# Life Cycle Analysis

Review the information pulled from the two water heater Energy Guide labels below and then answer questions 1-8.

Water Heater #1:

• Water heater: Electric
• Capacity (first hour rating): 58 gallons
• This model uses 4622 kwh/year
• Energy use range of all similar models (kwh/year): 4624 (uses least energy) – 5109 (uses most energy)
• This model’s estimated yearly operating cost is \$388
• Based on a 1994 U.S. Government national average cost of \$0.0841 per kWh for electricity. Your actual operating cost will vary depending on your local utility rates and your use of the product.
• Sold for \$380

Water Heater #2:

• Water heater: Electric
• Capacity (first hour rating): 58 gallons
• This model uses 4989kwh/year
• Energy use range of all similar models (kwh/year): 4624 (uses least energy) – 5109 (uses most energy)
• This model’s estimated yearly operating cost is \$419
• Based on a 1994 U.S. Government national average cost of \$0.0841 per kWh for electricity. Your actual operating cost will vary depending on your local utility rates and your use of the product.
• Sold for \$335

### Questions:

#### Compare the Two Models

1) Are the sizes and features similar for both water heaters?

a. Yes

b. No

2) Calculate the annual energy savings of the more energy efficient model (hint: the difference between the annual energy consumption values).

a. 136 kWh

b. 276 kWh

c. 326 kWh

d. 367 kWh

3) Calculate the price difference between the two models.

a. \$25

b. \$35

c. \$45

d. \$55

Next assuming that the cost of electricity in the State College area is \$0.092/kWh, and the life expectancy is 13 years, respond to the following questions about the EnergyGuide Labels.

#### Determine Savings

4) Calculate the annual monetary savings on energy for the model with lower energy consumption (Hint: annual energy savings x cost of electricity).

a. \$33.76

b. \$85.88

c. \$165.15

d. \$337.60

5) Calculate the amount of energy saved over the life time of the appliance (Hint: annual energy savings x life expectancy in years).

a. 4287 kWh

b. 4388 kWh

c. 46810 kWh

d. 4771 kWh

6) Calculate the cost of energy savings over the life time of the appliance (Hint: annual monetary savings x life expectancy in years).

a. \$348.96

b. \$386.48

c. \$438.93

d. \$496.84

#### Determine the Pay Back Period

7) Calculate the Pay Back Period or years to recover the additional investment (Hint: how many years will it take to get back the price difference between the two models based solely on your annual monetary saving?

a. 1 year

b. 1.33 years

c. 1.5 years

d. 2 years

8) Calculate the total monetary benefit for choosing the more energy efficient model (Hint: monetary savings over the appliance’s life time – price difference of two models).

a. \$343.76

b. \$393.96

c. 496.34

d. \$634.76

1. Yes
2. 367 kWh
3. \$45
4. \$33.76
5. 4771 kWh
6. \$438.93
7. 1.33 years
8. \$393.96