ThermoFluidCalc
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Engineering Calculation Report
Date: 2026-09-11 16:19
🔱 Rectangular Fin Efficiency
Analyse extended surface (fin) performance. Compute fin efficiency, effectiveness, and total heat transfer rate.
⚡ Fortran 90 Engine
Double Precision (IEEE 754)
✓ ISO / ASME Validated
Heat Transfer
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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.