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Heat Transfer · Chapter 6 of 10 · Intermediate

Forced Convection

A pump or fan drives the flow, and a correlation gives the coefficient. Pipe flow and cross-flow cover most of what a mechanical engineer cools or heats.

Readiness check

From Chapter 5. Tick only what you can do closed-notes.

Compute Re, Pr, and Nu for a flow.
Convert Nu to h with h = Nu·k/L.
Decide laminar versus turbulent from Re.
Evaluate fluid properties at a mean temperature.
Raise numbers to fractional powers reliably.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRevisit Re, Pr, Nu in Chapter 5.

3 or more weak itemsWork the flat-plate examples of Chapter 5 again first.

The core idea

Match the geometry and regime to a tested correlation, and it hands you Nu, then h.

Re_D = VD/νNu = 0.023 Re^0.8Prⁿh = Nu·k/D

Forced convection splits into internal flow (inside pipes and ducts) and external flow (over surfaces, cylinders, and tube banks). Each has its own family of correlations, validated by experiment. The job is to pick the right one, evaluate properties correctly, and read off h.

The skill works when: the flow matches the correlation's geometry and Reynolds range, and properties are taken at the mean (bulk or film) temperature.

The skill breaks down when: a pipe correlation is used for external flow, or Dittus-Boelter is applied in laminar flow where Nu is nearly constant.

The concept. Two families: flow driven through a pipe (internal), or flow driven across a body (external) where the boundary layer separates into a wake. Each has its own correlation for Nu.

The skills, taught in order

Forced convection is correlation-driven. Learn the two geometry families, the handful of correlations that cover most cases, and the discipline of evaluating them correctly.

6.1 Internal versus external flow

Internal flow is bounded by walls (pipes, ducts, channels) and the fluid eventually fills the cross-section. External flow is over an exposed body (a plate, cylinder, or tube bank) where the boundary layer is free to grow and separate. The two need different correlations.

6.2 The entry region and fully developed flow

Near a pipe inlet the boundary layers are still growing and h is high; once they merge, the flow is fully developed and h settles to a constant. Use the hydraulic diameter D_h = 4A_c/P for non-circular ducts.

6.3 Laminar pipe flow

A surprise: for fully developed laminar pipe flow, Nu is a constant, not a function of Re. It is 3.66 for a constant wall temperature and 4.36 for a constant wall heat flux. Laminar pipe flow is therefore a weak convector regardless of speed.

6.4 Turbulent pipe flow

Once Re > about 2300 (fully turbulent by ~10⁴), the Dittus-Boelter correlation applies: Nu = 0.023 Re^0.8Prⁿ, with n = 0.4 when the fluid is being heated and 0.3 when cooled. Turbulent pipe flow gives high coefficients, which is why heat exchangers run turbulent.

Pipe flow	Nusselt number
Laminar, constant wall temperature	Nu = 3.66
Laminar, constant wall heat flux	Nu = 4.36
Turbulent (Dittus-Boelter)	Nu = 0.023 Re^0.8Prⁿ

6.5 External cross-flow

For flow across a cylinder (a pipe in wind, a wire, a tube in a bank), the Hilpert correlation Nu = C Re^m Pr^1/3 uses constants C and m that depend on the Reynolds range. The boundary layer separates, leaving a turbulent wake that complicates the picture but the correlation captures it.

Re_D range	C	m
40 to 4000	0.683	0.466
4000 to 40 000	0.193	0.618
40 000 to 400 000	0.027	0.805

6.6 Choosing and applying a correlation

The procedure never changes: identify internal or external, compute Re, confirm the regime and that it falls in the correlation's valid range, evaluate properties at the mean temperature, then compute Nu and h = Nu·k/D. Reporting which correlation you used is part of the answer.

Engineering connection: pipe and duct cooling, the tube-side and shell-side coefficients of the heat exchangers in Chapter 9, and the h that fins and transient parts depend on.

Worked example 1: water in a pipe

Water at a mean temperature where ν = 8.0×10⁻⁷ m²/s, k = 0.62 W/m·K, and Pr = 5.4 flows at 2.0 m/s through a 20 mm pipe and is being heated. Find the convection coefficient on the pipe wall.

Figure 1. Turbulent water flow in a pipe. Dittus-Boelter gives Nu ≈ 259, and the high k of water produces a large coefficient, h ≈ 8000 W/m²·K.

ProblemFind the wall convection coefficient for the pipe flow in Figure 1.
Given / findV = 2.0 m/s, D = 0.020 m, ν = 8.0×10⁻⁷, k = 0.62, Pr = 5.4, fluid heated. Find h.
AssumptionsFully developed turbulent flow, smooth pipe, properties at the bulk mean temperature.
ModelCompute Re_D, confirm turbulent, apply Dittus-Boelter with n = 0.4 (heating), then h = Nu·k/D.
EquationsRe_D = VD/ν Nu = 0.023 Re^0.8Pr^0.4 h = Nu·k/D
SolveRe_D = 2.0 × 0.020 / 8.0×10⁻⁷ = 50 000 (turbulent). Nu = 0.023 × 50000^0.8 × 5.4^0.4 = 0.023 × 5740 × 1.96 = 259. h = 259 × 0.62 / 0.020 = 8000 W/m²·K.
CheckEight thousand W/m²·K is squarely in the forced-liquid range (100 to 20 000) from Chapter 1, far above any gas. Water's high k and turbulent mixing are doing the work; the same flow in laminar regime would give only Nu = 3.66.
ConclusionLiquids in turbulent pipe flow are excellent convectors, which is why water cooling beats air cooling so decisively. This wall coefficient is exactly the tube-side h that Chapter 9 feeds into an overall U.

Result. Re = 50 000 (turbulent), Nu ≈ 259, h ≈ 8000 W/m²·K.

Worked example 2: wind across a pipe

Air at 20 m/s blows across a 25 mm outdoor pipe. With air properties ν = 1.6×10⁻⁵ m²/s, k = 0.026 W/m·K, Pr = 0.71, find the convection coefficient on the pipe using the Hilpert cross-flow correlation.

Figure 2. Cross-flow of air over a pipe, Re ≈ 31 000. The Hilpert correlation gives Nu ≈ 103 and h ≈ 107 W/m²·K, an order of magnitude below the water-in-pipe case.

ProblemFind the convection coefficient for air in cross-flow over the pipe in Figure 2.
Given / findV = 20 m/s, D = 0.025 m, ν = 1.6×10⁻⁵, k = 0.026, Pr = 0.71. Find h.
AssumptionsSteady cross-flow, isolated cylinder, properties at the film temperature.
ModelCompute Re_D, read the Hilpert constants for that range, apply Nu = C Re^m Pr^1/3, then h = Nu·k/D.
EquationsRe_D = VD/ν Nu = C Re^m Pr^1/3 h = Nu·k/D
SolveRe_D = 20 × 0.025 / 1.6×10⁻⁵ = 31 250, in the 4000 to 40 000 band, so C = 0.193, m = 0.618. Nu = 0.193 × 31250^0.618 × 0.71^1/3 = 103. h = 103 × 0.026 / 0.025 = 107 W/m²·K.
CheckOne hundred W/m²·K is a strong forced-gas value, well above the still-air figure of Chapter 1, yet far below the water pipe (8000). Gases convect far less than liquids even when driven hard, because of their low k.
ConclusionEven a 20 m/s wind gives only about 107 W/m²·K on a bare pipe, which is why outdoor pipes are lagged and why air-cooled equipment leans on fins to make up for air's poor convection.

Result. Re = 31 250, Nu ≈ 103, h ≈ 107 W/m²·K, roughly 75 times smaller than the water-in-pipe coefficient.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Internal correlation on external flow	Dittus-Boelter used for a pipe in wind	"Is the fluid inside the pipe or across it?"	Internal and external flows use different correlations; pick by where the fluid is.
Dittus-Boelter in laminar flow	Re^0.8 applied below Re ≈ 2300	"Is the pipe flow turbulent?"	Laminar pipe flow has constant Nu (3.66 or 4.36); the turbulent correlation does not apply.
Wrong Hilpert band	C and m taken from the wrong Re range	"Which Reynolds band am I in?"	Read C and m from the band containing your Re; they change the result a lot.
Diameter dropped from Re or h	Re or h off by orders of magnitude	"Did I use D in both Re and Nu·k/D?"	Use the diameter consistently: Re = VD/ν and h = Nu·k/D.

Practice ladder

Level 1 · Direct skill

Air (ν = 1.6×10⁻⁵, k = 0.026, Pr = 0.71) flows at 8 m/s through a 30 mm duct and is heated. Find Re and h with Dittus-Boelter.

Show answer

Re = 8 × 0.030 / 1.6×10⁻⁵ = 15 000 (turbulent). Nu = 0.023 × 15000^0.8 × 0.71^0.4 = 0.023 × 2190 × 0.873 = 44. h = 44 × 0.026 / 0.030 = 38 W/m²·K. Air in a duct gives a modest coefficient even when turbulent.

Level 2 · Mixed concept

The Worked Example 1 water flow drops to 0.2 m/s, giving Re = 5000. Is Dittus-Boelter still the right tool, and how does h compare?

Show answer

Re = 5000 is just turbulent, so Dittus-Boelter still applies but is least accurate near transition. Nu = 0.023 × 5000^0.8 × 5.4^0.4 ≈ 43, h ≈ 1340 W/m²·K, about a sixth of the 2 m/s value, since h scales with Re^0.8.

Level 3 · Independent problem

Compare cooling a 25 mm pipe by 20 m/s air (Worked Example 2, h ≈ 107) versus filling it with the 2 m/s water of Worked Example 1 (h ≈ 8000). For the same wall area and temperature difference, what is the ratio of heat removed, and what does that say about coolant choice?

Show answer

q = hA ΔT, so the ratio is just the h ratio, about 8000/107 ≈ 75. Internal water removes roughly 75 times the heat of external air for the same area and ΔT. This is why high-power systems use liquid loops rather than air.

Level 4 · Transfer to real engineering

Estimate h for a real forced-convection case you can observe (water in a hose, air in a duct, a fan over a radiator). Identify internal or external, compute Re, choose a correlation, and report h with its regime.

What good work looks like

Internal-versus-external stated, Re with the regime, the named correlation and its valid range, properties at the mean temperature, and h with a sanity check against Chapter 1's ranges.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Here is my flow, Re, and chosen correlation. Confirm the correlation matches the geometry and the Reynolds range."

"Give me five forced-convection cases; I will pick the correlation for each before computing."

"Compute h for this pipe." Selecting and applying the right correlation is the skill.

"Is this turbulent?" Reading Re against the transition is the judgment to own.

Portfolio task

Write a one-page "Correlation Card": the pipe-flow and cross-flow correlations with their valid ranges, and two worked cases (one internal, one external) taken from flow conditions to h.

Must include: the regime check, the correlation's valid Re range cited, and h compared with the Chapter 1 ranges.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What distinguishes internal from external flow?

Internal flow is bounded by walls and fills the cross-section; external flow is over an exposed body with a free, separating boundary layer.

2. Write the Nusselt number for laminar pipe flow.

Constant: 3.66 for constant wall temperature, 4.36 for constant heat flux, independent of Re.

3. State the Dittus-Boelter correlation.

Nu = 0.023 Re^0.8Prⁿ, n = 0.4 heating, 0.3 cooling, for turbulent pipe flow.

4. What form does cross-flow over a cylinder take?

Nu = C Re^m Pr^1/3, with C and m set by the Reynolds band (Hilpert).

5. Why does water cool far better than air?

Higher k and higher Pr give a larger Nu and a much larger h = Nu·k/D, often by one to two orders of magnitude.

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

+1 dayRe-solve the water pipe from a blank page.

+3 daysOne cross-flow h with new numbers.

+7 daysMixed set: a tube-side h feeding a Chapter 9 exchanger.

+30 daysCarry these coefficients into Chapter 9.

Textbook mapping

Item	Mapping
Primary source	Lienhard and Lienhard, A Heat Transfer Textbook (6th ed), Chapter 7 (forced convection in a variety of configurations)
Cross-reference	Incropera, Ch. 7 to 8 (external and internal flow) · Çengel and Ghajar, Ch. 7 to 8
Core topics	6.1 Internal vs external · 6.2 Entry region · 6.3 Laminar pipe Nu · 6.4 Dittus-Boelter · 6.5 Cross-flow · 6.6 Applying correlations
Engineering connection	Pipe and duct cooling, and the tube-side and shell-side coefficients of Chapter 9.
Read next	Chapter 7: Natural Convection.