Engineering
Fourier Heat Conduction formula: composite wall R-Value, U-Value & heat loss (W)
Calculate SI area-normalized resistance, transmittance and signed steady heat flow through a planar layered wall.
Why U-Values NEVER add in series—sum layer R-Values (L / k) first!
By Jean-Baptiste Joseph Fourier's Law of Heat Conduction, each solid wall layer of thickness L (m) and thermal conductivity k (W/(m*K)) acts like an electrical resistor in series with thermal resistance R_i = L_i / k_i (m^2*K/W). Adding indoor and outdoor boundary air-film resistances gives total thermal resistance R_total = SUM(D2:D5).
Overall thermal transmittance (U-Value) is the reciprocal U = 1 / R_total (W/(m^2*K)), and steady-state conductive heat transfer across wall area A and temperature drop DeltaT = T_in - T_out is Q = U * A * DeltaT (Watts)!
Assume one-dimensional steady flow through planar layers of equal area with constant conductivity, no heat storage, no internal generation and no thermal bridges or parallel paths. The 0.15 air-film resistance is a supplied combined boundary assumption, not a universal value. These are SI area-normalized resistances, not US customary R-values.
| Envelope Layer (A) | Thickness L m (B) | Conductivity k (C) | Thermal Resistance R = L/k (D) |
|---|---|---|---|
| Gypsum Drywall | 0.016 | 0.16 | 0.10 m^2*K/W |
| Mineral Wool Batt | 0.144 | 0.04 | 3.60 m^2*K/W (90% of insulation!) |
| Brick Veneer | 0.120 | 0.80 | 0.15 m^2*K/W |
| Air Films (In + Out) | — | — | 0.15 -> R_tot = 4.00 | U = 0.25 | Q = 250 W |
Overall U-Value = 1 / SUM(R_layers) (W/(m^2*K))
Summing layer resistances D2:D5 (4.00 m^2*K/W) yields U = 0.25 W/(m^2*K).
=IF(OR(COUNT(B2:C4)<>6,MIN(B2:B4)<0,MIN(C2:C4)<=0,NOT(ISNUMBER(D5)),D5<0),"Invalid inputs",IF(OR(COUNT(D2:D5)<>4,SUM(D2:D5)<=0),"Invalid inputs",ROUND(1 / SUM(D2:D5), 4)))Steady-state conductive heat loss Q = (A * (T_in - T_out)) / SUM(R)
Across Area F2 = 40 m^2 with Indoor F3 = 21 C and Outdoor F4 = -4 C (DeltaT = 25 K), heat loss is 250 W—versus 2,500 W without the mineral wool layer!
Positive power means outward heat loss; negative power means inward heat gain. This is thermal power in watts, not energy in watt-hours. Setting insulation thickness to zero removes that layer mathematically; it does not model an actual replacement air gap.
=IF(OR(COUNT(B2:C4)<>6,MIN(B2:B4)<0,MIN(C2:C4)<=0,NOT(ISNUMBER(D5)),D5<0),"Invalid inputs",IF(OR(COUNT(D2:D5)<>4,SUM(D2:D5)<=0,COUNT(F2:F4)<>3,F2<0),"Invalid inputs",ROUND((F2 * (F3 - F4)) / SUM(D2:D5), 2)))Try the example
Use the original values or change them to test your understanding.