Heat escapes (or transfers) from inside to outside (high temperature to low temperature) by three mechanisms (either individually or in combination) from a home:

• Conduction
• Convection
• Radiation

Examples of Heat Transfer by Conduction, Convection, and Radiation

Click here to open a text description of the examples of heat transfer by conduction, convection, and radiation

• Conduction: heat moving through walls of a home from high temperature inside to low temperature outside.
• Convection: heat circulating within the rooms of a house.
• Radiation: Heat from the sun entering a home.

Credit: © Penn State is licensed under CC BY-NC-SA 4.0 [2]

Conduction

Conduction is a process by which heat is transferred from the hot area of a solid object to the cool area of a solid object by the collisions of particles.

In other words, in solids the atoms or molecules do not have the freedom to move, as liquids or gases do, so the energy is stored in the vibration of atoms. An atom or molecule with more energy transfers energy to an adjacent atom or molecule by physical contact or collision.

In the image below, heat (energy) is conducted from the end of the rod in the candle flame further down to the cooler end of the rod as the vibrations of one molecule are passed to the next; however, there is no movement of energetic atoms or molecules.

Click the play button to start the animation.

With regard to residential heating, the heat is transferred by conduction through solids like walls, floors, and the roof.

Click the play button to start the animation.

Convection

Convection is a process by which heat is transferred from one part of a fluid (liquid or gas) to another by the bulk movement of the fluid itself. Hot regions of a fluid or gas are less dense than cooler regions, so they tend to rise. As the warmer fluids rise, they are replaced by cooler fluid or gases from above.

In the example below, heat (energy) coming from candle flame rises and is replaced by the cool air surrounding it.

Click the play button to start the animation.

In residential heating, convection is the mechanism by which heat is lost by warm air leaking to the outside when the doors are opened, or cold air leaking into the house through the cracks or openings in walls, windows, or doors. When cold air comes in contact with the heater in a room, it absorbs the heat and rises. Cold air, being heavy, sinks to the floor and gets heated, and thus slowly heats the whole room air.

Instructions: Press the play button below and observe what happens to the cold air (blue arrows) as it enters the house and encounters the warm air (red arrows) coming from the heating vent:

Radiation

Radiation is the transfer of heat through electromagnetic waves through space. Unlike convection or conduction, where energy from gases, liquids, and solids is transferred by the molecules with or without their physical movement, radiation does not need any medium (molecules or atoms). Energy can be transferred by radiation even in a vacuum.

In the image below, sunlight travels to the earth through space, where there are no gases, solids, or liquids.

Click the play button to start the animation.

Test Yourself

First, identify the type of home heat loss pictured in images A-J as either: conduction, convection or radiation. Then click and drag each image down to the correct category at the bottom of the screen.

Reducing Energy Consumption

There are two ways in which we can reduce energy consumption.

1. The most cost-effective way is to improve the home’s “envelope”—the walls, windows, doors, roof, and floors that enclose the home—by improving the insulation (conduction losses) and sealing the air leaks with caulking (convection losses).
2. The second way to reduce the energy consumption is by improving the efficiency of the furnace that provides the heat.

Conduction and Convection

Click here to open a text description of the Conduction and Convection diagram

Line drawing of a house with arrows pointing out from the walls and roof showing conduction & arrows flowing in a circular motion inside the house showing convection.

Credit: © Penn State is licensed under CC BY-NC-SA 4.0 [2]

Main Factors of Heat Loss

What does a house's heat loss depend on? Complete the activity below to find out the three main factors leading to heat loss.

Most heat is lost through a house's walls through conduction. As you learned from the activity on the previous screen, the amount of heat loss depends on three factors:

• Size of the house (area through which the heat can escape)
• Local weather or climatic conditions:
  • The inside temperature is often constant at a comfortable temperature of 65°F.
  • As the outside temperature falls lower than 65°F, the heat is lost to the outside.
  • The higher the temperature difference, the higher the heat loss to outside.
  • By calculating the Heating Degree Days (HDD), we can determine how many degrees the mean temperature fell below 65ºF for the day.
• Wall's capacity to resist heat loss.
  • Insulation is rated in terms of thermal resistance, called R-value, which indicates the resistance to heat flow.
  • The higher the R-value, the greater is the insulating effectiveness.

Local weather or climatic conditions are one of three factors that affects the amount of heat loss through conduction. When examining weather conditions, we look at both the inside and outside temperature of a home.

The inside temperature is usually taken as a standard comfort temperature of 65ºF. The outside temperature varies by the hour. Knowing this information can help us to understand two concepts:

• Average outside temperature = Average of the maximum and minimum temperature during the day
• Heating Degree Day (HDD) = The temperature difference through which air has to be treated, or how many degrees the mean temperature fell below 65ºF for the day. It is also an index of fuel consumption.

Temperatures Inside and Outside a House

Credit: © Penn State is licensed under CC BY-NC-SA 4.0 [2]

Determining Heating Degree Day (HDD)

The formula for determining the Heating Degree Day (HHD) is:

$${\text{HDD = T}}_{\text{base}}-{\text{T}}_{\text{a}}$$

To calculate HDD:

1. Determine the base temperature or inside temperature: $${\text{T}}_{\text{base}}\text{ = usually 65}°\text{F}$$
2. Find the day's average outside temperature using this formula: $${\text{T}}_{\text{a}}\text{ = average outside temperature = }\frac{{\text{T}}_{\text{max}}+{\text{T}}_{\text{min}}}{\text{2}}$$
3. Use the HDD formula to solve:
   $${\text{T}}_{\text{base}}{\text{ - T}}_{\text{a}}$$

Try This!

Click the play button below, and observe the temperature changes. Then calculate the average temperature and the Heating Degree Day.

Heating Degree Day

For this problem, we need to calculate HDD for one full week. The data that is given is each day, what is the outside temperature -- average outside temperature. For example, Sunday, the average outside temperature is 49 degrees F. Monday 47 degrees Fahrenheit, Tuesday it’s 51 and Wednesday it is 60 degrees F and on Thursday it is 65, Friday, 67 and on Saturday it is 58.

So we need to calculate heating degree days (HDD) for each day. One day, that is Sunday, the outside temperature is 65 minus 49 the outside temperature will give you the degree days. Similarly, 65 minus 47° Fahrenheit, one day times 65 minus 51 in this case. One day times 65 minus 60 and one day 65 minus 65. This would be zero. One day 65 minus 67. Remember when it exceeds 65, the heating degree days would be zero. In this case, also it is zero. One day on Saturday it is 65 minus 58.

This case it is 7.

These two are zeros, this is 5 and this is 14 and this happens to be 18 and here it is 16. So the total sum is for one full week is 60 degree days.

This is equal to 60 degree days.

In previous examples, we are assuming that the outside temperature remains the same for all 150 heating days in a season. This is not realistic, but it explains the method to calculate the HDD. In a more realistic example, we need to find the temperature difference for each day and add all the temperature differences.

We will now look at Seasonal Heating Degree Days (HDD), which is the sum of temperature differences of ALL days - rather than just 1 day or 1 week - during which heating is required.

The table below provides Seasonal HDDs for selected places in the United States. The higher HDD indicates a higher heat loss and therefore, higher fuel requirements.

HDD is used to estimate the amount of energy required for residential space heating during a cool season, and the data are published in local newspapers or on the National Weather Service website.

Source: NOAA [3]

Calculating Seasonal Heating Degree Days

To calculate Seasonal Heating Degree Days, use this formula:

Remember, in months where the average temperature is equal to or greater than 65, there will be no heating degree days, so the value for the month will be 0.

Example 1

Given the following set of average temperatures, by month, for State College, PA, calculate the HDD for the heating season:

Please watch the following (2:32) presentation about problem #1:

Example 2

If Ms. S. Belle moves from Birmingham, AL (HDD=2,800) to State College, PA (HDD=6,000) how much can she expect her heating bill to increase?

Please watch the following 1:34 presentation about problem #2:

As we have learned, most heat is lost through a house's walls through conduction. One of the three factors that affect heat loss is a wall's capacity to resist heat loss.

We will now look at how to calculate the rate of heat loss of the walls of a house, using the following formula:

From the above equation, it can be seen that once the house is built, these two variables will NOT change:

• The area of the walls
• The R-value of the walls

The only variable that will change is the temperature difference between inside and outside.

Example

Calculate the heat loss for a 10 ft by 8 ft wall, insulated to R-value 22. The inside temperature is maintained at 70° F. The temperature outside is 43° F.

Please watch the following 2:25 presentation about Hourly Heat Loss:

Now that you know how to calculate hourly heat loss, how would you calculate daily heat loss?

Since there are 24 hours in a day, you would simply multiply the hourly heat loss by 24.

Example

What is the hourly and daily heat loss of a 15-ft by 15-ft room with an 8-ft ceiling, with all surfaces insulated to R13, with inside temperature 65°F and outside temperature 25°F?

$$HourlyHeatLossrate=\frac{Q(BTUs)}{t(hour)}=\frac{(480f{t}^2\times \left({65}^{o}F-{25}^{o}F\right)}{13\frac{f{t}^2{}^{o}Fh}{BTU}}=1,477\frac{BTUs}{h}$$

Once we know the heat loss rate per hour, we can determine the heat loss per day by multiplying by 24 (hours in a day).

In a 24-hour period or one day, the heat loss would be:

$$DailyHeatLossperday=\frac{1477BTUs}{h}x\frac{24h}{day}=35,448\frac{BTUs}{day}$$

$$HeatLoss\left(\frac{BTUs}{h}\right)=\frac{Area(f{t}^2)\times TemperatureDifference{(}^{o}F)}{Rvalue\left(\frac{f{t}^2{}^{o}Fh}{BTU}\right)}$$

Area (ft²) is the sum of the area of all four walls. Each wall is 8' x 15', or 120 sq. feet, so take 4 x 120 sq ft to get 480 ft² in the equation.

In the previous calculations, we determined hourly and daily heat loss. How do we calculate annual or seasonal heat loss?

Since the temperature outside the house may not remain the same day after day, the heat loss will vary by the day. Thus, to obtain the heat loss for a whole year, we do the following:

• Calculate heat loss for each day.
• Add the heat loss for all days in a year that needed heating.
• Leave the R-value of the wall and the area of the wall the same – they will not change.
• Determine the difference between inside and outside temperature, since it will change for each day.

Recall that the formula for daily heat loss is:

Thus, theoretically, we would need to perform this calculation for every day of the 365-day calendar year.

For example, if the average outside temperature were to be 35°F, 32°F, 28°F, and so on for each day, the heat loss for the whole year or the season can be calculated as follows:

$$\begin{array}{l}\text{Heat Loss = }\underset{\text{day 1 heat loss}}{\underbrace{\frac{480f{t}^2\times \left(65-\stackrel{\text{outside temp for day 1}}{\stackrel{︸}{35}}\right)\text{ °F}}{13\frac{f{t}^2\text{ °F h}}{\text{BTU}}}\times 24\frac{h}{day}}}+\underset{\text{day 2 heat loss}}{\underbrace{\frac{480f{t}^2\times \left(65-\stackrel{\text{outside temp for day 2}}{\stackrel{︸}{32}}\right)\text{ °F}}{13\frac{f{t}^2\text{ °F h}}{\text{BTU}}}\times 24\frac{h}{day}}}+\\ \underset{\text{day 3 heat loss}}{\underbrace{\frac{480f{t}^2\times \left(65-\stackrel{\text{outside temp for day 3}}{\stackrel{︸}{28}}\right)\text{ °F}}{13\frac{f{t}^2\text{ °F h}}{\text{BTU}}}\times 24\frac{h}{day}}}+\mathrm{...}\text{ and so on for all heating days}\text{.}\end{array}$$

Since the area (480 ft²), R-value $\frac{f{t}^2°FBTU}{h}$,and 24 h in a day are common for ALL heating days, we can bring those out and rewrite the equation as:

$$\begin{array}{l}=\frac{480f{t}^{2}x}{13\frac{f{t}^2{}^{o}Fh}{BTU}}\times 24\frac{h}{day}\left\{{\left(65-35\right)}^{o}F+{\left(65-32\right)}^{o}F+{\left(65-28\right)}^{o}F\mathrm{...}soonforallheatingdays\right\}\\ \\ =\frac{480f{t}^{2}x}{13\frac{f{t}^2{}^{o}Fh}{BTU}}\times 24\frac{h}{day}\left\{{30}^{o}F+{33}^{o}F+{37}^{o}F+soonforallheatingdays\right\}\\ \\ Where\left\{{30}^{o}F+{33}^{o}F+{37}^{o}F+soonforallheatingdays\right\}isthesumofHeatingDegreeDays\end{array}$$

This equation can be even further simplified. The formula for Annual or Seasonal heat loss can be written in general terms as:

$$HeatLossinaSeason=\frac{Area(f{t}^2)}{Rvalue\left(\frac{f{t}^2{}^{o}Fh}{BTU}\right)}\times 24\frac{h}{day}\times (HDDfortheseason)$$

Example 1

Please watch the following 3:29 presentation on Example Problem #1. For the 150-day heating season in Roanoke, VA, the average temperature was 47° F. How much heat is lost through a 176 ft² wall (R=16) during the entire season?

Example 2

Please watch the following 3:42 presentation on Example Problem #2. In Fargo, ND, the heating season lasts about 220 days and the average outside temperature is around 27° F. How much heat is lost through an 8 ft by 6 ft window (R=1) during the heating season?

As you may recall from the first part of this lesson, R-value is a wall's capacity to resist heat loss or its thermal resistance. Insulation materials are rated in terms of their R-value, with a higher R-value indicating better insulating effectiveness.

The R-value of thermal insulation depends on the type of material, its thickness, and its density. Generally, walls are not made up of just one material or one layer.

R-values for most commonly used building materials are given in the table below. From looking at the table below, we can see that natural materials like stone and bricks are not good insulation materials, but most of the synthetic insulation products such as polystyrene or polyurethane are very effective insulation materials.

If insulation materials have a low R-value per inch, then they will need to be thicker than those materials with a higher R-value per inch to achieve the same degree of effectiveness in resisting heat loss.

Look at the chart below to compare the thickness required of various insulation materials to achieve the same R-Value of 22 and then answer the questions below.

Thickness required of various insulation materials to achieve an R-Value of 22.

Credit: "Insulation Value Of CMAA Australia [4]." LAT Works Construction.

Look at the table below to learn about six types of insulation.

Types of Insulation

Insulation	What is it made of?	What does it look like?	Additional Information
Fiberglass	Molten glass spun into microfibers	Pink or yellow in the form of batts or rolled blankets.
Rock Wool	Rock	Gray or brown fibers in batts or blankets or as shredded loose-fill.	Manufactured in a similar way as fiberglass, but with molten rock instead of glass.
Cellulose	Recycled paper – newsprint or cardboard shredded into small bits of fiber.	Blown in as loose fill.	It is treated with fire- and insect-resistant chemicals.
Rigid Foam	Different types, but some made from post-consumer recycled content from fast food containers and cups.	Rigid sheets that are applied directly to framing.	Best where space is limited, but a high R-value is needed. Can be installed on the interior of a wall, but if installed inside, must be covered by a fire resistant material like wallboard. One drawback to foam is it deteriorates unless it is protected from prolonged exposure to sunlight and water. It is also more expensive than other insulation.
Synthetic Insulation	Usually polystyrene or polyurethane foam.	Polystyrene comes as rigid boards, and Polyurethane comes as rigid boards or sprayed in place systems.	Polystyrene is used for insulating basements, cathedral ceilings, or sidewalls. Polyurethane foams are high performance insulating materials.

Generally, walls are made up of several layers of different materials. The R-value of a composite wall is calculated by adding the effective R-values of each of the layers of the wall.

For example, the image below shows a wall made up of four layers—½ inch drywall inside for aesthetic purposes, real insulation in between the studs, ¾ inch plywood sheathing outside, and wood siding as the final exterior finish. Together, the layers of the wall are preventing heat loss.

Diagram of a wall

Credit: © Penn State is licensed under CC BY-NC-SA 4.0 [2]

We can calculate this wall’s composite R-value by adding the R-values of each layer.

We can find the R-values for the walls by using the table from the page about R-Values [5] (the table is repeated below):

Examples

Example 1

A ceiling is insulated with 0.75" plywood, 2" of polystyrene board, and a 3" layer of fiberglass. What is the R-Value for the ceiling?

Solution:

The ceiling consists of three layers, and all three layers together prevent the heat loss. So, we need to add the R-values of all three layers

$\raisebox{1ex}{$\mathrm{3/}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$" of plywood has an R-value of 0.94
2" of polystyrene at 5.0 per inch will have an R-value of 10.00
3" of fiberglass at 3.7 per inch will have an R-value of 11.10

So the R-value of the ceiling is 22.04 ft^{2 o}Fh / BTU.

Pink insulation in the ceiling

Credit: © Penn State is licensed under CC BY-NC-SA 4.0 [2]

Insulation Needs By Region

The United States map below shows insulation needs by region, as indicated by color and numbers.

Instructions: Click on the “zone” buttons below the map to see the U.S. Department of Energy’s Recommended Total R-Values for new construction of houses. Note that insulation R-values are different for the ceilings, walls, floor, etc.

U.S. Department of Energy recommended total R-Values for new construction houses, by regions and by various parts of the house.

Heat loss from the surface of a wall can be calculated by using any one of the three formulas we have covered in Part A of this Lesson.

Equations

The heat loss in an hour equation

The heat loss in a day equation

The heat loss in full heating season

The heat loss from walls, windows, roof, and flooring should be calculated separately, because of different R-Values for each of these surfaces. If the R-value of walls and the roof is the same, the sum of the areas of the walls and the roof can be used with a single R-value.

Example #1

Please watch the following 4:58 presentation about Example #1. A house in State College, PA has 580 ft² of windows (R = 1), 1920 ft² of walls, and 2750 ft² of roof (R = 22). The walls are made up of wood siding (R = 0.81), 0.75” plywood, 3” of fiberglass insulation, 1.5” of polyurethane board, and 0.5” gypsum board. Calculate the heating requirement for the house for the heating season.

Example #2

Please watch the following 9:00 presentation about Example #2. A single-story house in Anchorage, AK (HDD = 11,000) is 50 ft by 70 ft with an 8 ft high ceiling. There are six windows (R = 1) of identical size, 4 ft wide by 6 ft high. The roof is insulated to R-30. The walls consist of a layer of wood siding (R = 0.81), 2” of polyurethane board (R = 6.25 per inch), 4” of fiberglass (R = 3.70 per inch), and a layer of drywall (R = 0.45). Calculate the heat loss through the house (not counting the floor) for the season. (A good estimate for the area of the roof is 1.1 times the area of the walls, before you subtract out the area for the windows.)

Various fuels such as natural gas, electricity, fuel oil, and so on, are used to heat a house. Click on the graph below to see the percent of households that use each type of heating fuel.

As you see, about 50 percent of the households in the United States use natural gas as their main heating fuel, and about 35 percent of the households use electricity to heat their homes. Another 11% use fuel oil, and the last 6% use something other than natural gas, electricity, and fuel oil.

Capacity and Consumption

Capacity

The amount of heat a furnace can deliver is called its capacity. Heating units are manufactured and sold by their capacity. The heating capacities of Natural Gas, Propane and Fuel Oil are measured according to BTU/h, and the capacity of Electricity is measured in kilowatts.

Consumption

The amount of energy a furnace actually uses is called consumption. In other words, we pay monthly bills for the consumption of a particular heating fuel. Heating Fuels are sold to consumers in different units of measure. For example, Natural gas is sold by cubic feet (ft³).

Press play to see the difference between capacity and consumption in a gas-heated home. (36 seconds)

Fuel Comparison

Heating Values of various fuels

Each unit of fuel when burned gives different amounts of energy. The energy that is released when a unit amount of fuel is burned is called the heating value. The heating value of a fuel is determined under a standard set of conditions. A comparison of approximate heating values of various fuels is shown in the table below.

From the table above, it should be noted that if a gallon of fuel oil is burned, one would get 140,000 BTUs. Similarly, a CCF of natural gas would fetch 100,000 BTUs.

An assumption is made here that all the energy from the fuel is released, and all the heat is available to heat the place. However, generally, when a fuel is burned in a furnace, not all the energy (heat) is available for the final end user.

The energy efficiency of a furnace is not 100 percent. Not all the energy from the fuel is released, and not all the heat is available to heat the place. For example, if a furnace’s efficiency is, say, 50 percent, then twice as much fuel would be needed to heat a home.

Looking again at the table on the previous screen, we saw that the heating value of fuel oil is given as 140,000 BTUs. However, if the furnace’s efficiency is 50 percent, then the actual heating value of fuel oil is 140,000 BTUs x 0.5 (efficiency) = 70,000 BTUs. In other words, when a gallon of oil is burned, 70,000 BTUs of heat is actually available to the user.

It can be clearly seen that as the efficiency of the furnace increases, the amount of heat available increases proportionally.

The higher the efficiency, the less oil needs to be put into the furnace to get the same amount of heat output.

Most of the heating furnaces burn fuel and release hot combustion gases. The hot combustion gases heat the incoming cold air and go out through the chimney. In older furnaces, all the heat in the fuel is not released or not transferred to the cold air (or water, in the case of heat registers and water heaters), and therefore is lost through the chimney. The air or water that is heated distributes the heat throughout the house. Newer models of furnaces have gotten better at getting more of the heat into the cold air and, therefore, into your house.

Click the play button and observe how a wood burning furnace operates at 75% efficiency.

Furnaces are usually not as efficient when they are first firing up as they are running at steady-state. It is sort of like a car getting better mileage in steady highway driving than in stop-and-go city traffic.

What matters over the course of the year is the total useful heat the furnace delivers to your house versus the heat value of the fuel it consumes. This is kind of like measuring the gas mileage your car gets by asking how many miles you drove this year and dividing it by how many gallons of gas you bought.

For furnaces, they call this measure the AFUE (Annual Fuel Utilization Efficiency). The federal minimum-efficiency standards for furnaces and boilers took effect in 1992, requiring that new furnace units have an AFUE of at least 78 percent and new boiler units have an AFUE of at least 80 percent. In comparison, many old furnaces and boilers have AFUE ratings of only 55 to 65 percent.

To find out how efficient your furnace is, look for an energy guide label like this:

Energy Guide Label.

Credit: Union Gas: Wise Energy Guide [8]

The table below gives the efficiencies of most efficient furnaces that were available in 2002–2003.

When buying a new furnace, make sure its heating capacity (output) is appropriate for your home. If the insulation and/or windows in your home have been upgraded since the old heating equipment was installed, you can probably use a much less powerful furnace or boiler. Oversized furnaces operate less efficiently because they cycle on and off more frequently; in addition, larger furnaces are more expensive to buy.

It is clear now that when a unit of fuel is burned not all of it is available to the end user, and that as the furnace efficiency increases, higher amounts of heat will be available. An important question that needs to be addressed is how much it costs to buy the energy or heat to heat a place.

Fuel is usually sold in gallons or CCF or kWh. Comparing the actual cost of energy to produce a certain amount of heat for the end user would be easy if the comparison is made on an energy basis rather than on a unit basis. That is, \$/BTUs rather than \$/gal or CCF or kWh.

We can use the following formula to calculate Actual Energy Cost:

Using the formula below, we can calculate the Actual Energy Cost.

$$\text{Actual Energy Cost}=\frac{\text{Fuel Cost}\left(\frac{\$}{\text{Unit of Fuel}}\right)}{\text{HeatingValue}\left(\frac{\text{MMBtus}}{\text{Unit of Fuel}}\right)\text{×Efficiency}}$$

$$\text{Oil(in central heating system) cost}=\frac{\frac{\$1.25}{\bcancel{Gal}}}{\frac{0.14MMBtus}{\bcancel{Gal}}\times 0.85\text{(Efficiency)}}=\$10.50/MMBtu$$

$$\text{Electrical Resistance Heat Cost}=\frac{\frac{\$0.082}{\bcancel{kWh}}}{\frac{0.003412MMBtus}{\bcancel{kWh}}\times 0.97\text{(Efficiency)}}=\$24.77/MMBtus$$

Example 1

Please watch the following 1:26 presentation about Example #1. Your old oil furnace runs at about 68% efficiency. If you buy your oil for $1.02/gal, calculate your actual cost on an MM BTU basis.

Example 2

Please watch the following 2:44 presentation about Example #2. Natural gas costs 9.74 dollars/MCF. Heating oil costs 0.99 cents/gal. The natural gas furnace runs at 90% efficiency and the oil furnace runs at 80% efficiency. Which fuel is cheaper?

In the example on the page 12, we see that the heat loss from the house (walls, windows, and the roof) was 116.53 MM BTUs. We also know that it costs $24.77 for 1MM BTUs if electrical resistance heating is used (see Example 17 on page 26). The total cost for the heating can be calculated as follows:

$$CostofHeating=\left(116.53MMBTUs\right)\times \frac{\$24.77}{MMBTUs}=\$2,886.44$$

The price of fuel oil is $10.50 per MMBtu. The annual heating cost would be:

$$CostofHeating=\left(116.53MMBTUs\right)\times \frac{\$10.50}{MMBTUs}=\$1,1223.57$$

Example 1

Please watch the following 6:01 presentation about Example Problem #1. A house in Bismark, ND (HDD = 9,000) has 860 ft² of windows (R = 1), 2,920 ft² of walls (R = 19), and 3,850 ft² of roof (R = 22). Calculate how much heating oil is required to heat this house for the heating season. The furnace efficiency is 80%.

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.

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

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:

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

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

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

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.

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.

R₁ = Initial R-value of section

R₂ = Final R-value of section

R₂ – R₁ = 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 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?

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?