Library / Heat transfer and thermal systems / HTS-12
Heat transfer and thermal systems
Distance to a specified temperature fraction on a long fin
Distance from the base at which the excess temperature of a very long fin has fallen to the fraction f = (T − T∞)/(Tb − T∞).
- Ref
- HTS-12
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- x [m]
On a very long fin the excess temperature decays exponentially, θ/θb = exp(−mx). Solving for the distance at which the excess has fallen to a fraction f gives x = −ln(f)/m. The decay length is 1/m. It is valid when the real tip is much farther than the distance found, so the tip has no influence on the profile.
Distance from the base at which the excess temperature of a very long fin has fallen to the fraction f = (T − T∞)/(Tb − T∞). It uses the semi-infinite fin solution, so the real tip must be far away.
Calculator
- Decay length, 1/m0.05 m
- Distance to half the base excess, ln 2/m0.03466 m
- Distance to 1 % of the base excess, ln 100/m0.2303 m
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The fin excess falls to 10 % of its base value at 0.1151 m, which is 2.303 decay lengths.
- Each decay length of 0.05 m cuts the excess by a factor of e, so the fin is effectively finished after about 4 to 5 decay lengths.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| m Fin parameter √(hP/(kAc)) | 0.1 to 1,000 1/m | 0.002303 | 23.03 | 23.02 | +10% | −9.09 % |
| f Temperature fraction (T−T∞)/(Tb−T∞) | 0.001 to 1 | 0 | 0.3454 | 0.3454 | +10% | −4.14 % |
Sweep of Temperature fraction (T−T∞)/(Tb−T∞)
| Temperature fraction (T−T∞)/(Tb−T∞) | Distance from the base (m) |
|---|---|
| 0.001 | 0.3454 |
| 0.002371 | 0.3022 |
| 0.005623 | 0.259 |
| 0.01334 | 0.2159 |
| 0.03162 | 0.1727 |
| 0.07499 | 0.1295 |
| 0.1778 | 0.08635 |
| 0.4217 | 0.04317 |
| 1 | 0 |
Sweep one variable
Variables and how each one contributes
Distance from the base
What the equation gives you. Everything else feeds it.
Temperature fraction (T−T∞)/(Tb−T∞)
The temperature fraction (T − T∞)/(Tb − T∞), between 0 and 1. A smaller fraction lies farther along the fin. Raising it by 10 % (to 0.11) moves the result by −4.14 %, from 0.1151 to 0.1104 m.
Fin parameter √(hP/(kAc))
The fin parameter √(hP/(kAc)), in 1/m. A larger m means faster decay, so the distance is shorter in inverse proportion. Raising it by 10 % (to 22) moves the result by −9.09 %, from 0.1151 to 0.1047 m.
Worked example with the values above
- Natural log of the fraction: ln(0.1) = −2.3026.
- Negate and divide by m = 20 1/m.
- x = 2.3026 / 20 = 0.11513 m.
Assumptions and limits
- Uniform cross-section, constant k and h, and one-dimensional conduction along the fin.
- Negligible radiation, with the base held at a uniform temperature.
Common mistakes
- Using the cross-section perimeter wrongly when finding m.
- Mixing absolute temperatures and excess temperatures above ambient.
- Using an insulated-tip formula when the tip convects, which matters for short fins.
Statistics and uncertainty
Fin results depend strongly on the heat-transfer coefficient h, which is usually known to only 20 % to 30 %. Because m varies as √h, the temperature and heat-rate errors are somewhat smaller, but still significant.
Where it is used
- Pin, rib and rod fins on heat sinks, engines and pipes.
- Rating finned tubes and checking tip temperatures.
Maths you need
- Fin parameter: m = √(hP/(kAc))
- Excess temperature: θ = T − T∞
- Fin efficiency: actual heat rate divided by the rate if the whole fin were at the base temperature