on average, home heating uses more energy than any other system in a home about 45% of total energy...
TRANSCRIPT
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Heating and Cooling a Home
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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.
Energy Use to Heat Homes
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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
Heating Losses
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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
Heating Systems
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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
Reducing Heating Losses
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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
Saving Energy
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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)
Saving Energy
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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
Modeling Heating a Home
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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
Modeling Heating a Home
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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
Modeling Heating a Home
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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
Modeling Heating a Home
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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
Modeling Heating a Home
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Construction: Heat loss is driven by the temperature
difference between the inside and outside of a house.
The thermal resistance (R-value) reduces the amount of heat lost.
Increasing the surfaced area of a house exposed to the outside increases the heat loss
Modeling Heating a Home
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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
Modeling Heating a Home
𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 h𝐶 𝑎𝑛𝑔𝑒=𝐻𝑒𝑎𝑡 𝐿𝑜𝑠𝑠 ∗ time 𝐻𝑒𝑎𝑡𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦
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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
Sample Calculation