Designing#Your#Energy#Efficient#Home# !Name ANSWERKEY · AK#EnergySmart:#Designing!Your!Energy!Efficient!House!Part!1:!The!Heat!Loss!Equation! 2! RECOMMENDED!R>VALUES! Alaska!is!a!large!state!withmany!different!climates

AK EnergySmart: Designing Your Energy Efficient House Part 1: The Heat Loss Equation

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Designing Your Energy Efficient Home

Name______ANSWER KEY___________________________

When building a house, a builder must determine if he or she is using the correct materials and building techniques for the location. This worksheet will walk you through the types of calculations that builders must make in order to construct a house that protects the occupants from the outside climate.

R-‐VALUE

The R-‐value of an insulating material tells you how well that material insulates (or its ability to resist heat from going through; the “R” stands for “resistance”). The larger the R-‐value, the better the material insulates. For example:

The R-‐value of wood is 1.25 per 1-‐inch of thickness while sawdust has an R-‐value of 2.2 per 1-‐inch of thickness.

1. Which is a better insulator—wood or sawdust? _____sawdust_______________________

Why? _________ Sawdust traps air in spaces; these pockets of air are better at insulating than wood.________

Total R-‐value = (R-‐value per inch) X (thickness of material)

2. What is the R-‐value of a 5 inch thick wood wall? _______5 x 1.25 = 6.25________

3. What is the R-‐value of sawdust insulation in a wall that is 5 inches thick? __5 x 2.2 = 11


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RECOMMENDED R-‐VALUES

Alaska is a large state with many different climates. The Building Energy Efficiency Standards (BEES) give minimum R-‐values for different components (ceiling, walls, floors, etc.) for builders to meet when constructing a residential or commercial building. These minimum R-‐values depend on geographic location and Alaska is divided into 4 different zones (See Figure 1).

Figure 1. Climate zones in Alaska

4. The following table lists the BEES’s minimum R-‐value for walls

Climate Zone Region of Alaska Minimum wall R-‐value Zone 6

Southeast

21

Zone 7

Southcentral and Aleutians

21

Zone 8

Interior, Southwestern, and Western

30

Zone 9

North Slope

35


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Use the same equation to calculate insulation thickness for the different climate zones in Alaska using wood or sawdust.

Total R-‐value = (R-‐value per inch) X (thickness of material)

A.1) How many inches thick would a log wood wall need to be to meet this minimum requirement in

Southeast Alaska? 21 ÷ 1.25 = 16.8 inches

A.2) Southcentral and Aleutians? 21 ÷ 1.25 = 16.8 inches

A.3) Interior, Southwestern, and Western? 30 ÷ 1.25 = 24 inches

A.4) North Slope? 35 ÷ 1.25 = 28 inches

B.1) How many inches thick would sawdust insulation need to be to meet this minimum requirement in

Southeast Alaska? 21 ÷ 2.2 = 9.5 inches

B.2) Southcentral and Aleutians? 21 ÷ 2.2 = 9.5 inches

B.3) Interior, Southwestern, and Western? 21 ÷ 2.2 = 13.6 inches

B.4) North Slope? 21 ÷ 2.2 = 15.9 inches

5. Fiberglass is a popular insulation material and it comes in different thicknesses and R-‐values. If a 12-‐inch-‐thick batt of fiberglass typically has an R-‐value of 38, what would be the R-‐value per inch for this

piece of fiberglass? _____38 ÷ 12 = 3.2 per inch____

U FACTOR

Glass is not a good insulator, so heat is easily lost from a house through windows. Instead of measuring their R-‐values (the ability to resist heat transfer), windows are measured by their U Factor. The U Factor tells us how well heat is transferred through the material. Therefore, the U Factor is the inverse of the R-‐value.

! = !! or ! = !

!

Because of this inverse relationship, a high R-‐value equals a low U Factor.


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Windows come in different styles with different numbers of panes of glass. Additionally, to further decrease a window’s U Factor the glass can be treated with a “low emissive” or “low e” coating to reduce heat loss. Or, the space between panes of glass can be filled with a gas such as argon which can also reduce heat loss by preventing heat from moving through the window. Improving the insulation on the window frame will also decrease heat loss through the window.

6. Calculate the R-‐value for different types of windows based on their U Factor

Type of window Approximate U Factor Calculated R-‐value Single paned

1

1

Double paned

0.5

2

Triple paned

.33

3

Triple paned with low e coating and insulated frame

0.16

6.25

APPLYING R-‐VALUES AND U FACTORS TO CALCULATE HEAT LOSS

Knowing the R-‐value of the walls, doors, and roof and the U Factor of the windows, we can calculate the amount of the heat being lost from a house. The British Thermal Unit (BTU) is a common way to measure heat. A BTU is approximately the amount of heat released by a single match.


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7. Calculate the total surface areas of the house—you can use graph paper to draw out each side of the house to help calculate the different surface areas. Assume that the two walls that you can’t see are identical to the two that you can see.

Door—total surface area of two doors: a front door and a back door (each door is 3 ft. wide by 7 ft. tall)

Sq. ft. 2 ( 3 x 7 ) = 42

Windows—total surface area of all windows (eight windows: two 3 ft. x 3 ft.; six 3 ft. x 4 ft.

Sq. ft. 2 ( 3 x 3 ) + 6 ( 3 x 4 ) = 90

Exterior walls—total surface area (minus the doors and windows; use the formula for the area of a triangle to calculate the exterior wall sections under the pitched roof: A = ½ x b x h)

Sq. ft. [2 ( 24 x 8 ) + 2 ( 32 x 8 ) + 2 ( ½ x 24 x 7)] – [42 + 90] = 932

Roof—total surface area

Sq. ft. 2 ( 14 x 32 ) = 896


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The following equation determines how much heat loss occurs from each type of surface (windows, doors, walls, and roof)

Δ T is the temperature difference in Fahrenheit between inside the house and outside the house.

For example, if it is 65°F inside the house and 25°F outside

Δ T = 65°F -‐ 25°F = 40°F

8. Calculate Δ T for the following situations

Inside temperature Outside temperature Δ T 70°F 50°F 20°F 65°F 32°F 33°F 65°F 0°F 65°F 60°F -‐40°F 100°F

Once we know the R-‐value, surface area, and Δ T, we can calculate heat loss. For example, a wall that is

160 sq. ft. with an R-‐value of 30 and Δ T of 45°F would have a heat loss of

9. Using the total surface areas calculated in #7, determine the total heat loss for a house when the inside temperature is 70 and the outside temperature is 20. The surfaces have the following R-‐values (if given the U Factor, then calculate the R-‐Value).

Doors R-‐10 Window R-‐4.5 U-‐0.22

Exterior walls R-‐24 Roof R-‐38

!"#$ !"## !"# ℎ!"# = !"#$%&' !"#! X ∆!

!-‐!"#$%

!"#$ !"## !"# ℎ!"# = 160 !" !" X 45°F

!-‐30 = 240 !"#$ !"# ℎ!"#


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Doors

Windows

Exterior walls

Roof

___!""#_____!"#$% !"#$ !"# ℎ!"# !"#$ !"" !"#$%&'!

__!"___ !" !" X __!"___°F!-‐_!"___ = _____!"#_____ !"#$ !"# ℎ!"#

__!"___ !" !" X __!"___°F!-‐__!. !__ = ___!"""_______ !"#$ !"# ℎ!"#

___!"#__ !" !" X __!"___°F!-‐__!"__ = ____!"#$______ !"#$ !"# ℎ!"#

__!"#___ !" !" X ___!"__°F!-‐__!"__ = ____!!"#______ !"#$ !"# ℎ!"#


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10. From what surface is most of the heat lost? If you could pick one surface to improve the R-‐value in order to prevent heat loss, which surface would it be and why?

The exterior walls lose the most heat because of their large surface are, so this would be the best surface to improve the R-‐value on.

11. In order to lower your Δ T and reduce heat loss, you turn down the thermostat to 60°F (and put on

a sweater). What is the new amount of heat loss?

Doors

Windows

Exterior walls

Roof

___!"__ !" !" X __!"___°F!-‐__!"__ = ___!"#_______ !"#$ !"# ℎ!"#

__!"___ !" !" X __!"___°F!-‐_!. !___ = ___!""_______ !"#$ !"# ℎ!"#

___!"#__ !" !" X __!"___°F!-‐__!"__ = ___!""#_______ !"#$ !"# ℎ!"#

___!"#__ !" !" X _!"____°F!-‐__!"__ = ___!"#______ !"#$ !"# ℎ!"#


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___!"#"_____!"#$% !"#$ !"# ℎ!"# !"#$ !"" !"#$%&'!

Calculate the amount of energy that you conserved by turning down the thermostat.

CALCULATED VERSUS ACTUAL VALUES

Measurements were taken and you discovered that heat loss was much higher than what you calculated it would be. You double checked your math and it was all correct—how could this be?

After talking to a friend, you discover that your walls are not a solid R-‐24 because of the wooden framing studs that provide structure to the wall but also interrupt the fiber glass insulation:

Figure 3: A 3-‐dimensional look into a wall

Since wood has a lower R-‐value than fiber glass, more heat is lost through the wall where the studs are located.

____!""#_____!"#$ !"#$ !" !" !!__!"#"____!"#$ !"#$ !" !" !____!"#_____!"#$ !"#$%&'%(


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12. You calculate that 11% of the exterior walls are wooden studs with an R-‐8. What is the average R-‐value of you exterior walls?

A)

B) Δ T of 60°F inside temperature and 20°F outside temperature = _____40_______________°F

C) Heat loss of stud section

Heat loss of fiberglass section

D) Total heat loss for exterior walls

____!"#. !______ !"#$ !"# ℎ!"# (!"#$) + ____!"#$. !______ !"#$ !"# ℎ!"#(!"#$%&'())) =____!"#$______ !"!#$ !"#$ !"#$ !"# ℎ!"#

E) Calculate average R-‐Value for the exterior walls

102.5

829.5

__!"#. !___ !" !" X __!"___°F!-‐8 = ___!"#. !____ !"#$ !"# ℎ!"#

__!"#. !___ !" !" X __!"___°F!-‐24 = ____!"#$. !___ !"#$ !"# ℎ!"#

___!"#______!"!#$ !" !" !" !"#!$%&$ !"##$ ! __!"___°F__!"#$_______!"!#$ !"#$ !"#$ !"# ℎ!"! = !-‐___!". !____

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Designing#Your#Energy#Efficient#Home# !Name ANSWERKEY · AK#EnergySmart:#Designing!Your!Energy!Efficient!House!Part!1:!The!Heat!Loss!Equation! 2! RECOMMENDED!R>VALUES! Alaska!is!a!large!state!withmany!different!climates