ThermoFluidCalc
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Engineering Calculation Report
Date: 2026-05-30 00:17
Steady Heat Conduction in Plane Walls
Analyze heat transfer through single and multilayer plane walls using thermal resistance networks.
Calculation Domain Inputs
Define the geometry and boundary conditions of the 1D composite wall:
- Wall Area ($A$): Cross-sectional area perpendicular to heat transfer.
- Inner Temperature ($T_1$): Surface temperature on the hot (left) face.
- Outer Temperature ($T_2$): Surface temperature on the cold (right) face.
- Layer Thickness ($L_i$): Width of each individual material layer.
- Thermal Conductivity ($k_i$): Material's ability to conduct heat.
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.