Library / Heat Transfer / HTR-06
Heat Transfer
Nusselt number
Dimensionless heat transfer coefficient comparing convection with conduction in the fluid.
- Ref
- HTR-06
- Rev
- v1
- Field
- Heat Transfer
- Result
- Nu
Use the fluid conductivity kf and the characteristic length specified by the correlation (diameter for pipes, length for plates). Correlations give Nu from Re and Pr, from which h follows.
The Nusselt number compares heat carried by fluid motion at the surface with what pure conduction through the same fluid layer would carry.
Calculator
- Equivalent conduction layer kf/h1.04 mm
- Heat flux at ΔT = 10 K250 W/m²
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.
- Nu of 1 would mean conduction only; larger values show convection is boosting the heat transfer.
- Correlations such as Dittus–Boelter give Nu from Re and Pr, from which h = Nu kf/Lc.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| h Heat transfer coefficient | 2 to 5,000 W/(m²·K) | 3.846 | 9,615 | 9,612 | +10% | +10 % |
| Lc Characteristic length | 0.001 to 2 m | 0.9615 | 1,923 | 1,922 | +10% | +10 % |
| kf Fluid conductivity | 0.01 to 1 W/(m·K) | 1.25 | 125 | 123.8 | +10% | −9.09 % |
Sweep of Heat transfer coefficient
| Heat transfer coefficient (W/(m²·K)) | Nusselt number |
|---|---|
| 2 | 3.846 |
| 5.318 | 10.23 |
| 14.14 | 27.2 |
| 37.61 | 72.32 |
| 100 | 192.3 |
| 265.9 | 511.4 |
| 707.1 | 1,360 |
| 1,880 | 3,616 |
| 5,000 | 9,615 |
Sweep one variable
Variables and how each one contributes
Nusselt number
What the equation gives you. Everything else feeds it.
Heat transfer coefficient
Surface convection coefficient. Directly proportional.
Characteristic length
Length scale used by the correlation. Directly proportional.
Fluid conductivity
Fluid thermal conductivity (air 0.026, water 0.6). Inversely proportional.
Worked example with the values above
- h Lc = 25 × 0.05 = 1.25 W/K per m.
- Divide by kf = 0.026.
- Nu = 48.08.
Assumptions and limits
- Fluid properties at the specified reference temperature.
- Characteristic length consistent with the correlation.
Common mistakes
- Using the solid’s conductivity.
- Using the wrong characteristic length.
Statistics and uncertainty
Correlations are good to ±15–25%, so h from Nu inherits this uncertainty.
Where it is used
- Applying convection correlations.
- Scaling model tests to full size.
Maths you need
- Convection coefficient: h
- Thermal conductivity: k