ThermoFluidCalc
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Engineering Calculation Report
Date: 2026-05-30 00:16
Rectangular Fin Heat Conduction
Analyze fin efficiency, effectiveness, and heat dissipation for straight rectangular fin geometries.
Calculation Domain Inputs
Specify geometry parameters and boundary conditions to calculate the thermal distribution and heat loss of a rectangular fin:
- Fin Length ($L$): The extension length from the base wall.
- Fin Thickness ($t$): The profile height of the fin.
- Fin Width ($w$): The lateral dimension of the fin.
- Thermal Conductivity ($k$): Heat conduction capacity of the material.
- Convection Coefficient ($h$): Heat transfer coefficient of the surrounding fluid.
Configuration
Rectangular Fin Theory:
Q = √(hPkA) × (T₀ - T∞) × tanh(mL)
Where:
• m = √(hP/kA)
• P = 2(w + t) = Perimeter
• A = w × t = Cross-sectional area
• η = tanh(mL)/(mL) = Fin efficiency
• ε = Q/(hA_total(T₀ - T∞)) = Effectiveness
• Q = Heat transfer rate [W]
• h = Convection coefficient [W/m²·K]
• k = Thermal conductivity [W/m·K]
• L = Fin length [m]
Q = √(hPkA) × (T₀ - T∞) × tanh(mL)
Where:
• m = √(hP/kA)
• P = 2(w + t) = Perimeter
• A = w × t = Cross-sectional area
• η = tanh(mL)/(mL) = Fin efficiency
• ε = Q/(hA_total(T₀ - T∞)) = Effectiveness
• Q = Heat transfer rate [W]
• h = Convection coefficient [W/m²·K]
• k = Thermal conductivity [W/m·K]
• L = Fin length [m]
Results & Visualization
Results and visualizations will be displayed here upon completion of the computation.
Calculation Methodology
Mathematical Model & Theory
Fins are used to enhance heat transfer from a surface by increasing the surface area. The steady conduction along a rectangular fin with convection from its outer surfaces is modeled using the fin equation:
$$Q_{fin} = M \tanh(m L_c)$$
$$m = \sqrt{\frac{h P}{k A_c}}, \quad M = \sqrt{h P k A_c} (T_b - T_{\infty})$$
$$L_c = L + t/2 \quad \text{(corrected fin length for convective tip)}$$
Variable Definitions & Units:
- $Q_{fin}$: Fin heat transfer rate [W]
- $P$: Fin perimeter ($2w + 2t$) [m]
- $A_c$: Cross-sectional area ($w \times t$) [m²]
- $T_b$: Base temperature [°C]
- $T_{\infty}$: Ambient temperature [°C]
- $k$: Fin thermal conductivity [W/m·K]
- $h$: Convection coefficient [W/m²·K]
Validity & Bounds:
- Valid for thin fins where the Biot number of the fin $Bi_{fin} = h t / k < 0.1$ ensures a uniform temperature across the fin thickness.
Assumptions & Boundary Conditions:
- Steady-state one-dimensional conduction along the fin length $x$.
- Constant thermal conductivity $k$ of the fin material.
- Uniform convection heat transfer coefficient $h$ over the entire fin surface.
- Constant ambient temperature $T_\infty$.
- Convective boundary condition at the fin tip (modeled via corrected length $L_c$).
- No heat generation within the fin.
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:
An aluminum rectangular fin ($k = 200$ W/m·K), 50 mm long, 2 mm thick, and 100 mm wide, is attached to a base at 100°C. Convection occurs to ambient air at 20°C with $h = 25$ W/m²·K. Calculate the fin heat transfer rate.
Step-by-step Solution:
1. Calculate parameters:
$$P = 2(w + t) = 2(0.1 + 0.002) = 0.204 \text{ m}$$ $$A_c = w \times t = 0.1 \times 0.002 = 0.0002 \text{ m}^2$$ $$m = \sqrt{\frac{h P}{k A_c}} = \sqrt{\frac{25 \times 0.204}{200 \times 0.0002}} = 11.29 \text{ m}^{-1}$$ $$L_c = L + t/2 = 0.05 + 0.001 = 0.051 \text{ m}$$ $$mL_c = 11.29 \times 0.051 = 0.5759$$ 2. Calculate base parameter $M$:
$$M = \sqrt{25 \times 0.204 \times 200 \times 0.0002} \times (100 - 20) = 36.13 \text{ W}$$ 3. Calculate heat transfer rate:
$$Q_{fin} = M \tanh(mL_c) = 36.13 \times \tanh(0.5759) = 18.76 \text{ W}$$
Final Result:
Fin heat transfer is 18.76 W.
An aluminum rectangular fin ($k = 200$ W/m·K), 50 mm long, 2 mm thick, and 100 mm wide, is attached to a base at 100°C. Convection occurs to ambient air at 20°C with $h = 25$ W/m²·K. Calculate the fin heat transfer rate.
Step-by-step Solution:
1. Calculate parameters:
$$P = 2(w + t) = 2(0.1 + 0.002) = 0.204 \text{ m}$$ $$A_c = w \times t = 0.1 \times 0.002 = 0.0002 \text{ m}^2$$ $$m = \sqrt{\frac{h P}{k A_c}} = \sqrt{\frac{25 \times 0.204}{200 \times 0.0002}} = 11.29 \text{ m}^{-1}$$ $$L_c = L + t/2 = 0.05 + 0.001 = 0.051 \text{ m}$$ $$mL_c = 11.29 \times 0.051 = 0.5759$$ 2. Calculate base parameter $M$:
$$M = \sqrt{25 \times 0.204 \times 200 \times 0.0002} \times (100 - 20) = 36.13 \text{ W}$$ 3. Calculate heat transfer rate:
$$Q_{fin} = M \tanh(mL_c) = 36.13 \times \tanh(0.5759) = 18.76 \text{ W}$$
Final Result:
Fin heat transfer is 18.76 W.