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Heat Transfer · Chapter 7 of 10 · Intermediate

Natural Convection

No fan, just buoyancy. Warm fluid rises and drives its own gentle flow, giving the small but free heat transfer that passively cools so much equipment.

Readiness check

From Chapters 5 and 6. Tick only what you can do closed-notes.

Use Nu = hL/k to get h.
Recall the Prandtl number and the film temperature.
Understand that warm fluid is less dense and rises.
Handle quarter-power and cube terms in a formula.
Read temperatures in kelvin for property work.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRevisit Nu and the film temperature in Chapter 5.

3 or more weak itemsReread the convection fundamentals of Chapter 5 first.

The core idea

A hot surface warms the fluid touching it; the warm fluid is lighter and rises, dragging cooler fluid in behind it.

Ra = gβ(T_s−T_∞)L³/(να)Nu = C·Raⁿh = Nu·k/L

With no pump or fan, the flow is created by the heat transfer itself, through buoyancy. The Rayleigh number Ra (the product of the Grashof and Prandtl numbers) plays the role Reynolds played in forced convection: it sets the regime and the correlation. Natural-convection coefficients are small, but they cost nothing to run.

The skill works when: there is no significant forced flow, so buoyancy dominates and a Rayleigh-based correlation applies.

The skill breaks down when: even a light breeze is present: once Gr/Re² is well below 1, forced convection takes over and natural-convection formulas understate h.

The concept. A hot vertical surface heats the adjacent fluid, which becomes buoyant and rises in a thin layer, pulling cooler fluid up from below. The heat transfer creates the very flow that carries it away.

The skills, taught in order

Natural convection reuses the Nusselt machinery, swapping Reynolds for Rayleigh. Six skills cover what drives the flow, the number that scales it, and when buoyancy still matters.

7.1 What drives natural convection

Heat lowers the density of the fluid near a hot surface, and gravity lifts that lighter fluid. The strength of the effect is set by the volumetric thermal expansion coefficient β; for an ideal gas, β = 1/T with T in kelvin.

7.2 The Grashof and Rayleigh numbers

The Grashof number Gr = gβΔT L³/ν² compares buoyancy to viscous damping. Multiplying by Pr gives the Rayleigh number Ra = Gr·Pr, the single group that governs natural convection. It replaces Reynolds as both the regime indicator and the correlation variable.

Micro-example. For air at 313 K, β = 1/313 = 3.2×10⁻³ K⁻¹, so a 40 K difference over a 0.3 m wall already gives Ra ≈ 8×10⁷, comfortably laminar.

7.3 The vertical-plate correlation

For a vertical isothermal wall, Nu = C·Raⁿ: roughly Nu = 0.59 Ra^1/4 in the laminar range (10⁴ to 10⁹) and Nu = 0.10 Ra^1/3 when turbulent (above ~10⁹). The Churchill-Chu correlation covers the whole range more precisely.

Vertical plate	Nusselt number
Laminar (10⁴ < Ra < 10⁹)	Nu ≈ 0.59 Ra^1/4
Turbulent (Ra > 10⁹)	Nu ≈ 0.10 Ra^1/3

7.4 Other geometries

Horizontal plates, cylinders, spheres, and enclosed layers each have their own C and n and their own characteristic length (a heated plate facing up convects more than one facing down). The form Nu = C·Raⁿ persists; only the constants change.

7.5 Natural versus forced

Natural-convection coefficients are small, typically 2 to 25 W/m²·K for gases. The ratio Gr/Re² tells you which mechanism rules: much greater than 1 means buoyancy dominates, much less than 1 means a forced flow has taken over, and near 1 both matter (mixed convection).

7.6 Why it matters

Natural convection is free and silent, so it cools transformers, baseboard heaters, fanless electronics, and building surfaces, and it drives the heat loss from any warm object left in still air. Its small h is exactly why such surfaces are often finned.

Engineering connection: passive and fanless cooling, building heat loss, and the still-air h that the fins of Chapter 3 are designed to overcome.

Worked example 1: a hot vertical panel in still air

A 0.3 m tall electronics panel sits at 60 °C in still air at 20 °C. Using air properties at the film temperature (ν = 1.7×10⁻⁵ m²/s, α = 2.4×10⁻⁵ m²/s, k = 0.027 W/m·K, Pr = 0.71, β = 1/313 K⁻¹), find the natural-convection coefficient.

Figure 1. A warm vertical panel cooling by buoyant air. Ra ≈ 8×10⁷ is laminar, giving Nu ≈ 56 and a modest h ≈ 5 W/m²·K, typical of free convection in air.

ProblemFind the natural-convection coefficient on the vertical panel in Figure 1.
Given / findL = 0.3 m, T_s = 60 °C, T_∞ = 20 °C, ν = 1.7×10⁻⁵, α = 2.4×10⁻⁵, k = 0.027, Pr = 0.71, β = 1/313. Find h.
AssumptionsStill surroundings, isothermal vertical wall, ideal-gas β, properties at the film temperature.
ModelCompute Ra, confirm the laminar range, apply the vertical-plate correlation, and convert Nu to h.
EquationsRa = gβ(T_s−T_∞)L³/(να) Nu = 0.59 Ra^1/4 h = Nu·k/L
SolveRa = 9.81 × (1/313) × 40 × 0.3³ / (1.7×10⁻⁵ × 2.4×10⁻⁵) = 8.3×10⁷ (laminar). Nu = 0.59 × (8.3×10⁷)^1/4 = 0.59 × 95.4 = 56.3. h = 56.3 × 0.027 / 0.3 = 5.1 W/m²·K.
CheckFive W/m²·K sits in the free-convection-gas range (2 to 25) from Chapter 1, so it is reasonable. Buoyancy alone is a weak convector, which is why panels like this run warm unless finned or fanned.
ConclusionFree convection gives only about 5 W/m²·K here. To shed serious heat passively you must add area (fins) or accept a large surface; to do better still you need a fan, which is the next example's question.

Result. Ra = 8.3×10⁷ (laminar), Nu ≈ 56, h ≈ 5.1 W/m²·K.

Worked example 2: does a breeze take over?

A 3 m/s breeze now blows across the same panel. Find the forced-convection coefficient, compute Gr/Re² to judge which mechanism dominates, and compare with the natural-convection result.

Figure 2. With a 3 m/s breeze, Gr/Re² falls to about 0.04, so forced convection dominates and lifts h to roughly 12 W/m²·K, more than double the still-air value.

ProblemFor the panel in a 3 m/s breeze, find the forced h, the ratio Gr/Re², and the comparison with natural convection.
Given / findSame panel and properties as Worked Example 1; V = 3 m/s. Find h_forced, Gr/Re², and h_forced/h_natural.
AssumptionsFlat-plate flow, laminar (Re below 5×10⁵), same film properties.
ModelUse the laminar flat-plate correlation for the forced h, then form Gr/Re² with Gr = Ra/Pr from Worked Example 1.
EquationsRe = VL/ν, Nu = 0.664 Re^1/2Pr^1/3 Gr = Ra/Pr, criterion Gr/Re²
SolveRe = 3 × 0.3 / 1.7×10⁻⁵ = 52 900. Nu = 0.664 × 52900^1/2 × 0.71^1/3 = 136, so h_forced = 136 × 0.027/0.3 = 12.3 W/m²·K. Gr = Ra/Pr = 8.3×10⁷/0.71 = 1.17×10⁸; Gr/Re² = 1.17×10⁸/(52900²) = 0.042, far below 1, so forced convection dominates. The ratio h_forced/h_natural = 12.3/5.1 ≈ 2.4.
CheckGr/Re² ≪ 1 and h_forced > h_natural agree: once even a light breeze exists, buoyancy is a side effect. A stronger wind would push the ratio further down and the gap wider.
ConclusionBuoyancy is easily overwhelmed: a gentle 3 m/s breeze more than doubles the coefficient and makes the natural-convection estimate an underprediction. Always check Gr/Re² before trusting a free-convection number outdoors or near any airflow.

Result. h_forced ≈ 12.3 W/m²·K; Gr/Re² = 0.042 (forced dominates); about 2.4 times the still-air value.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Reynolds used for natural convection	Re sought where there is no forced flow	"What is driving the flow here?"	Buoyancy drives it; use the Rayleigh number, not Reynolds.
β in Celsius	Expansion coefficient an order of magnitude off	"Is my temperature in kelvin for β = 1/T?"	β for a gas is 1/T with T absolute; use kelvin.
Natural formula in a draft	Free-convection h used outdoors with wind	"Is Gr/Re² near or below 1?"	If a forced flow exists, check Gr/Re²; below 1, switch to a forced correlation.
Same correlation for every geometry	Vertical-plate constants used for a horizontal plate	"Does this geometry have its own C, n, and L?"	Each orientation has its own correlation and characteristic length; pick the right one.

Practice ladder

Level 1 · Direct skill

Find β for air at 350 K, and explain why warm air rises faster than cooler air for the same temperature difference.

Show answer

β = 1/T = 1/350 = 2.86×10⁻³ K⁻¹. A higher β means a given ΔT produces a larger density difference and stronger buoyancy, so hotter ambient conditions actually weaken β slightly while the driving ΔT matters most.

Level 2 · Mixed concept

Recompute Worked Example 1 for a taller, 0.9 m panel at the same temperatures. How do Ra and h change, given Ra ∝ L³ and Nu ∝ Ra^1/4?

Show answer

Ra scales with L³, so it rises 27-fold to about 2.2×10⁹ (now just turbulent). h = Nu·k/L: Nu ∝ Ra^1/4 ∝ L^3/4, and dividing by L gives h ∝ L^−1/4, so h actually falls slightly with height, to about 4.2 W/m²·K. Taller walls convect a touch less per unit area.

Level 3 · Independent problem

A transformer relies on natural convection from a 1.2 m tall side at 70 °C in 25 °C air. Estimate Ra and h, then decide whether fins or a fan would help more.

Show answer

With film-temperature air properties, Ra ≈ 10¹⁰ (turbulent), Nu ≈ 0.10 Ra^1/3 ≈ 215, h ≈ 0.10 × 215 × 0.028/1.2 ≈ 5 W/m²·K. The coefficient stays small, so fins (more area) help, but a fan (raising h severalfold) helps more per unit cost if noise and reliability allow.

Level 4 · Transfer to real engineering

Find a real passively cooled object (a power brick, a radiator, a fanless amplifier). Estimate its Ra and natural h, and judge whether it is area-limited (needs fins) or could benefit from forced flow.

What good work looks like

β from 1/T, Ra with the right characteristic length, h from a Ra-based correlation, and a Gr/Re² check if any airflow is present, ending with a fins-versus-fan recommendation.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Here is my Rayleigh number and correlation. Check the characteristic length and that β is in 1/K."

"Give me five warm surfaces; I will judge natural versus forced from Gr/Re² before computing."

"Compute h for this panel." Building Ra and choosing the regime is the skill.

"Does the breeze matter?" The Gr/Re² judgment is what this chapter trains.

Portfolio task

Write a one-page natural-convection note for a passively cooled surface: Ra, the correlation, h, and a Gr/Re² check, ending with whether fins or a fan would do more.

Must include: β = 1/T in kelvin, Ra with its characteristic length, and the natural-versus-forced verdict.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What drives natural convection, and what coefficient measures it?

Buoyancy from density differences; the volumetric expansion coefficient β (1/T for a gas) sets its strength.

2. Write the Rayleigh number and its role.

Ra = gβΔTL³/(να) = Gr·Pr; it replaces Reynolds as the regime indicator and correlation variable.

3. Give the laminar vertical-plate correlation.

Nu ≈ 0.59 Ra^1/4 for 10⁴ < Ra < 10⁹.

4. How do you tell natural from forced convection?

The ratio Gr/Re²: much greater than 1 is buoyancy-dominated, much less than 1 is forced, near 1 is mixed.

5. Why are natural-convection surfaces often finned?

Because the coefficient is small (a few W/m²·K), so extra area is the cheap way to move more heat without a fan.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-solve the vertical panel from a blank page.

+3 daysOne Gr/Re² check on a new surface.

+7 daysMixed set: a natural h feeding a Chapter 3 fin design.

+30 daysContrast with the phase-change coefficients of Chapter 8.

Textbook mapping

Item	Mapping
Primary source	Lienhard and Lienhard, A Heat Transfer Textbook (6th ed), Chapter 8 (natural convection)
Cross-reference	Incropera, Ch. 9 (free convection) · Çengel and Ghajar, Ch. 9
Core topics	7.1 Buoyancy and β · 7.2 Grashof and Rayleigh · 7.3 Vertical plate · 7.4 Other geometries · 7.5 Natural vs forced · 7.6 Why it matters
Engineering connection	Passive and fanless cooling, building heat loss, and the still-air h that Chapter 3 fins fight.
Read next	Chapter 8: Boiling and Condensation.