If every household lowered its average house temperature by 6 degrees every day, we’d save the equivalent of 500,000 gallons of oil a day.
Problem Restatement
How many gallons of oil are saved per winter day by lowering the average U.S. house temperature 6 degrees?
Governing Equations and Input
Heat loss from infiltration/exfiltration:
Heat loss through walls, windows, doors, and roof:
Heat loss through a slab:
For calculating U-values:
ΔT [°F] = Indoor Temperature [°F] – Outdoor Temperature [°F]
[BTU] = [BTU/hr] * [hr]
1 BTU = 0.0000072 gallons of oil
Information Sources, Data, and Measurement
This analysis assumes that all houses in the U.S. are single-story, brick houses with no basements, with wooden studs and asphalt roofing. All houses are assumed to operate according to the input parameters, to have negligible heat gains from occupancy and equipment, and to be heated with natural gas. Actual results will vary depending on occupancy, appliances, type of heating system, and other factors.
Data concerning the construction, occupancy, and heating equipment of U.S. housing can be obtained from the American Housing Survey1.
The National Oceanic and Atmospheric Administration (NOAA)2 provides climatic data including average, high, and low temperatures by month for several hundred U.S. cities.
Reference data including equations, U-values, and load calculation procedures are detailed in the 2001 ASHRAE Fundamentals Handbook3.
Procedure for a calculating a solution [Sources in brackets]:
Define housing materials and find U and F values. [ASHRAE]
Determine dimensions (surface area) for each material. [ASHRAE]
Determine indoor and outdoor temperatures. [NOAA]
Calculate heating load using formulas listed above, considering infiltration/exfiltration and conduction through walls, windows, doors, roof, and slab. Find a load for each of the three outdoor temperatures. Repeat for the adjusted indoor temperature. [ASHRAE]
Determine energy use by multiplying load by the corresponding duration of time at a given temperature. (For this analysis, it is assumed over the course of a day that the high and low outdoor temperatures last for 6 hours each and the average temperature lasts for 12 hours.) Also consider types of and efficiency of heating equipment. [ASHRAE]
Subtract energy use at the lower indoor temperature from energy use at the higher indoor temperature.
Convert the amount of saved energy to units of gallons per hour.
Multiply energy saved per house by total number of U.S. houses. [American Housing Survey]
Find your city's annual high, low and mean temperature.
Final Answer
This analysis shows that the amount of energy saved from lowering thermostats in every U.S. home greatly exceeds the claim’s value. The result is highly dependent on the assumptions made in the analysis and the level of detail considered by the analysis. The fact that the claim was originally formulated circa 1970 may account for some of the discrepancy between the two results.
Deeper Issues
One of the most important insights gained from this analysis is the relative importance of certain building components on the overall heating load of the building. Factors such as heat loss from a slab are usually not as important as losses from windows and roofs. Energy conservation measures should be directed towards those building components that most significantly affect energy use.
Other Related Questions
How much energy is saved during a summer day by raising the thermostat 6 degrees? (Analysis of a cooling load involves additional load considerations such as latent heat, internal gains from people and appliances, solar radiation, and increased heat gain through opaque surfaces.)
How much energy is saved by adjusting the thermostat in a commercial building?
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