Natural convection
Buoyancy-driven convection: β, Grashof and Rayleigh numbers, vertical and horizontal plate correlations, orientation effects, mixed convection and why radiation matters, with worked numericals.
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Why it matters
When a car is parked, idling in traffic or soaking after a hot run, there is no ram air: the engine bay, the exhaust, the battery pack and the electronics cool largely by natural (free) convection and radiation. Natural-convection coefficients for air are small, only a few W/m²·K, so these "heat-soak" conditions often set the worst-case component temperatures. The same physics drives the thermosiphon in old cooling systems and the airflow in cabin vents.
Key ideas
Mechanism. Fluid next to a hot surface warms, becomes less dense and rises under buoyancy; cooler fluid replaces it. No fan or pump is needed. The velocity is zero at the wall and far from it, with a maximum in between, so the velocity and thermal boundary layers are coupled.
Volume expansion coefficient. β = −(1/ρ)(∂ρ/∂T)_P. For an ideal gas β = 1/T, with T the absolute film temperature in kelvin. For liquids, take β from property tables.
Grashof number. Gr = gβΔT L³/ν² is the ratio of buoyancy to viscous forces. It plays the role that Re² plays in forced convection.
Rayleigh number. Ra = Gr·Pr. Natural-convection correlations are written as Nu = C Ra^n. For a vertical plate the boundary layer becomes turbulent at about Ra ≈ 10⁹. Laminar: n ≈ 1/4; turbulent: n ≈ 1/3, which makes h independent of L in turbulent free convection.
Characteristic length. Vertical plate: its height L. Horizontal cylinder: its diameter D. Horizontal plate: L_c = A/P (area over perimeter).
Orientation matters. For a horizontal hot plate facing up (or cold plate facing down), the buoyant plume rises freely and heat transfer is good. For a hot plate facing down (or cold facing up), the warm fluid is trapped beneath the plate and must escape around its edges, so h is much lower. Vertical plates fall in between.
Properties are evaluated at the film temperature T_f = (T_s + T_∞)/2.
Mixed convection. The ratio Gr/Re² measures the relative importance of buoyancy and forced flow: Gr/Re² ≪ 1 means forced convection dominates; ≫ 1 means natural convection dominates; around 1, both matter. A car at low road speed is often in the mixed regime.
Enclosures (double glazing, battery-module gaps, door cavities) use correlations in terms of the gap width; below a critical Ra the air in the gap is still and heat crosses by conduction only (Nu = 1).
Radiation is comparable. With h of 3–10 W/m²·K, radiation from a surface with high emissivity is of similar size, so in parked-car and heat-soak problems the two are usually added.
Formulas
Gr_L = g β (T_s − T_∞) L³ / ν² — g = 9.81 m/s²; β: 1/K (= 1/T_f for gases); L: m; ν: m²/s.
Ra_L = Gr_L · Pr
Nu = h L / k — k of the fluid at T_f.
Nu = 0.59 Ra^(1/4) (10⁴ < Ra < 10⁹), Nu = 0.10 Ra^(1/3) (10⁹ < Ra < 10¹³) — vertical plate.
Nu = {0.825 + 0.387 Ra^(1/6) / [1 + (0.492/Pr)^(9/16)]^(8/27)}² — vertical plate, all Ra (Churchill–Chu).
Nu = 0.54 Ra^(1/4) (10⁴–10⁷), Nu = 0.15 Ra^(1/3) (10⁷–10¹¹) — upper surface of a hot horizontal plate (or lower surface of a cold one); L_c = A/P.
Nu = 0.27 Ra^(1/4) (10⁵–10¹¹) — lower surface of a hot plate (or upper surface of a cold one).
Gr/Re² — mixed-convection parameter.
Constants vary slightly between textbooks; use the set given in the question or your data book.
Worked examples
Example 1 (standard, vertical plate). The 0.5 m high, 0.4 m wide vertical side of a battery pack is at 60 °C in still air at 20 °C. Air properties at T_f = 40 °C: ν = 1.702 × 10⁻⁵ m²/s, k = 0.02662 W/m·K, Pr = 0.7255. Find h and the convective heat loss.
- β = 1/T_f = 1/313.15 = 3.193 × 10⁻³ K⁻¹.
Gr = g β ΔT L³/ν²= 9.81 × 3.193 × 10⁻³ × 40 × 0.5³/(1.702 × 10⁻⁵)² = 0.1566/(2.897 × 10⁻¹⁰) = 5.41 × 10⁸.Ra = Gr Pr= 5.41 × 10⁸ × 0.7255 = 3.92 × 10⁸ (< 10⁹, laminar).Nu = 0.59 Ra^(1/4)= 0.59 × 140.7 = 83.0.h = Nu k/L= 83.0 × 0.02662/0.5 = 4.42 W/m²·K.Q = h A ΔT= 4.42 × (0.5 × 0.4) × 40 = 35.4 W.
(Churchill–Chu gives Nu ≈ 92, h ≈ 4.9 W/m²·K; correlations agree within about 10–15%, typical for free convection.)
Example 2 (GATE level, horizontal plate). A car roof, treated as a flat horizontal plate 1.5 m × 1.2 m, is heated by the sun to 60 °C in still air at 30 °C. Air at T_f = 45 °C: ν = 1.750 × 10⁻⁵ m²/s, k = 0.02699 W/m·K, Pr = 0.7241. Find the convective loss from the upper surface. How would h change if the same plate were hot on its lower face?
L_c = A/P= (1.5 × 1.2)/(2 × (1.5 + 1.2)) = 1.8/5.4 = 0.333 m.- β = 1/318.15 = 3.143 × 10⁻³ K⁻¹.
- Gr = 9.81 × 3.143 × 10⁻³ × 30 × 0.333³/(1.750 × 10⁻⁵)² = 1.12 × 10⁸; Ra = 1.12 × 10⁸ × 0.7241 = 8.10 × 10⁷.
- Hot surface facing up, Ra in 10⁷–10¹¹:
Nu = 0.15 Ra^(1/3)= 0.15 × 432.7 = 64.9. h = Nu k/L_c= 64.9 × 0.02699/0.333 = 5.26 W/m²·K.Q = h A ΔT= 5.26 × 1.8 × 30 = 284 W.- Hot surface facing down: Nu = 0.27 Ra^(1/4) = 0.27 × 94.9 = 25.6, h = 2.07 W/m²·K, less than half, because the warm air is trapped under the plate.
Common mistakes
- Using β in °C⁻¹ as 1/t(°C); for gases β = 1/T in kelvin.
- Forgetting to square ν in Gr.
- Using plate width instead of height for a vertical plate, or the long side instead of A/P for a horizontal plate.
- Applying the laminar correlation above Ra ≈ 10⁹.
- Swapping the hot-up and hot-down correlations for horizontal surfaces.
- Ignoring radiation, which in still air is often as large as convection.
For GATE ME
Expect calculations of Gr and Ra, identification of laminar or turbulent free convection, h from a supplied correlation, the effect of plate height on h (h ∝ L^(−1/4) in laminar, independent of L in turbulent), and conceptual MCQs on the meaning of Gr, Ra and the mixed-convection parameter Gr/Re². Practise checking the Ra range before choosing the correlation.
Quick check
- Gr = 1.2 × 10⁶ and Pr = 0.7. What is Ra?
- For an ideal gas at a film temperature of 300 K, what is β?
- In laminar free convection on a vertical plate, how does average h change if the plate height increases 16 times?
- Which has the higher h: a hot horizontal plate facing up or facing down?
Answers: 1. 8.4 × 10⁵. 2. 1/300 = 3.33 × 10⁻³ K⁻¹. 3. h ∝ L^(−1/4), so it halves. 4. Facing up.
Interview questions
All Thermodynamics and Heat Transfer interview questionsTry answering each one aloud before you open it.
1.What is natural convection in the context of heat transfer?Concept
Natural convection is a mode of heat transfer where fluid motion is generated by buoyancy forces that result from density variations due to temperature gradients in the fluid. Unlike forced convection, it does not require any external mechanical devices like fans or pumps.
2.Explain how natural convection occurs in a heated room.Concept
In a heated room, the air near the heat source (like a radiator) becomes warmer and less dense. This causes it to rise, and as it rises, cooler, denser air moves in to take its place. This cycle creates a natural circulation pattern, distributing heat throughout the room.
3.What are the key factors affecting natural convection?Concept
The key factors affecting natural convection include the temperature difference between the surface and the fluid, the properties of the fluid (such as viscosity and thermal conductivity), the geometry of the surface, and the gravitational field strength.
4.Where does natural convection matter in a vehicle's thermal management?Application
Natural convection matters whenever there is little forced airflow: when the vehicle is parked after a hot run (heat soak), crawling in traffic, or for components shielded from the airflow. Then the engine bay, exhaust, battery pack and electronics shed heat mainly by free convection and radiation, with h for air only a few W/m²·K, so these conditions often set peak component temperatures. That is why electric fans and coolant pumps may run after shutdown. Historically, thermosiphon cooling systems used buoyancy alone to circulate coolant, but at normal operating loads modern engines rely on forced convection.
5.What happens if the temperature difference in a natural convection system is very small?Application
If the temperature difference in a natural convection system is very small, the buoyancy forces will be weak, leading to minimal fluid motion. This can result in inefficient heat transfer and may require additional methods, such as forced convection, to achieve the desired cooling or heating effect.
6.How does the orientation of a surface affect natural convection heat transfer?Application
Orientation changes how easily the buoyant fluid can move away. A hot horizontal surface facing up sheds a rising plume freely and has a relatively high h; a hot surface facing down traps warm fluid underneath, which must escape around the edges, so h can be less than half as large. A vertical surface lies in between, with a boundary layer that grows up the plate. For cold surfaces the cases swap: a cold plate facing down behaves like a hot plate facing up. Correlations therefore differ by orientation, and horizontal plates use the characteristic length A/P.
7.Why are fins used in heat exchangers to enhance natural convection?Application
Fins are used in heat exchangers to increase the surface area available for heat transfer. By increasing the surface area, fins enhance the rate of heat dissipation through natural convection, making the heat exchanger more efficient without requiring additional energy input.
8.Calculate the heat transfer rate by natural convection from a vertical plate 1 m high and 0.5 m wide, with a surface temperature of 60°C in air at 20°C. Assume a heat transfer coefficient of 5 W/m²·K.Numerical
The heat transfer rate (Q) can be calculated using the formula: Q = h·A·ΔT. Here, h = 5 W/m²·K, A = 1 m × 0.5 m = 0.5 m², and ΔT = 60°C - 20°C = 40 K. Therefore, Q = 5 W/m²·K × 0.5 m² × 40 K = 100 W.
9.A horizontal cylinder with a diameter of 0.1 m and length of 1 m is maintained at 80°C in air at 25°C. If the heat transfer coefficient is 6 W/m²·K, calculate the heat loss due to natural convection.Numerical
Lateral surface area A = πDL = π × 0.1 × 1 = 0.3142 m², ignoring the ends. Temperature difference ΔT = 80 − 25 = 55 K. Then Q = hAΔT = 6 × 0.3142 × 55 ≈ 103.7 W. In practice h would come from a horizontal-cylinder correlation using Ra based on diameter, and radiation from the surface would add a comparable amount.
10.Explain the role of Grashof number in natural convection.Concept
The Grashof number, Gr = gβΔT L³/ν², is the ratio of buoyancy forces to viscous forces, and in natural convection it plays the role that Re² plays in forced convection. Multiplied by the Prandtl number it gives the Rayleigh number, Ra = Gr·Pr, which is what the correlations Nu = C·Ra^n actually use and which decides the regime: for a vertical plate the flow is laminar below about Ra = 10⁹ and turbulent above. The ratio Gr/Re² tells you whether buoyancy matters when there is also a forced flow.
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