A homeowner asks an energy consultant to find the heat loss

A homeowner asks an energy consultant to find the heat loss from his home. On one wall, measuring 15 ft by 9 ft (without windows), the consultant measures a temperature of 66°F on the inside surface of the wall and 18°F on the outside surface. The wall has a thermal resistance of 0.35 ft2 -°F-hr/BTU (air films included). What is the rate of heat loss through the wall?

Solution

Given that,

The height of the wall = 15ft

The width of the wall = 9ft

The temperature inside the wall surface = 66 degree.F

The temperature outside the wall surface = 18 deg.F

The thermal resistance, R= 0.35 ft^2-hr/BTU

The rate of the heat loss = [Surface Area*(Temperature A-Temperature B)]/Thermal Resistance

Surface Area = height*width

=15ft * 9ft

=135 square ft

Heat Loss = [135 * (66-18)]/0.35

= (135*48)/0.35

= 18,514.28 BTU/HR