Natural convection correlations

Buoyancy-driven convection: Grashof and Rayleigh numbers, standard correlations for plates and cylinders, scaling of h, and mixed convection.

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Why it matters

Heat loss from hot pipes, tanks and reactor shells to still air, cooling of transformers and electronics without fans, heating of liquids in unstirred storage tanks and the outside coefficient in many insulation problems are all natural (free) convection. Its coefficients are small — a few W/m²·K in air — so it often controls the overall resistance and must be estimated properly before an insulation thickness or a tank heater is chosen.

Key ideas

Mechanism. No pump or fan drives the flow. Fluid next to a hot surface is heated, its density falls, and buoyancy lifts it; cooler fluid moves in to replace it. The velocity is therefore caused by the temperature field, so the momentum and energy equations are coupled. The velocity is zero both at the wall and far away, peaking inside the boundary layer.

Volumetric expansion coefficient β. β = −(1/ρ)(∂ρ/∂T)_p (1/K). For an ideal gas β = 1/T with T in kelvin (absolute film temperature). For liquids take β from tables (water at 25 °C ≈ 2.6 × 10⁻⁴ 1/K).

Grashof number Gr = g·β·ΔT·L³/ν². Ratio of buoyancy to viscous forces; it plays the role that Re² plays in forced convection. ΔT = |T_s − T∞|.

Rayleigh number Ra = Gr·Pr = g·β·ΔT·L³/(ν·α). Correlations are written Nu = C·Raⁿ. For a vertical plate the boundary layer is laminar up to Ra ≈ 10⁹ and turbulent beyond.

Exponent tells you how h scales.

  • Laminar (n = 1/4): h ∝ (ΔT/L)^(1/4). A taller plate has a lower average h.
  • Turbulent (n = 1/3): h ∝ ΔT^(1/3) and is independent of L, because L³ inside Ra^(1/3) cancels the L in Nu = hL/k. This makes turbulent free convection easy to scale up.

Characteristic length. Vertical plate or vertical cylinder (if not too thin): height L. Horizontal cylinder: diameter D. Sphere: D. Horizontal plate: L = A_s/P (area divided by perimeter).

Orientation matters for horizontal plates. The hot surface facing up (or cold facing down) lets the plume rise freely and gives a higher h than the hot surface facing down, where warm fluid is trapped beneath the plate.

Properties at the film temperature T_f = (T_s + T∞)/2, including β = 1/T_f for gases.

Mixed convection. When a forced flow is also present, compare Gr/Re²: much less than 1 → forced convection dominates; about 1 → mixed; much greater than 1 → natural convection dominates.

Enclosures. Fluid between two plates or in double glazing is described by an effective conductivity k_eff/k = f(Ra) based on the gap width; below Ra ≈ 1000 the fluid is effectively stagnant and k_eff = k.

Correlations. The simple McAdams-type forms below are widely taught; the Churchill–Chu correlations cover the whole Ra range in one equation and are more accurate. Constants differ slightly between textbooks, so use the set your syllabus or data book specifies and its validity range.

Formulas

Gr = g·β·(T_s − T∞)·L³/ν² g = 9.81 m/s², β (1/K), L characteristic length (m), ν kinematic viscosity (m²/s).

Ra = Gr·Pr, Nu = h·L/k = C·Raⁿ Rayleigh number; average Nusselt number with k of the fluid (W/m·K).

Nu = 0.59·Ra^(1/4) (10⁴ < Ra < 10⁹), Nu = 0.10·Ra^(1/3) (10⁹ < Ra < 10¹³) Vertical plate or large vertical cylinder, L = height.

Nu = 0.53·Ra^(1/4) (10⁴ < Ra < 10⁹) Horizontal cylinder, L = D (McAdams).

Nu = {0.6 + 0.387·Ra^(1/6)/[1 + (0.559/Pr)^(9/16)]^(8/27)}² Horizontal cylinder, Churchill–Chu, Ra < 10¹².

Nu = 0.54·Ra^(1/4) (10⁴–10⁷), Nu = 0.15·Ra^(1/3) (10⁷–10¹¹) Horizontal plate, hot surface facing up; L = A_s/P.

Nu = 0.27·Ra^(1/4) (10⁵–10¹⁰) Horizontal plate, hot surface facing down.

β = 1/T_f Ideal gas; T_f absolute film temperature (K).

Gr/Re² Mixed-convection criterion.

Worked examples

Example 1 (standard). A vertical plate 0.5 m high is held at 70 °C in quiescent air at 30 °C. Air properties at the film temperature 50 °C (323 K): ν = 1.798 × 10⁻⁵ m²/s, k = 0.02735 W/m·K, Pr = 0.723. Find the average h and the heat loss per metre width from one side.

  1. β = 1/T_f = 1/323 K⁻¹.
  2. Gr = g·β·ΔT·L³/ν² = 9.81 × (1/323) × 40 × 0.125/(1.798 × 10⁻⁵)² = 4.70 × 10⁸.
  3. Ra = Gr·Pr = 4.70 × 10⁸ × 0.723 = 3.40 × 10⁸ < 10⁹ → laminar.
  4. Nu = 0.59·Ra^(1/4) = 0.59 × 135.7 = 80.1.
  5. h = Nu·k/L = 80.1 × 0.02735/0.5 = 4.38 W/m²·K.
  6. q = h·A·ΔT = 4.38 × (0.5 × 1) × 40 = 87.6 W.

Answer: h ≈ 4.4 W/m²·K; q ≈ 88 W per metre width.

Example 2 (GATE level). A bare horizontal pipe of 80 mm outside diameter at 90 °C runs through still air at 30 °C. Properties at 60 °C (333 K): ν = 1.896 × 10⁻⁵ m²/s, k = 0.02808 W/m·K, Pr = 0.720. (a) Find h and the heat loss per metre. (b) For a vertical plate in the laminar regime, by what factors do the average h and the heat loss per unit width change if the height is doubled?

(a)

  1. β = 1/333 K⁻¹; ΔT = 60 K; L = D = 0.08 m.
  2. Ra = g·β·ΔT·D³·Pr/ν² = 9.81 × (1/333) × 60 × 5.12 × 10⁻⁴ × 0.720/(1.896 × 10⁻⁵)² = 1.81 × 10⁶.
  3. Nu = 0.53·Ra^(1/4) = 0.53 × 36.7 = 19.4.
  4. h = 19.4 × 0.02808/0.08 = 6.83 W/m²·K.
  5. q = h·π·D·ΔT = 6.83 × π × 0.08 × 60 = 103 W per metre.
  6. Cross-check: Churchill–Chu gives Nu = 17.2, h = 6.05 W/m²·K — correlations differ by about 10 %, typical of free convection. (b)
  7. Laminar: h ∝ L^(−1/4), so h changes by 2^(−1/4) = 0.84.
  8. Heat loss ∝ h·L, so it changes by 2^(3/4) = 1.68.

Answer: (a) h ≈ 6.8 W/m²·K, q ≈ 103 W/m (radiation, often similar in size, is extra); (b) h × 0.84, q × 1.68.

Common mistakes

  • Using β = 1/T with T in °C. It must be absolute temperature (K).
  • Using diameter for a vertical cylinder or height for a horizontal one.
  • Forgetting to check Ra against the correlation's range (laminar vs turbulent constants).
  • Ignoring radiation: from a surface at 70–100 °C in air, radiation is comparable to free convection.
  • Using hot-facing-up constants for a hot plate facing down.
  • Treating a forced-convection problem as free when Gr/Re² ≪ 1, or vice versa.

For GATE CH

Expect calculations of Gr, Ra, Nu and h from a given correlation, ratio questions (how h or q changes with height, ΔT or diameter for laminar Ra^(1/4) or turbulent Ra^(1/3) scaling), the identification of the controlling dimensionless group, and conceptual questions on β for ideal gases and mixed convection. Practise expressing h ∝ ΔTᵃ·Lᵇ directly from the exponent of Ra.

Quick check

  1. What is β for air at a film temperature of 27 °C?
  2. Which dimensionless group replaces Re in free convection?
  3. In turbulent free convection on a vertical plate, does h depend on plate height?
  4. A horizontal plate is 2 m × 1 m. What characteristic length do horizontal-plate correlations use?
  5. Gr/Re² = 0.01. Forced, mixed or natural convection?

Answers: 1. 1/300 K⁻¹ ≈ 3.33 × 10⁻³ K⁻¹. 2. Grashof number (or Ra = Gr·Pr). 3. No. 4. A/P = 2/6 = 0.333 m. 5. Forced convection dominates.

Interview questions

All Heat Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What drives natural convection, and how is it different from forced convection?Concept

    In natural convection the flow is created by the heat transfer itself: fluid near a hot surface becomes less dense and rises under buoyancy, while cooler, denser fluid replaces it. No pump or fan is involved, so velocities and heat transfer coefficients are small (typically 2–25 W/m²·K in gases). Because the velocity field depends on the temperature field, the momentum and energy equations are coupled, and correlations are in terms of Gr or Ra instead of Re.

  2. 2.Define the Grashof and Rayleigh numbers and explain their physical meaning.Concept

    Gr = g·β·ΔT·L³/ν² is the ratio of buoyancy forces to viscous forces; it plays the role of Re² in forced convection. Ra = Gr·Pr = g·β·ΔT·L³/(ν·α) also brings in the ratio of momentum to thermal diffusion and is the parameter used in most correlations, Nu = C·Raⁿ. For a vertical plate the boundary layer turns turbulent at about Ra = 10⁹.

  3. 3.How do you evaluate β for air in a natural convection calculation?Concept

    For an ideal gas at constant pressure ρ = p/(RT), so β = −(1/ρ)(∂ρ/∂T)_p = 1/T. T must be the absolute film temperature, (T_s + T∞)/2 in kelvin, the same temperature at which ν, k and Pr are read. Using °C is a common error that can make β wrong by an order of magnitude. For liquids β is taken from property tables.

  4. 4.Why does the average heat transfer coefficient in turbulent natural convection not depend on the height of the plate?Concept

    In the turbulent regime Nu = C·Ra^(1/3). Since Ra ∝ L³, Ra^(1/3) ∝ L, and Nu = hL/k ∝ L, so the L cancels and h ∝ (gβΔT/να)^(1/3)·k, independent of height. Physically the turbulent layer's resistance is set by a thin near-wall region whose thickness does not grow with height. This makes laboratory results on small surfaces easy to scale up to large walls and tanks.

  5. 5.Why does a hot horizontal plate facing upward lose more heat by natural convection than the same plate facing downward?Concept

    When the hot surface faces up, heated fluid can rise freely off the surface in plumes and is continuously replaced by cooler fluid, so the boundary layer stays thin. When it faces down, the heated, lighter fluid is trapped beneath the plate and must creep sideways to the edges before it can rise, giving a thick, sluggish layer and a lower h. Correlations reflect this, e.g. Nu ≈ 0.54Ra^(1/4) for hot-up versus 0.27Ra^(1/4) for hot-down in the laminar range. The opposite holds for a cold plate.

  6. 6.When a fan blows over a heated surface, how do you decide whether natural convection also needs to be considered?Concept

    Compare buoyancy with inertia using Gr/Re² (the Richardson number). If Gr/Re² ≪ 1 (say below 0.1) forced convection dominates and natural convection can be ignored; if it is ≫ 1 (say above 10) natural convection dominates; around 1 it is mixed convection and both must be combined, often as Nuⁿ = Nu_forced ⁿ ± Nu_natural ⁿ with n ≈ 3. The sign depends on whether buoyancy assists or opposes the forced flow.

  7. 7.A vertical plate cooled by laminar natural convection has its height doubled. How do the average h and the heat loss per unit width change?Concept

    In laminar free convection Nu = C·Ra^(1/4) with Ra ∝ L³, so h ∝ L^(−1/4). Doubling the height multiplies h by 2^(−1/4) ≈ 0.84. The heat loss per unit width is proportional to h·L, so it rises by 2^(3/4) ≈ 1.68, assuming the flow stays laminar (Ra < 10⁹).

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