Convective heat transfer coefficients and correlations

Newton's law of cooling, boundary layers, Nusselt, Prandtl, Grashof and Reynolds numbers, and forced, natural and in-tube convection correlations.

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Why it matters

Convection decides how fast a casting cools in air, how quickly a quench tank hardens steel, how much coolant a machining centre needs and how big a heat exchanger must be. Since h cannot be looked up as a material property, engineers estimate it from dimensionless correlations; choosing the right correlation for the geometry and flow regime is the core skill of this topic.

Key ideas

Newton's law of cooling. The heat transfer between a surface and a moving fluid is written Q = h·A·(T_s − T∞). This is a definition of the heat transfer coefficient h, not a law of nature: all the physics is hidden in h, which depends on fluid properties, velocity, geometry, flow regime and sometimes temperature difference. Typical values (W/m²·K): free convection in air 2–25; forced convection in air 25–250; forced convection of water 100–20 000; boiling and condensation 2500–100 000.

Mechanism. At the wall the fluid is at rest (no-slip), so heat passes into the fluid by conduction across the innermost layer; fluid motion then carries it away. This gives h = −k_f·(∂T/∂y)_wall/(T_s − T∞). Anything that steepens the temperature gradient at the wall (higher velocity, turbulence, thinner boundary layer) raises h.

Boundary layers. Over a surface a velocity boundary layer and a thermal boundary layer grow from the leading edge. Their relative thickness is set by the Prandtl number: δ_t/δ ≈ Pr^(−1/3) for Pr not too small. Gases (Pr ≈ 0.7) have thermal and velocity layers of similar thickness, the thermal one slightly thicker; oils (Pr in the hundreds) have a much thinner thermal layer; liquid metals (Pr ≪ 1) have a much thicker one. On a flat plate, flow is laminar up to a critical Reynolds number of about 5 × 10⁵ and turbulent beyond.

Forced versus natural convection. In forced convection, a pump, fan or relative motion drives the flow; the correlation is Nu = f(Re, Pr). In natural (free) convection, buoyancy from density differences drives the flow; the correlation is Nu = f(Gr, Pr) or f(Ra), with Rayleigh number Ra = Gr·Pr. When both are important (Gr/Re² ≈ 1) the convection is mixed.

Key dimensionless groups.

  • Nusselt number Nu = hL/k_f: dimensionless heat transfer coefficient; ratio of convective to conductive heat transfer across the fluid layer. (Do not confuse it with the Biot number, which uses the solid's k.)
  • Reynolds number Re = VL/ν: flow regime.
  • Prandtl number Pr = ν/α = μc_p/k: fluid property; momentum versus thermal diffusivity.
  • Grashof number Gr = gβΔT·L³/ν²: buoyancy versus viscous forces. β = 1/T (absolute) for an ideal gas.
  • Stanton number St = Nu/(Re·Pr) = h/(ρVc_p).

Internal flow in tubes. Characteristic length is the diameter. Laminar (Re < 2300) fully developed flow has a constant Nu: 3.66 for constant wall temperature and 4.36 for constant wall heat flux. Turbulent flow uses the Dittus–Boelter correlation. Fluid properties are evaluated at the mean bulk temperature (for external flow, at the film temperature, the mean of surface and free-stream temperatures).

Reynolds analogy. For Pr ≈ 1, heat and momentum transfer are similar: St = C_f/2. The Chilton–Colburn form, St·Pr^(2/3) = C_f/2, extends it to other Prandtl numbers. It lets you estimate h from a known friction factor.

Correlation constants (0.664, 0.037, 0.023 and the like) are empirical and have validity limits on Re and Pr; quote the ranges from your data book when using them.

Formulas

Q = h·A·(T_s − T∞)

  • Q (W); h (W/m²·K); A (m²); temperatures (°C or K, difference only).

Nu = h·L/k_f

  • k_f fluid conductivity (W/m·K); L characteristic length (m).

Nu_avg = 0.664·Re_L^0.5·Pr^(1/3)

  • Laminar flat plate, average over length L, Re_L < 5 × 10⁵, Pr ≥ 0.6.

Nu_avg = 0.037·Re_L^0.8·Pr^(1/3)

  • Turbulent over the whole plate (tripped at the leading edge).

Nu = 0.023·Re^0.8·Pr^n

  • Dittus–Boelter, turbulent tube flow (Re > 10 000, 0.6 < Pr < 160), n = 0.4 for heating, 0.3 for cooling of the fluid.

Nu = 3.66 (constant T_s), Nu = 4.36 (constant q″)

  • Laminar fully developed tube flow.

Re = 4·ṁ/(π·D·μ)

  • Tube Reynolds number from mass flow (kg/s).

Gr = g·β·(T_s − T∞)·L³/ν², Ra = Gr·Pr

  • Natural convection; β (1/K).

Worked examples

Example 1 (standard): laminar flow over a flat plate. Given: air flows at 3 m/s over a 0.5 m long, 1 m wide plate held at 80 °C; free-stream air at 20 °C. Air properties at the film temperature (50 °C), as data: ν = 1.6 × 10⁻⁵ m²/s, k = 0.0263 W/m·K, Pr = 0.71. Find h and the heat loss from one side.

  1. Re_L = V·L/ν = 3 × 0.5/(1.6 × 10⁻⁵) = 93 750 → laminar (below 5 × 10⁵).
  2. Nu = 0.664 × 93 750^0.5 × 0.71^(1/3) = 0.664 × 306.2 × 0.892 = 181.4
  3. h = Nu·k/L = 181.4 × 0.0263/0.5 = 9.54 W/m²·K
  4. Q = h·A·ΔT = 9.54 × (0.5 × 1) × 60 = 286 W Answer: h ≈ 9.5 W/m²·K; Q ≈ 286 W

Example 2 (GATE level): water heated in a tube. Given: 0.2 kg/s of water flows through a 25 mm tube and is heated from 20 °C to 40 °C by a uniform wall heat flux. Properties at the mean bulk temperature (data): μ = 8.0 × 10⁻⁴ Pa·s, k = 0.615 W/m·K, Pr = 5.4, c_p = 4180 J/kg·K. The wall is 15 K hotter than the water along the tube. Find h and the tube length required.

  1. Re = 4ṁ/(πDμ) = 4 × 0.2/(π × 0.025 × 8.0 × 10⁻⁴) = 12 732 → turbulent.
  2. Heating, so n = 0.4: Nu = 0.023 × 12 732^0.8 × 5.4^0.4 = 86.8
  3. h = 86.8 × 0.615/0.025 = 2136 W/m²·K
  4. Heat duty: Q = ṁ·c_p·ΔT = 0.2 × 4180 × 20 = 16 720 W
  5. Q = h·(πDL)·ΔT_wall → L = 16 720/(2136 × π × 0.025 × 15) = 6.65 m Answer: h ≈ 2140 W/m²·K; L ≈ 6.6 m

Example 3 (quick): laminar tube. For laminar, fully developed flow with constant wall temperature in a 10 mm tube with k = 0.6 W/m·K: h = 3.66 × 0.6/0.01 = 220 W/m²·K — independent of velocity.

Common mistakes

  • Using the solid's conductivity in Nu (that gives Biot number).
  • Applying a laminar correlation without checking Re, or Dittus–Boelter for Re < 10 000.
  • Using n = 0.4 when the fluid is being cooled.
  • Using the plate length for an internal-flow Re (use diameter) or diameter for a plate.
  • Evaluating properties at the wrong temperature (film temperature for external flow, bulk temperature for internal flow).
  • Forgetting that h for natural convection depends on ΔT, so Q is not simply proportional to ΔT.

For GATE PI

Expect questions on the meaning of Nu, Pr, Gr and Re, choosing the right correlation, short numericals using flat-plate or Dittus–Boelter correlations (constants given), laminar tube values 3.66 and 4.36, energy balances for fluid heated in a tube, and boundary-layer concepts (δ_t/δ and Pr). Practise the chain Re → Nu → h → Q with careful units.

Quick check

  1. Water (Pr ≈ 6) and air (Pr ≈ 0.7): which has a thermal boundary layer thinner than its velocity boundary layer?
  2. What is Nu for laminar, fully developed flow in a tube with uniform heat flux?
  3. Which dimensionless number replaces Re in natural convection?
  4. A surface at 70 °C in air at 30 °C has h = 12 W/m²·K. Heat flux?

Answers: 1. Water 2. 4.36 3. Grashof number (or Rayleigh number) 4. 480 W/m²

Interview questions

All Thermal and Fluids Engineering interview questions

Try answering each one aloud before you open it.

  1. 1.What is convective heat transfer and how does it differ from conductive heat transfer?Concept

    Convective heat transfer is the transfer of heat between a solid surface and a fluid (liquid or gas) in motion. It involves the movement of the fluid, which enhances the heat transfer process. In contrast, conductive heat transfer occurs through a solid or stationary fluid without any bulk movement, relying solely on molecular interactions. Convection can be natural or forced, depending on whether the fluid motion is due to buoyancy forces or external means like a pump or fan.

  2. 2.Explain the significance of the convective heat transfer coefficient.Concept

    The convective heat transfer coefficient (h) is a measure of the heat transfer rate per unit area and per unit temperature difference between the surface and the fluid. It is significant because it quantifies the efficiency of heat transfer in convective processes. A higher value of h indicates more effective heat transfer. The coefficient depends on factors such as fluid velocity, fluid properties, surface geometry, and temperature difference.

  3. 3.What are some common correlations used to estimate the convective heat transfer coefficient?Concept

    Common correlations for estimating the convective heat transfer coefficient include the Dittus-Boelter equation for turbulent flow in pipes, the Nusselt number correlations for laminar flow, and the Churchill-Bernstein equation for external flow over cylinders. These correlations are empirical and are derived from experimental data, taking into account factors like flow regime, fluid properties, and geometry.

  4. 4.Why is the Nusselt number important in convective heat transfer analysis?Concept

    The Nusselt number (Nu) is a dimensionless number that represents the ratio of convective to conductive heat transfer across a boundary. It is important because it helps in determining the convective heat transfer coefficient. A higher Nusselt number indicates a more efficient convective heat transfer process. It is used in various correlations to estimate the heat transfer coefficient for different flow conditions and geometries.

  5. 5.How does fluid velocity affect the convective heat transfer coefficient?Application

    Fluid velocity significantly affects the convective heat transfer coefficient. As the velocity of the fluid increases, the convective heat transfer coefficient generally increases due to enhanced mixing and reduced thermal boundary layer thickness. This results in more efficient heat transfer. However, the relationship can vary depending on whether the flow is laminar or turbulent.

  6. 6.What happens to the convective heat transfer coefficient if the fluid viscosity increases?Application

    If the fluid viscosity increases, the convective heat transfer coefficient typically decreases. Higher viscosity fluids have a thicker boundary layer, which reduces the rate of heat transfer. This is because the fluid's resistance to flow increases, leading to less effective mixing and slower heat transfer rates.

  7. 7.Why is forced convection often used in industrial heat exchangers?Application

    Forced convection is often used in industrial heat exchangers because it allows for greater control over the heat transfer process. By using pumps or fans to move the fluid, the flow rate can be increased, leading to higher convective heat transfer coefficients and more efficient heat transfer. This is particularly important in applications where large amounts of heat need to be transferred quickly and efficiently.

  8. 8.Calculate the convective heat transfer coefficient for air flowing over a flat plate with a Nusselt number of 50, thermal conductivity of air as 0.026 W/m·K, and characteristic length of 0.5 m.Numerical

    To calculate the convective heat transfer coefficient (h), use the formula: h = Nu * k / L. Here, Nu = 50, k = 0.026 W/m·K, and L = 0.5 m. Therefore, h = 50 * 0.026 / 0.5 = 2.6 W/m²·K.

  9. 9.A fluid with a Prandtl number of 0.7 flows over a heated cylinder. Explain how the Prandtl number affects the heat transfer process.Application

    Pr = ν/α compares how fast momentum and heat diffuse, and it sets the relative thickness of the velocity and thermal boundary layers, roughly δ_t/δ ≈ Pr^(−1/3). At Pr = 0.7 (air and most gases) heat diffuses slightly faster than momentum, so the thermal boundary layer is a little thicker than the velocity layer and the two are of similar size. Correlations include Pr as a factor, typically Nu ∝ Pr^(1/3) to Pr^0.4, so at the same Reynolds number air gives a lower Nu than water (Pr ≈ 5–7) or oil (Pr in the hundreds).

  10. 10.Determine the heat transfer rate from a surface with an area of 2 m², a surface temperature of 80°C, and an ambient air temperature of 25°C, given a convective heat transfer coefficient of 10 W/m²·K.Numerical

    The heat transfer rate (Q) can be calculated using the formula: Q = h * A * ΔT. Here, h = 10 W/m²·K, A = 2 m², and ΔT = 80°C - 25°C = 55°C. Therefore, Q = 10 * 2 * 55 = 1100 W.

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