Published on *EGEE 102: Energy Conservation and Environmental Protection* (https://www.e-education.psu.edu/egee102)

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.

- Heat loss through ceilings.
- Heat loss through windows.
- Heat loss through doors.
- Heat loss through frame walls.
- Heat loss through cracks in walls, windows, and doors.
- Heat loss through basement walls.
- Heat loss through basement floor.

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:

- Conduction
- Convection
- 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.

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 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.

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:

- Heat escaping through the roof of a house
- A hot stove burner
- Boiling water
- A torch halogen lamp producing light and heat
- A door hanging wide open, letting in cold air
- A fire creating heat
- Heat escaping through a wall
- A mirror reflecting sunlight
- Heat escaping through a window
- Heat escaping through a chimney

**Answers:**

- Conduction
- Radiation
- Convection
- Radiation
- Convection
- Radiation
- Conduction
- Radiation
- Conduction
- Radiation

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

- 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).
- The second way to reduce the energy consumption is by improving the efficiency of the furnace that provides the heat.

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.

- Which house loses more heat?
- House A
- House B

- Why do you think this house loses more heat?
- More people in it
- More appliances and lights are used
- Larger size/more area

**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.

- Which house loses more heat?
- House A
- House B

- Why do you think this house loses more heat?
- People skiing need more heat to keep warm
- Snow on the roof is good insulation
- Outside temperature

**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 R-value of house B is .63 and the R-value of house A is unknown.

- Which house loses more heat?
- House A
- House B

- Why do you think this house loses more heat?
- Less insulation
- It's only one color
- It's thicker

**Example 1:**

- B: House B
- C: Larger size/more area

**Example 2:**

- B: House B
- C: Outside temperature

**Example 3:**

- B: House B
- A: Less insulation

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.

- Insulation is rated in terms of thermal resistance, called

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.

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

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

To calculate HDD:

- Determine the
**base temperature**or inside temperature: $${\text{T}}_{\text{base}}\text{=usually65}\xb0\text{F}$$ - Find the day's
**average outside temperature**using this formula: $${\text{T}}_{\text{a}}\text{=averageoutsidetemperature=}\frac{{\text{T}}_{\text{max}}+{\text{T}}_{\text{min}}}{\text{2}}$$ - Use the
**HDD formula**to solve:

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

**Note:** If the T_{a} is equal to or above 65 ºF, there are no heating degree days for that 24-hour 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:**

- What is the average temperature?

Hint: T_{a}= (max temperature + minimum temperature) divided by 2. - What is the heating degree days?

Hint: HDD = T_{base}- T_{a}

**Answers:**

- 50 degrees
- 15 degrees

**Calculate the HDD for one day when the average outside temperature is 13º F.**

**Calculate the HDD for one day when the average outside temperature is 2º C.**

**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:

$$\begin{array}{c}\text{SeasonalHDD=}\left({\text{(T}}_{\text{b}}{\text{-T}}_{\text{a}}\text{)}\times \text{No}\text{.DaysinMonth1}\right)\\ \text{+}\left({\text{(T}}_{\text{b}}{\text{-T}}_{\text{a}}\text{)}\times \text{No}\text{.DaysinMonth2}\right)\\ +\left({\text{(T}}_{\text{b}}{\text{-T}}_{\text{a}}\text{)}\times \text{No}\text{.DaysinMonth3}\right)\end{array}$$

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:

$$\text{HeatLoss}\left(\frac{\text{BTUs}}{\text{h}}\right)\text{=}\frac{\text{Area}\left({\text{ft}}^{\text{2}}\right)\times \text{TemperatureDifference}\left(\xb0\text{F}\right)}{\text{R-Value}\left(\frac{{\text{ft}}^{\text{2}}\text{}\xb0\text{Fh}}{\text{BTUs}}\right)}$$

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.

**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.

$$Heat\text{}Loss\left(\frac{BTUs}{h}\right)=\frac{Area(f{t}^{2})\times Temperature\text{}Difference{(}^{o}F)}{R\text{}value\left(\frac{f{t}^{2}{}^{\text{}o}F\text{}h}{BTUs}\right)}\text{}\times \text{}24hr/day$$

**What is the hourly and daily heat loss of a 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?**

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:

$$Daily\text{}Heat\text{}Loss\text{}per\text{}day=\frac{1477\text{}BTUs}{h}\text{}x\frac{24\text{}h}{day}=35,448\text{}\frac{BTUs}{day}$$
$$Heat\text{}Loss\left(\frac{BTUs}{h}\right)=\frac{Area(f{t}^{2})\times Temperature\text{}Difference{(}^{o}F)}{Rvalue\left(\frac{f{t}^{2}{}^{\text{}o}F\text{}h}{BTU}\right)}$$

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:

- Calculate heat loss for each day.
- Add the heat loss for all days in a year that needed heating.
- Leave the 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:

$$Heat\text{}Loss\text{}\left(\frac{BTUs}{h}\right)=\frac{Area\text{}(f{t}^{2})\times Temperature\text{}Difference\text{}{(}^{o}F)}{Rvalue\left(\frac{f{t}^{2}{\text{}}^{o}F\text{}h}{BTUs}\right)}\text{}\times \frac{24\text{}h}{day}\text{}$$

Thus, theoretically we would need to perform this calculation for every day of the 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:

Since the area (480 ft2), R-value $\frac{f{t}^{2}\text{}\xb0F\text{}Btu}{h}$ , and 24 h in a day are common for ALL heating days, we can bring those out and rewrite the equation as:

$$\begin{array}{l}=\frac{480\text{}f{t}^{2\text{}}x}{13\frac{f{t}^{2}{\text{}}^{o}F\text{}h}{BTU}}\times 24\frac{h}{day}\left\{{\left(65-35\right)}^{o}F+{\left(65-32\right)}^{o}F+{\left(65-28\right)}^{o}F\mathrm{...}so\text{}on\text{}for\text{}all\text{}heating\text{}days\right\}\\ \\ =\frac{480\text{}f{t}^{2\text{}}x}{13\frac{f{t}^{2}{\text{}}^{o}F\text{}h}{BTU}}\times 24\frac{h}{day}\left\{{30}^{o}F+{33}^{o}F+{37}^{o}F+so\text{}on\text{}for\text{}all\text{}heating\text{}days\right\}\\ \\ Where\left\{{30}^{o}F+{33}^{o}F+{37}^{o}F+so\text{}on\text{}for\text{}all\text{}heating\text{}days\right\}is\text{}the\text{}sum\text{}of\text{}Heating\text{}Degree\text{}Days\end{array}$$

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 150-day 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?

**Links**

[1] http://www.noaa.gov/