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 On average, home heating uses more energy than any other system in a home  About 45% of total energy use  More than half of homes use natural gas.

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Presentation on theme: " On average, home heating uses more energy than any other system in a home  About 45% of total energy use  More than half of homes use natural gas."— Presentation transcript:

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2  On average, home heating uses more energy than any other system in a home  About 45% of total energy use  More than half of homes use natural gas.

3  In the winter homes lose heat to the outside through conduction, convection, radiation, and infiltration  These losses can be reduced by good home design, but there is always some loss of heat  To keep the inside of a home warm the lost heat needs to be replaced

4  Heating systems replace lost heat to keep your home warm  Typical kinds of heating systems are: ◦ Gas furnaces ◦ Oil-fired furnaces ◦ Heat pumps ◦ Electrical heating  Passive solar features can help capture the sun’s energy for heating

5  Good Construction Techniques ◦ Making sure joints are sealed and tight ◦ Installing insulation uniformly and at sufficient depths ◦ Windows and doors function properly and shut well ◦ Ensuring penetrations for plumbing and electricity are well sealed ◦ Radiant and infiltration barriers ◦ Well constructed ventilation ductwork  Good materials ◦ High efficiency windows reduce conductive, convective, and radiant heat losses ◦ Ceiling and wall insulation that performs better than standard materials

6 Typically, the best ways to save energy in existing homes is:  Sealing holes between the house and the attic and then adding attic insulation  Sealing the ductwork that moves heated air throughout the home in some systems  Tuning or replacing the heating and cooling equipment.  Incorporating passive solar systems when possible

7  More efficient heating systems ◦ Some types of systems, such as electric heating, use more energy than needed to heat a home. In mild climates heat pumps can use less energy to warm the house by the same amount. ◦ The efficiency of all systems has increased over time. Replacing older systems (20 years old or more) can save energy. Look for Energy Star systems. (www. energystar.gov)

8  Modeling a home is a good way to determine the least expensive ways to save energy  Over the next several lessons you will explore a simplified model of a building  The initial model calculates heat lost from a home assuming the heat does not come on  Models from other lessons will use more advance programming functions to make the model more realistic

9 Scope:  A model that estimates the rate of heat loss from a home based on the inside and outside temperatures. Output:  The rate of heat loss from a home, assuming the heat loss is uniformly distributed around the house.  The temperature change over 30 minutes if that rate of heat loss continued

10 Relationships:  Heat loss is higher the greater the difference between the inside and outside temperatures  Heat loss is higher with more wall surface area separating the inside and outside

11 Relationships (continued):  Heat loss is lower the greater the wall’s resistance to heat loss ◦ The resistance to heat loss of a building material is often called the R-value ◦ Different materials have different R-values  Glass is low  Wood is moderate  Insulation is high ◦ Heat loss is reduced by using materials with a high R-value and by using more than one layer

12 Relationships (continued):  A building with more mass (more things) inside takes longer to cool down than one with less mass (fewer things) inside  As time passes, more heat is lost

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14 Construction (continued):  The greater the heat loss, the greater the temperature change  The greater the heat capacity, the smaller the temperature change  The more time that passes, the more heat is lost and so the greater the temperature change

15  Inside is 22 °C, outside is 0 °C  R-value is 5 m²·°C/W  Surface Area: 650 m²  Heat capacity inside house: 5,000,000 joule/ °C  Calculation of temperature drop over 30 minutes (time = 1800 seconds): ◦ Heat loss = (22 – 0)* 650/5 = 2,860 joule/second ◦ Temp. change = 2,860*1800/5,000,000 = 1.0 °C ◦ Final temperature = 22°C – 1.0° C = 21°C


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