The House as a System

1

Air Temperature and Water Vapor

Simple Psychrometric Chart100% Relative Humidity Line = Dew Point Line

Simple Psychrometric ChartAs temperatures rise, relative humidity decreases

Simple Psychrometric ChartAs temperatures decrease, relative humidity increases

1. Indoor winter air at 70 deg F and 40% RH leaks into wall

2. Slide across to 100% RH line to find dew point

3. If inside surface of wall’s exterior sheathing is below 44 deg F dew point, condensation may occur

1. Indoor winter air at 70 deg F and 40% RH leaks into wall

2. Slide across to 100% RH line to find dew point

3. If inside surface of wall’s exterior sheathing is below 44 deg F dew point, condensation may occur

Winter Concerns

2

3

Summer Concerns1. Outdoor

summer air at 88 deg F and 50% RH leaks into wall

2. Slide across to 100% RH line to find dew point

3. If wall’s interior finish is below 66 deg F condensation may occur

1. Outdoor summer air at 88 deg F and 50% RH leaks into wall

2. Slide across to 100% RH line to find dew point

3. If wall’s interior finish is below 66 deg F condensation may occur

The Real Psychrometric Chart (for later)

Crawl Space Moisture Problems

9

92/ 70%RH

72/ 100%RH

92/ 70%RH

High Performance Recommendations (Test for Results)

Air leakage – blower door

Gauges for pressure problems

Duct leakage – duct testing fan

Indoor air quality

You don’t know ‘til you test

Testing for Pressure Imbalances

• Pressure imbalances occur when exhaust fans (including dryers and attic vent fans) or the HVAC blower cause high negative or positive pressures in the home or in areas with combustion appliances

• Combustion appliances -- those with flues/ vents such as fuel-fired water heaters, furnaces, vented space heaters, vented dryers

Pressure Imbalance Testing

Pressure gauge

Channel A; Top Port Reference to Inside PressureChannel A; Bottom Port Tube Goes Outside

-4.5 Pa

Pressure Imbalance Testing Protocol

• Set up gauge to measure pressure difference between inside and outside

• Turn on exhaust fans, including dryer; note pressure • Turn on HVAC; note pressure• Close doors to interior rooms without exhaust fans;

note pressure• If house pressure is -5 Pascals or lower, you could

backdraft combustion appliances

Pressure Imbalance Testing Protocol (cont.)

• Also test areas with combustion appliances (Combustion-Air Zone -- CAZ)

• Set up gauge to measure difference in pressure between zone and outside

• Turn on all blowers in combustion zone and measure negative pressure

• Try combinations of other fans and door closings to see if you can create negative pressures

Wood or metal frame

Air tight fabric

Gauges

Blower

Blower Door Components

Blower Door Components

16

CFM50

Blower Door - 50

Pascals

Blower Door Test for Air Leakage

• To compare homes• To check for compliance with ASHRAE indoor air

quality guidelines• To find ACH50, use CFM50 from blower door test

• Find volume of building• Find ACH50 = CFM50 x 60/ House Volume

• Many feel a better measure is CFM50/ square foot of exterior envelope

• To compare homes• To check for compliance with ASHRAE indoor air

quality guidelines• To find ACH50, use CFM50 from blower door test

• Find volume of building• Find ACH50 = CFM50 x 60/ House Volume

• Many feel a better measure is CFM50/ square foot of exterior envelope

Purposes of Finding ACH50(air changes per hour at 50 Pascals)

• Blower door test reveals 5,000 CFM50.• The volume is:

• Area of each floor = 28ft x 48ft = 1,344 sq ft

• First floor volume = 1,344 sq ft x 9 ft =12,096 cu ft

• Second floor volume = 1,344 sq ft x 8 ft =10,752 cu ft

• Total volume = 22,848 cu ft

8’9’

Calculating ACH50

• ACH50 = CFM50 x 60 Volume

• ACH50 = 5,000 CFM50 x 60 / 22,848 cu ft

• ACH50 = 13.1 ACH50

• ACH50 = CFM50 x 60 Volume

• ACH50 = 5,000 CFM50 x 60 / 22,848 cu ft

• ACH50 = 13.1 ACH50

Finding ACH50 (continued)

Calculating CFM50/sq ft

Area of walls = Perimeter x height = Perimeter = 2 * (28 + 48) = 152 lin ft

Wall height = 17 feet Area = 152 * 17 = 2,584

Floor area = 1,344 sq ft

Ceiling area = 1,344 sq ft

Total area = 5,272 sq ft

CFM50/ sq ft = 5,000/ 5,272 = 0.95

Very tight new home 1-3Tight new home 3-5Moderate new home 5-8Standard new home 8-22Older home 12-24Very leaky older home 24-50

Very tight new home 1-3Tight new home 3-5Moderate new home 5-8Standard new home 8-22Older home 12-24Very leaky older home 24-50

CFM50/ square foot of exterior surface area:

0.30 for fairly tight home

CFM50/ square foot of exterior surface area:

0.30 for fairly tight home

Range of ACH50 – Southeastern Homes

Southface Study -- Early 90’s

0

5

10

15

20

25

30

0-4 4-6 6-8 8-10 10-12 12-14 14-16 16-21

ACH50

Nu

mb

er o

f H

omes

New home buyers (less than 2 years old) received flyer offering a blower door test. Results are for 100 homes.

• 0.35 air changes per hour, or • At least 15 cfm per occupant• New ASHRAE 62 Residential

7.5 cfm per person + 0.01 * floor area• Kitchen exhaust fans

– 100 cfm intermittent or – 25 cfm of continuous fresh air or operable window

• Bathroom exhaust fans – 50 cfm intermittent or – 20 cfm continuous fresh air or operable window

ASHRAE 62 (not revision): Minimum Ventilation Requirements for Indoor Air Quality

1. Conduct blower door test - find ACH50 = CFM50 * 60/ volume

2. Find factor in table:3. ach = ACH50/ factor

in table

Number Degree of Protectionof

Floors Exposed Average Good1 18.9 21 25.2

1.5 17 18.9 22.72 15.1 16.8 20.2

2.5 14.2 15.8 18.93 13.2 14.7 17.6

over 3 12.3 13.7 16.4

Estimating Natural ACH

• 2-story home with 4 bedrooms– Area = 2,000 sq ft; Volume = 16,000 cubic feet– Good protection from wind– 10 ACH50 on blower door test

• Natural air changes per hour– From table, ach Factor = 20.2– ach = ACH50/ach Factor = 10 / 20.2 = 0.5

(often used in computer load sizing and energy estimating calculations)

Example - Natural ACH

Example : ASHRAE 62 - cfm Requirement

• Required minimum cfm by ASHRAE 62 = (# of bedrooms + 1) x 7.5 cfm + (0.01 * area) =

((4+1) x 7.5) + (0.01 * 2000) = 57.5 cfm• Measured cfm for home = ach x volume / 60

= 0.5 x 16,000/ 60 = 133 cfm, which complies since it exceeds required minimum of 57.5 cfm

Blower Door/ ASHRAE 62 Homework

• A home has a bottom floor measuring 24 feet x 48 feet and having 10’ ceilings. The second floor measures 18 feet x 48 feet, but has a sloped ceiling that is 8 feet high at its lowest point and 16 feet high at its apex. The ceiling runs this way along the entire 48-foot length.

• During a blower door test, you have a house pressure of 50 Pa, fan pressure of 225 Pa, and 4,500 cfm of air flow out the blower.

• What is the CFM50 and the ACH50?• Does the home comply with the old ASHRAE 62?

(assume average shielding) – factor from table is 18.9

• Sealing air leaks is important:• Saves energy• Can reduce moisture problems• Increases comfort

• Blower door measures leakage rates: ACH50 or CFM50/sq ft

• Most important leaks are to attic and crawl space -- usually larger and air is undesirable

• Select durable sealants

• Sealing air leaks is important:• Saves energy• Can reduce moisture problems• Increases comfort

• Blower door measures leakage rates: ACH50 or CFM50/sq ft

• Most important leaks are to attic and crawl space -- usually larger and air is undesirable

• Select durable sealants

Air Leakage Summary