Thermal engineering · Heat transfer · conduction

Multilayer wall calculator — thermal resistance

Calculate steady conduction through 2 to 4 plane layers in series and see each layer's resistance and temperature drop.

Created v1.0

By Thibaut Grzelak, Mechanical Analysis Engineer

Heat transfer · conduction
01Calculation inputs
k₁k₂…kₙTₕT𝒸q″R″tot = Σ(eᵢ/kᵢ)
R″tot = Σ(eᵢ/kᵢ)Calculate steady conduction through 2 to 4 plane layers in series and see each layer's resistance and temperature drop.
1
W/(m·K)
2
W/(m·K)
3
W/(m·K)
02

Results

ΣR″
Heat flux q″
—W/m²

Heat-transfer rate per unit wall area through the series layers.

Total area resistance R″tot
—m²·K/W

Sum of the active layers' conductive resistances eᵢ/kᵢ.

Total thermal resistance Rtot
—K/W

Total conductive resistance of the wall for the entered area.

Area conductance U
—W/(m²·K)

Inverse of the total conductive area resistance, excluding surface films.

R″tot = —

Steady-state, one-dimensional conduction through plane layers in series. Each layer is homogeneous, has constant area and constant thermal conductivity.

01Formulas and symbols

Formulas used

R″ᵢeᵢ / kᵢArea thermal resistance of layer i
R″totΣ(eᵢ / kᵢ)Sum of the layer resistances in series
U1 / R″totConductive area conductance of the wall
q″ΔT / R″totHeat flux through all layers
Q̇A · q″Total heat-transfer rate
eᵢe₁ … eₙm
kᵢk₁ … kₙW/(m·K)
AArea Am²
ΔTTemperature difference ΔTK
R″ᵢArea thermal resistance of layer im²·K/W
q″Heat flux q″W/m²
Q̇Heat-transfer rate Q̇W
UArea conductance UW/(m²·K)
#ekR″ᵢΔTᵢ
02Assumptions and limits

Scope of validity

  • Steady-state, one-dimensional conduction through plane layers in series.
  • Each layer is homogeneous, has constant area and constant thermal conductivity.
  • Select 2 to 4 layers; every selected layer must have strictly positive thickness.
  • No contact resistance, surface convection or radiation is included.
  • Thermal bridges, parallel paths, moisture and cylindrical geometries are outside this model.
  • The displayed U value represents conduction through the entered layers only and is not a regulatory whole-assembly U-value.
03Validation example

Reference numerical case

  1. Reference case: A = 10 m² and ΔT = 20 K with three active layers.
  2. e₁/k₁ = 0.10/0.04 = 2.50; e₂/k₂ = 0.05/0.50 = 0.10; e₃/k₃ = 0.10/1.00 = 0.10 m²·K/W.
  3. R″tot = 2.70 m²·K/W; U = 0.37037 W/(m²·K).
  4. q″ = 7.4074 W/m² and Q̇ = 74.074 W for A = 10 m².
04References

Technical references

  1. MIT OpenCourseWare — Thermal Resistance Circuits: composite slab with resistances in series.
  2. NIST Special Publication 811 — SI units and temperature intervals.