Heating and cooling load calculation
Domain 3 tests heating and cooling load calculation: the inputs, the difference between load and equipment capacity, and the ACCA/ASHRAE workflow.
What the calculation includes
A load calculation accounts for design outdoor and indoor conditions, envelope U-values and solar heat gain, orientation, infiltration, ventilation, internal gains from people and appliances, and duct gain or loss. Heating load is mainly conduction and infiltration. Cooling load has sensible and latent components and is not simply outdoor dry-bulb minus indoor temperature. Use Manual J or an equivalent ASHRAE method, then select equipment with Manual S.
Worked example
A wall has U = 0.10 Btu/h·ft²·°F, area 400 ft², and design ΔT = 30°F: 0.10 × 400 × 30 = 1,200 Btu/h. A window has U = 0.50, area 80 ft², same ΔT: 0.50 × 80 × 30 = 1,200 Btu/h. Add internal sensible gain 2,400 Btu/h and infiltration 1,000 Btu/h: sensible total = 5,800 Btu/h. Add latent gain 800 Btu/h: total cooling load = 6,600 Btu/h. Equipment is selected at design conditions, not by the old unit's tonnage.
Traps that cost marks
Do not use a square-foot rule, size from the existing unit, ignore duct losses or ventilation, use only dry-bulb temperature for cooling, or forget altitude and envelope changes. Oversizing causes short cycling, poor humidity control, and higher energy use.