Method and interpretation

Nusselt Number

See how to enter values, calculate a result, and check the limits of its interpretation.

Convection measured against conduction

The Nusselt number divides the convective coefficient h by k/L, the coefficient a stagnant fluid layer of thickness L would give by conduction alone. Nu therefore says how many times more heat the moving fluid carries than that baseline. The Michigan Tech CM3120 notes call it the dimensionless heat-transfer coefficient in convection.

Nu is used in two directions. Design goes from a correlation Nu to h = Nu·k/L; review of tests or CFD goes from a measured h to Nu for comparison across diameters and fluids.

Choosing L and k in Nu = hL/k

Nu = hL / k

h is the convective coefficient in W/(m²·K), L the characteristic length in m and k the fluid conductivity in W/(m·K); W/(m²·K) × m ÷ W/(m·K) = 1. The calculator does not predict h; it must come from a measurement or a correlation.

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Reference values: laminar 3.66 and 4.36, and turbulent correlations

Once laminar flow in a circular tube is thermally fully developed, Nu is a constant independent of Re and Pr. RELAP-7 uses Nu = 4.36 for uniform wall heat flux (eqs 45 and 591) and Nu = 3.66 for constant wall temperature (eq 591), eq (45) being the exact solution for uniform heat flux and constant properties. Here velocity does not raise h; since h = Nu·k/D, smaller tubes do.

In turbulent flow Nu grows with Re and Pr. The widely used Dittus–Boelter form is below; RELAP-7 eq (46) uses C = 0.023 with n = 0.4 when the fluid is heated and n = 0.3 when it is cooled.

NuD = 0.023 ReD0.8 Prn

It is for fully developed turbulent flow in smooth round tubes with moderate wall-to-fluid temperature differences. Commonly quoted limits are roughly Re ≳ 10⁴, Pr about 0.6 to 160 and L/D ≳ 10, with the exact bounds varying between textbooks. RELAP-7's wall heat-transfer section (eqs 592 and 593) uses the Gnielinski correlation and takes the largest of the laminar, turbulent and natural-convection Nu so h does not jump at transition (eq 590). Liquid metals with Pr below 0.1 get a separate Nu = 5.0 + 0.025Pe0.8.

Working back from Nu to pipe heat-transfer coefficients

Example 1. Laminar 20 °C water in a 4 mm tube: turning Nu 3.66 and 4.36 into h

Given: 20 °C water from IAEA Table 3.3, saturated liquid (ρ = 998.2 kg/m³, μ = 1.0000×10⁻³ Pa·s, k = 0.5995 W/(m·K), Pr = 6.981). The 4 mm inner diameter and 0.2 m/s mean velocity are assumed operating conditions.

Flow check: Re = 998.2 × 0.2 × 0.004 ÷ 1.0000×10⁻³ = 798.6, well below the typical pipe transition of about 2300, so laminar.

Calculation: constant wall temperature gives h = 3.66 × 0.5995 ÷ 0.004 = 548.5 W/(m²·K); uniform heat flux gives h = 4.36 × 0.5995 ÷ 0.004 = 653.5 W/(m²·K). Doubling the diameter to 8 mm halves the constant-temperature value to 274.3 W/(m²·K).

Interpretation: the boundary condition alone moves h by about 19%. The constants hold only once the flow is fully developed. The common textbook estimate of laminar thermal entry length, 0.05RePrD, gives 0.05 × 798.6 × 6.981 × 0.004 = 1.11 m, about 279 diameters; shorter tubes have a mean Nu above 3.66. RELAP-7 likewise notes that the fully developed assumption may be questionable in a short pipe with strong entrance effects.

Example 2. Turbulent 60 °C water in a 25 mm pipe: Dittus–Boelter against Gnielinski

Given: 60 °C water from IAEA Table 3.3 (ρ = 983.2 kg/m³, μ = 4.677×10⁻⁴ Pa·s, k = 0.6510 W/(m·K), Pr = 3.005). Assumed: 25 mm inner diameter, 1.0 m/s mean velocity, the wall heating the water, and a 10 K wall-to-fluid difference.

Re: Re = 983.2 × 1.0 × 0.025 ÷ 4.677×10⁻⁴ = 52,555, inside the usual Dittus–Boelter range along with Pr.

Nu and h: for heating (n = 0.4), Nu = 0.023 × 52,5550.8 × 3.0050.4 = 213.5 and h = 213.5 × 0.6510 ÷ 0.025 = 5559 W/(m²·K). Cooling (n = 0.3) gives Nu = 191.2 and h = 4980 W/(m²·K), about 10% lower. With 10 K, the heat flux is q = h·ΔT = 5559 × 10 = 55.6 kW/m².

Second correlation: Gnielinski as in RELAP-7 eqs (592) and (593) gives f = (1.58 ln 52,555 − 3.28)⁻² = 0.00518, Nu = 236.1 and h = 6148 W/(m²·K); Dittus–Boelter is 9.6% lower.

Interpretation: two validated correlations differ by nearly 10%; treat a correlation h as an estimate and size with margin. Against the laminar 549 to 654 W/(m²·K) of Example 1, turbulence raised h by a factor of 8.5 to 10.

Example 3. One h, two numbers: Nu and Bi

Given: h = 5559 W/(m²·K) from Example 2. The tube wall is assumed to be 2 mm of type 316 stainless steel, with conductivity approximated by the 300 K value in IAEA Table 6.20, 14.0 W/(m·K).

Calculation: fluid side, Nu = hD/kfluid = 5559 × 0.025 ÷ 0.6510 = 213.5. Wall side, Bi = ht/kwall = 5559 × 0.002 ÷ 14.0 = 0.79. Putting the wall's k into Nu by mistake gives 5559 × 0.025 ÷ 14.0 = 9.93.

Interpretation: the MIT 3.185 notes separate the two: in Nu conduction and convection both happen in the fluid, while Bi compares conduction resistance in the solid with the fluid boundary-layer resistance. At Bi = 0.79 the wall resistance is comparable to the convective one, so a single wall temperature is a poor assumption.

What the size of Nu says about the thermal layer

Substituting h ~ k/δT, the MIT 3.185 notes rewrite Nu as Nu ~ L/δT, with δT the thermal boundary-layer thickness. A large Nu means a thin thermal layer relative to L. The Nu of 213.5 in Example 2 indicates that the resistance in the 25 mm pipe is concentrated in a wall layer of order 0.1 mm (25 mm ÷ 213.5 = 0.117 mm), a scale estimate rather than a computed thickness.

Conditions to check before using a Nu correlation

Nusselt number questions

I doubled the velocity in a laminar tube. Why did h not change?

In thermally fully developed laminar flow Nu is fixed at 3.66 or 4.36 and h = Nu·k/D contains no velocity. A higher velocity lengthens the entry region, though, so in a short tube the mean h can rise.

Constant wall temperature or constant heat flux: which do I pick?

Whichever is closer to the real heating. Electric heaters approximate constant flux (4.36); tubes surrounded by condensing steam or boiling liquid approximate constant wall temperature (3.66). If unsure, compute both and report the range.

Dittus–Boelter and Gnielinski disagree. Which one is right?

Treat the gap, nearly 10% in Example 2, as correlation uncertainty; check validity ranges and compare with data for the same geometry where possible.

Nu and the Biot number have the same formula. What is the difference?

Nu uses the fluid's k, Bi the solid's. The Michigan Tech CM3120 notes warn that the two have completely different purposes despite confusingly similar definitions.

These are inputs to Nu correlations or help interpret the result.

References