Energy Conservation Physics /24/03

Reducing energy consumption may help alleviate environmental problems: Conserve fossil fuel resources Reduce CO2 emissions Reduce extraction impacts Reduce use of water and other resources.

There are two ways to reduce energy consumption. Alter lifestyle – e.g. public transportation v.s. private cars. Increase the efficiency of energy use.

Efficiency – General Definition Efficiency = useful output / energy input

Example 1 – Fluorescent Bulbs A 20 watt Fluorescent bulb produces the same luminous flux as a 100 watt incandescent bulb. So, the Fluorescent bulb is 5 times more efficient. Work the Lighting Efficiency problem.

Example 2 – Space Heating

Space Heating The amount of heat that flows through a wall or window may be calculated by the following formula: Heat Loss (in Btu/hour) = Wall Area  in - Tout)  R-value

How do we lower energy consumption? Lower Tin. (Lifestyle change) Decrease wall size (Lifestyle change) Add insulation to the walls. Increase efficiency of heater.

Example 3 – Air-Conditioner Efficiency The efficiency of Central A/C units is governed by U.S. law and regulated by the U.S. Department of Energy. Every A/C unit is assigned an efficiency rating known as its seasonal energy efficiency ratio (SEER). The SEER is defined as the total cooling output (in Btu-British thermal units) provided by the unit during its normal annual usage period divided by its total energy input (in Watt- hours) during the same period.

Air-Conditioner Efficiency - SEER The SEER is displayed on a yellow label affixed to the A/C unit. Higher SEERs are better. The minimum SEER allowed by law for a central A/C is 10 for a split system or 9.7 for a single-package unit. The best available SEER is about 18, while many older units have SEER ratings of 6 or less. Most consumers should look for a SEER of 12 or higher when buying a new A/C system.

Air Conditioner Efficiency Calculation

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