Composite Wall Heat Flux Calculator
Series thermal resistance of a multi-layer wall: resistance per layer, total, heat flux for an overall temperature difference, and the share of the temperature drop across each layer. The fields start with the DOE worked example; replace any value, and leave a layer row empty to skip it.
Model: steady one-dimensional conduction through flat layers in perfect contact, with no surface films and no contact resistance. Ri = (thickness ÷ 12) ÷ k, Rtotal = ΣRi, flux = ΔT ÷ Rtotal, ΔTi = flux × Ri, heat rate = flux × area. Method and worked example: DOE Fundamentals Handbook, Thermodynamics, Heat Transfer, and Fluid Flow, Volume 2 of 3, Module 2 Heat Transfer, DOE-HDBK-1012-92, Rev. 0, Conduction Heat Transfer, example on HT-02 pp. 10–11 (energy.gov). For preliminary estimation only, not a substitute for review by a qualified or licensed engineer.
Where the temperature actually drops in a layered wall
A wall assembly is rarely one material. A metal sheet, a thin coating or air-tight liner, and a layer of insulation sit in series, and the same heat flux passes through every one of them. What differs is the resistance of each layer, and the temperature drop across a layer is simply that flux times its resistance. For anyone checking a construction, the useful question is not only how much heat gets through, but which layer is doing the work.
This calculator is built for HVAC and building envelope engineers who need that picture before sizing equipment or checking a build-up. It computes the series thermal resistance of a multi-layer wall: resistance per layer, the total, the heat flux for an overall temperature difference, and the share of the temperature drop across each layer. Resistances are printed to enough significant digits that a thin, highly conductive layer never collapses to a misleading zero.
The DOE worked example shows how lopsided the split can be. A 1 in metal layer with k = 240 takes only 0.022 F of a 500 F overall drop, a 0.125 in layer with k = 0.048 takes 13.92 F, and a 2 in layer with k = 0.022 takes 486.05 F. The thin, low-conductivity layer carries nearly all of the resistance, so it is where the temperature falls.
How to use the calculator
- In the first layer row, type the layer thickness in inches and its conductivity k in Btu/hr-ft-F.
- Repeat for each additional layer, up to five; leave a row blank to skip it.
- Enter the overall temperature difference across the wall in F.
- Optionally enter the wall area in ft² to also get the heat rate.
- Read each layer's R and temperature drop, the total R and the heat flux from the results as they update.
FAQ
How is heat flux through a composite wall calculated?
Each layer's resistance is R_i = (thickness in inches ÷ 12) ÷ k, the layer resistances add in series to R_total, and the flux is the overall temperature difference divided by R_total, in Btu/hr-ft^2.
Which units does this calculator use?
US customary: thickness in inches, conductivity in Btu/hr-ft-F, temperature difference in F, area in ft^2, resistance in hr-ft^2-F/Btu and flux in Btu/hr-ft^2.
How much of the temperature drop falls across the insulation?
The drop across a layer is flux × R_i. In the DOE worked example, a 2 in layer with k = 0.022 takes 486.05 F of a 500 F overall drop, while the 0.125 in layer at k = 0.048 takes 13.92 F and the 1 in metal layer at k = 240 takes only 0.022 F.
Does it include surface film coefficients or contact resistance?
No — it models steady one-dimensional conduction through flat layers in perfect contact, with no surface films and no contact resistance, so the result is for preliminary estimation only and not a substitute for review by a qualified or licensed engineer.