Energy Efficient Buildings
Heat Gain/Loss through Infiltration
Infiltration is air leakage into and out of a building. Buildings may also have forced
ventilation. Neglecting air storage and changes in air density (which are almost always
quite small), the total volume flow rates of air moving into and out of a building are
almost always nearly identical. Heating or cooling this air requires significant amounts
of energy. Thus, energy-efficient buildings are designed and built using construction
practices the reduce air leakage to the minimum level acceptable for adequate
ventilation. To reduce energy usage further, energy-efficient buildings may also employ
forced ventilation with a heat exchanger between the incoming and exhaust air streams
to capture heat that would otherwise be lost. This chapter discusses the drivers of
infiltration, how to calculate average and hourly rates of infiltration, the heating and
cooling energy required to heat and cool infiltration air, and the heat recovery
ventilators.
Infiltration Drivers
The primary drivers of infiltration in buildings are stack effect, wind effect, combustion
effect, exterior ducts, unbalanced supply and return air. These effects are discussed
below.
Stack Effect
Stack effect occurs when warm air rises and pressurizes the upper part of a room or
building. As this pressurized air leaks out the top, cool air is drawn into the lower part of
the room or building. The pressure difference depends on vertical height and
indoor/outdoor air temperature difference. Hence, stack-driven infiltration is higher in
multi-story buildings and during winter.
Z
ΔP 
zgP  1
1

 
2R  Toa Tia 
Wind Effect
Wind effect occurs as wind impinges on the windward side of a building then moves
over the top and around the sides of the building. The moving air increases the pressure
on the windward side of the building. As the air passes over and around the building it
entrains air on the leeward side into the air stream, creating a negative pressure on the
leeward side of the building. This pressure difference between the windward and
leeward sides of the building drives infiltration. The pressure difference is proportional
1
to the square of wind velocity V; hence wind-driven infiltration increases significantly
during windy weather. Wind shading by terrain, vegetation and other buildings
significantly reduces wind-driven infiltration.
V
PH
PL
V
ρv
ΔP 
2
2
Combustion Effect
Combustion effect occurs when combustion devices such as furnaces use indoor air for
combustion, and then expel exhaust gasses through stacks to the outdoors. Air must
leak into the building to make up the air lost through the exhaust stack. Combustion
effect is greatest during winter when combustion furnaces, boilers, and hot water
heaters burn more fuel.

mex


m
a

minf
mag
Exterior Ducts
In warm climates it is common to run air supply ducts through unconditioned spaces
such as attics. Air supply ducts are rarely tightly sealed and typically leak air. Air leaked
from supply ducts into unconditioned spaces is replaced through infiltration. Air lost
from leaky ducts can be a major driver of infiltration; the average infiltration rate in 50
houses in Florida with exterior ducts increased from 0.46 when the air conditioner was
off to 2.10 when the air conditioner was on.

ma
50 houses in Florida:
N ac, off  2.10
N ac, on  .46

minf
AC
2
Unbalanced Supply & Return Air
Furnaces and air conditioners supply conditioned air to rooms and draw return air back
to the furnace or air conditioner. When supply and return ducts are not located in the
same room, rooms with supply ducts are over-pressurized and rooms with return ducts
are under-pressurized. These pressure differences increase air infiltration. Thus,
recommended practice is for each room to have equally sized supply and return ducts.

minf
Return
 return in each room  0
N
 central return  1.5N
 otherwise
N
Calculating Energy Loss/Gain Due to Infiltration
Furnaces and boilers increase air temperature, but do not add or remove moisture from
the air. Hence, infiltration heat loss during winter includes only sensible heat loss. Air
conditioners cool warm and humid air below the dew point temperature and condense
moisture out of the air. Hence, the air conditioners remove both energy sensible and
latent energy from the air. Thus, infiltration heat gain during summer must include both
sensible and latent cooling.
Winter Heating
 2h2  m
 1h 1
Q net,out  Q inf  m
 (h2  h1 )
m
 V ρ (h2  h1 )
Toa
 V N ρ (h2  h1 )
m1

Tia
m2

In winter, furnaces & heat pumps add sensible heat only. Thus Δh = cpΔT
 inf  V N
 ρc p (T2  T1 )
Q
 ρc p (Tia  Toa )
Q inf  V N
where
3
V  volume
of house
ft 
3
N  air changes per hour
1hr 
 Btu 
ρc p  0.018  3
 ft  F 
Example
Calculate Q inf for a 1,500 ft2 house with 8 ft ceilings if N  1.0, Tia  72F, Toa  30F .
Q inf  VN ρc p (Tia  Toa )
 
1
 Btu 
 Btu 
 (1500  8) ft 3  1.0   .018 3
 (72  30)F  9,072 

 hr 
 ft  F 
 hr 
Summer Air Conditioning
 1h1  m
 2 h2
Q net,in  Q inf  m
 (h1  h2 )
m

 ρ (h2  h1 ) but h includes both Q sensible
VN

& Q latent )
In summer, A/C removes sensible & latent heat (dries air), Thus
Q inf  Q inf,sensible  Q inf,latent
Q inf,sensible  V N ρc p (T1  T2 )  V N ρc p (Toa  Tia )
Q inf,latent  energy required to condense water from cooling coil
 w h fg  m
 a ω 1  ω 2 h fg
m
 V N ρ ω 1  ω 2 h fg
where
 
V  volume of house ft 3
 hr 
N  air changes per hour 1

ρ  density of air  .075 lba
ft 3


ω 1  specific humidity of outside air from psychromet ric chart lbw

ω 2  specific humidity of inside air from psychromet ric chart lbw

h fg  enthalpy of evaporatio n  1,076 Btu
lbw

lba
lba


4
Example
Calculate the sensible & latent infiltration loads on a 1,500 ft2 house with 8 ft ceilings if
 hr ,
  1.0 1
N
Tia  70F, RHin  50%, To a  90F, RHoa  80%
From the psychrometric chart:
 lba
T  90F, RH  80%  .0248 lbw lba
ω ia T  70F, RH  50%  .0078 lbw
ω oa
Q inf,sensible  V N ρc p Toa  Tia 
 
1
 Btu 
 Btu 
 1500  8 ft 3  1.0   .018 3
 90  70F  4,320  

 hr 
 ft  F 
 hr 
5
Q inf,latent  V N ρ ω oa  ω ia h fg
 
1
 lba 
 lbw 
 Btu 
Q inf,latent  1500  8  ft 3  1.0    .075 3   .0248  .0078 
 1,076 

 hr 
 ft 
 lba 
 lbw 
 Btu 
Q inf,latent  16,463
 hr 
 Btu 
Q inf  Q inf,sensib le  Q inf,latent  4,320  16,463  20,783
 hr 
Calculating Average Air Infiltration
Air infiltration is typically measured in terms of the number of times the entire volume
of air in a building is replaced each hour. This unit is called “air changes per hour” N .
The actual number of air changes per hour in a building varies continuously over time
with drivers of infiltration such as wind velocity. Instantaneous measurements of
infiltration can be made by releasing a tracer gas in a house and recording the rate of
decay in the concentration of the gas. However, this method is somewhat impractical
for wide spread use. Two methods for estimating the average number of air changes
per hour are described below.
ASHRAE Method
The ASHRAE method provides a simple way to estimate air infiltration based on the
tightness of construction, seasonal variations in wind speed, and the indoor/outdoor air
temperature difference that drives the stack effect.
  a  b Tia  Toa
N
Time of year Construction
Tight
Winter
Medium
Loose
Tight
Summer
Medium
Loose
a
0.280
0.408
0.483
0.210
0.310
0.310
b
0.00630
0.00873
0.01224
0.00720
0.00840
0.01400
Blower Door Method
The blower door method estimates the average air infiltration by depressurizing a house
with a blower and measuring the quantity of air that must be removed from the house
to maintain a constant pressure difference between indoor and outdoor air. Leaky
houses will require more air to maintain a constant pressure difference than tight
houses. To use the blower door method, install a blower door in a door way and seal
the door against the door jam to minimize leakage. Next, depressurize house until
6
ΔP = Pia – Poa = 50 Pascals
The air flow required to maintain a 50 Pa pressure difference is called V 50 or “CFM50.”
Use the following relation to convert from V 50 (CFM50) to N 50 (ÄCH50)
3
1
 min 
N50  V 50  ft  
 60
3
 hr 
 min  Vhouse ft
 
CFM50  60 

 or ACH50 

V ft 3


 
The “Princeton” Method to estimate the average N from N 50 is:

  N50
N
20
ACH50 

 or ACH 

20 

The “Sherman” Method to estimate the average N from N 50 is:
N 
N 50
C  H S  L
ACH50 

 or ACH 

C  H S  L 

where the following coefficients can be estimated from the graph and tables that
follow.
C = leakage infiltration ratio: 26 (hot climate) < C < 14 (cold climate)
H = height correction factor
S = wind shielding correction factor
L = leakiness correction factor
7
Coefficients for Sherman Method
Example
Use the Sherman method to calculate the average infiltration rate for a 1,500 ft2 house
in Dayton, OH with 8 ft ceilings, a single story, average shielding, and average cracks, if
the measured CFM50 = 1,800 cfm.
CFM50  ft 3 
 min  1800  60
1
N 50  ACH50 

 9.00  
  60 
3 

V ft  min 
 hr  1500  8
 hr 
 
N  ACH 
ACH50
9.00
1

 .45  
C  H  S  L 20  1.0  1.0  1.0
 hr 
LBNL Effective Air Leakage Area
The LBNL method calculates an effective air leakage area, ELA at 0.016 inwc from a
blower door test. The air flow due to infiltration, Q, is then calculated as:
Q = ELA (Cs (Tia-Toa) + Cw U2)0.5
Where Cs is stack coefficient, Cw is wind coefficient and U is wind speed. Values of
these coefficients are listed in the tables below.
8
Source: ASHRAE Fundamentals, 2005 p. 27.21
Estimating Hourly Air Infiltration
Researchers at Lawrence Berkeley National Laboratory have developed methods to
estimate hourly infiltration based on “effective leakage area”, building characteristics,
and weather drivers.
The effective leakage area at a typical pressure difference of 4 Pa (ELAat4Pa) can be
estimated from air density (p) and the volume of air displaced from a building at 50 Pa
(CFM50) during a blower door test;
p = density (kg/m3) = 1.2
Q50 (m3/s) = CFM50 (ft3/m) / (35.314 ft3/m3 x 60 s/m)
ELAat4Pa (m2) = (p/8)^.5 Q50 (4/50)^.65 0.106188433
The effective velocity of air (s) through the effective leakage area (ELAat4Pa) can be
calculated from a stack effect coefficient (fs) and the indoor outdoor air temperature
(dT), and from a wind effect coefficient (fw) and wind speed (ws). Values for fs and fw
for typical residential buildings are shown below.
dT = [Tindoor – Toudoor]+
fs (m/s-K^0.5) = 0.120
9
fw = 0.132
s = (fs^2 dT + fw^2 ws^2)^0.5
The volume flow rate of infiltration (V) is the product of the effective velocity (s) and the
effective leakage area (ELAat4Pa). The number of air changes per hour (ACH) is the ratio
of volume flow rate and the volume of the conditioned space (Vol).
V = s ELAat4Pa
N = ACH = V / Vol
Example
Calculate the number of air changes per hour for a residential building in which
the volume of air displaced at 50 Pa (CFM50) during a blower door test is 3,000 cfm.
The floor area is 1,500 ft2, the ceiling height is 8 ft. The temperature difference
between the indoor and outdoor air is 10 C and the wind speed is 4 m/s.
Input Data
CFM50 (cfm)
A = floor area (ft2)
H = height (ft)
dT = Indoor outdoor temperature difference (deg C)
ws = wind speed (m/s)
3000
1500
8
10
4
Constants
p = density (kg/m3)
n
fs (m/s-K^0.5)
fw
1.2
0.65
0.12
0.132
Calculations
A = floor area (m) (1 m2 = 10.76 ft2)
H = height (m) (1 m = 3.28 ft)
Q50 (m3/s) = CFM50 / (35.314 x 60)
ELA(at 4 Pa) (m2) = (p/8)^.5 Q50 (4/50)^.65
s (m/s) = (fs^2 dT + fw^2 ws^2)^0.5
Q (m3/s) = ELA s = air flow
Vol = A H (m3)
ACH (1/hr) = Q/ Vol x 3600 s/hr
139.4052
2.439024
1.415869
0.106188
0.650218
0.069046
340.0127
0.731045
10
The chart below shows hourly air change per hour rates for a residential building
calculated using this procedure and TMY3 temperature and wind speed data from
Dayton, Ohio. Hourly infiltration rates are higher during winter when the indooroutdoor air temperature is larger. Hourly variations in wind speed cause significant
variations in Infiltration rates throughout the year.
Hourly and average air changes per hour for a residential building in Dayton, Ohio.
Typical Air Infiltration Rates
An independent study of air infiltration rates in new Minnesota houses found average
rates of ACH50 = 4.2 and ACH = 0.27. ASHRAE research shows that 75% of newly
constructed homes have infiltration rates between 0.25 and 0.75 air changes per hour.
Older homes have higher infiltration rates.
Source: ASHRAE Fundamentals, 2005 p. 27.15
11
Recommended Air Ventilation/Infiltration Rates
U.S. E.P.A. Energy Star
The U.S. Environmental Protection Agency awards “Energy Star” designation to highly
energy efficient houses. One of the criteria for the Energy Star award is based on a
blower door test:
Energy Star House:
CFM50 [cfm]
 0.35
A ft 2
 
Example
The measured infiltration rate at 50 Pa indoor/outdoor pressure difference in a 1,500 ft 2
house is 1,800 cfm. Determine whether the house meets the U.S. Energy Star criteria
for air leakage.
CFM50 1,800 cfm

 1.2  0.35
A ft 2
1,500 ft 2
 
 
The house does not meet the U.S. Energy Star criteria for air leakage.
ASHRAE Standard 62-1989
ASHRAE Standard 62-1989 recommends that all houses have ventilation rates exceeding
0.35 ACH to remove air contaminants. Thus, in very tight houses in which the natural
infiltration rate is less than 0.35 ACH, recommended practice is to install an air-to-air
heat exchanger, which continually brings in and exhausts outdoor air but recovers
energy between the two air streams.
12
Blower Door Instructions
Blower doors measure infiltration rate for a house. Infiltration is the amount of air that
is moved from within a house to outside. Because this air carries with it heat/cooling put
into it by the boiler/air conditioner, that energy must then be replaced. Reducing
infiltration is an important part of conserving energy for a house. Infiltration rate is
typically measured in Air Changes per Hour (ACH), or the amount of house volumes
drafted out in a one-hour period. A good ACH for a tightly sealed house is about 0.5.
Blower doors measure approximate ACH by depressurizing the house.
Blower Door Operation
1. Make sure house is tightly sealed.
2. Select Time Mode of 5 seconds. (Turn dial to Time Mode, and push down on
Select).
3. Fan Select 3-0 (Turn dial to Fan Mode. Type 3 (select up), Configuration 0 (select
down)).
4. Range – Low.
5. Channel-Set to channel A.
6. Gauge Setup (refer to diagram next page)
 Connect Channel A reference knob with a tube to outside of the house.
 Leave Channel A input knob open
 Connect Channel B input knob with a tube to the blower door fan.
 Leave Channel B reference knob open
7. Remove all 3 rings from fan.
8. Keep gauge level.
9. Raise fan speed until pressure reads about 50. Then switch to cfm, and record
CFM.
13
ACH calculation
1. Calculate the volume of the house. You can find area of the floor on
housing.udayton.edu. Multiply the area by about 8 to get the volume of the
house in cubic feet.
CFM(measured)  60
2. ACH at 50 Pascals, or ACH50 
.
Volume
3. Find N  C  H  S  L .
4. Let C = 19, S = 1, L = 1 (normal house), 0.7 (drafty house), H = 1 (one story), 0.8 (2
stories).
House Pressure
CRF
ACH50
45
1.1
5. ACH 
.
40
1.2
N
35
1.3
If you can’t reach 50 Pa, then your house is either big or very
30
1.4
leaky. Use the Can’t Reach Fifty (CRF) Factor on the speed dial
25
1.6
of the fan. Then multiply your ACH50 by this CRF.
20
1.8
15
2.2
Example EEB Infiltration Measurements
229 Lowes
Built ~1930
Vol. = 14,120 ft3
CFM50 = 5704
dP = 45 Pa (therefore factor of 1.1)
Fan dP = 143 Pa;
CFM50  60
ACH50 
 24.21
Volume
ACH50
N 
 1.21 (Princeton Method)
20
435 Kiefaber
(Built 1999)
Vol. = 15,683 ft3
CFM50 = 3255
Total dP = 50 Pa
Fan dP = 46 Pa;
CFM50  60
ACH50 
 12.45
Volume
ACH50
N 
 .62 (Princeton Method)
20
14
Reducing Infiltration Losses in Energy Efficient Buildings
Create Wind Shields
Build an earthen berm, locate the garage, or plant trees and shrubs on the windward
side of a building to shield it from the wind.
Supply Outside Air to Combustion Devices
Use PVC piping to directly supply outside air for combustion to natural gas furnaces,
boilers and water heaters. This will reduce drafts of cold air directly into the
conditioned space.
Seal Air Leaks
Probably the single most cost effective measure to improve energy efficiency is sealing
leaks.
In new construction, the goal is to create a continuous air barrier. This entails sealing
joints between the foundation, sill plate, floor joist header, sub-floor, wall framing
plates and upper rim joist with caulk and/or other materials. In addition, seal all
penetrations in the wall and ceiling drywall including electrical boxes and plumbing.
Finally, tape and seal all joints in the house wrap and moisture barriers. One study
found that at 50 P pressure difference, air leakage in a sealed house averaged 1.24 air
changes per hour compared with 2.41 air changes per hour in an identical but unsealed
house (1).
House wrap to reduce infiltration.
Close-up of taped joints.
In existing buildings, use weatherstripping to seal doors and windows. Felt and open
cell foam are inexpensive. However, vinyl and metal weatherstripping tends to last
15
longer and be more effective at reducing air flow. Use caulk to seal joints around
plumbing and electrical fixtures and between walls, floors, ceilings, windows and doors.
Also,
Ducts
Locate ducts within the insulated space and use mastic to seal duct joints. The photo
below shows a duct located within the insulated space and sealed with mastic. A new
technology allows contractors to seal duct leaks by injecting polymer laden air through
the ducts. The company reports that the cost for a typical residence is about $800 to
$1,500 with about a one year payback. In addition, balance supply and return air in
every room to minimize over/under pressurization. The second photo shows a
technician measuring the flow rate from a supply duct.
Photos: NREL. http://www.nrel.gov/data/pix/
Air-to-Air Energy Recovery Units
To minimize heating and cooling energy losses associated with exhausting and
replenishing outdoor air, heat can be exchanged between the entering and exiting air
streams. All air-to-air energy recovery units transfer sensible heat from the warm air
stream to the cool air stream, raising the temperature of the cool air stream and
lowering the temperature of the hot air stream. Some air-to-air energy recovery units
can also transfer latent heat, in the form of water vapor, between the air steams. In
these units, moisture is transferred from the moist air stream to the dry air stream. In
an air-conditioned building when the indoor air is dryer than the outdoor air, this
decreases the humidity of air entering the building.
16
The schematic below shows an energy recovery unit that uses a heat exchange wheel.
As the wheel slowly rotates through the hot side, fins within the wheel absorb heat from
air and cool the exit air. When the fins rotate to the cool side, they give up this heat and
heat the exit air. The performance of a typical sensible+latent air-to-air heat exchanger
is depicted in the schematic. The picture below shows a different type an energy
recovery unit with a fixed plate heat exchanger. One air stream is drawn diagonally
from upper left to lower right. The other air stream is drawn diagonally from lower left
to upper right. Sensible heat is transferred through plates between the two air streams.
SUMMER CONDITIONS
WINTER CONDITIONS
SUPPLY AIR
81°F DB
68°F WB
OUTDOOR AIR
95°F DB
78°F WB
SUPPLY AIR
53°F DB
40°F WB
OUTDOOR AIR
7°F DB
6°F WB
RETURN AIR
75°F DB
63°F WB
EXHAUST AIR
89°F DB
73°F WB
RETURN AIR
72°F DB
54°F WB
EXHAUST AIR
27°F DB
73°F WB
Source of data: http://www.airxchange.com/
Photo: NREL. http://www.nrel.gov/data/pix/
Energy savings from air-to-air heat exchangers can be modeled using the heat
exchanger effectiveness method as:
Sensible only:
Sensible + Latent:
Q = e m cp (Th1 – Tc1) = e V (p cp) (Th1 – Tc1)
Q = e m (hh1 – hc1) = e V p (hh1 – hc1)
where e is heat exchanger effectiveness, m is mass flow rate, V is volume flow rate, p is
density, cp is specific heat, T is temperature, h1 is the entering hot stream and c1 is the
entering cold stream.
17
Example
Consider a sensible air-to-air heat exchanger moving 100 cfm into and out of a building
with effectiveness = 0.70. The inside air temperature is 72 F and the outside air
temperature is 30 F. The product of air density and specific heat, (p cp), is 0.018 Btu/ft3F. Determine the energy reclaimed (Btu/hr) and the temperature of the pre-heated as it
leaves the heat exchanger and enters the building.
The rate of heat reclaimed by the heat exchanger is:
Q = e V (p cp) (Th1 – Tc1) = 0.70 100 ft3/min 60 min/hr 0.018 Btu/ft3-F (72 – 30)
Q = 3,175 Btu/hr
From an energy balance on the incoming cold air stream:
Q = V (p cp) (Tc2 – Tc1)
Tc2 = Tc1 + Q / [ V (p cp) ] =30 F + 3,175 Btu/hr/[100 ft3/min 60 min/hr 0.018 Btu/ft3-F]
Tc2 = 59.4 F
References :
(1) (http://www.energysavers.gov/your_home/insulation_airsealing/index.cfm/myt
opic=11310).
18