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Heat Transfer · Chapter 5 of 10 · Intermediate

Convection and Boundary Layers

The convection coefficient h is not a material property; it is set by the flow. Three dimensionless numbers, Reynolds, Prandtl, and Nusselt, turn that messy fact into usable correlations.

01

Readiness check

From Chapter 1 and basic fluid mechanics. Tick only what you can do closed-notes.

  • State Newton's law of cooling q = hA ΔT.
  • Recall the Reynolds number and laminar-versus-turbulent flow.
  • Work with fluid properties: density, viscosity, conductivity.
  • Evaluate powers and roots on a calculator.
  • Keep units consistent across a correlation.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsReview the Reynolds number in Fluid Mechanics first.
3 or more weak itemsRevisit the convection law in Chapter 1.
02

The core idea

Convection is conduction across a thin boundary layer that the flow keeps stirring away. The thinner the layer, the higher h.

Re = ρVL/μPr = ν/αNu = hL/k = f(Re, Pr)

Near any surface the fluid slows to rest, forming a boundary layer; heat must cross this layer, mostly by conduction, before the bulk flow carries it off. Faster flow (higher Reynolds number) thins the layer and raises h. Correlations express the dimensionless coefficient Nu as a function of Re and Pr, and then h = Nu·k/L.

The skill works when: you pick the correlation matching the geometry and flow regime, evaluate properties at the right temperature, and read h from Nu.
The skill breaks down when: h is treated as a fixed property, or a laminar correlation is used in turbulent flow (or the reverse).
The concept. Flow slows to zero at the surface, building a boundary layer that thickens downstream. Heat crosses this slow layer by conduction; thinning it (faster flow) is how convection is enhanced.
03

The skills, taught in order

Convection looks fluid-mechanical and messy, but three dimensionless numbers tame it. Learn what each measures, and the route from flow conditions to h becomes a recipe.

5.1 The convection coefficient is not a property

Unlike k, the coefficient h is not a fixed property of the fluid; it depends on velocity, geometry, and whether the flow is laminar or turbulent. That is why convection is handled by correlations rather than a lookup table.

5.2 Boundary layers

At a surface, viscosity drags the fluid to rest, forming a velocity boundary layer; a thermal boundary layer forms likewise between the surface and fluid temperatures. Heat crosses the thermal layer mostly by conduction, so a thinner layer means a steeper gradient and a higher h.

5.3 The Reynolds number

Re = ρVL/μ = VL/ν compares inertial to viscous forces and decides the flow regime. On a flat plate the boundary layer trips from laminar to turbulent near Re ≈ 5×10⁵; in a pipe, near Re ≈ 2300. Turbulent layers mix vigorously and carry far more heat.

5.4 The Prandtl number

Pr = ν/α = μcp/k is a fluid property comparing how fast momentum and heat diffuse. It sets the relative thickness of the velocity and thermal layers, which is why it appears in every convection correlation.

FluidPrandtl number Pr
Liquid metals0.004 to 0.03
Gases (air)≈ 0.7
Water≈ 2 to 7
Oils50 to 100 000

5.5 The Nusselt number

Nu = hL/k is the dimensionless convection coefficient: the ratio of convective to pure-conduction heat transfer across the layer. Correlations give Nu as a function of Re and Pr; once you have Nu, the coefficient is h = Nu·k/L.

GroupDefinitionCompares
ReynoldsRe = ρVL/μinertia vs viscosity (regime)
PrandtlPr = ν/αmomentum vs thermal diffusion
NusseltNu = hL/kconvection vs conduction

5.6 Using a correlation

The recipe: identify geometry and regime, evaluate fluid properties (usually at the film temperature, the average of surface and fluid), compute Re and Pr, apply the matching Nu correlation, and convert Nu to h. For a laminar flat plate, Nu = 0.664 Re1/2 Pr1/3; turbulent, Nu = 0.037 Re0.8 Pr1/3.

Engineering connection: the foundation under every forced- and natural-convection correlation in Chapters 6 and 7, and the h that fins and transient problems need.

04

Worked example 1: laminar flow over a plate

Air at 10 m/s flows along a 0.2 m circuit board (treat as a flat plate). Using air properties ν = 1.6×10⁻⁵ m²/s, k = 0.026 W/m·K, Pr = 0.71, find the Reynolds number, confirm the flow is laminar, and find the average convection coefficient.

Figure 1. Air in laminar flow over a short board. Re = 1.25×10⁵ is below the 5×10⁵ transition, so the boundary layer stays laminar and h ≈ 27 W/m²·K.
  1. ProblemFind Re, the flow regime, and the average h for the plate in Figure 1.
  2. Given / findV = 10 m/s, L = 0.2 m, ν = 1.6×10⁻⁵ m²/s, k = 0.026 W/m·K, Pr = 0.71. Find Re, regime, h.
  3. AssumptionsSteady incompressible flow, smooth flat plate, constant properties at the film temperature.
  4. ModelCompute Re = VL/ν, compare with the 5×10⁵ transition, apply the laminar flat-plate correlation, and convert Nu to h.
  5. EquationsRe = VL/ν Nu = 0.664 Re1/2 Pr1/3 h = Nu·k/L
  6. SolveRe = 10 × 0.2 / 1.6×10⁻⁵ = 1.25×10⁵, below 5×10⁵, so laminar. Nu = 0.664 × (1.25×10⁵)1/2 × 0.711/3 = 0.664 × 354 × 0.892 = 209. h = 209 × 0.026 / 0.2 = 27.2 W/m²·K.
  7. Check27 W/m²·K sits squarely in the forced-gas range (25 to 250) from Chapter 1, so the answer is physically reasonable. The dimensionless groups did the work: pick the regime, get Nu, scale by k/L.
  8. ConclusionThe board sheds about 27 W per m² of board per kelvin of surface-to-air difference. Doubling the air speed would raise Re and h by roughly √2 in laminar flow, which is why fans help but only modestly until the flow turns turbulent.
Result. Re = 1.25×10⁵ (laminar), Nu = 209, h = 27.2 W/m²·K.
05

Worked example 2: turbulence changes everything

Now air at 40 m/s flows along a 1.0 m surface (same properties). Find Re, confirm it is turbulent, and find h with the turbulent correlation. Compare with what a laminar formula would have predicted.

Figure 2. Once Re passes 5×10⁵ the boundary layer becomes turbulent and mixes vigorously, raising h far above the laminar value: here about 113 versus 24 W/m²·K.
  1. ProblemFind Re, the regime, and h for the faster, longer plate in Figure 2, and compare with the laminar estimate.
  2. Given / findV = 40 m/s, L = 1.0 m, ν = 1.6×10⁻⁵, k = 0.026, Pr = 0.71. Find Re, regime, h.
  3. AssumptionsFully turbulent layer over the plate, constant properties, smooth surface.
  4. ModelCompute Re, confirm it is past transition, and apply the turbulent flat-plate correlation Nu = 0.037 Re0.8 Pr1/3.
  5. EquationsRe = VL/ν Nu = 0.037 Re0.8 Pr1/3 h = Nu·k/L
  6. SolveRe = 40 × 1.0 / 1.6×10⁻⁵ = 2.5×10⁶, well past 5×10⁵, so turbulent. Nu = 0.037 × (2.5×10⁶)0.8 × 0.711/3 ≈ 4340. h = 4340 × 0.026 / 1.0 = 113 W/m²·K.
  7. CheckThe laminar formula at the same Re would give Nu ≈ 935 and h ≈ 24 W/m²·K, so turbulence multiplies h by about 4.6. That jump (the Re0.8 versus Re0.5 scaling) is exactly why turbulators and rough surfaces are used to boost cooling.
  8. ConclusionTurbulence is the single biggest lever on convection: the same air, faster, moves several times the heat. Knowing the regime before picking a correlation is therefore the most important step in any convection estimate.
Result. Re = 2.5×10⁶ (turbulent), Nu ≈ 4340, h ≈ 113 W/m²·K, about 4.6 times the laminar value.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
h treated as a propertyOne h looked up for a fluid regardless of flow"What is the velocity and geometry?"h depends on the flow; compute it through Re, Pr, and a correlation each time.
Wrong regime correlationLaminar formula used in turbulent flow"Is Re above or below transition?"Check Re first: 5×10⁵ on a plate, 2300 in a pipe; pick the matching correlation.
Properties at the wrong temperatureRe and Pr from room-temperature data on a hot surface"Did I use the film temperature?"Evaluate properties at the film temperature, the mean of surface and fluid.
Nu mistaken for hNusselt number reported in W/m²·K"Did I multiply by k/L?"Nu is dimensionless; h = Nu·k/L carries the units.
07

Practice ladder

Level 1 · Direct skill

Water (ν = 1.0×10⁻⁶ m²/s) flows at 1.5 m/s along a 0.3 m plate. Find Re and state the regime.

Show answer

Re = VL/ν = 1.5 × 0.3 / 1.0×10⁻⁶ = 4.5×10⁵, just below 5×10⁵, so still laminar but near transition. Water's low viscosity gives a high Re even at modest speed.

Level 2 · Mixed concept

Why does water give a much higher h than air in the same laminar flow? Use the roles of k and Pr.

Show answer

Nu depends on Pr1/3, and water's Pr (≈ 7) far exceeds air's (≈ 0.7), so its Nu is higher; on top of that h = Nu·k/L scales with k, and water's k (≈ 0.6) is over 20 times air's (≈ 0.026). The two effects together make liquid cooling far more effective.

Level 3 · Independent problem

For Worked Example 1, recompute h if the air speed rises to 30 m/s (still laminar over the 0.2 m board). By what factor does h change, and why is the gain modest?

Show answer

Re triples to 3.75×10⁵ (still < 5×10⁵). Since Nu ∝ Re1/2, h rises by √3 ≈ 1.73, to about 47 W/m²·K. Laminar convection improves only with the square root of speed, so big fan increases buy modest cooling until the flow turns turbulent.

Level 4 · Transfer to real engineering

Estimate h for a real cooled surface (a heat sink face, a car panel, a window in wind). Identify the geometry, estimate V and the film properties, compute Re and Nu, and report h with the regime.

What good work looks like

A stated geometry and characteristic length, properties at the film temperature, Re with the regime named, the matching correlation, and h with a sanity check against the Chapter 1 ranges.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Here is my Re and chosen correlation. Check whether the regime and the geometry match the formula I used."
"Give me five flows; I will call each laminar or turbulent from Re before you confirm."
"Compute h for me." Picking the regime and correlation is the judgment being trained.
"Which correlation do I use?" Matching geometry and Re to a correlation is the core skill.

Portfolio task

Write a one-page "Convection Recipe Card": the definitions of Re, Pr, Nu, the flat-plate correlations, and one worked surface where you go from flow conditions to h, with the regime justified.

Must include: properties at the film temperature, Re compared with transition, and h = Nu·k/L shown explicitly.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Why is h not a material property?

It depends on the flow velocity, geometry, and regime, not just the fluid, so it must be correlated rather than looked up.

2. Define Re, Pr, and Nu and what each compares.

Re = ρVL/μ (inertia vs viscosity, the regime); Pr = ν/α (momentum vs thermal diffusion); Nu = hL/k (convection vs conduction).

3. How does h follow from a correlation?

Find Nu = f(Re, Pr) for the geometry and regime, then h = Nu·k/L.

4. What happens to h at transition to turbulence?

It jumps, because Nu scales as Re0.8 turbulent versus Re1/2 laminar; mixing greatly enhances heat transfer.

5. At what temperature are properties evaluated?

Usually the film temperature, the average of the surface and free-stream temperatures.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-solve the laminar plate from a blank page.
+3 daysOne regime call plus h for a new flow.
+7 daysMixed set: a convection h feeding a Chapter 3 fin.
+30 daysCarry Re, Pr, Nu into the forced-convection correlations of Chapter 6.
10

Textbook mapping

ItemMapping
Primary sourceLienhard and Lienhard, A Heat Transfer Textbook (6th ed), Chapter 6 (laminar and turbulent boundary layers)
Cross-referenceIncropera, Ch. 6 to 7 (convection fundamentals, external flow) · Çengel and Ghajar, Ch. 6
Core topics5.1 h is not a property · 5.2 Boundary layers · 5.3 Reynolds · 5.4 Prandtl · 5.5 Nusselt · 5.6 Using correlations
Engineering connectionThe basis of all convection correlations and the h needed by fins and transient problems.
Read nextChapter 6: Forced Convection.