Library / Heat Transfer / HTR-11
Heat Transfer
Dittus–Boelter correlation
Nusselt number for turbulent flow heated in a smooth pipe.
- Ref
- HTR-11
- Rev
- v1
- Field
- Heat Transfer
- Result
- Nu
Re is the Reynolds number based on the pipe diameter and Pr the Prandtl number of the fluid. Nu = hD/k, so the heat transfer coefficient is h = Nu k/D. The 0.4 exponent on Pr is for heating the fluid; for cooling it is 0.3. The correlation is accurate to about ±25 %.
Nusselt number for turbulent flow heated in a smooth pipe. Multiply by k/D to get the heat transfer coefficient.
Calculator
- Heat transfer coefficient h for water (k = 0.6 W/m·K) in a 25 mm tube6,905 W/m²·K
- Nusselt number if Re is doubled500.9
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 defaults suit water at Re = 50 000: Nu is 288, which for a 25 mm tube of water gives h of about 6900 W/m²·K.
- Doubling the flow speed raises h by 74 % but pumping power by about 8 times, so there is a limit beyond which more flow is not worth it.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| Re Reynolds number | 10,000 to 1,000,000 | 79.39 | 3,161 | 3,081 | +10% | +7.92 % |
| Pr Prandtl number | 0.6 to 160 | 107.7 | 1,006 | 898.2 | +10% | +3.89 % |
Sweep of Reynolds number
| Reynolds number | Nusselt number |
|---|---|
| 10,000 | 79.39 |
| 133,800 | 632.1 |
| 257,500 | 1,068 |
| 381,300 | 1,461 |
| 505,000 | 1,830 |
| 628,800 | 2,180 |
| 752,500 | 2,518 |
| 876,300 | 2,844 |
| 1,000,000 | 3,161 |
Sweep one variable
Variables and how each one contributes
Nusselt number
What the equation gives you. Everything else feeds it.
Reynolds number
The Reynolds number of the flow in the pipe. Power 0.8: doubling the flow speed raises Nu by 74 %. A 10 % rise in Re raises Nu by about 8 %.
Prandtl number
The Prandtl number of the fluid (about 0.7 for air, 7 for water at 20 °C, 100 or more for oils). Power 0.4: a more viscous fluid with the same conductivity carries heat from the wall less well, but only gently. A 10 % rise in Pr raises Nu by about 4 %.
Worked example with the values above
- Powers: Re0.8 = 5,743, Pr0.4 = 2.1779.
- Multiply: 0.023 × 5,743 × 2.1779.
- Nu = 287.7.
Assumptions and limits
- Fully developed turbulent flow (Re above about 10 000) in a smooth, long (L/D above 10) circular tube.
- Moderate property variation across the pipe and Prandtl number between about 0.6 and 160.
Common mistakes
- Using it for laminar flow (Re below 2300), where Nu is nearly constant.
- Evaluating the properties at the wrong temperature. Use the mean bulk temperature.
Statistics and uncertainty
Correlations like this one scatter by ±25 % from real data, which is larger than any input uncertainty here, so the result should not be quoted to more than two figures. The Re exponent of 0.8 means a 5 % error in flow rate gives about 4 % in h.
Where it is used
- Estimating convection in coolant lines and heat exchanger tubes.
- Sizing the flow needed to reach a required coolant coefficient.
Maths you need
- Dimensionless numbers: Re, Pr and Nu
- Powers: decimal exponents