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Wall Thermal Resistance Calculator

Calculate nominal R, nominal U, bridge-adjusted U, effective R, and steady heat transfer through a wall assembly from surface films, three material layers, wall area, and temperature difference.

Nominal assembly R-value (m²·K/W)-
Nominal U-value (W/(m²·K))-
U-value after entered bridge allowance (W/(m²·K))-
Effective R-value (m²·K/W)-
Steady conductive heat transfer (kW)-
Nominal unbridged heat transfer (kW)-
Added heat transfer from bridge allowance (kW)-
Adjusted U-value increase-

Decision view

Layered wall section and thermal-bridge bypass

Layered wall section and thermal-bridge bypassThe five series resistances form the nominal path while a separate bridge arrow exposes the increase from nominal to adjusted heat flow.
Exact scenario comparisonThermal-bridge U-value increase (%) changes while all other entered assumptions remain constant.
Thermal-bridge U-value increase (%)Nominal assembly R-value (m²·K/W)Nominal U-value (W/(m²·K))U-value after entered bridge allowance (W/(m²·K))Effective R-value (m²·K/W)Steady conductive heat transfer (kW)Nominal unbridged heat transfer (kW)Added heat transfer from bridge allowance (kW)Adjusted U-value increase

How to use Wall Thermal Resistance Calculator

  1. Enter surface-film and material-layer R-values in the same unit.
  2. Enter the thermal-bridge U-value increase.
  3. Enter wall area and the design temperature difference.
  4. Compare nominal and adjusted paths in the layered wall diagram.

Calculator guide

Understanding Wall Thermal Resistance Calculator

A wall's nominal layer resistance is not its whole-building performance. This calculator adds the five entered resistances, converts that series total to U-value, applies the explicit thermal-bridge increase, and reconciles both nominal and adjusted heat flow.

R adds in series Every layer contributes to nominal resistance.
U is reciprocal Transmittance equals one divided by resistance.
Bridge acts on U The entered penalty raises transmittance.
Heat follows UAΔT Area and temperature difference scale the load.

Calculation method

How the calculation works

Sum entered series thermal resistances, invert to nominal U-value, and apply a transparent U-value bridge allowance before calculating steady heat transfer. Add all series resistances, invert the sum for nominal U, multiply U by the bridge factor, invert again for effective R, and apply Q = UAΔT.

Detailed calculation process

Move from layer resistance to bridge-adjusted wall heat flow

The defaults model 0.12 and 0.04 m²·K/W films around three layers of 0.18, 2.80, and 0.25 m²·K/W, with an 18% U-value bridge increase.

General formula: R_nom = R_si + R_1 + R_2 + R_3 + R_se; U_nom = 1/R_nom; U_eff = U_nom(1+b); R_eff = 1/U_eff; Q = U_eff A ΔT / 1000 Series resistances add. U-value is their reciprocal, so a bridge penalty is applied to U rather than R. Multiplying adjusted U by area and temperature difference gives watts; division by 1,000 gives kilowatts.

What each symbol means

R_si / R_se Inside and outside surface-film resistances, in m²·K/W.
R_1, R_2, R_3 Entered material-layer resistances, each in m²·K/W.
U_nom / U_eff Nominal and bridge-adjusted thermal transmittance, in W/(m²·K).
b Entered fractional increase in U caused by thermal bridging, unitless.
A Wall area, in square metres (m²).
ΔT / Q Indoor-outdoor temperature difference in K and resulting heat flow in kW.

Worked substitution with the default inputs

1. Add the five series resistances: R_nom = 0.12 + 0.18 + 2.80 + 0.25 + 0.04 = 3.390 m²·K/W Surface films and material layers share one series heat path.
2. Invert nominal resistance: U_nom = 1 / 3.390 = 0.294985 W/(m²·K) Resistance and transmittance are reciprocals for the same assembly.
3. Apply the bridge allowance: b = 18/100 = 0.18; U_eff = 0.294985 × 1.18 = 0.348083 W/(m²·K) The percentage is converted to a decimal before multiplying U.
4. Recover effective resistance: R_eff = 1 / 0.348083 = 2.872881 m²·K/W The bridge-adjusted wall therefore has less effective resistance than the nominal layer stack.
5. Check heat transfer: Q = 0.348083 × 120 × 25 / 1000 = 1.044248 kW; nominal Q = 0.884956 kW The 0.159292 kW difference is exactly the modeled bridge increment.

The default 3.390 m²·K/W nominal assembly becomes 2.8729 m²·K/W effective resistance and passes 1.04425 kW across 120 m² at a 25 K temperature difference.

Assembly anatomy

See the insulation path and bridge bypass together

The cross-section distinguishes the five series layers from the parallel bridge penalty.

Interior film The room-side boundary contributes R_si.
Layer stack Three entered materials add in series.
Exterior film The outdoor boundary contributes R_se.
Bridge bypass The U-value increase shows the extra heat path.

Worked situations

Practical examples

  • The five default resistances add to 3.390 m²·K/W.
  • An 18% bridge allowance raises U from 0.294985 to 0.348083 W/(m²·K).
  • The adjusted steady heat flow is 1.044248 kW.

Better inputs

Useful tips

  • Use assembly-specific layer data at the intended temperature and moisture condition.
  • Keep thermal-bridge effects separate from series-layer R-values.
  • Use net opaque wall area if windows and doors are modeled elsewhere.

Before relying on the result

Limitations and common mistakes

  • This is one-dimensional steady-state heat flow with a single aggregate bridge factor.
  • Air leakage, solar gain, moisture, thermal mass, framing geometry, and dynamic weather are excluded.
  • Code compliance and certified assembly values require local documentation.

Reference

Key terms

R-value
Thermal resistance of a layer or assembly.
U-value
Heat transfer rate per area and temperature difference.
Thermal bridge
A more conductive path that raises assembly heat transfer.

Important note

Calculated from the entered measurements and stated coverage or quantity rules. Confirm field dimensions, waste, product requirements, structural conditions, and local codes before purchasing or building.

Frequently asked questions

Why is the bridge percentage applied to U?

Bridging increases heat transmission, so the model multiplies transmittance directly.

Can I add window area to the wall area?

Only if the entered wall U-value represents the combined assembly; otherwise model glazing separately.

Why divide by 1,000?

U × area × temperature difference produces watts, and 1,000 W equals 1 kW.

Is the effective R a material R-value?

No. It is the reciprocal of the adjusted whole-assembly U-value.