Department of Natural Sciences
Clayton State University
Morrow, Georgia

SCI 1901E
Example
Payback Time Calculations

1. Examine the summary table on page 3 of the your heating audit forms (also shown below) and determine the major causes for heat loss in your home. Using the class example, this summary table revealed:

The two major heat loss problems with the example house are the floor and infiltration. At a minimum, these two problems should be addressed in a heating audit report. Recall the floor was uninsulated on an unheated crawl space.

2. Floor:

a. Remedy. One way of addressing this problem would be to add insulation to the floor.

b. New surface factor. If 6" of fiberglass insulation is added to the floor, the new surface factor would be 2.6 Btu / ft²/ hr (from Appendix C of your text book).

c. New conductive heat loss rate. From the example the total floor area was:

A_floor,total = A_floor,Rm1 + A_floor,Rm2

A_floor,total = 750 ft² + 450 ft² = 1200 ft²

Then, for the floor,

Q_c/ t_new = ( 2.6 Btu / ft²/ hr ) ( 1200 ft²) = 3120 Btu / hr

d. Change in conductive heat loss rate. Comparing the old and new conductive heat loss rates, the rate saved by adding the insulation is:

Q_c/ t_save = Q_c/ t_old - Q_c/ t_new

Q_c/ t_save = 18000 Btu / hr - 3120 Btu/hr = 14880 Btu / hr

e. Energy savings over the entire heating season. Use the calculation method of line E of the third page of the heating audit forms:

E_save = (14880 Btu / hr) (3120°F)(24 hr /day) / (1 x 10⁶ Btu / MBtu) / (65°F)

E_save = 17.14 MBtu

f. Cost savings over the entire heating season. To find the cost savings multiply the energy savings (in MBtu) by the cost per MBtu (calculated on page 1 of the heating audit forms):

Cost_save/year = (17.14 MBtu)($17.25 / MBtu) = $295.66 / year

g. Determine the cost of the remedy. Visit a store that sell the materials required to make the change and determine their price (you will need the floor dimensions when you go to the store). In our example, I determined that the cost would be $220.00. This assumes that I will do the installation myself so that there are no labor costs.

h. Determine the payback time. The payback time is cost of the materials divided by the cost saving per year:

t_payback = ($220.00) / ($295.66 / year) = 0.744 years = 8.93 months

(This calculation overestimates the payback time since cost saving will also occur during the cooling season as a result of the addition of the insulation.)

3. Infiltration.

a. Remedy. Your text book indicates that the infiltration loss rate can typically be reduced by a factor of one-half by caulking and weather-stripping.

b. New infiltration heat loss rate. From the example, the infiltration heat loss rate was 14040 Btu / hr. The new infiltration heat loss rate is then:

Q_infil / t_new = (0.5)(14040 Btu / hr) = 7020 Btu / hr

and

Q_infil / t_save = 7020 Btu / hr

c. Energy savings over the entire heating season. Use the calculation method of line E of the third page of the heating audit forms:

E_save = (7020 Btu / hr) (3120°F)(24 hr /day) / (1 x 10⁶ Btu / MBtu) / (65°F)

E_save = 8.09 MBtu

d. Cost savings over the entire heating season. To find the cost savings multiply the energy savings (in MBtu) by the cost per MBtu (calculated on page 1 of the heating audit forms):

Cost_save/year = (8.09 MBtu)($17.25/MBtu) = $139.55 / year

e. Determine the cost of the remedy. Visit a store that sell the materials required to make the change and determine their price (you will need the door and window dimensions when you go to the store). In our example, I determined that the cost would be $52.00. This assumes that I will do the work myself so that there are no labor costs.

f. Determine the payback time. The payback time is cost of the materials divided by the cost saving per year:

t_payback = ($52.00) / ($139.55 / year) = 0.373 years = 4.47 months

(This calculation overestimates the payback time since cost saving will also occur during the cooling season as a result of the addition of the insulation.)

Last update: February 11, 2009