Question

A single room building has dimensions of 110ft x 60ft x 12ft high the walls have a R value of 20, the ceiling R value is 50 Heat loss through the floor slab on grade is 3btu/hr/ft assume 2 air change per hour. Inside temperature is 70degF and the outside temperature is 30degF Size the heating unit in terms of btu/hr. Express this in horsepower

Answer 1

Given in example

Room dimension = 110ft x 60ft x 12ft

Area of wall = 2*110*12 + 2*60*12 = 4080 ft^2

Area of roof = 110*60 = 6600 ft^2

perimeter of slab = 340 ft

Now we have to find the size the heating unit in terms of btu/h = Total heat Losses = Transmission Losses + Air Change Loss

Transmission Losses = Loss due to wall + Loss due to celling + Loss due to slab

Loss due to wall = Area/R factor = 4080/20 = 204 BTU/Hr/Degrees F

Loss due to celling = Area/R factor = 6600/50 = 132 BTU/Hr/Degrees F

Loss due to slab = (Heat loss through the floor slab on grade * perimeter)/(T inside - T outside) = (3*(110+60)*2)/40 = 1020/40 = 25.5 BTU/Hr/Degrees F

Transmission Losses = 361.5 BTU/Hr/Degrees F

Now we will find air change loss

Air Infiltration Heat Loss (BTU/Hr/Degrees F)= Room Volume X AIR CHANGES PER HOUR X .O18.

where = 0.018 is a coefficient its depends on conditions of room ( very tight house, you make this number very low)

Air Infiltration Heat Loss = 79200 * 2 * 0.018 = 2851.2 BTU/Hr/Degrees F

Total loss = 2851.2 + 361.5 = 3212.7 BTU/Hr/Degrees F

Total Heating Requirement = Total loss * change in temperature = 3212.7 * 40 = 128508 BTU/Hr

as we know that 1 Watt = 3.412142 BTU/hr

So 128508 BTU/Hr = 128508/3.412142 = 37661.97 Watt