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Thermal Conductivity Calculator

Physics

Or start from a worked scenario:

Layers run inside → outside. Reversing the order reverses every interface temperature.

Layer 1
Layer 2
Layer 3

3 of 8 layers

ISO 6946 convention, editable.
Both systems are always printed; this only sets the order.
Set to zero to hide the seasonal panel.
Values above 1 are heat-pump COPs.

Wall heat loss

Heat rate Q
90.944 W
Heat flux q″
9.094 W/m²
Temperature difference
26.000 K
Direction
Inside → outside

Resistance and transmittance

QuantitySIUS
Total resistance RR-2.859 (SI) m²·K/WR-16.234 (US) h·ft²·°F/BTU
Transmittance U0.350 W/(m²·K)0.062 BTU/(h·ft²·°F)
Reference area10.000 m²107.639 ft²

The two systems differ by exactly 5.678263×, which is why a US "R-13" batt is only R-2.289 in SI units.

Surface films contribute 5.946366 % of the total resistance.

Layer breakdown

LayerR (m²·K/W)ΔT (K)% of ΔTCold face (°C)

Inside surface film (R_si)

convention
0.1301.1824.547 %19.818

Plasterboard

0.0500.4551.749 %19.363

Mineral wool

2.50022.73687.447 %-3.373

Brick

0.1391.2634.858 %-4.636

Outside surface film (R_se)

convention
0.0400.3641.399 %-5.000

Condensation check

Interfaces below the dew point
Mineral wool at -3.373 °C; Brick at -4.636 °C; Outside surface film (R_se) at -5.000 °C. A vapour control layer belongs on the warm side of the insulation.

Heat rate in other units

UnitValue
W90.944
kW0.091
BTU/h310.315
hp0.122

Seasonal fabric loss

Heat through the fabric
209.872 kWh
Delivered energy
233.191 kWh
Running cost
65.29
Thermal diffusivity of Plasterboard (gypsum)
α = k / (ρ · c) = 2.8670e-7 m²/s. Conductivity says how much heat flows; diffusivity says how fast a temperature front travels, and it is why a metal bench feels colder than a wooden one at the same temperature.

Working

R_total = R_si + Σ(Lᵢ/kᵢ) + R_se, then q″ = ΔT / R_total and Q = q″ · A. Each layer drops ΔTᵢ = q″ · Rᵢ.

Units cancel as W/(m·K) · m² · K / m → W. R_total = 2.859 m²·K/W, ΔT = 26.000 K, A = 10.000 m² → Q = 90.944 W.

Model limit
An interface sits below the dew point, so moisture can condense inside the build-up. Water conducts 0.598 W/(m·K) against still air's 0.026 — 23× more — and damp insulation loses most of its rated value. A vapour control layer belongs on the warm side.
Model limit
Surface films R_si = 0.13 and R_se = 0.04 m²·K/W are ISO 6946 tabulated conventions for horizontal heat flow, not derived constants. They move with wind speed, emissivity and flow direction, and standards disagree at roughly the ±30 % level.
Model limit
This is a one-dimensional result. Studs, joists, ties, lintels and fixings are parallel high-conductivity paths the series model cannot see; a timber-framed wall's effective U-value is typically 10–30 % worse, and a steel-framed one far worse.
Model limit
Fourier's law here assumes steady state — temperatures that have stopped changing. A lightweight wall settles in hours and a masonry wall over a day or more, so this is not the heat flow immediately after a temperature step.
Model limit
Degree-day estimates assume a fixed base temperature and ignore solar gain, internal gains, air infiltration and thermal mass. Infiltration alone often exceeds fabric conduction in older buildings, so read this as the fabric conduction component only.

About This Tool

Thermal Conductivity Calculator – Fourier’s Law, R-Values and U-Values

Heat leaks through everything, and how fast it leaks is set by one material property and one piece of geometry. A thermal conductivity calculatorputs the two together with Fourier’s law of steady-state conduction, Q = k · A · ΔT / L, and then rearranges it for whichever quantity you do not know. Enter a wall, a pipe or a lab sample and you get the heat rate, the heat flux, the thermal resistance and the transmittance — the four ways engineers describe the same physics.

A single wall, and then a real one

Take 10 m² of common brick 100 mm thick, with k = 0.72 W/(m·K), holding 20 °C against 0 °C. The conduction heat loss is 0.72 × 10 × 20 / 0.1 = 1440 W, a heat flux of 144 W/m². Add insulation and the picture changes completely. A plasterboard–mineral wool–brick build-up stacks its resistances in series, R_total = Σ Lⁱ/kⁱ, giving 2.689 m²·K/W and only 96.694 W across the same area at a larger 26 K difference. The 100 mm of wool carries 92.975 % of the temperature drop across 3.7 % of the wall’s thickness, because at equal thickness its resistance is 0.72 / 0.040 = 18×the brick’s.

Interface temperatures and condensation

Because every layer sees the same flux, the calculator can walk the temperature across the build-up and report the value at each interface. In that wall the plasterboard/wool face sits at 20.517 °C while the wool/brick face is at −3.657 °C. Compare those against the dew point and the reason a vapour control layer belongs on the warm side becomes obvious: the cold face of the insulation is far below any indoor dew point, and condensation inside insulation is expensive. Water conducts 0.598 W/(m·K)against still air’s 0.026— 23 times more — so damp insulation is barely insulation at all.

R-13 is not R-13 everywhere

The R-value is resistance per unit area and it comes in two incompatible systems. One h·ft²·°F/BTU equals 0.17611018368230588 m²·K/W, so the two scales differ by 5.678263×. A US batt sold as “R-13” is only R-2.289in SI units. This calculator always prints both with the system in the label, and it also gives the per-inch figure US insulation is actually sold by — mineral wool at k = 0.040 works out at R-3.6 per inch. The U-value is simply U = 1 / R_total, in W/(m²·K) or BTU/(h·ft²·°F).

Never quote a bare conduction U-value
Surface films matter. ISO 6946 assigns conventional values of R_si = 0.13 and R_se = 0.04 m²·K/W for the boundary layers. On a well-insulated wall they change the answer by about 6 %, but on 6 mm single glazing conduction alone predicts U = 166.7 W/(m²·K) against a realistic 5.68— 29.3 times too much heat.

Pipes are not flat

Radial area grows with radius, so the flat-plate formula has no single correct area to use. The cylindrical result is Q = 2πkLΔT / ln(r₂/r₁). On 25 mm of lagging over a 60 mm pipe at 80 °C in a 25 °C room, a 10 m run loses 228.051 W, or 22.805 W/m. Approximating it as a slab of the outer area overstates the loss by 33.35 %; using the inner area understates it by 27.26 %. Only the log-mean area, 2πL(r₂ − r₁)/ln(r₂/r₁), reproduces the exact answer.

Working backwards: identify, size and cost

Rearranged for k, the same law identifies a sample: 12 W through a 20 mm, 0.1 m² specimen across 15 K gives k = 0.16 W/(m·K), which the calculator ranks against its material library. Rearranged for L, it sizes insulation to hit a target R-value — always round up to the next stocked thickness. Multiply the U-value by area and heating degree-days and you get seasonal energy, E = U · A · HDD · 24 / 1000kWh, which shows why doubling insulation never halves the bill: the other layers and the films do not change, so a 100 mm to 200 mm upgrade cuts that wall’s loss by 46.65 %, not 50 %.

What the model cannot see

The arithmetic is exact; the assumptions are the limit. This is steady-state, one-dimensional conduction. Studs, ties and fixings are parallel thermal bridgesthat typically make a framed wall’s real U-value 10–30 % worse. Tabulated k values drift with temperature, so each preset carries the temperature it was measured at. Degree-day costs ignore solar gain, internal gains and infiltration, and are best read as the fabric conduction component rather than a heating bill.

Frequently Asked Questions

Is the Thermal Conductivity Calculator free?

Yes, Thermal Conductivity Calculator is totally free :)

Can I use the Thermal Conductivity Calculator offline?

Yes, you can install the webapp as PWA.

Is it safe to use Thermal Conductivity Calculator?

Yes, any data related to Thermal Conductivity Calculator only stored in your browser (if storage required). You can simply clear browser cache to clear all the stored data. We do not store any data on server.

How does this thermal conductivity calculator work?

It applies Fourier's law of steady-state conduction, Q = k·A·ΔT/L, and rearranges it for whichever quantity you leave blank — heat rate, conductivity, thickness, area or temperature difference. A 10 m² brick wall 100 mm thick with k = 0.72 W/(m·K) across a 20 K difference passes 0.72 × 10 × 20 / 0.1 = 1440 W. Multi-layer walls are solved as resistances in series, R_total = Σ L/k, and pipes use the logarithmic form Q = 2πkL·ΔT / ln(r₂/r₁).

Why is R-13 insulation not R-13 everywhere?

Because the two R-value systems use different units. One h·ft²·°F/BTU equals 0.17611018368230588 m²·K/W, so a US batt labelled R-13 is only R-2.289 in SI units — reading the US number as an SI one overstates the resistance by 5.678263×. This calculator always prints both values with the system in the label, and it never infers the system from the magnitude of a number.

What are surface film resistances and should I include them?

R_si = 0.13 and R_se = 0.04 m²·K/W are the ISO 6946 conventional values that stand in for convection and radiation at the inside and outside faces. They are tabulated conventions, not derived constants, and they vary with wind speed, emissivity and flow direction. Include them for any real building element: leaving them off a 6 mm pane of glass gives U = 166.7 W/(m²·K) instead of a realistic 5.68 — 29.3 times too much heat.

Why does the pipe result use a logarithm?

Because the area heat crosses grows with radius, so a flat-plate formula has no single correct area to use. On 25 mm of lagging over a 60 mm pipe, using the outer area overstates the loss by 33.35 % and the inner area understates it by 27.26 %. The log-mean area, 2πL(r₂−r₁)/ln(r₂/r₁), is the value that makes a slab calculation agree exactly with the correct cylindrical solution.

How accurate is a one-dimensional wall calculation?

The arithmetic is exact, but the model is the limit. Studs, joists, wall ties and fixings are parallel high-conductivity paths this series-resistance calculation cannot see, and they typically make a timber-framed wall's real U-value 10–30 % worse than the plain result. Tabulated k values also drift with temperature — mineral wool gains 15–20 % between 0 °C and 100 °C — and damp insulation can lose most of its rated value because water conducts 23 times better than still air.

Can it tell me whether condensation will form inside a wall?

It flags the risk. Every interface temperature is computed from the same heat flux and compared against the dew point you enter, so you can see exactly where in the build-up the temperature falls below it. In a plasterboard–wool–brick wall at 21 °C inside and −5 °C outside, the warm face sits at 20.52 °C but the wool/brick face at −3.66 °C is far below a typical 12 °C dew point, which is why a vapour control layer belongs on the warm side of the insulation.