Earlier sections illustrated that adding more insulation and improving the R-value of a wall would help in cutting heat loss. Less heat loss reduces the amount of fuel that needs to be burned, thereby reducing the heating costs and protecting the environment. However, adding insulation often involves additional investment.

The money invested into insulation can be recovered or paid back using the money saved because of the reduction of fuel usage. The time it takes to recover the additional cost through savings is called the **pay-back period**. A simple pay back is the initial investment divided by annual savings after taxes.

A simple calculation illustrates this term. If the R-value of the wall used in an earlier example is improved to R-23 by adding additional insulation, which costs \$254, the heat loss can be reduced. The new heat loss after improvement can be calculated using the equation below.

$$\text{New Heat Loss from Walls}=\frac{1,248\text{}\overline{)f{t}^{2}}\times 10,500\overline{){}^{o}F}-\overline{)\text{days}}\times \frac{24\overline{)h}}{\overline{)day}}}{23.0\frac{\overline{)f{t}^{2}}\overline{){}^{o}F\text{}}\overline{)h}}{Btu}}=\frac{13.7MMBtu}{Year}$$

Heat loss from the roof remains the same and is equal to 9.99 MM BTUs. Therefore, new annual total heat loss is only 13.7 + 9.99 =23.69 MM BTUs. The annual cost of heating after this improvement would be:

$$=(23.69\text{}\overline{)MMBtus})\times \frac{\$9.80}{\overline{)MMBtus}}=\$232.16$$

The savings is \$334.96 - \$232.16 = \$102.80 every year. Remember that to get this savings, an investment of \$254 was made. So if this investment was paid off by the savings, it would take

$$\frac{\$254.00}{\$102.80/year}\text{}=\text{}2.47\text{}years$$

The pay-back period is 2.47 years. Shorter pay-back periods indicate that the additional investment can be paid off quickly and the homeowner can start saving money after that.

The formula below will help you to estimate the cost effectiveness of adding insulation in terms of the "years to payback" for savings in heating costs.

$$Years\text{}to\text{}Payback\text{}=\text{}\frac{{C}_{i}\times {R}_{1}\times {R}_{2}\times E}{{C}_{e}\times \left({R}_{2}-{R}_{1}\right)\times HDD\times 24}$$

Where:

**C _{i} **=

**Cost of insulation in $/square feet.**Collect insulation cost information; include labor, equipment, and vapor barrier if needed.

**C _{e}** =

**Cost of energy, expressed in $/BTUs**. To calculate this, divide the actual price you pay per gallon of oil, kilowatt-hour (kWh) of electricity, gallon of propane, or therm (or per one hundred cubic feet [CCF]) of natural gas by the BTU content per unit of fuel.

**E** = Efficiency of the heating system. For gas, propane, and fuel oil systems, this is the Annual Fuel Utilization Efficiency, or AFUE.

**R _{1}** = Initial R-value of section

**R _{2}** = Final R-value of section

**R _{2} – R_{1}** = R-value of additional insulation being considered

**HDD** = Heating degree days/year. This information can usually be obtained from your local weather station, utility, or oil dealer.

**24** = Multiplier used to convert HDD to heating hours (24 hours/day).

The formula above works only for uniform sections of the home. For example, you can estimate years to pay back for a wall or several walls that have the same R-values, if you add the same amount of insulation everywhere. Ceilings, walls, or sections of walls with different R-values must be figured separately.

### Example

Mr. Energy Conscious (who lives in East Lansing, MI with an HDD of 7,164) wants to know how many years it will take to recover the cost of installing additional insulation in his attic. He renovated his attic and increased the level of insulation from R-19 to R-30 by adding additional insulation. He has a gas furnace with an AFUE of 0.88 and pays \$0.95/CCF for natural gas. The attic insulation costs \$340 to cover 1,100 sq. ft.

The pay-back period is given by the equation 5.5:

$$Years\text{}to\text{}Payback\text{}=\text{}\frac{{C}_{i}\times {R}_{1}\times {R}_{2}\times E}{{C}_{e}\times \left({R}_{2}-{R}_{1}\right)\times HDD\times 24}$$R_{1} = 19; R_{2} = 30; and R_{2 }- R_{1} = 30 - 19 = 11

HDD = 7,164 and E = 0.88

The most important part of this problem is to determine the cost of insulation per one sq. ft (C_{i}) and cost of energy per one BTU (C_{e}).

$$And,{C}_{e}=\frac{\$0.95}{\overline{)1\text{}CCF}\times \frac{100,000\text{}Btus}{\overline{)1\text{}CCF}}}=\$0.0000095/Btu$$

Note: The cost for one BTU is a very small number.

Substituting the values in the equation,

$$Years\text{}to\text{}Payback\text{}=\text{}\frac{0.31\times 19\times 30\times 0.88}{0.0000095\text{}\times \text{}\left[11\right]\text{}\times \text{}7,164\text{}\times \text{}24}\text{}=\text{}8.65\text{}years$$**After 8.65 years, Mr. Energy **Conscious** can start saving money for himself for the rest of the period that he lives in that home. During the entire period the energy that he is not using can help the environment.**