Calculating Wall Heat Loss

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.

Equations

The heat loss in an hour equation

$H e a t L o s s (\frac{B T U s}{h}) = \frac{A r e a (f t^{2}) \times T e m p . D i f f e r e n c e (^{o} F)}{R - V a l u e \frac{f t^{2}^{o} F h}{B T U}}$

The heat loss in a day equation

$H e a t L o s s (\frac{B T U s}{h}) = \frac{A r e a (f t^{2}) \times T e m p . D i f f e r e n c e (^{o} F)}{R - V a l u e \frac{f t^{2}^{o} F h}{B T U}} \times 24$

The heat loss in full heating season

$S e a s o n a l H e a t L o s s = \frac{A r e a}{R - V a l u e} \times 24 \times H D D$

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

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.

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 ft², HDD = 6,100, and the composite R- value of the wall needs to be calculated.

Materials and their R-Value
Material	R-Value
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 R-Value of the walls	20.40

Heat  Loss from Walls = \frac{1,920 {ft}^{2} \times 6,100 ° F - days \times \frac{24 h}{day}}{20 .4 \frac{{ft}^{2} ° F h}{Btu}} = 13 .78 MMBtu

Heat Loss from Windows = \frac{580 {ft}^{2} \times 6,100 ° F - days \times \frac{24 h}{day}}{1 \frac{{ft}^{2} ° F h}{Btu}} = 84.91 MMBtu

Heat Loss from Roof = \frac{2,750 {ft}^{2} \times 6,100 ° F - days \times \frac{24 h}{day}}{22 \frac{{ft}^{2} ° F h}{Btu}} = 18.30 MMBtu

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.

Equations

The heat loss in an hour equation

The heat loss in a day equation

The heat loss in full heating season

Example

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