ThermoFluidCalc
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Engineering Calculation Report
Date: 2026-09-11 15:08
🧱 Steady Conduction — Plane Walls
Multi-layer thermal resistance networks for flat walls. Compute heat flux, interface temperatures, and overall U-value.
⚡ Fortran 90 Engine
Double Precision (IEEE 754)
✓ ISO / ASME Validated
Conduction
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Wall Configuration
Thermal Resistances:
$$Q = A(T_1 - T_2) / R_{total}$$ $$R_{total} = \sum (L_i / k_i)$$ • $Q$ = Heat flux [W]
• $A$ = Area [m²]
• $R_{total}$ = Total thermal resistance [m²K/W]
• $L_i$ = Thickness of Layer i [m]
• $k_i$ = Thermal Conductivity of Layer i [W/mK]
$$Q = A(T_1 - T_2) / R_{total}$$ $$R_{total} = \sum (L_i / k_i)$$ • $Q$ = Heat flux [W]
• $A$ = Area [m²]
• $R_{total}$ = Total thermal resistance [m²K/W]
• $L_i$ = Thickness of Layer i [m]
• $k_i$ = Thermal Conductivity of Layer i [W/mK]
Results & Visualization
Results and visualizations will be displayed here upon completion of the computation.
Calculation Methodology
Mathematical Model & Theory
Conduction through a plane wall is governed by Fourier's Law of Heat Conduction. For steady-state 1D heat flow with no heat generation, the heat transfer rate is constant and defined as:
$$Q = \frac{A(T_1 - T_2)}{R_{total}}$$
$$R_{total} = \sum_{i=1}^{n} R_i = \sum_{i=1}^{n} \frac{L_i}{k_i}$$
Variable Definitions & Units:
- $Q$: Heat transfer rate [W]
- $A$: Heat transfer area [m²]
- $T_1, T_2$: Inner and outer surface temperatures [°C]
- $R_i$: Thermal resistance of layer $i$ [K/W or °C/W]
- $L_i$: Thickness of layer $i$ [m]
- $k_i$: Thermal conductivity of layer $i$ [W/m·K]
Assumptions & Boundary Conditions:
- One-dimensional heat flow (perpendicular to wall surfaces).
- Steady-state conditions (temperatures do not vary with time).
- Constant material properties (thermal conductivity $k$ is uniform).
- No internal heat generation ($q = 0$).
Validity & Bounds:
- Applicable to flat surfaces where height and width are significantly larger than total thickness.
Academic References:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer.
- Çengel, Y. A. (2015). Heat and Mass Transfer: Fundamentals and Applications.
Worked Engineering Example
Problem Statement:
A multilayer cold-room wall consists of a 100 mm concrete layer ($k = 1.7$ W/m·K) and a 50 mm polyurethane insulation board ($k = 0.03$ W/m·K). The wall area is 20 m². The inner surface of the concrete is at 20°C and the outer surface of the insulation is at -10°C. Calculate the heat transfer rate.
Step-by-step Solution:
1. Calculate the thermal resistances of individual layers:
$$R_1 = \frac{L_1}{k_1} = \frac{0.10}{1.7} = 0.0588 \text{ m}^2\text{K/W}$$ $$R_2 = \frac{L_2}{k_2} = \frac{0.05}{0.03} = 1.6667 \text{ m}^2\text{K/W}$$ 2. Calculate total thermal resistance:
$$R_{total} = R_1 + R_2 = 0.0588 + 1.6667 = 1.7255 \text{ m}^2\text{K/W}$$ 3. Calculate the heat transfer rate $Q$:
$$Q = \frac{A(T_1 - T_2)}{R_{total}} = \frac{20 \times (20 - (-10))}{1.7255} = 347.72 \text{ W}$$
Final Result:
Heat transfer rate is 347.7 W.
A multilayer cold-room wall consists of a 100 mm concrete layer ($k = 1.7$ W/m·K) and a 50 mm polyurethane insulation board ($k = 0.03$ W/m·K). The wall area is 20 m². The inner surface of the concrete is at 20°C and the outer surface of the insulation is at -10°C. Calculate the heat transfer rate.
Step-by-step Solution:
1. Calculate the thermal resistances of individual layers:
$$R_1 = \frac{L_1}{k_1} = \frac{0.10}{1.7} = 0.0588 \text{ m}^2\text{K/W}$$ $$R_2 = \frac{L_2}{k_2} = \frac{0.05}{0.03} = 1.6667 \text{ m}^2\text{K/W}$$ 2. Calculate total thermal resistance:
$$R_{total} = R_1 + R_2 = 0.0588 + 1.6667 = 1.7255 \text{ m}^2\text{K/W}$$ 3. Calculate the heat transfer rate $Q$:
$$Q = \frac{A(T_1 - T_2)}{R_{total}} = \frac{20 \times (20 - (-10))}{1.7255} = 347.72 \text{ W}$$
Final Result:
Heat transfer rate is 347.7 W.