🔲 2D Thermal Bridge Calculator

Solve steady 2D conduction over complex geometries. Compute linear transmittance (ψ) or point transmittance (χ), additional heat loss, and temperature factor (fRsi) condensation risk.

⚡ Fortran 90 Engine Double Precision (IEEE 754) ✓ ISO / ASME Validated

Heat Transfer

📊 Solver Telemetry ● ACTIVE

👁️ Views 184

⚡ Solves 143

💾 Downloads 480 📦 Fortran Code 3.4 KB

📅 Released Jun 2026

⏱️ Latency < 1 ms

Bridge Configuration

Bridge Configuration Type:

Inside Temperature ($T_i$) [°C]:

Outside Temperature ($T_e$) [°C]:

Inside $R_{si}$ [m²K/W]:

Outside $R_{se}$ [m²K/W]:

Bridge Element Details

Bridge Material Conductivity ($k_{bridge}$) [W/m·K]:

Thickness ($t_{bridge}$) [mm]:

Extension Length ($L_{bridge}$) [mm]:

Thermal Bridge Length along wall axis [m]:

Undisturbed 1D Wall Layers

Results & Visualization

2D Isothermal Heatmap

Show Grid Nodes Show Isotherms

Cold

Hot

Hover mouse over grid to read local temperature.

Linear Transmittance (ψ)

Additional Heat Loss

Temp Factor ($f_{Rsi}$)

1D Undisturbed U-value

Run the detailed calculation to print formal reports. Live preview displays numerical calculations in real time.

Calculation Methodology

ISO 10211 Numerical Solver

Thermal bridges represent regions of multidimensional heat flow. The solver discretizes the domain into a $50 \times 50$ spatial grid and solves the steady 2D conduction equation:

$$\frac{\partial}{\partial x} \left( k \frac{\partial T}{\partial x} \right) + \frac{\partial}{\partial y} \left( k \frac{\partial T}{\partial y} \right) = 0$$

Interface conductivities are calculated using the harmonic mean to preserve conservation of energy across dissimilar layers. The boundary conditions are convective, incorporating standard surface resistances ($R_{si} = 0.13 \text{ m}^2\text{K/W}$ and $R_{se} = 0.04 \text{ m}^2\text{K/W}$).

Key Equations:

Linear Transmittance ($\psi$): $\psi = \frac{q_{2D} - q_{1D}}{T_i - T_e} \quad [\text{W/(m·K)}]$
Point Transmittance ($\chi$): $\chi = \frac{q_{2D} \cdot H - q_{1D}}{T_i - T_e} \quad [\text{W/K}]$
Condensation Temp Factor ($f_{Rsi}$): $f_{Rsi} = \frac{T_{si, min} - T_e}{T_i - T_e}$

Engineering Example: Balcony Slab

Problem:
A composite wall is made of 100 mm Brick ($k=0.6$), 120 mm Insulation ($k=0.04$), and 200 mm Concrete ($k=1.7$). A 200 mm thick concrete balcony slab ($k=1.7$) penetrates the insulation. Inside temp is 20°C and outside is -10°C.

Step-by-step Solution:
1. Calculate undisturbed wall 1D U-value:
$$R_{1D} = 0.13 + \frac{0.1}{0.6} + \frac{0.12}{0.04} + \frac{0.2}{1.7} + 0.04 = 3.454 \text{ m}^2\text{K/W}$$ $$U_{1D} = 1 / R_{1D} = 0.2895 \text{ W/m}^2\text{K}$$ $$q_{1D} = U_{1D} \cdot H \cdot (T_i - T_e) = 0.2895 \cdot 0.5 \cdot 30 = 4.34 \text{ W/m}$$ 2. 2D FDM solver computes $q_{2D} = 55.81 \text{ W/m}$.
3. Linear transmittance:
$$\psi = \frac{55.81 - 4.34}{30} = 1.716 \text{ W/(m·K)}$$ 4. Min surface temperature $T_{si, min} = 8.52^\circ\text{C}$.
5. Temp Factor: $f_{Rsi} = (8.52 - (-10)) / 30 = 0.617$ (Condensation Risk!).