__Energy & Cost
saving__

Life Time = 10 years Electricity = 0.08$/kwh Gas = 11$/MMBTU

[REFRIGERATOR]

( Lifetime Cycle Analysis)

This is a step for calculating the best efficient model between comparison model 1 and 2

Comparison 1= $549.99 + 482kwh x $0.08kwh x 10years = $935.59

Comparison 2= $589.99 + 417kwh x $0.08kwh x 10years = $923.59

=> Comparison model 2 is more efficient than comparison model 1 over a period of 10 years

1. How much money (if you ) you need to invest to make your house appliance more energy efficient?

= There is no need additional money to make my house appliance more energy efficient because model 2 that I chosen is more cheaper than the given model.

2. What will the total energy savings (kwh) per year if you replace the appliances with the energy efficient appliances?

= 442kwh – 417kwh= 25kwh/yr

= I can save 25kwh/yr by replacing given model to new efficient model.

3. how much money will you save per year if you replace the appliance with the energy efficient appliances?

= 25kwh x $0.08/kwh = $2

4. What will be the overall (all appliances together) pay back period?

= There is no payback for period model because the cost of the model and power consumption both are lower than the given model.

5. How much money will you save over a period of 10 years.

= $2 x 10years = $20

= I will save $20 over a period of 10 years when I replace the given model to the new efficient model

[DISHWASHER]

(Life Cycle Analysis)

Comparison 1= $509 + 361kwh x 0.08$/kwh x 10year = $797.8

Comparison 2= $649.99 + 361wh x 0.08$/kwh x 10year =$ 938.79

=> Comparison Model 1 is more efficient than model 2 over a period of 10 years

1. How much money (if you ) you need to invest to make your house appliance more energy efficient?

= $509 - $500 = $9

= I need to invest $9 to make my house appliance more energy efficient

2. What will the total energy savings (kwh) per year if you replace the appliances with the energy efficient appliances?

= 386kwh – 361kwh = 25kwh/yr

= I can save 25khw/yr by replacing given model to new efficient model

3. how much money will you save per year if you replace the appliance with the energy efficient appliances?

25kwh x $0.08/kwh = $2/yr

4. What will be the overall (all appliances together) pay back period? (Initial investment/Annual saving)

= 386kwh – 361kwh = 25kwh/yr

= 25kwh x $0.08/kwh = $2/yr

= 9($) / 2($/yr) = 4.5 years

= It take 4.5years to recover additional money

5. How much money will you save over a period of 10 years.

= 25kwh/yr x 0.08$/yr x 10yr = 20$ - 9$= 11$

= I will save 11$ over a period of 10 years by replacing given model to new efficient model.

[WATER HEATER]

(Life Cycle Analysis)

Comparison 1 $279.99 + 4721khw/yr x 0.08$/yr x 10years = $4056.79

Comparison 2 $439.99 + 4622khw/yr x 0.08$/yr x 10years = $4137.59

=>Comparison model 1 is more efficient than comparison model 2 over a period of 10 years

1. How much money (if you ) you need to invest to make your house appliance more energy efficient?

= $279.99 - $250 = $29.99

= I need to invest $29.99 to make my house appliance more energy efficient

2. What will the total energy savings (kwh) per year if you replace the appliances with the energy efficient appliances?

= 4879khw/yr - 4721khw/yr = 158kwh/yr

= I can save 158khw/yr by replacing given model to new efficient model

3. how much money will you save per year if you replace the appliance with the energy efficient appliances?

= 158kwh/yr x 0.08$/kwh = 12.64$

4. What will be the overall (all appliances together) pay back period?

= 279.99$ - 25$ = 29.99$

= 158kwh/yr x 0.08$.kwh = 12.64($/yr)

= 29.99$ / 12.64($/yr) = 2.37 years

= It take 2.37years to recover additional money

5. How much money will you save over a period of 10 years.

= 158kwh/yr x 0.08$/kwh x 10years = 126.4$ - 29.99$ = 96.41$

= I will save 96.41$ over a period of 10 years by replacing given model to new efficient model.

[CLOTHES WASHER]

(Life Cycle Analysis)

Comparison1 729$ + 259kwh/yr x 0.08$/kwh x 10years = 936.2$

Comparison2 999$ + 176kwh/yr x 0.08$/kwh x 10years = 1139.8$

=>Comparison model 1 is more efficient than comparison model 2 over a period of 10 years

1. How much money (if you ) you need to invest to make your house appliance more energy efficient?

= There is no need additional money to make my house appliance more energy efficient because model 2 that I chosen is

more cheaper than the given model.

= 260kwh/yr - 142kwh/yr = 118kwh/yr

= I can save 118khw/yr by replacing given model to new efficient model

= 118kwh/yr x 0.08$/kwh = 9.44$/yr

4. What will be the overall (all appliances together) pay back period?

= There is no payback for period model because the cost of the model and power consumption both are lower than the given model.

5. How much money will you save over a period of 10 years.

= 118kwh/yr x 0.08$/kwh x 10years = 94.4$

= I will save 94.4$ over a period of 10 years by replacing given model to new efficient model.