Energy Efficient Housing

Energy Efficient Housing Construction

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Heating Systems

An energy efficient home by definition has very little heat loss because of high insulation levels and airtight construction. This leads to two problems: finding a properly-sized heat source and providing adequate ventilation to maintain indoor air quality.

Except for extremely cold periods, a properly designed and constructed energy efficient home can sometimes gain almost enough daily heat from 'waste' sources such as the heat given off by lights, people and appliances. During sunny, cold days solar energy gains also contribute to reducing the heating load. These heat sources are often called 'internal gains'.

Modern control systems such as programmable thermostats can further help to reduce heating energy consumption.

Sizing the Heating System

Heating equipment in a home must be capable of maintaining an interior temperature of 68 to 72 F (20 to 22) during the heating season. Heating equipment is generally oversized for most homes, but is even worse if the home is an energy efficient home. This leads to frequent on/off operation reducing both efficiency of fuel use and service life of heating equipment.

A heat loss calculation is required to determine the heating system required. It should have a capacity of no more than 10% in excess of the calculated requirement. Installing the smallest capacity heating equipment to meet the loads will save both energy and money. Calculation of total home heat loss is generally done by heating contractors. Contractors inexperienced in understanding low energy house design and heat loss, however, may still result in drastic over sizing. A simple heat loss calculation method is provided in this section.

Isolating The Heating System

If a fuel burning furnace, boiler and/or hot water heater is required, building an airtight enclosure (mechanical room) around the appliances can help control chimney heat loss. Separate combustion and fresh air supplies feed into this room. No previously heated air is used by the fuel burning appliances and cold outside air is prevented from entering other areas of the home. This isolated room must be insulated and sealed from the rest of the home. Water pipes and heat supply ducts should also be insulated.

Isolating The Heating System

Calculating Heat Loss

Calculating the heat loss from a home is quite simple. The heating requirement will be highest when the outside temperature is lowest and there is no solar gain. A cold winter night is when the heating load will be greatest. Heat flows out through all the building surfaces including walls, ceilings, floors, windows and doors.

Building Heat Loss Areas

Heat loss through each surface can be calculated using the following equation:

Heat Loss Formula

Heat is also lost through infiltration and exfiltration - air leakage. This heat loss can be calculated using:

Air Leakage Formula

Note: change the constant 0.36 to 0.018 Imperial units.

There are also a number of sources of heat gain in a typical house. Not only do the occupants give off heat, appliances and lights contribute significantly to home heating. Each person can provide about 75 watts of heating energy while 200 or 300 watts are available from appliances (like freezers, ranges, refrigerators, etc). The average home therefore provides 500 or more watts daily of the total energy required for space heating.

An example heat loss calculation is shown using Plan 13 from the REED Energy Efficient House Plans section. Plan 13 is a 1,920 square foot (178 sq. metres) two level, rectangular bungalow. If this house was to be built in the Red Deer (Alberta, Canada) area, the outside heating design day temperature is - 26 F (-33 C). A common inside temperature is 68 F (20 C) - the difference between them is 94 F (53 C). Design day temperatures and heating degree days information for your locale is usually available from your local weather office. A short list is provided as an example of January design temperatures and degree days for various Canadian cities.

The first set of equations is used to calculate the heat loss from each of the seven surfaces (substitute the areas, R-values and the temperature difference for each surface from your plan). The air leakage heat loss is next calculated using the second equation. The building air volume is 16,775 cubic feet (475 cubic metres) and an air change rate of one-third (1/3 of the house air is replaced every hour with fresh air) can be used in the equation (a typical rate for a well-built, energy efficient home). Substitute the house volume from your plan into the equation. Using the same temperature difference, the total air leakage heat loss is calculated and added to the surface heat loss figures. Subtracting the heat gain average of 500 W results in a total space heat requirement of 8200 W for this example (8.2 kW or about 28,000 btu/hr).

There are a wide variety of computer software programs available which can be used to more accurately calculate building heat loss. These programs require a detailed breakdown of each building component and complete area weather data. Most of the programs available require a considerable learning curve and are often not practical unless you do a lot of heat loss calculations, are a house designer or are designing a complex solar building.

The efficiency of the heat source must be taken into account when selecting it. In the example, an 8.2 kW heat source would be needed (28,000 btu/hr). If one chooses a 100% efficient electric heating source, the exact figure calculated above can be used to size equipment. Gas furnaces range from 70% to 80% efficient (measured seasonally - over an entire year of operation). Divide the heat load (8.2kW) by the system efficiency (0.70) to obtain the 'bonnet' size of 11.7kW (40,000 btu/hr) necessary to provide 8.2kW. Gas-fired furnace and boiler units with efficiencies of 90% to 95% are also available but are usually produced in large output sizes and are more expensive.

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