HEAT LOSS

HEAT LOSS

Q = U × A × ΔT (per surface)
ft²
ft²
ft²
ft²
°F
°F
RESULT
FILL IN ABOVE
Conductive loss only. Add infiltration + duct losses for full Manual J. Estimate only — verify with a licensed HVAC contractor running full ACCA Manual J before equipment purchase.

About this calculator

This heat loss calculator computes the conductive heat loss through a building envelope using Q = U × A × ΔT per surface, where U-value is the inverse of R-value. Enter the area and U-value for walls, windows, ceiling/roof, and floor, then the indoor design temp and outdoor 99% design temperature for your location. Add infiltration losses (typical 0.35 ACH × volume × 0.018) for a full envelope figure. The result is BTU/hr at design conditions — multiply by hours of heating season for annual energy. ESTIMATE ONLY — full Manual J adds duct losses, internal gains, and solar to get installed equipment size.

Common questions

What is the 99% design temp?
The outdoor temperature your area is colder than only 1% of the heating season hours. ACCA Manual J Table 1 lists it for every US city. Sizing equipment to the absolute lowest temperature on record means oversized equipment that short-cycles 99% of the year. Examples: Chicago -3°F, Boston 6°F, NYC 11°F, Atlanta 23°F, Phoenix 31°F.
Are R-value and U-value the same thing?
They are inverses. R-value is resistance to heat flow (insulation rating). U-value is the rate of heat flow (window rating). U = 1/R. A wall at R-19 has a U-value of 0.053. Windows are sold by U-value because their assemblies (glass + frame + spacer) have non-uniform heat flow — single R-value doesn't describe them well.
Why does conductive loss not equal Manual J?
Manual J adds three pieces this calculator doesn't: (1) infiltration loss — air leaking through the envelope, typically 0.35 ACH × volume × 0.018 BTU per CFH-°F, (2) duct losses — ducts in unconditioned attics or crawlspaces lose 15-25% of system output, (3) internal gains — bodies, lights, appliances offsetting load. Add 25-40% for full envelope; reduce by internal gains for net heating load.