outlet cover savings calculations

These calculations display your estimated savings based on assumptions below

For customized calculations enter your own data in the green fields

size of house sq ft
ceiling height ft
region



cooling degree days CDD
heating degree days HDD
type of heater

cost per kilowatt hour
cost per therm
cost per CCF
cost per cubic meter
1 CCF = 103,100 BTU = 1.031 therm
1 m3 = 35.31 ft3 = 0.3531 CCF
infiltration through uncovered outlets %
number of outlets on exterior walls
Heating Seasonal Performance Factor (HSPF) BTU/(W hr)
Seasonal Energy Efficiency Rating (SEER) BTU/(W hr)
Annual Fuel Utilization Efficiency (AFUE) %
air changes per hour ACH
0.5 ACH - House is a new, energy-efficient construction, or has been thoroughly weatherized by professionals. All doors and windows seal tightly, and no drafty areas are noticed within.
0.9 ACH - House is a recent construction (~2000), or an older home that has weatherstripping and seals in good repair. There may be some small gaps around doors and windows, or a few noticeable drafty places.
1.5 ACH - House is an older construction (~1990), or has several noticeable gaps around doors and windows. There will be several noticeably drafty areas that do not maintain temperature well.
2.0 ACH - House is much older, likely built without a central conditioning system. Remote rooms are uncomfortable during extreme temperature conditions (i.e. hot summer/cold winter conditions).

How your savings are calculated

Step 1 determine rate of infiltration

2,000 sq ft x 8 ft ceilings = 16,000 cubic feet volume inside the house

16,000 cu ft x 1.5 volumes/hour (ACH) = 24,000 cubic feet per hour

24,000 / 60 minutes per hour = 400 CFM of infiltration

Step 2 determine infiltration due to uncovered outlets

400 CFM x 2% air leakage waste due to uncovered outlets / 20 outlets on exterior walls = 0.40 CFM

Step 3 determine cooling energy

0.081 density of air (lb/cu ft) x 0.240 heat capacity of air (BTU/lb °F) x 2008 CDD (°F day) x 0.40 CFM x 24 hrs/day x 60 min/hr = 22,484 BTUs

Step 4 determine heating energy

0.081 density of air (lb/cu ft) x 0.240 heat capacity of air (BTU/lb °F) x 2914 HDD (°F day) x 0.40 CFM x 24 hrs/day x 60 min/hr = 32,629 BTUs

Step 5 determine cooling utility used cool air

22,484 BTUs / 15 SEER (BTU/W hr) / 1,000 W/kW = 1.5 kWh per year

Step 6 determine heating utility used heat air

32,629 BTUs / 8.2 HSPF (BTU/W hr) / 1,000 W/kW = 4.0 kWh per year

Step 7 determine cooling savings

1.5 kWh x $0.118 = $0.18 per year

Step 8 determine heating savings

4.0 kWh x $0.118 = $0.47 per year

Step 9 determine total savings

1.5 kWh cooling + 4.0 kWh heating = 5.5 kWh total savings per year

$0.18 cooling + $0.47 heating = $0.65 total savings per year

5.5 kWh x 15 = 82.2 kWh lifetime savings

$0.65 x 15 = $9.70 lifetime savings
Savings calculations are presented as an example of potential savings; actual savings may vary based on use, utility rates and other factors. Calculations are based on national averages and assumptions and may be adjusted for a more accurate estimate of individual savings. While EnergyEarth strives to provide the most accurate information, we cannot guarantee the savings displayed.