The links below provide an outline of the material for this lesson. Be sure to carefully read through the entire lesson before returning to Canvas to submit your assignments.
We are into Lesson Seven now: home heating basics. This is a very interesting and very important chapter. Let me warn you about one thing: this chapter has a lot of calculations, and in the entire course this is the most, I would say, calculationoriented chapter. So I want you to pay attention to this one.
Basically, if you have a house, you will be losing heat from various places like windows, like the ceiling, the roof, the walls, the doors, through the floor, everywhere. So in this chapter, we will be looking at how much heat loss occurs and by what means you will be losing it. Let's say we have this wonderful mansion and we want to keep the interior at, let's say, 70 degrees all the time. And we have a little furnace that puts out heat and that keeps the interior at 70 degrees even when outside it's, let's say, 10 degrees Fahrenheit. Heat always tends to get out. It can actually go through those solid walls, or if we have windows, it can go through windows or even the roof. And when you open doors and windows, air might leak in and out, so that's another way of losing heat. So there are several ways in which heat is lost to the outside.
How much heat the furnace has to put out to keep the temperature constantly at 70 degrees inside can easily be calculated or known when we know how much heat is escaping. If somehow, ideally, we can seal off this entire place without any heat loss, then once you bring this house to 70 degrees you can turn on the furnace and it will remain at 70 degrees for the rest of the season. But you and I both know that is not the case. So in this chapter, you're going to learn how much heat is escaping through various surfaces here. And obviously, whatever is escaping, that is the amount of energy that the furnace has to put out. Which means we have to buy energy and put it into the furnace in the form of fuel. So if we are losing a million BTUs like this through all these areas, we have to get a million BTUs made up from this furnace. And we also know that the furnaces are not 100% efficient, either. If we need to get 100,000 BTUs, or 1 million BTUs from the furnace, we cannot expect a furnace to put out if we put in 1 million BTUs. So if we want 1 million BTUs as output, obviously we have to put more in in the form of fuel. So to know how much fuel we have to put in, we have to know how much heat loss we actually have in the house.
So we will be looking at residential heat loss, how to calculate that, and in what ways we lose heat in this first chapter of Lesson Seven. We will understand the mechanisms of heat transfer and also calculate the heating degree days for a heating season, which is basically the number of degrees that we have to heat our air. If the outside temperature is low, obviously we have to heat our air to a higher temperature or higher number of degrees. That's more degrees of heating. So we will calculate how many degrees we have to heat per day and per season, and so on and so forth. And if we know that, we can calculate the heat loss from a solid wall. That is conduction heat loss. We will talk about conduction, convection, and radiation  those three mechanisms of heat loss.
And also, once we have a wall, not all walls are the same. Some walls resist more heat loss than the others. So we define a property of a wall, a property of a solid material that tells us how much the material resists the heat loss. That's called Rvalue. So we will understand and articulate the concept of Rvalue here. And we can also increase the Rvalue of a wall by increasing its thickness or by going to a different material, and so on. And if we were to have a wall with lower Rvalue, which means lower resistance, we would be losing more heat to the outside. And we would be requiring more heat or more fuel to heat that place. So we'll talk about the cost of various fuels for a given heat loss. Seeing 1 million BTUs that we lose, if we were to heat with natural gas to get that same 1 million BTUs, what would it cost? Or if we use electricity, what would it cost to get 1 million BTUs? Or if we use propane, what would it cost? But the heat loss is always 1 million BTUs. For the same amount of heat loss, which fuel happens to be cheaper fuel or more expensive fuel? That's what we will be determining here.
We will also understand that if we have to install more insulation, we will need to borrow more money. And if we borrow more money and put in more insulation, can we save enough through heat loss to pay back that money that we borrowed? That is payback period. You already know the concept. We will do some calculations to see whether it is wise to borrow money and put more into insulation. All right, that's basically part A here. And we will also look at part B, which is insulation and home heating fuels. And in this part, we will talk about various types of insulation materials and how we can improve a wall's performance by adding more and more layers to that wall. So if we have a wall that has four different materials together, we'll see that, for example, inside we would have drywall that we could paint easily. And behind the drywall we have the framework, like a wooden frame with wooden studs that are used for structural integrity. And between those studs you always pack some insulation material. And outside the insulation material is not visible because we'll have a plywood or sheathing outside. And even sheathing doesn't look good outside, so on top of that we put a siding, a vinyl siding or brick or whatever it is.
So when we have different materials together, one right behind the other, they will all together resist the heat loss. So we will calculate the heat loss resistivity or Rvalue of a wall that has different layers of insulation. And if we have that, we can find how much energy we can save or how we can cut the heat loss, et cetera. And also we will talk about the efficiencies, the furnaces, and how we can distribute the heat in the house into various rooms, et cetera, of different heating systems. We'll talk about that later on in the next chapter. So this is basically a calculationoriented and problembased kind of chapter or lesson. And if we have any more difficulties in this one, I also added on top of this the practice questions, a bunch of practice problems which explain you how to do numerical problems. For example, I have given here a bunch of problems, actually, using the formulas that we'll be looking at here for practice. There are a bunch of problems  about 30 problems or so.
Those of you who are comfortable with the material can do these problems yourselves and see if you got them correct by going to this answers page, where the same problems are given with answers. So you can actually look at whether you got the same answer or not. If you got it right, you are happy. If you don't, you may feel lost. If that's the case, what you do is click on this third one here, where for every problem we have a solution, actually, like audio that I'm speaking to you right now. I have made some movies again for each of these problems that will explain to you like a blackboard, or white board rather.
So there are a bunch of problems like this that you can look at. And most of the problems have solutions like this. So I've tried to make your life simpler. Lesson Seven is a twoweek lesson, which means you will have a quiz at the end of two weeks after the lesson is assigned. And you take the quiz at the end of the lesson.
Good luck.
Home heating is the single highest energy expense for a household.
Therefore, reduction of energy consumption in home heating results in substantial monetary savings and reduction in air pollution. With appropriate improvements, average home heating costs can be reduced by 30 percent (i.e., about $500 a year).
Appliances  Percentage 

Space Heating  47% 
Electric Air Conditioning  8% 
Water Heating  17% 
Refrigerators  5% 
Other Appliances and Lighting  23% 
Upon completing this lesson, you will be able to:
Step  Activity  Access / Directions 

1  Read the online lesson  Lesson 7a  Residential Heat Loss Lesson 7b  Insulation and Home Heating Fuels 
2  Watch  Lesson 7  Guided Review (Flash movie). (A printable Review Sheet is also provided.) 
3  Read  Lesson 7  Questions for Review and Discussion. 
4  Review  Lesson 7  Resources (supplemental materials that are optional...but informative!). 
5  Complete  Lesson 7  StudyMate Activities. (You will obtain feedback for these exercises, but they will not count toward your final course grade.) 
6  Take  Lesson 7  Quiz (graded). The quiz is available in Canvas. 
7  Complete  Home Activity 4: Lighting

See the Calendar tab in Canvas for due dates/times.
If you have any questions, please post them to the General Course Questions forum in located in the Discussions tab in Canvas. I will check that discussion forum daily to respond. While you are visiting the discussion board, feel free to post your own responses to questions posted by others  this way you might help a classmate!
Houses are heated to keep the temperature inside at about 65°F when the outside temperature is lower. A house requires heat continuously because of the heat loss. Heat can escape from a house through various places; some are well known and some are not noticeable. Heat can escape from the roof, walls, doors, windows, basement walls, chimney, vents, and even the floor.
The more heat the house leaks, the more the furnace has to put out to make up for the loss. For the furnace to generate more heat to compensate the heat loss, more fuel needs to be put into the furnace, hence higher fuel or heating costs.
As you will recall from Lesson 3, not all energy conversion devices are efficient. Thus, it is important to note that furnaces are not 100 percent efficient. When a furnace’s efficiency is lower, the fuel consumption for the same amount of heat output will be even higher.
Participate in the following activity to find out where the most heat loss occurs.
Below are the most common places for heat loss in a typical home.
Question: Which one has the most heat loss compared to the others?
Answer: # 5 Heat loss through cracks in walls, windows, and doors
Actual percentages of heat loss for each:
Heat loss through ceilings = 5%
Heat loss through windows = 16%
>Heat loss through doors = 3%
Heat loss through frame walls = 17%
Heat loss through cracks in walls, windows, and doors = 38%
Heat loss through basement walls = 20%
Heat loss through basement floor = 1%
Heat escapes (or transfers) from inside to outside (high temperature to low temperature) by three mechanisms (either individually or in combination) from a home:
Click here to open a text description of the examples of heat transfer by conduction, convection, and radiation
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 play to start animation.
Click here to open a text description of the Conduction Candle animation
A hand holds a metal rod above a lit candle. The molecules quickly heat up at the point where the flame touches the rod. The heat then spreads across the entire metal rod and the heat is then able to be felt by the hand.
With regard to residential heating, the heat is transferred by conduction through solids like walls, floors, and the roof.
Click here to open a text description of Conduction in Regard to Residential Heating example
Picture the cross section of a wall in a house. Inside the house it is 65°F and outside it is 30°F. Two arrows point from inside the house to the outside to show how heat is transferred from the inside of the house to the outside through the wall via conduction.
Heat loss across a solid wall by conduction
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 here to open a text description of the Convection Candle animation
A hand is held above a lit candle. As the candle heats the air, the heat rises to the hand. Eventually, it gets too hot and the hand pulls away from the candle.
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:
Click here to open a text description of the Convection in a Room animation
Picture a room with an open door letting in cool air on the left and a radiator creating heat on the right. As the radiator heats the air around it, the air rises and is replaced by cool air. Once the warm air hits the ceiling, it travels left towards the open door, cooling as it moves. The cool air from the open door travels to the right across the floor towards the radiator to be heated. The overall effect is a circular convection current of air within the room.
Radiation is 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 here to open a text description of the Radiation Example animation
Picture the Sun and the Earth with arrows traveling from the Sun to the Earth through space. The arrows represent the energy that travels to the Earth via radiation, which does not require any medium (atoms or molecules) to do so.
First, identify the type of home heat loss pictured in images AJ as either: conduction, convection or radiation. Then click and drag each image down to the correct category at the bottom of the screen.
Click here to open a text description of the Test Yourself activity
Identify the type of heat loss (conduction, convection, or radiation) for each of the following examples:
Answers:
There are two ways in which we can reduce energy consumption.
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.
What does a house's heat loss depend on? Complete the activity below to find out the three main factors leading to heat loss.
Click here to open a text description of the conduction heat loss activity.
What are the three main factors a house's heat loss depends on?
Example 1:
House A sits next to house B. Though both houses have the same basic design, house B is significantly larger than house A.
Example 2:
House A and house B are the exact same size and design. House A sits on the beach in a warm, tropical area while house B sits by a ski resort in the mountains up north, surrounded by snow.
Example 3:
House A and house B are the same size and sit next to each other. The design for both houses is the same except house A has a thick layer of pink insulation installed. The Rvalue of house B is .63 and the Rvalue of house A is unknown.
Example 1:
Example 2:
Example 3:
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:
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:
The formula for determining the Heating Degree Day (HHD) is:
$${\text{HDD = T}}_{\text{base}}{\text{T}}_{\text{a}}$$
To calculate HDD:
Note: If the T_{a} is equal to or above 65 ºF, there are no heating degree days for that 24hour period, or HDD = 0.
Click the play button below, and observe the temperature changes. Then calculate the average temperature and the Heating Degree Day.
Click here to open a text description of the Heating Degree Days activity.
Picture a house with a thermometer on the porch. The current outside temperature is 60 degrees. As thick clouds start to move in, the temperature begins to drop in 5 degree increments, first to 55 degrees, then 50 degrees, etc. When the clouds have completely blocked the Sun, the thermometer reads 40 degrees.
Questions:
Answers:
Calculate the HDD for one day when the average outside temperature is 13º F.
$$\begin{array}{ccc}\text{HeatingDegreeDay}& =& {T}_{base}{T}_{a}\\ \text{}& =& 65\xbaF13\xbaF\\ \text{}& =& 52\xbaF\end{array}$$Calculate the HDD for one day when the average outside temperature is 2º C.
$$\text{ConvertfromCelsiustoFahrenheit:}2\xba\text{}C\text{}=\text{}35.6\xba\text{}F$$ $$\begin{array}{ccc}\text{HeatingDegreeDay}& =& \text{}{T}_{base}\text{}{T}_{a}\\ \text{}& =& \text{}65\xba\text{}F\text{}\text{}35.6\xba\text{}F\\ \text{}& =& \text{}29.4\xba\text{}F\end{array}$$Given the following data, calculate the HDD for the week:
Day  Average Temperature 

Sunday  49° F 
Monday  47° F 
Tuesday  51° F 
Wednesday  60° F 
Thursday  65° F 
Friday  67° F 
Saturday  58° F 
Please watch the following (2:21) presentation about Heating 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.
Place  Degree Days 

Birmingham, AL  2,823 
Anchorage, AK  10,470 
Barrow, AK  19,893 
Tucson, AZ  1,578 
Miami, FL  155 
Pittsburgh, PA  5,829 
State College, PA  6,345 
Source: NOAA [1]
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.
Given the following set of average temperatures, by month, for State College, PA, calculate the HDD for the heating season:
Jan.  Feb.  Mar.  Apr.  May  Jun.  Jul.  Aug.  Sep.  Oct.  Nov.  Dec. 

25°F  28°F  37°F  48°F  59°F  67°F  71°F  70°F  62°F  51°F  41°F  31°F 
Please watch the following (2:32) presentation about problem #1:
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 only variable that will change is the temperature difference between inside and outside.
Calculate the heat loss for a 10 ft by 8 ft wall, insulated to Rvalue 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.
$$Heat\text{}Loss\left(\frac{BTUs}{h}\right)=\frac{Area(f{t}^{2})\times Temperature\text{}Difference{(}^{o}F)}{R\text{}value\left(\frac{f{t}^{2}{}^{\text{}o}F\text{}h}{BTUs}\right)}\text{}\times \text{}24hr/day$$
What is the hourly and daily heat loss of a 15ft by 15ft room with an 8ft ceiling, with all surfaces insulated to R13, with inside temperature 65°F and outside temperature 25°F?
$$Hourly\text{}Heat\text{}Loss\text{}rate=\frac{Q(BTUs)}{t(hour)}\text{}=\text{}\frac{(480f{t}^{2}\times \left({65}^{o}F{25}^{o}F\right)}{13\frac{f{t}^{2}{\text{}}^{o}Fh}{BTU}}=1,477\frac{BTUs}{h}$$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 24hour period or one day, the heat loss would be:
$$Daily\text{}Heat\text{}Loss\text{}per\text{}day=\frac{1477\text{}BTUs}{h}\text{}x\frac{24\text{}h}{day}=35,448\text{}\frac{BTUs}{day}$$Area (ft^{2}) 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 ft^{2} 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:
Recall that the formula for daily heat loss is:
$$Heat\text{}Loss\text{}\left(\frac{BTUs}{h}\right)=\frac{Area\text{}(f{t}^{2})\times Temperature\text{}Difference\text{}{(}^{o}F)}{Rvalue\left(\frac{f{t}^{2}{\text{}}^{o}F\text{}h}{BTUs}\right)}\text{}\times \frac{24\text{}h}{day}\text{}$$
Thus, theoretically we would need to perform this calculation for every day of the 365day 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:
Since the area (480 ft2), Rvalue $\frac{f{t}^{2}\text{}\xb0F\text{}Btu}{h}$ , and 24 h in a day are common for ALL heating days, we can bring those out and rewrite the equation as:
$$\begin{array}{l}=\frac{480\text{}f{t}^{2\text{}}x}{13\frac{f{t}^{2}{\text{}}^{o}F\text{}h}{BTU}}\times 24\frac{h}{day}\left\{{\left(6535\right)}^{o}F+{\left(6532\right)}^{o}F+{\left(6528\right)}^{o}F\mathrm{...}so\text{}on\text{}for\text{}all\text{}heating\text{}days\right\}\\ \\ =\frac{480\text{}f{t}^{2\text{}}x}{13\frac{f{t}^{2}{\text{}}^{o}F\text{}h}{BTU}}\times 24\frac{h}{day}\left\{{30}^{o}F+{33}^{o}F+{37}^{o}F+so\text{}on\text{}for\text{}all\text{}heating\text{}days\right\}\\ \\ Where\left\{{30}^{o}F+{33}^{o}F+{37}^{o}F+so\text{}on\text{}for\text{}all\text{}heating\text{}days\right\}is\text{}the\text{}sum\text{}of\text{}Heating\text{}Degree\text{}Days\end{array}$$
This equation can be even further simplified. The formula for Annual or Seasonal heat loss can be written in general terms as:
$$Heat\text{}Loss\text{}in\text{}a\text{}Season\text{}=\text{}\frac{Area(f{t}^{2})}{Rvalue\left(\frac{f{t}^{2}{\text{}}^{o}F\text{}h}{BTU}\right)}\times 24\frac{h}{day}\times (HDD\text{}for\text{}the\text{}season)$$
$$Heat\text{}Loss\text{}in\text{}a\text{}Season\text{}=\text{}\frac{Area}{Rvalue}\times 24\times HDD\text{}$$
Please watch the following 3:29 presentation on Example Problem #1. For the 150day heating season in Roanoke, VA, the average temperature was 47° F. How much heat is lost through a 176 ft^{2} wall (R=16) during the entire season?
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?
The following section of Lesson 7 wil discuss insulation and home heating fuels.
As you may recall from the first part of this lesson, Rvalue is a wall's capacity to resist heat loss or its thermal resistance. Insulation materials are rated in terms of their Rvalue, with a higher Rvalue indicating better insulating effectiveness.
The Rvalue 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.
Rvalues 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.
Material  RValue (ft^{2 o} Fh / BTU) 

Plain glass, 1/8 inch  0.03 
Stone per inch  0.08 
Common Brick per inch  0.20 
Asphalt Roof Shingles  0.44 
1/2 inch Gypsum Board (Drywall or plasterboard)  0.45 
Wood Siding, 1/2 inch  0.81 
Plywood, 3/4 inch  0.94 
Insulating sheathing, 3/4 inch  2.06 
Fiberglass, per inch (battens)  3.50 
Polystyrene per inch  5.00 
Polyurethane Board  6.25 
If insulation materials have a low Rvalue per inch, then they will need to be thicker than those materials with a higher Rvalue 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 RValue of 22 and then answer the questions below.
Look at the table below to learn about six 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 loosefill.  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 insectresistant chemicals. 
Rigid Foam  Different types, but some made from postconsumer recycled content from fast food containers and cups.  Rigid sheets that are applied directly to framing.  Best where space is limited, but a high Rvalue 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 Rvalue of a composite wall is calculated by adding the effective Rvalues 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.
We can calculate this wall’s composite Rvalue by adding the Rvalues of each layer.
Plasterboard (1/2 inch)
+ Fiberglass (3.5 inches @ 3.70 per inch)
+ Plywood (3/4 inches)
+ Wood Siding (1/2 inch)

Total RValue of Composite Wall
We can find the Rvalues for the walls by using the table from the page about RValues [2] (the table is repeated below):
Material  RValue (ft^{2 o} Fh / BTU) 

Plain glass, 1/8 inch  0.03 
Stone per inch  0.08 
Common Brick per inch  0.20 
Asphalt Roof Shingles  0.44 
1/2 inch Gypsum Board (Drywall or plasterboard)  0.45 
Wood Siding, 1/2 inch  0.81 
Plywood, 3/4 inch  0.94 
Insulating sheathing, 3/4 inch  2.06 
Fiberglass, per inch (battens)  3.50 
Polystyrene per inch  5.00 
Polyurethane Board per inch  6.25 
Cinder Block (12 inches)  1.89 
Plasterboard (1/2 inch) = 0.4524
+ Fiberglass (3.5 inches @ 3.70 per inch) = 12.95
+ Plywood (3/4 inches) = 0.94
+ Wood Siding (1/2 inch) = 0.81

Total RValue of Composite Wall = 15.15 $\frac{{\text{ft}}^{\text{2}}\text{}\xb0\text{Fh}}{\text{Btu}}$
A ceiling is insulated with 0.75" plywood, 2" of polystyrene board, and a 3" layer of fiberglass. What is the RValue for the ceiling?
Solution:
The wall consists of three layers, and all three layers together prevent the heat loss. So, we need to add the Rvalues of all three layers
$\raisebox{1ex}{$3$}\!\left/ \!\raisebox{1ex}{$4$}\right.$ " in Plywood has an Rvalue of 0.94
2" of polystyrene at 5.0 per inch will have an Rvalue of 10.00
3" of fiberglass at 3.7 per inch will have an Rvalue of 11.10
So the Rvalue of the composite wall is 22.04 ft^{2 o}Fh / BTU.
Please watch the following 1:31 video presentation about Example #2. A wall consists of 0.5” wood siding (R = 0.81), 0.75” plywood (R = 0.94), 3.5” of fiberglass (R = 13.0), and 0.5” plasterboard (R = 0.45). What is the composite Rvalue of the wall?
Please watch the following 1:18 video presentation about Example #3. What is the Rvalue of a wall that is made up of wood siding (R = 0.81), 5” of fiberglass (R = 3.70 per inch), and a layer of 0.5” drywall (R = 0.45)?
Please watch the following 1:44 video presentation about Example #4. A wall is made up of 8” of stone, 3” of polyurethane board, and 0.75” of plywood. Calculate the composite Rvalue for the wall.
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 RValues for new construction of houses. Note that insulation Rvalues are different for the ceilings, walls, floor, etc.
Click here to open a text description of the Insulation Needs activity
The U.S. Department of Energy released recommended total Rvalues for new construction houses. The info is based on regional zone and covers various parts of the house. The states within each zone are listed below, followed by a data table containing the Rvalues for each part of the house. The Rvalues are dependent on the type of heating system being used and may vary for each. Some states may lie in multiple zones.
States:
Zone 1  Gas Heat Pump Fuel Oil 
Electric Furnace 

Attic  R49  R49 
Cathedral Ceiling  R38  R60 
Wall  R18  R28 
Floor  R25  R25 
Crawl Space  R19  R19 
Slab Edge  R8  R8 
Basement Interior  R11  R19 
Basement Exterior  R10  R15 
States:
Zone 2  Gas Heat Pump Fuel Oil 
Electric Furnace 

Attic  R49  R49 
Cathedral Ceiling  R38  R38 
Wall  R18  R22 
Floor  R25  R25 
Crawl Space  R19  R19 
Slab Edge  R8  R8 
Basement Interior  R11  R19 
Basement Exterior  R10  R15 
States:
Zone 3  Gas Heat Pump Fuel Oil Electric Furnace 

Attic  R49 
Cathedral Ceiling  R38 
Wall  R18 
Floor  R25 
Crawl Space  R19 
Slab Edge  R8 
Basement Interior  R11 
Basement Exterior  R10 
States:
Zone 4  Gas Heat Pump Fuel Oil 
Electric Furnace 

Attic  R38  R49 
Cathedral Ceiling  R38  R38 
Wall  R13  R18 
Floor  R13  R25 
Crawl Space  R19  R19 
Slab Edge  R4  R8 
Basement Interior  R11  R11 
Basement Exterior  R4  R10 
States:
Zone 5  Gas  Heat Pump Fuel Oil 
Electric Furnace 

Attic  R38  R38  R49 
Cathedral Ceiling  R30  R38  R38 
Wall  R13  R13  R18 
Floor  R11  R13  R25 
Crawl Space  R13  R19  R19 
Slab Edge  R4  R4  R8 
Basement Interior  R11  R11  R11 
Basement Exterior  R4  R4  R10 
States:
Zone 6  Gas  Heat Pump Fuel Oil 
Electric Furnace 

Attic  R22  R38  R49 
Cathedral Ceiling  R22  R30  R38 
Wall  R11  R13  R18 
Floor  R11  R11  R25 
Crawl Space  R11  R13  R19 
Slab Edge  (c)  R4  R8 
Basement Interior  R11  R11  R11 
Basement Exterior  R4  R4  R10 
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.
$$Heat\text{}Loss\text{}\left(\frac{BTUs}{h}\right)=\frac{Area(f{t}^{2})\times Temp.\text{}Difference\text{}{(}^{o}F)}{R\text{}Value\text{}\frac{f{t}^{2}{\text{}}^{o}Fh}{BTU}}$$
$$Heat\text{}Loss\text{}\left(\frac{BTUs}{h}\right)=\frac{Area\text{}(f{t}^{2})\text{}\times \text{}Temp.\text{}Difference\text{}{(}^{o}F)}{R\text{}Value\text{}\frac{f{t}^{2}{\text{}}^{o}Fh}{BTU}}\text{}\times \text{}24$$
$$Seasonal\text{}Heat\text{}Loss\text{}=\text{}\frac{Area\text{}}{R\text{}Value}\text{}\times \text{}24\text{}\times \text{}HDD$$
The heat loss from walls, windows, roof, and flooring should be calculated separately, because of different RValues for each of these surfaces. If the Rvalue 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 Rvalue.
A house in Denver, CO has 580 ft^{2} of windows (R = 1), 1920 ft^{2} of walls and 2750 ft^{2} of roof (R = 22). The walls are made up of wood siding (R = 0.81), 0.75” plywood, 3.5” of fiberglass insulation, 1.0” of polyurethane board, and 0.5” gypsum board. Calculate the heating requirement for the house for the heating season, given that the HDD for Denver is 6,100.
Solution:
Heating requirement of the house = Heat loss from the house in the whole year. To calculate the heat loss from the whole house, we need to calculate the heat loss from the walls, windows, and roof separately, and add all the heat losses.
Heat loss from the walls:
Area of the walls = 1,920 ft^{2}, HDD = 6,100, and the composite R value of the wall needs to be calculated.
Material  RValue 

Wood Siding  0.81 
3/4 inch plywood  0.94 
3.5 inches of fiberglass 3.5 in x 3.7 / in  12.95 
1.0 inch of polyurethane board = 1.0 in x 5.25 / in  5.25 
1/2 inch Gypsum board  0.45 
Total RValue of the walls  20.40 
Total heat loss from the house = 13.78 + 84.91 + 18.30 =116.99 MMBTU in a year or heating requirement is 116.99 million BTUs per year.
Please watch the following 4:58 presentation about Example #1. A house in State College, PA has 580 ft^{2} of windows (R = 1), 1920 ft^{2} of walls, and 2750 ft^{2} 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.
Please watch the following 9:00 presentation about Example #2. A singlestory 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 R30. 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.
Fuel Type  Percentage used 

Natural Gas  56% 
Electricity  26% 
Fuel Oil  11% 
Other  10% 
As you see, more than 50 percent (56% to be exact) of the households in the United States use natural gas as their main heating fuel, and about 26 percent of the households use electricity to heat their homes. Another 11% use fuel oil, and the last 10% use something other than natural gas, electricity and fuel oil.
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.
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 (ft^{3}).
Press play to see the difference between capacity and consumption in a gasheated home.
Fuel  Capacity  Consumption  Additional Information 

Natural Gas  Measured in British thermal units per hour (BTU/h). Most heating appliances for home use have heating capacities of between 40,000 and 150,000 BTU/h. In the past, gas furnaces were often rated only on heat input; today the heat output is given.  Consumption of natural gas is measured in cubic feet (ft^{3}). This is the amount that the gas meter registers and the amount that the gas utility records when a reading is taken. One cubic foot of natural gas contains about 1,000 BTU of energy.  Utility companies often bill customers for CCF (100 cu. ft) or therms of gas used: one therm equals 100,000 BTUs. Some companies also use a unit of MCF, which is equal to 1,000 cu. ft One MCF equals 1,000,000 BTUs (1 MM BTUs). 
Propane or Liquefied Petroleum Gas (LPG)  Measured according to BTU/h.  Consumption of propane is usually measured in gallons; propane has an energy content of about 91.300 BTUs per gallon.  Can be used in many of the same types of equipment as natural gas. It is stored as a liquid in a tank at the house, so it can be used anywhere, even in areas where natural gas hookups are not available. 
Fuel Oil  The heating (bonnet) capacity of oil heating appliances is the steadystate heat output of the furnace, measured in BTU/h. Typical oilfired central heating appliances sold for home use today have heating capacities of between 56,000 and 150,000 BTU/h.  Oil use is generally billed by the gallon. One gallon of #2 fuel oil contains about 140,000 BTU of potential heat energy.  Several grades of fuel oil are produced by the petroleum industry, but only #2 fuel oil is commonly used for home heating. 
Electricity  The heating capacity of electric systems is usually expressed in kilowatts (kW); 1 kW equals 1,000 W. A kilowatthour (kWh) is the amount of electrical energy supplied by 1 kW of power over a 1hour period. Electric systems come in a wide range of capacities, generally from 10 kW to 50 kW.  Electricity is sold in kWh (kilowatts per hour).  The watt (W) is the basic unit of measurement of electric power. 
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.
Fuel  Unit  Heating Value (BTU's) 

Natural Gas  CCF (100 Cu. ft) or Therm  100,000 
Natural Gas  MCF (1,000 Cu.ft)  1,000,000 
Fuel Oil  Gallon  140,000 
Electricity  kWh  3,412 
Propane  Gallon  91,300 
Bituminous Coal  Ton  23,000,000 
Anthracite Coal  Ton  26,000,000 
Hardwood  Cord  24,000,000 
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.
Fuel  Unit  Heating Value (BTU's)  If Efficiency = 50 % 

Natural Gas  CCF (100 Cu. ft) or Therm  100,000  50,000 
Natural Gas  MCF (1,000 Cu.ft)  1,000,000  500,000 
Fuel Oil  Gallon  140,000  70,000 
Electricity  kWh  3,412  1706 
Propane  Gallon  91,300  45,650 
Bituminous Coal  Ton  23,000,000  11,500,000 
Anthracite Coal  Ton  26,000,000  13,000,000 
Hardwood  Cord  24,000,000  12,000,000 
The table below contains the data from a line plot showing the efficiency of a furnace. The percent efficiency is dependent on the heating value of fuel oil (measured in BTU) and shows a strong positive correlation.
Efficiency of a Furnace  Heating Value of Fuel Oil (BTUs) 

50%  70000 
55%  77000 
60%  84000 
65%  90000 
70%  100000 
75%  105000 
80%  111000 
85%  122000 
90%  125000 
95%  131000 
100%  140000 
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.
Click here to open a text description of How a Wood Furnace Operates
Cold combustion air from outside is pulled into the furnace combustion chamber by an electric blower. The wood fire in the chamber heats the air to 1 million BTUs. The hot combustion gasses heat the cold, outside air to 750,000 BTUs, which will circulate through the house. The combustion gasses (250,000 BTUs) travel out through the chimney.
Furnaces usually are not as efficient when they are first firing up as they are running at steadystate. It is sort of like a car getting better mileage in steady highway driving than in stopandgo 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 minimumefficiency 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:
The table below gives the efficiencies of most efficient furnaces that were available in 2002–2003.
Fuel  Furnace Type  Efficiencies (%) 

Natural Gas  Hot air  93.0  96.6 
Natural Gas  Hot Water  83.0  95.0 
Natural Gas  Steam  81.0  82.7 
Fuel Oil  Furnace  83.8  86.3 
Fuel Oil  Hot Water  86.0  87.6 
Fuel Oil  Steam  82.5  86.0 
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:
$$Actual\text{}Energy\text{}Cost\text{}=\text{}\frac{Fuel\text{}Cost\text{}\left(\frac{\$}{Unit\text{}of\text{}Fuel}\right)}{Heating\text{}Value\text{}\left(\frac{MMBTUs}{Unit\text{}of\text{}Fuel}\right)\text{}\times \text{}Efficiency}$$
Let’s say we need one million BTUs to keep a place warm at a certain temperature. What would it cost to get those million BTUs from oil or gas or electricity? Let’s assume that:
Material  Cost per unit  Efficiency  Heating Value 

Natural Gas  $6.60/MCF  90%  1,000,000 BTUs or 1.0 MM BTU/MCF 
Oil  $1.25/Gallon  85%  140,000 BTUs or 0.14 MM BTUs/Gallon 
Electricity  $0.082/kWh  97%  3412 BTUs or .003412 MM BTUs/kWh 
Using the formula below, we can calculate the Actual Energy Cost.
$$\text{Actual Energy Cost}=\frac{\text{Fuel Cost}\left(\frac{\$}{\text{Unit of Fuel}}\right)}{\text{HeatingValue}\left(\frac{\text{MMBtus}}{\text{Unit of Fuel}}\right)\text{\xd7Efficiency}}$$
$$\text{Oil(in central heating system) cost}=\frac{\frac{\$1.25}{\overline{)Gal}}}{\frac{0.14\text{}MMBtus}{\overline{)Gal}}\text{}\times \text{}0.85\text{}\text{(Efficiency)}}\text{}=\text{}\$10.50\text{}/\text{}MMBtu$$
$$\text{Electrical Resistance Heat Cost}=\text{}\frac{\frac{\$0.082}{\overline{)kWh}}}{\frac{0.003412\text{}MMBtus}{\overline{)kWh}}\times 0.97\text{(Efficiency)}}=\$24.77/MMBtus$$
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.
Click here to open a transcript of the Energy Cost 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.
Ok, the old furnace runs at about 68% efficiency in this problem. This is 5.6 and the old furnace runs at an efficiency of E=0.68 and we have the actual cost per unit price or unit fuel which is \$1.02 per gallon. And we also know the calorific value or heating value in millions of BTUs. When you burn one gallon of oil, we get .13 million BTUs so we apply the same formula to get the actual cost. Which is cost per unit fuel which is \$1.02/gallon divided by the heating value which is 0.13 MMBTUs per gallon. We have to have the same units here. And times the efficiency here. Efficiency is 0.68. Gallons and Gallons are canceled and we get this one as \$11.50 / Million BTUs.
$$\begin{array}{l}\text{E=0}\text{.68}\\ \text{Fuel=\$1}\text{.02/gal}\\ \text{1Gal=0}\text{.13MMBTUs}\\ \\ =\text{}\frac{\text{\$1}\text{.02/gal}}{\text{0}\text{.13MMBTUs/gal}\times \text{0}\text{.68}}\\ \\ =\frac{\text{\$11}\text{.50}}{\text{MMBTUs}}\end{array}$$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?
Click here to open a transcript of the Energy Cost Example #2
Natural gas costs \$9.74/MCF. Heating oil costs \$0.99/gal. The natural gas furnace runs at 90% efficiency and the oil furnace runs at 80% efficiency. Which fuel is cheaper?
Ok. This 5.7 is an interesting problem here. We are trying to compare the prices of two fuels – Natural Gas which sells for \$9.74/MCF and we also have oil that sells at \$0.99/gallon. We are trying to compare the prices of these two and choose which one is the best fuel or cheapest fuel. So we need to calculate the price per million BTUs so that we can compare these two fuels. And we also know the furnace efficiencies of each of these. Natural gas furnace efficiency is 0.9 and we know the oil furnace efficiency is 0.8; it is given. So we need to calculate the actual cost and compare the cost.
Natural gas actual cost will be cost per unit fuel which is \$9.74/MCF divided by the heating value per unit fuel. Heating value for this one happens to be 1.0 Million BTUs per MCF and we have to multiply by the efficiency here in the denominator which is 0.9 so the Natural Gas price turns out to be \$10.83 or \$10.83 per Million BTUs (MMBTUs).
$$\begin{array}{l}\text{NaturalGas}\raisebox{1ex}{$\text{\$9}\text{.74}$}\!\left/ \!\raisebox{1ex}{$\text{MCF}$}\right.\\ \\ \text{0}\text{.9efficiency}\\ \\ \text{NaturalGas=}\frac{\raisebox{1ex}{$\text{\$9}\text{.74}$}\!\left/ \!\raisebox{1ex}{$\text{MCF}$}\right.}{\raisebox{1ex}{$\text{1}\text{.0MMBTU}$}\!\left/ \!\raisebox{1ex}{$\text{MCF}\times \text{0}\text{.9}$}\right.}\\ \\ \text{=}\frac{\text{\$10}\text{.82}}{\text{MMBTUs}}\end{array}$$When you do similar calculation for oil here, the actual price is, per unit is \$0.99 per gallon here and how many million BTUs do we get per gallon? 0.13 Million BTUs (0.13 MMBTUs). We have done this before. We have to have the same units here. Gallons and gallons and MCF and MCF here in this case (natural gas) and times the efficiency is 0.8. So the price works out to be \$9.50 per Million BTUs. Same million BTUs would cost \$10.82 for Natural Gas and oil would be \$9.50 so oil 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:
$$Cost\text{}of\text{}Heating\text{}=\text{}\left(116.53\text{}MMBTUs\right)\text{}\times \text{}\frac{\$24.77}{MMBTUs}\text{}=\text{}\$2,886.44$$
The price of fuel oil is \$10.50 per MMBtu. The annual heating cost would be:
$$Cost\text{}of\text{}Heating\text{}=\text{}\left(116.53\text{}MMBTUs\right)\text{}\times \text{}\frac{\$10.50}{MMBTUs}\text{}=\text{}\$1,1223.57$$
A house in International Falls, MN (HDD = 10,500) consists of 1248 ft^{2} of walls with an Rvalue of 13 and 1150 ft^{2} of roof with an R value of 29. The home is heated with natural gas. The AFUE is 0.90 and the price of natural gas is \$0.88/CCF. What is the annual heating cost?
Energy cost per million BTUs from natural gas can be calculated using the following equation.
$$Actual\text{}Energy\text{}Cost\text{}=\text{}\frac{Fuel\text{}Cost\text{}\left(\frac{\$}{Unit\text{}of\text{}Fuel}\right)}{Heating\text{}Value\text{}\left(\frac{MMBTUs}{Unit\text{}of\text{}Fuel}\right)\text{}\times \text{}Efficiency}$$$$\text{Actual Energy Cost}=\frac{\frac{\$0.88}{\overline{)CCF}}}{\frac{0.1\text{}MMBtus}{\overline{)CCF}}\times 0.90\text{}\text{(Efficiency)}}=\$9.80/MMBtu$$
Heat required can be calculated from the heat loss. Heat loss from the house is from two sources: walls and the roof. Heat loss from each of these sources for a year (season) can be calculated by using the following equation.
$$\text{Heat Loss from Walls}=\frac{1,248\text{}\overline{)f{t}^{2}}\times 10,500\text{}\overline{){}^{o}f}\overline{)days}\times \frac{24\overline{)h}}{\overline{)day}}}{13.0\frac{\overline{)f{t}^{2}}\text{}\overline{){}^{o}f}\overline{)h}}{Btu}}=24.19\text{}MMBtu$$
$$\text{Heat Loss from Roof}=\frac{1,150\text{}\overline{)f{t}^{2}}\times 10,500\text{}\overline{){}^{o}f}\overline{)days}\times \frac{24\overline{)h}}{\overline{)day}}}{29.0\frac{\overline{)f{t}^{2}}\text{}\overline{){}^{o}f}\overline{)h}}{Btu}}=9.99\text{}MMBtu$$
Total heat loss = sum of heat loss from the walls and the roof
$$=\text{}24.19\text{}+\text{}9.99\text{}=\text{}34.18\text{}MMBTUs$$Annual heating cost = Annual heat loss (MMBTUs) x Actual energy cost ($/MMBTU)
$$=(34.18\text{}\overline{)MMBtus})\times \frac{\$9.80}{\overline{)MMBtus}}=\$334.96$$
Please watch the following 6:01 presentation about Example Problem #1. A house in Bismark, ND (HDD = 9,000) has 860 ft^{2 }of windows (R = 1), 2,920 ft^{2} of walls (R = 19), and 3,850 ft^{2} of roof (R = 22). Calculate how much heating oil is required to heat this house for the heating season. The furnace efficiency is 80%.
Click here to open a text description of the Annual Heating Cost Example 1.
A house in Bismark, ND (HDD = 9000) has 860 ft^{2} of windows (R = 1), 2920 ft^{2} of walls (R = 19), and 3850 ft^{2} of roof (R = 22). Calculate how much heating oil is required to heat this house for the heating season. The furnace efficiency is 80%.
Ok, this problem is very similar to what we have done in before. Which is calculating the heat loss from various surfaces and adding all of those to get the total heat loss from the house. And to calculate the oil requirement based on the furnace efficiency. The house is located in Bismark, ND and HDD there is given as 9,000 so HDD is given. This HDD remains the same for all different surfaces. And we also know the surfaces of each of these components like windows. Let’s do the calculation first for windows.
Windows area is given as 860 ft and we also know the Rvalue of this. Rvalue is given as 1 – this is ft °F h/BTUs. So we can calculate heat loss through windows. Where all we need is the area, 860 ft times HDD which is 9,000 in this case times 24 hours in a day divided by, we get the Rvalue here. That is 1 ft°F h/BTU. So ft/ ft cancel, °F/°F, 24 hours/24 hours, days and days so the total heat loss through windows is the number that we get here. It happens to be 185,760,000 BTUs.
$$\begin{array}{l}{\text{Windows=860ft}}^{\text{2}}\\ {\text{Rvalue=1ft}}^{\text{2}}\text{}\xb0\text{Fh/BTUs}\\ \text{HeatLoss=}\frac{{\text{860ft}}^{\text{2}}\text{}\times {\text{9,000ft}}^{\text{2}}\text{}\xb0\text{Fday}\times \text{24hrs/day}}{{\text{1ft}}^{\text{2}}\text{}\xb0\text{Fh/BTUs}}\\ \\ \text{=185,760,000BTUs}\end{array}$$Similarly we can calculate the heat loss  this is through windows. We can calculate heat loss through the walls. Wall area is 2,920 ft^{2} times 9,000 °F days times 24 hours per day divided by Rvalue of 19 for walls. And we can cancel out the units and make sure that everything makes sense and this turns out to be 33,195,789 BTUs.
$$\begin{array}{l}\text{HeatLoss=}\frac{{\text{2,920ft}}^{\text{2}}\text{}\times {\text{9,000ft}}^{\text{2}}\text{}\xb0\text{Fday}\times \text{24hrs/day}}{{\text{19ft}}^{\text{2}}\text{}\xb0\text{Fh/BTUs}}\\ \\ \text{=33,195,789BTUs}\end{array}$$Now heat loss through the roof. You can calculate it again separately and the area of the roof happens to be 3850 ft^{2}, and 9,000 degree days times 24 hours over a day divided by, the roof generally has higher Rvalue, 22 ft^{2} °F h/BTUs, ok? Now let’s cancel out these units and the heat loss is to be 37,800,000 BTUs. So the total heat loss is the sum of all these three and when you add these up you get 256,755,789 BTUs.
$$\text{HeatLoss=256,755,789BTUs}$$Now, the furnace efficiency is given and we are using in this case, heating oil. Heating oil is 130,000 BTUs so how much is required? How much heating oil is required? We need 256,755,789 BTUs. When we buy heating oil, we get 130,000 BTUs per every gallon and although we get 130,000, this is theoretical, the efficiency is given as 0.8 so only 80% will get really available as heat. So, that makes it, actually, the requirement little bit higher, so that is equal to now, if we do this calculation, it will be 2,468.8 gallons or roughly 2,469 gallons of oil is required to heat this place.
$$\frac{\text{256,755,789BTUs}}{\text{130,000BTUs/gal}\times \text{0}\text{.8}}\text{=2,468}\text{.8gallonsor2,469gallons}$$Earlier sections illustrated that adding more insulation and improving the Rvalue 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 Rvalue of the wall used in an earlier example is improved to R23 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 payback period is 2.47 years. Shorter payback 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, kilowatthour (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 Rvalue of section
R_{2} = Final Rvalue of section
R_{2} – R_{1} = Rvalue 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 Rvalues, if you add the same amount of insulation everywhere. Ceilings, walls, or sections of walls with different Rvalues must be figured separately.
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 R19 to R30 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 payback 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}).
$${C}_{i}\text{}=\text{}\frac{\$340}{1,100\text{}sq.\text{}ft.}\text{}=\text{}\$0.31/sq.\text{}ft.$$
$$And,{C}_{e}=\frac{\$0.95}{\overline{)1\text{}CCF}\times \frac{100,000\text{}Btus}{\overline{)1\text{}CCF}}}=\$0.0000095/Btu$$
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.
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 R13 to R22 is \$0.60/ft^{2}. 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.
For a house in Hackensack, NJ (HDD = 4,600), the installed cost to upgrade from R13 to R22 is \$0.60/ft^{2}. 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 R times R times efficiency divided by cost of energy times the difference in Rvalue times HDD times 24. That is the formula that we need to use.
$$\text{Payback=}\frac{{\text{C}}_{\text{i}}\text{}\times {\text{R}}_{\text{1}}\text{}\times {\text{R}}_{\text{2}}\text{}\times \text{E}}{{\text{C}}_{\text{e}}\text{}\times \text{}\left[{\text{R}}_{\text{2}}\text{}{\text{R}}_{\text{1}}\right]\text{}\times \text{HDD}\times \text{24}}$$We have all the data. HDD is given as 4,600 and cost of insulation C_{i} is given as \$0.60 per ft^{2} and the efficiency of the furnace is given as 78% or 0.78. We know the initial Rvalue, R_{1} = 13 and we know the final Rvalue; the improved Rvalue 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.
$$\begin{array}{l}\text{HDD=4,600}\xb0\text{Fdays}\\ {\text{C}}_{\text{i}}\text{=\$0}{\text{.61/ft}}^{\text{2}}\\ \text{E=0}\text{.78}\\ {\text{R}}_{\text{1}}\text{=13}\\ {\text{R}}_{\text{2}}\text{=22}\\ {\text{Difference=R}}_{\text{2}}\text{}{\text{R}}_{\text{1}}\text{=9}\\ \\ \text{Costofoil=\$1}\text{.13/gal}\\ \\ \frac{\text{\$1}\text{.13}}{\text{130,000BTUs}}\text{=\$0}\text{.0000869perBTU}\end{array}$$So, for every BTU we pay \$0.000869. So we can use the formula now. It consists of C_{i}, C_{i} is cost of insulation, which is 0.06 per ft^{2}, dollars per ft^{2} times R_{1} which is 13, again units, you have to make sure ft^{2}, °F, hr/BTU. And here R_{2} 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.
$$\begin{array}{l}=\frac{\text{\$0}{\text{.60/ft}}^{\text{2}}\text{}\times \text{1322}\times \text{0}\text{.78}}{\text{\$0}\text{.0000869/BTU}\times \text{9}\times \text{4,600}\times \text{24}}\\ \\ \text{PaybackPeriod=15}\text{.5years}\end{array}$$Please watch the following 4:10 presentation about Example Problem #2. Lt. Dave Rajakovich has 1200 ft^{2} of roof in his home in Pittsburgh, PA (HDD = 6,000). He is considering upgrading the insulation from R16 to R22. 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.
Lt. Dave Rajakovich has 1200 ft^{2} of roof in his home in Pittsburgh, PA (HDD = 6,000). He is considering upgrading the insulation from R16 to R22. 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 R_{1} times R_{2} times efficiency divided by cost of energy times again R_{2} minus R_{1}, this is the final Rvalue minus the initial Rvalue times HDD times 24.
$$\text{Payback=}\frac{{\text{C}}_{\text{i}}\text{}\times {\text{R}}_{\text{1}}\text{}\times {\text{R}}_{\text{2}}\text{}\times \text{E}}{{\text{C}}_{\text{e}}\text{}\times \text{}\left[{\text{R}}_{\text{2}}\text{}{\text{R}}_{\text{1}}\right]\text{}\times \text{HDD}\times \text{24}}$$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 ft^{2}. To cover this area with insulation it costs \$775 so we can calculate the cost per ft . This happens to be \$0.65 per ft^{2}. That’s what we want, per ft^{2}.
Do we know R_{1}? Yes we know, R_{1} is equal to 16, R_{2} 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.
$$\begin{array}{l}\frac{\text{\$775}}{{\text{1,200ft}}^{\text{2}}}\text{=\$0}{\text{.65/ft}}^{\text{2}}\\ \\ \text{R1=16}\\ \text{R2=22}\\ \text{E=0}\text{.86}\\ \text{HDD=6,000}\end{array}$$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.
$$\begin{array}{l}{\text{C}}_{\text{e}}\text{=\$9}\text{.86/MCF=9}\text{.86/1,000,000BTUs}\\ \\ \text{=0}\text{.00000986}\end{array}$$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 R_{2} – R_{1} 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.
$$\begin{array}{l}=\frac{\text{0}\text{.65}\times \text{16}\times \text{22}\times \text{0}\text{.86}}{\text{\$0}\text{.0000986}\times \text{6}\times \text{6,000}\times \text{24}}\\ \\ \text{PaybackPeriod=23years}\end{array}$$The questions below are your chance to test and practice your understanding of the content covered in this lesson. In other words, you should be able to answer the following questions if you know the material that was just covered! If you have problems with any of the items, feel free to post your question on the unit message board so your classmates, and/or your instructor, can help you out!
For more information on topics discussed in Lesson 7, see these selected references:
The following practice exercises are designed to help you assess your readiness for the Lesson 7 quiz. You can attempt these activities as many times as you like and they will not count for your course grade.
Try them out and see how well you do!
You must complete a short quiz that covers the reading material in lesson 7. The Lesson 7 Quiz, can be found in the Lesson 7: Home Heating Basics module in Canvas. Please refer to the Calendar in Canvas for specific timeframes and due dates.
You must complete the Home Activity 4: Lighting. The activity can be found under the Home Activities [12] link in the Lessons menu. You have a limited window of time to access and complete the Home Activity, please refer to the Calendar in Canvas for specific timeframes and due dates.
Note: Your grade for the Home Activity will be posted one week after the due date.
Links
[1] http://www.noaa.gov/
[2] http://www.eeducation.psu.edu/egee102/node/2062
[3] http://www.eere.energy.gov
[4] https://www.uniongas.com/~/media/residential/savemoneyenergy/energysavingresources/wiseenergyguide/WEG_Booklet_Web_Final.pdf?la=en
[5] http://www.aceee.org/consumerguide/topfurn.htm
[6] http://www.eere.energy.gov/consumerinfo/factsheets/ea3.html
[7] http://www.nrel.gov/docs/fy00osti/27835.pdf
[8] http://web.ornl.gov/sci/roofs+walls/facts/Insulation%20Fact%20Sheet%202008.pdf
[9] http://www.naima.org/pages/resources/library/pdf/BI403.PDF
[10] http://www.naima.org/pages/resources/library/html/BI409.HTML
[11] http://www.eere.energy.gov/consumerinfo/energy_savers/insulation.html
[12] https://www.eeducation.psu.edu/egee102/2217