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.
The heat loss in an hour equation
The heat loss in a day equation
The heat loss in full heating season
The heat loss from walls, windows, roof, and flooring should be calculated separately, because of different R-Values for each of these surfaces. If the R-value of walls and the roof is the same, the sum of the areas of the walls and the roof can be used with a single R-value.
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.
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 ft2, HDD = 6,100, and the composite R- value of the wall needs to be calculated.
|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 R-Value 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.