Heating and Cooling a Home

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

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

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

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

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

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

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

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

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

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

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

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

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

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