Fin Heat Transfer Calculator
Compute heat transfer, fin efficiency, effectiveness, and temperature profile for straight fins with convection (rectangular and pin fins).
Fin geometry & thermal conditions
Straight fin of uniform cross‑section attached to a wall.
Example: Aluminum ≈ 205 W/m·K, Copper ≈ 385 W/m·K, Steel ≈ 50 W/m·K.
Used for fin effectiveness (bare area = wb × 1 m depth).
0 at the base, L at the tip. Used to compute local temperature.
Results
Quick interpretation
Enter parameters and click “Compute fin performance” to see whether your fin is efficient and worthwhile.
How this fin heat transfer calculator works
This tool implements the standard one‑dimensional fin theory used in heat transfer textbooks (e.g. Incropera, Holman, Çengel). It supports straight fins of uniform cross‑section with convection to a surrounding fluid.
1. Geometry and thermal properties
For a straight fin of length \(L\), cross‑sectional area \(A_c\), and perimeter \(P\):
Rectangular fin (width \(b\), thickness \(t\), unit depth):
\(A_c = b\,t\)
\(P = 2(b + t)\)
Circular pin fin (diameter \(D\), unit depth):
\(A_c = \dfrac{\pi D^2}{4}\)
\(P = \pi D\)
The fin is made of a material with thermal conductivity \(k\) and is exposed to a fluid with convection coefficient \(h\). The base is at temperature \(T_b\) and the ambient fluid at \(T_\infty\).
2. Fin parameter \(m\)
The governing differential equation for a straight fin leads to the fin parameter:
\[ m = \sqrt{\frac{h P}{k A_c}} \]
3. Heat transfer from the fin
The heat transfer rate depends on the tip condition.
Adiabatic (insulated) tip
\[ Q_{\text{fin}} = \sqrt{h P k A_c}\,(T_b - T_\infty)\,\tanh(mL) \]
Convective tip
For a convective tip, the exact solution is:
\[ Q_{\text{fin}} = \sqrt{h P k A_c}\,(T_b - T_\infty)\, \frac{\sinh(mL) + \dfrac{h}{m k}\cosh(mL)} {\cosh(mL) + \dfrac{h}{m k}\sinh(mL)} \]
For long fins with large \(mL\), the adiabatic‑tip formula is often a good approximation.
4. Fin efficiency and effectiveness
The maximum possible heat transfer from the fin (if the entire fin were at \(T_b\)) is:
\[ Q_{\text{max}} = h\,A_{\text{fin}}\,(T_b - T_\infty) \]
where \(A_{\text{fin}}\) is the total fin surface area (sides + tip).
Then:
Fin efficiency
\[ \eta_f = \frac{Q_{\text{fin}}}{Q_{\text{max}}} \]
Fin effectiveness (compared to bare base area \(A_b\))
\[ \varepsilon_f = \frac{Q_{\text{fin}}}{h A_b (T_b - T_\infty)} \]
A fin is usually considered worthwhile if \(\varepsilon_f > 2\), i.e. it at least doubles the heat transfer compared to the same base area without a fin.
5. Temperature distribution along the fin
The dimensionless temperature is defined as:
\[ \theta(x) = \frac{T(x) - T_\infty}{T_b - T_\infty} \]
Adiabatic tip
\[ \theta(x) = \frac{\cosh\big(m(L - x)\big)}{\cosh(mL)} \]
\[ T(x) = T_\infty + \theta(x)\,(T_b - T_\infty) \]
Convective tip
\[ \theta(x) = \frac{\cosh\big(m(L - x)\big) + \dfrac{h}{m k}\sinh\big(m(L - x)\big)} {\cosh(mL) + \dfrac{h}{m k}\sinh(mL)} \]
\[ T(x) = T_\infty + \theta(x)\,(T_b - T_\infty) \]
Typical values and design tips
- Convection coefficient h: natural convection in air ≈ 5–15 W/m²·K; forced air ≈ 30–200 W/m²·K; boiling or condensing fluids can be much higher.
- Fin materials: aluminum and copper are common due to high thermal conductivity and low weight.
- Thin, long fins increase area but may have low efficiency if too long (temperature drops quickly along the fin).
- Use efficiency to judge material usage and effectiveness to judge whether adding a fin is beneficial at all.
FAQ
What assumptions does this calculator make?
The model assumes steady‑state, one‑dimensional conduction along the fin, constant thermal conductivity, uniform convection coefficient, negligible radiation, and a straight fin of uniform cross‑section.
Can I model annular or triangular fins?
Not directly. However, you can approximate them by an equivalent straight fin with an effective cross‑sectional area and perimeter. For accurate design of complex fins, use numerical methods (FEM/CFD).
How do I know if my fin is “too long”?
If increasing \(L\) further does not significantly increase \(Q_{\text{fin}}\) or if the efficiency \(\eta_f\) becomes very low (e.g. < 0.4), the extra length is not being used effectively.
Why is fin effectiveness sometimes less than 1?
This happens when the fin has low conductivity, is very short or thick, or the convection coefficient is low. In that case, the fin does not improve heat transfer compared to the bare surface and is not worth using.