tests are done to estimate
aggregate envelope leakage and to
locate and fix leaks:
•Device consists of a door-insert
with rubber edge.
•Variable speed fan and measurements for
flow and pressure difference
•Tests conducted till fairly high pressures
(about 50 Pa) in 10 Pa incremental steps
HCB-3: Chap 6 Infiltration
47
The effective leakage area (ELA) of the house is subsequently determined as
ELA  cV r [  / (2p)]0.5
(6.29)
with reference volume flow rate V r determined from the identified power law equation with p = 4 Pa
taken as reference. Note that though the blower-door testing and estimation of model parameters by
regression is done with pressure differences in the range of 10-50 Pa, extrapolation to the lower values
of 4 Pa is required which may result in inaccurate predictions of V r due to the somewhat imprecise form
of the flow-pressure relation assumed (ASHRAE Fundamentals, 2013).
HCB-3: Chap 6 Infiltration
48
Presentation and Analysis of Test Data of Blower Door Tests
Q  K ( p )n
log Q  log K  n log p
HCB-3: Chap 6 Infiltration
49
Outcomes
• Familiarity with the three causes of pressure difference across building
envelopes resulting in infiltration
• Understanding the energy implications of air infiltration
• Familiarity with different pathways/locations and types of air leakage:
component perforations, openings, and background or fabric leakage
• Familiarity with the scientific background for analyzing wind and stack effects
and engineering methods of estimating the associated pressure difference
• Be able to analyze situations involving wind and stack effects on buildings
• Understanding of the ELA concept and be able to use it along with the LBNL
model to solve simple problems
• Be able to apply engineering models for estimating leakage through various
types of envelope components
• Familiarity with multi-zone modeling methods
• Familiarity with lab testing of components and with the blower door test
HCB-3 Chap 4: Solar Radiation
50