The amount of heat that flows through a wall or window may be calculated by the following formula: Heat Loss (Btu/hour) = Wall Area in - Tout) R-value R-value of insulation indicates its resistance to the flow of heat. R-20 has twice the resistance to heat loss as R-10.
How do we lower energy consumption for space heating? Increase R-value of walls, ceiling, and windows. Increase efficiency of furnace. Use heat pump. Use natural gas rather than electric heater. Better home design and construction. Lower Tin. (Easy lifestyle change) Decrease wall size. (Major lifestyle change)
Insulation Increasing ceiling and wall insulation is the cheapest technique for reducing heating and cooling loads for a building.
Windows Double or triple pane windows have a much higher R-value than single pane windows. Low emissivity glass, heat absorbing glass, and reflective glass depending on climate. Using better insulated frames and casements and sealing leaks.
Furnace Efficiency Annualized Fuel Utilization Efficiency Older units have 50% or 60% values. Newer units have 80% efficiency. Payback period for new units is about a year.
Geothermal Heat Pump (GHP) Heating GHP is very efficient and can be used for heating and cooling.
Geothermal Heat Pump (GHP) Cooling Payback Period – 2 to 10 years. High initial cost.
How do we lower energy consumption for Air-Conditioning? Increase R-value of walls, ceiling, and windows. Increase efficiency of Air-Conditioner. Use heat pump. Better home design and construction. Raise Tin. (Easy lifestyle change) Decrease wall size. (Major lifestyle change)
Air-Conditioner Efficiency Regulated by the U.S. DOE. Efficiency rating -SEER (seasonal energy efficiency). SEER is defined as the annual cooling output (Btus) divided by its total energy input (Watt-hours) during the same period.
Air-Conditioner Efficiency - SEER The minimum SEER allowed for a central A/C is 9.7. The best available SEER is 18. Older units have SEER ratings of 6 or less. Consumers should look for a SEER of 12 or higher when buying a new A/C system.
Retrofit and Payback Suppose an older home in San Antonio is equipped with a central air-conditioning unit with SEER rating equal to 6 and that this home consumes 36,000 kWh of electrical power each year to power this system. Suppose this system is replaced by a central unit with a SEER rating equal to 18.
1. How much energy (in kWh) is consumed annually by the central A/C unit after the upgrade? Energy Consumed = = 36,000 kWh Old SEER / New SEER = 36,000 kWh 6 /18 = 36,000 kWh 1 /3 = 12,000 kWh
2. How much energy is saved annually by switching to the A/C unit with the higher SEER rating? Energy Saved = = 36,000 kWh – 12,000 kWh = 24,000 kWh
3. How much money is saved annually, assuming that the cost of electricity equals $0.10/kWh? Money Saved = = 24,000 kWh $0.10/kWh = $2,400
4. What is the payback period if the cost of purchasing and installing the new A/C unit equals $4,800? Cost = $4,800 Savings= $2,400/year Payback = Cost / Savings = 2 years