Forced and natural convection

Newton's law of cooling, boundary layers, Nu, Re, Pr, Gr and Ra, flat-plate, pipe-flow and natural-convection correlations, with plate, Dittus–Boelter and vertical-plate examples.

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Why it matters

Convection sets the heat flow at almost every surface a mechatronics engineer designs: a heat sink in a fan duct, a motor casing in still air, coolant in a battery-pack channel, an enclosure wall on a sunny roof. The convection coefficient h is not a material property; it depends on the flow, the fluid and the geometry, and choosing the right correlation for it is most of the work.

Key ideas

Newton's law of cooling. Heat flow from a surface is Q = hA(T_s − T∞). All the physics is hidden in h, which ranges from about 2–25 W/m²·K for natural convection in gases, 25–250 for forced convection in gases, 50–20 000 for liquids, and far higher for boiling and condensation.

Boundary layers. Near the wall the fluid velocity falls to zero (no slip), so heat crosses the wall–fluid interface by conduction through a thin thermal boundary layer. A thinner boundary layer means a larger h. The velocity layer grows along a flat plate; the flow is laminar up to a critical Reynolds number of about 5 × 10⁵ and turbulent beyond it, and turbulence greatly increases h.

Dimensionless groups.

  • Nusselt number Nu = hL/k_f — the convection coefficient made dimensionless with the fluid conductivity; it is the ratio of convective to purely conductive heat transfer across the fluid layer.
  • Reynolds number Re = ρVL/μ = VL/ν — inertia versus viscous forces; decides laminar or turbulent in forced convection.
  • Prandtl number Pr = ν/α = μc_p/k — momentum diffusivity versus thermal diffusivity; about 0.7 for air, 2–7 for water, hundreds for oils, ≪ 1 for liquid metals.
  • Grashof number Gr = gβΔT L³/ν² — buoyancy versus viscous forces; plays the role of Re² in natural convection.
  • Rayleigh number Ra = Gr·Pr — the single group used in most natural-convection correlations.

Forced convection correlations take the form Nu = f(Re, Pr); natural convection correlations take Nu = f(Gr, Pr) = f(Ra). When Gr/Re² is near 1, both matter (mixed convection).

Properties. Evaluate fluid properties at the film temperature T_f = (T_s + T∞)/2 for external flow, and at the bulk mean temperature for internal flow. For an ideal gas β = 1/T_f in kelvin.

Internal flow. In a pipe the flow is laminar below Re_D ≈ 2300. For fully developed laminar flow Nu_D is constant (3.66 for uniform wall temperature, 4.36 for uniform heat flux), independent of velocity. For turbulent flow use Dittus–Boelter (Re_D > 10 000, 0.6 < Pr < 160, L/D > 10).

Natural convection. Hot fluid rises along a heated vertical wall. A hot surface facing up (or a cold surface facing down) gives stronger plumes and higher h than a hot surface facing down. The exponent in Nu = C·Raⁿ is about ¼ for laminar and ⅓ for turbulent flow; the transition on a vertical plate is near Ra ≈ 10⁹. Take C and n for each geometry from your data book.

Formulas

Q = h A (T_s − T∞) — Q (W), h (W/m²·K), A (m²). Nu = h L / k_f — L: characteristic length (m), k_f: fluid conductivity (W/m·K). Re = V L / ν — V (m/s), ν: kinematic viscosity (m²/s). For a pipe, Re_D = 4ṁ / (π D μ). Pr = μ c_p / k Gr = g β (T_s − T∞) L³ / ν² — g = 9.81 m/s², β (1/K). Ra = Gr · Pr Nu_L = 0.664 Re_L^0.5 Pr^(1/3) — laminar flat plate, average, Re_L < 5 × 10⁵, Pr ≥ 0.6. Nu_x = 0.332 Re_x^0.5 Pr^(1/3) — laminar flat plate, local; local h falls as x^(−1/2) and the average over L is twice the local value at L. δ ≈ 5 x / √Re_x — laminar velocity boundary-layer thickness (m). Nu_L = 0.037 Re_L^0.8 Pr^(1/3) — fully turbulent flat plate (average). Nu_D = 3.66 (uniform T_s) or 4.36 (uniform q) — fully developed laminar pipe flow. Nu_D = 0.023 Re_D^0.8 Prⁿ — Dittus–Boelter; n = 0.4 when the fluid is heated, 0.3 when cooled. Nu_L = 0.59 Ra_L^(1/4) (10⁴ < Ra < 10⁹), Nu_L = 0.10 Ra_L^(1/3) (10⁹ < Ra < 10¹³) — vertical plate, L = height (typical data-book constants).

Worked examples

Example 1 (standard, flat plate). Air at 20 °C flows at 3 m/s over a plate 0.5 m long (in the flow direction) and 1 m wide, held at 60 °C. Air at the film temperature 40 °C: ν = 16.97 × 10⁻⁶ m²/s, k = 0.02735 W/m·K, Pr = 0.7255 (data book). Find h and the heat loss from one side.

  1. Re_L = V L / ν = 3 × 0.5 / 16.97 × 10⁻⁶ = 88 390 < 5 × 10⁵, laminar throughout.
  2. Nu_L = 0.664 Re^0.5 Pr^(1/3) = 0.664 × 297.3 × 0.8985 = 177.4.
  3. h = Nu k / L = 177.4 × 0.02735 / 0.5 = 9.70 W/m²·K.
  4. Q = h A ΔT = 9.70 × (0.5 × 1) × 40 = 194 W.

h ≈ 9.7 W/m²·K, Q ≈ 194 W

Example 2 (GATE level, pipe flow). Water flows at 0.2 kg/s through a 20 mm bore tube and is being heated. At the bulk mean temperature: μ = 6.53 × 10⁻⁴ Pa·s, k = 0.631 W/m·K, Pr = 4.32. Find h.

  1. Re_D = 4ṁ / (π D μ) = 4 × 0.2 / (π × 0.02 × 6.53 × 10⁻⁴) = 19 500 > 10 000, turbulent.
  2. Nu_D = 0.023 Re^0.8 Pr^0.4 (heating, n = 0.4) = 0.023 × 2704 × 1.796 = 111.7.
  3. h = Nu k / D = 111.7 × 0.631 / 0.02 = 3520 W/m²·K.
  4. Since Nu ∝ Re^0.8, doubling the flow rate at the same diameter raises h by 2^0.8 = 1.74 times.

h ≈ 3.5 kW/m²·K

Example 3 (natural convection). The same 0.5 m tall plate (1 m wide) at 60 °C hangs vertically in still air at 20 °C. Use the properties of Example 1 and β = 1/313 K⁻¹.

  1. Gr = g β ΔT L³ / ν² = 9.81 × (1/313) × 40 × 0.125 / (16.97 × 10⁻⁶)² = 5.44 × 10⁸.
  2. Ra = Gr Pr = 3.95 × 10⁸ < 10⁹, laminar.
  3. Nu = 0.59 Ra^(1/4) = 0.59 × 141.0 = 83.2; h = 83.2 × 0.02735 / 0.5 = 4.55 W/m²·K.
  4. Q = 4.55 × 0.5 × 40 = 91 W, less than half of the forced-convection value.

h ≈ 4.5 W/m²·K, Q ≈ 91 W

Common mistakes

  • Using the solid's conductivity in Nu (that belongs to Bi).
  • Taking ν in the Grashof number as dynamic viscosity, or forgetting to square it.
  • Using °C for β = 1/T; it must be in kelvin.
  • Applying a turbulent correlation to laminar flow, or Dittus–Boelter below Re ≈ 10 000.
  • Using the local Nusselt number where the average is asked (the laminar average is twice the local value at the trailing edge).
  • Mixing heating and cooling exponents in Dittus–Boelter.

For GATE ME

Expect calculation of Re, Nu and h from a given correlation, ratios of h when velocity or length changes (h ∝ V^0.5 laminar, V^0.8 turbulent; laminar plate h̄ ∝ L^(−1/2)), boundary-layer thickness growth, local versus average h, the physical meaning of Pr and Gr, and identifying which groups govern forced versus natural convection. Practise proportionality reasoning; correlations are given in the question.

Quick check

  1. Which dimensionless groups appear in forced-convection correlations?
  2. If velocity doubles over a laminar flat plate, by what factor does h̄ change?
  3. What is β for air at a film temperature of 300 K?
  4. In laminar fully developed pipe flow with uniform wall temperature, what is Nu_D?
  5. Which fluid conductivity goes in Nu: the solid's or the fluid's?

Answers: 1. Nu, Re and Pr. 2. √2 ≈ 1.41. 3. 1/300 = 3.33 × 10⁻³ K⁻¹. 4. 3.66. 5. The fluid's.

Try answering each one aloud before you open it.

  1. 1.What is forced convection?Concept

    Forced convection is a mechanism where fluid motion is generated by an external source like a pump, fan, or a mixer. This external force enhances the heat transfer between the surface and the fluid. It is commonly used in applications where natural convection is insufficient to transfer heat effectively.

  2. 2.What is natural convection?Concept

    Natural convection is a heat transfer process where fluid motion is caused by buoyancy forces that result from density variations due to temperature differences in the fluid. This type of convection occurs without any external force, relying solely on the natural movement of the fluid.

  3. 3.Explain the difference between forced and natural convection.Concept

    The primary difference between forced and natural convection is the source of fluid motion. In forced convection, an external device like a fan or pump induces fluid movement, while in natural convection, the fluid motion is due to buoyancy forces arising from temperature-induced density differences. Forced convection generally results in higher heat transfer rates compared to natural convection.

  4. 4.Why is forced convection used in car radiators?Application

    Forced convection is used in car radiators to enhance the heat dissipation from the engine coolant to the air. A fan is typically used to increase the airflow over the radiator fins, which improves the heat transfer rate and helps maintain the engine at an optimal temperature, especially when the vehicle is stationary or moving slowly.

  5. 5.What happens if a fan in a forced convection system fails?Application

    If a fan in a forced convection system fails, the rate of heat transfer will significantly decrease because the fluid motion is no longer enhanced by the fan. This can lead to overheating of the system, as natural convection alone may not be sufficient to dissipate the heat generated, especially in high-power applications.

  6. 6.How does the orientation of a heated surface affect natural convection?Application

    Orientation decides how freely the buoyant warm fluid can move away. On a vertical heated plate a boundary layer rises along the surface and is well described by Nu = C·Raⁿ with the height as length. A hot surface facing upward (or a cold surface facing down) lets plumes rise freely and gives a relatively high h, while a hot surface facing downward traps warm fluid under it and gives a much lower h. Correlations, constants and characteristic lengths therefore differ by orientation and are taken from the data book; this is why heat-sink fins are usually mounted vertically.

  7. 7.Why is natural convection often used in electronic cooling systems?Application

    Natural convection is often used in electronic cooling systems because it is a passive cooling method that does not require additional energy input or moving parts, reducing the risk of mechanical failure. It is suitable for low-power devices where the heat generated is not excessive, and the ambient conditions allow for sufficient heat dissipation.

  8. 8.Calculate the heat transfer rate for a flat plate with an area of 2 m², a surface temperature of 80°C, and an ambient air temperature of 20°C, using a heat transfer coefficient of 25 W/m²·K.Numerical

    The heat transfer rate (Q) can be calculated using the formula: Q = h·A·ΔT, where h is the heat transfer coefficient, A is the area, and ΔT is the temperature difference. Here, Q = 25 W/m²·K × 2 m² × (80°C - 20°C) = 25 × 2 × 60 = 3000 W.

  9. 9.What factors affect the heat transfer coefficient in forced convection?Concept

    In forced convection, the heat transfer coefficient is influenced by several factors, including the velocity of the fluid, the properties of the fluid (such as viscosity and thermal conductivity), the surface geometry, and the type of flow (laminar or turbulent). Higher fluid velocities and turbulent flow generally increase the heat transfer coefficient.

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