Pay Back Period Examples

Example 1

Please watch the following 3:04 presentation about Example Problem #1. 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?

Click here to open a transcript of the Payback Period Example Problem #1.

Payback Period Example 1

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?

Ok, this is 5.10. Heating Degree Days are given. In this problem, basically we are trying to calculate the pay back period again using the formula that we have. That we derived basically and payback period for adding insulation is cost of insulation times R₁ times R₂ times efficiency divided by cost of energy times the difference in R-value times HDD times 24. That is the formula that we need to use.

$Payback = \frac{C_{i} \times R_{1} \times R_{2} \times E}{C_{e} \times [R_{2} - R_{1}] \times HDD \times 24}$

We have all the data. HDD is given as 4,600 and cost of insulation C_i is given as $0.60 per ft² and the efficiency of the furnace is given as 78% or 0.78. We know the initial R-value, R₁ = 13, and we know the final R-value; the improved R-value which is 22, the difference is 9. And we know the price of energy or cost of energy. We are paying $1.13 to buy a gallon of oil and when we buy a gallon of oil by paying$ 1.13 we get 130,000 BTUs.

$\begin{array}{l} HDD = 4,600 ° F days \\ C_{i} = $0 {.61 / ft}^{2} \\ E = 0 .78 \\ R_{1} = 13 \\ R_{2} = 22 \\ {Difference = R}_{2} - R_{1} = 9 \\ Cost of oil = $1 .13/gal \\ \frac{$1 .13}{130,000 BTUs} = $0 .0000869 per BTU \end{array}$

So, for every BTU we pay $0.0000869. So we can use the formula now. It consists of C_i, C_i is cost of insulation, which is 0.06 per ft², dollars per ft² times R₁ which is 13, again units, you have to make sure ft², °F, hr/BTU. And here R₂ is 22 and times the efficiency which is .78 divided by the cost of energy which is dollars per BTUs. $0.0000869/1 single BTU times the difference between these two R values here and that is 22 minus 13 (9) times days, (HDD) 4,600 times 24 and when you do this calculation it turns out to be 15.5 years is the payback period.

$\begin{array}{l} = \frac{$0 {.60/ft}^{2} \times 13 22 \times 0 .78}{$0 .0000869 / BTU \times 9 \times 4,600 \times 24} \\ Payback Period = 15 .5 years \end{array}$

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Example 2

Please watch the following 4:10 presentation about Example Problem #2. 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?

Click here to open a text description of the Payback Period Example Problem #2.

Payback Period Example #2

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?

Ok, this is 5.11. Here Dave has a house, and he is trying to improve the insulation and the contractor comes up with an estimate of $775. He is using natural gas to heat the home, and we are trying to calculate how long will it take Dave to recover his investment? And this is a typical problem that uses an equation for payback period. Equal to C_i, cost of insulation, times R₁ times R₂ times efficiency divided by cost of energy times again R₂ minus R₁, this is the final R-value minus the initial R-value times HDD times 24.

$Payback = \frac{C_{i} \times R_{1} \times R_{2} \times E}{C_{e} \times [R_{2} - R_{1}] \times HDD \times 24}$

Do we have all the pieces of information for this formula?

Cost of insulation: Cost of insulation should be in dollars per square foot. So Dave has got an estimate of $775 to cover an area of 1200 square foot. Roof area happens to be 1200 ft². To cover this area with insulation it costs $775, so we can calculate the cost per ft. This happens to be$ 0.65 per ft². That’s what we want, per ft².

Do we know R₁? Yes we know, R₁ is equal to 16, R₂ is equal to 22 and the furnace efficiency, E, is equal to 0.86. And we also know HDD. HDD is 6,000, and we know 24 hours.

$\begin{array}{l} \frac{$775}{{1,200 ft}^{2}} = $0 {.65 / ft}^{2} \\ R1 = 16 \\ R2 = 22 \\ E = 0 .86 \\ HDD = 6,000 \end{array}$

Only thing here is we need to calculate the cost, C_e, cost of energy. And we are paying here, or Dave is paying $9.86 per MCF. Remember, MCF is basically one million BTUs. So the price would be$ 9.86 per million BTUs. So this will be equal to 0.00000986.

$\begin{array}{l} C_{e} = $9 .86 / MCF = 9 .86 / 1,000,000 BTUs \\ = 0 .00000986 \end{array}$

And if we plug this into this number here, into this formula here, 0.65 times 16 times 22 times 0.86 and divided by this is 0.00000986 and R₂ – R₁ happens to be 6 and 6,000 is the degree days and 24. This should come out to be, I guess 23 years roughly. This is kind of a long time period to recover the investment.

$\begin{array}{l} = \frac{0 .65 \times 16 \times 22 \times 0 .86}{$0 .00 00986 \times 6 \times 6,000 \times 24} \\ Payback Period = 23 years \end{array}$

Example 1

Payback Period Example 1

Example 2

Payback Period Example #2

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