7.9: Insulation and Home Heating Fuels (V)

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Payback Period

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 payback 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.

\[ New \, Heat \, Loss \, from \, Walls = \dfrac{1,248 ft^2 * 10,500°F days * \tfrac{24 h}{day}}{23 \tfrac{ft^2 °F h}{BTU}} = 13.7 \, \dfrac{MMBTU}{Year} \nonumber\]

Heat loss from the roof remains the same and is equal to 9.99 MM BTUs (from section 7.9, example 2). 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

\[ Annual \, Cost \, of \, Heating = 23.69 MMBTUs * \dfrac{\$9.80}{MMBTUs} = \$232.16 \nonumber\]

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

\[ Time = \dfrac{\$254}{\$102.80/year} = 2.47 \, years \nonumber\]

The payback 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.

\[ Years \, to \, Payback = \dfrac{C_i * R_1 * R_2 * E}{C_e * (R_2 - R_1) * HDD * 24} \]

where

$ C_i $ is the cost of insulation in $/ft²
- Collect insulation cost information, and include labor, equipment, and vapor barrier if needed
$ C_e $ is the cost of energy in $/BTUs
- To calculate this, divide the actual price you pay per gallon of oil, kWh of electricity, gallon of propane, or therm (100 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 (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 per year
- This information can be obtained from your local weather station, utility, or oil dealer
$ 24 $ = multipler to convert HDD to heating hours

Equation 7.9.1 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 1

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 ft². What is his years to payback?

Answer

Let R₁ = 19 and R₂ = 30.

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).

\[ C_i = \dfrac{\$340}{1100 ft^2} = \$0.31/ft^2 \nonumber\]

\[ C_e = \dfrac{\$0.95}{1 \, CCF * \tfrac{100000 \, BTUs}{CCF}} = \$0.0000095/BTU \nonumber\]

Substituting the values into equation 7.9.1,

\[ Years \, to \, Payback = \dfrac{0.31 * 19 * 30 * 0.88}{0.0000095 * 11 * 7164 * 24} = 8.65 \, years \nonumber\]

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.

Example 2

For a house in Hackensack, NJ (HDD = 4,600), the installed cost to upgrade from R-13 to R-22 is $0.60/ft². The AFUE for the oil furnace is 0.78 and heating oil costs $1.13/gal. How long will it take to recover the initial investment?

Answer: The following video covers the solution.

Example 3

Lt. Dave Rajakovich has 1200 ft² of roof in his home in Pittsburgh, PA (HDD = 6,000). He is considering upgrading the insulation from R-16 to R-22. The estimate from the contractor was $775. His home is heated with natural gas. Last year, the average price he paid for natural gas was $9.86/MCF. Assuming an AFUE of 86%, how long will it take Dave to recover his investment?

Answer: The following video covers the solution.