7.6: Insulation and Home Heating Fuels (II)

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Calculating Wall Heat Loss

Heat loss from the surface of a wall can be calculated by using equations 7.6.1, 7.6.2, and/or 7.6.3.

Equations

Hourly Heat Loss

\[ Hourly \, Heat \, Loss \, (\dfrac{BTUs}{h}) = \cfrac{Area \, (ft^2) * Temperature \, Difference \, (°F)}{R-Value \, (\cfrac{ft^2 \, °Fh}{BTUs})} \]

Daily Heat Loss

\[ Daily \, Heat \, Loss \, (\dfrac{BTUs}{day}) = \cfrac{Area \, (ft^2) * Temperature \, Difference \, (°F)}{R-Value \, (\cfrac{ft^2 \, °Fh}{BTUs})} * 24 \, \dfrac{h}{day} \]

Full Heating Season Heat Loss

\[ Seasonal \, Heat \, Loss = \cfrac{Area \, (ft^2)}{R-Value \, (\cfrac{ft^2 \, °Fh}{BTUs})} * 24 \, \dfrac{h}{day} * HDD \]

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

Example Problems

Example 1

A house in Denver, CO has 580 ft² of windows (R = 1), 1920 ft² of walls and 2750 ft² of roof (R = 22). The walls are made up of wood siding (R = 0.81), 0.75” plywood, 3.5” 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.

Answer

The heating requirement for the house is equal to the 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 ft², HDD = 6,100, and the composite R-value of the wall needs to be calculated.

Materials and their R-values

Material	Total R-value
Wood siding	0.81
3/4 inch plywood	0.94
3.5 inches of fiberglass (R-value = 3.7 per inch)	12.95
1 inch of polyurethane	5.25
1/2 inch of Gypsum board	0.45
Total R-value of the walls	20.40

\[ Heat \, Loss \, from \, Walls = \cfrac{1920 ft^2 * 6100°F}{20.40 \, \cfrac{ft^2 °Fh}{BTUs}} * 24 \, \dfrac{h}{day} * 1 \, day = 13.78 \, MMBTUs \nonumber\]

\[ Heat \, Loss \, from \, Windows = \cfrac{580 ft^2 * 6100°F}{1 \, \cfrac{ft^2 °Fh}{BTUs}} * 24 \, \dfrac{h}{day} * 1 \, day = 84.91 \, MMBTUs \nonumber\]

\[ Heat \, Loss \, from \, Roof = \cfrac{2750 ft^2 * 6100°F}{22 \, \cfrac{ft^2 °Fh}{BTUs}} * 24 \, \dfrac{h}{day} * 1 \, day = 18.30 \, MMBTUs \nonumber\]

Total heat loss from the house = 13.78 + 84.91 + 18.30 =116.99 MMBTUs in a year

Therefore, the heating requirement is 116.99 MMBTUs per year.

The following video also provides an explanation for the problem.

Example 2

A single-story house in Anchorage, AK (HDD = 11,000) is 50 ft by 70 ft with an 8 ft high ceiling. There are six windows (R = 1) of identical size, 4 ft wide by 6 ft high. The roof is insulated to R-30. The walls consist of a layer of wood siding (R = 0.81), 2” of polyurethane board (R = 6.25 per inch), 4” of fiberglass (R = 3.70 per inch), and a layer of drywall (R = 0.45). Calculate the heat loss through the house (not counting the floor) for the season.

Hint: 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.

Answer: The following video provides the solution.