Forced Convection: Boundary Layers and Nusselt Correlations
Every chapter on conduction treated the film coefficient $h$ as a given. This chapter explains where $h$ comes from, what controls its value, and how to predict it from flow conditions. The key idea is that convective heat transfer is really conduction across a thin, slow-moving fluid layer at the surface, and the thickness of that layer is set by fluid dynamics.
Close your notes. Try each question from memory, then reveal the answer. The ones you cannot recall are exactly what this chapter will reinforce.
Write Newton’s law of cooling. Define every symbol with units. What physical mechanism does the film coefficient $h$ represent?
Name three physical factors that determine the magnitude of $h$. For each, state whether increasing it raises or lowers $h$ and why.
Write the Biot number formula. What does Bi ≤ 0.1 allow you to assume, and why is that useful?
The Part-Cooling Fan
Look at any FDM printer running at speed. A small fan is aimed directly at the nozzle exit, blasting the freshly deposited bead. Without it, PLA would sag before the next layer arrived; with it, the bead solidifies in under a second. The difference between a failed print and a crisp one is a few watts of convective heat removal.
The fan controls $h$, the film coefficient. But until now, $h$ has just been a number we assumed. A cooling fan running at 30% speed gives a different $h$ than one at full blast. Air gives a different $h$ than water. A narrow nozzle gives a different $h$ than a wide one. This chapter gives you the tools to calculate $h$ from first principles for the most common geometries in polymer processing.
The central idea is simple. No real fluid is perfectly still at a solid surface. Viscosity creates a thin layer of slow-moving fluid right at the wall, called the boundary layer. Heat must conduct across this layer before the bulk flow carries it away. The thinner the layer, the higher $h$ is, and the thicker the layer, the lower $h$ is. Everything else follows from understanding what controls the boundary layer thickness.
Typical values in FDM printing: part-cooling fan over a PLA bead, $h \approx 30$–80 W/m²K depending on fan speed and duct geometry. The heated bed surface in still air, $h \approx 5$–10 W/m²K. A water-cooled cold-end block, $h \approx 1{,}000$–5{,}000 W/m²K. Matching the right $h$ to each surface is the thermal design problem of the printer.
Before reading on — you already know the Biot number: $\mathrm{Bi} = hL_c/k_s$. This chapter introduces the Nusselt number: $\mathrm{Nu} = hL/k_f$. The two formulas look identical. But Bi uses the solid's conductivity and asks whether the solid's interior can keep up with the surface. Nu uses the fluid's conductivity and asks how much better real flow is than pure stagnant conduction. Same algebraic shape — completely opposite physical questions. If you can hold that distinction clearly in your mind, you will never confuse the two groups on an exam.
What Is Forced Convection?
Convection is always a combination of two mechanisms: bulk fluid motion (which carries energy from one place to another) and molecular conduction (which transfers energy at the fluid-solid interface where the velocity is zero). You cannot have convective heat transfer without conduction; you just benefit from flow making the effective conduction path much shorter.
Forced convection means the flow is driven externally: a pump, a fan, a moving belt, or a pressure difference. The key distinction from Chapter 10 (natural convection) is that the flow rate is set by the process, not by buoyancy forces arising from the heat transfer itself. In a cooling water line, a quench bath, or a die-exit air knife, the flow is forced and you know its velocity.
The no-slip condition of viscous fluid mechanics says that fluid touching a solid surface is stationary relative to that surface. Right at the wall, there is no bulk transport of energy at all. Heat leaves the wall by pure conduction through the nearly-still fluid immediately adjacent to it. The boundary layer picture makes this precise: it identifies the thin zone over which velocity rises from zero to the free-stream value, and within which conduction dominates.
The Velocity Boundary Layer
When a uniform flow at velocity $u_\infty$ encounters a flat solid surface, viscosity slows the fluid near the wall. The velocity boundary layer thickness $\delta$ is defined as the distance from the wall at which the fluid velocity reaches 99% of the free-stream value $u_\infty$. It starts at zero at the leading edge and grows as the fluid travels along the surface.
For laminar flow over a flat plate, the boundary layer thickness at position $x$ from the leading edge is:
Physically, the boundary layer grows because viscosity keeps "recruiting" more fluid as the flow travels further along the plate. The longer the plate, the thicker the layer at the trailing edge, and the lower the average $h$ — which is why shorter surfaces or smaller-diameter pipes tend to have higher heat transfer coefficients per unit area.
The Thermal Boundary Layer and the Prandtl Number
Alongside the velocity boundary layer grows a thermal boundary layer of thickness $\delta_T$, within which the fluid temperature transitions from the wall temperature $T_s$ to the free-stream temperature $T_\infty$. The two layers need not be the same thickness, and their ratio is set by the Prandtl number.
When Pr is large (momentum diffuses much faster than heat), the thermal boundary layer is much thinner than the velocity boundary layer. When Pr is small (heat diffuses fast), the thermal layer extends beyond the velocity layer. The relationship for laminar flow over a flat plate is:
Pr = ν/α = momentum diffusivity / thermal diffusivity. Polymer melts have extremely high viscosities (μ = 10–1000 Pa·s vs water’s 0.001 Pa·s) while their thermal conductivity stays low (k ≈ 0.1–0.5 W/mK). The result is Pr values of 103 to 106. Physically, a polymer melt resists flow far more than it resists heat conduction (relatively speaking). This means the thermal boundary layer is much thinner than the velocity one, heat transfer is concentrated in a very thin zone near the wall, and small changes in die geometry can dramatically affect cooling uniformity.
The Reynolds Number: Laminar or Turbulent?
The single most important dimensionless group in forced convection is the Reynolds number. It compares inertial forces (which want to keep the fluid moving in straight streamlines) to viscous forces (which want to damp out disturbances). When inertia wins, the flow becomes turbulent: chaotic, mixing rapidly, and dramatically more effective at heat transfer.
For flow over a flat plate, transition from laminar to turbulent occurs at $Re_x \approx 5 \times 10^5$. For flow inside a pipe, transition occurs at $Re_D \approx 2{,}300$. Fully turbulent pipe flow is established by $Re_D \approx 10{,}000$. Turbulent flow enhances $h$ by a factor of 5–50 compared to laminar flow at the same flow rate, because eddies actively mix hot fluid away from the wall.
High-viscosity polymer melts in processing equipment typically have Re << 1. A polymer melt at 200°C with μ = 500 Pa·s flowing at 0.01 m/s through a 10 mm die gives Re = 1000 × 0.01 × 0.01 / 500 = 0.0002. Turbulence is physically impossible. All forced convection correlations for polymer melts in dies and extruders assume laminar flow.
The Nusselt Number: Dimensionless h
The Nusselt number is the dimensionless form of the heat transfer coefficient. It tells you how much better actual convection is than pure conduction through a stagnant fluid layer of the same thickness:
Once you know Nu, you recover $h$ directly: $h = \mathrm{Nu} \cdot k_f / L$. The practical workflow is therefore: (1) compute Re and Pr from known fluid properties and flow conditions, (2) use a correlation to get Nu as a function of Re and Pr, (3) recover $h$ from $h = \mathrm{Nu} \cdot k_f / L$.
Both look like $hL/k$ but mean completely different things. The Nusselt number $\mathrm{Nu} = hL/k_f$ uses the fluid conductivity and characterises convective enhancement in the fluid. The Biot number $\mathrm{Bi} = hL_c/k_s$ uses the solid conductivity and tells you whether the solid's interior is isothermal. Never mix the two $k$ values.
Definition (always true)
$$\mathrm{Nu} = \frac{hL}{k_f}$$Nu = 1 means h equals what pure conduction would give across a stagnant layer of thickness L. Nu = 187 means flow does the job 187× faster.
What determines Nu (from scaling)
$$\mathrm{Nu} \sim Re^{1/2}\, \mathrm{Pr}^{1/3}$$Re controls the BL thickness; Pr controls how the thermal layer relates to the velocity layer. The constant 0.664 comes from the exact Blasius solution.
Step 1: Link h to the thermal boundary layer thickness. At the wall the velocity is zero (no-slip), so heat leaves the wall by pure conduction through the nearly-still fluid adjacent to it. The local heat flux is:
$$q = -k_f \left.\frac{\partial T}{\partial y}\right|_{y=0} \approx k_f \frac{\Delta T}{\delta_T}$$From Newton's law $q = h\,\Delta T$, it follows that $h \sim k_f / \delta_T$. Therefore:
$$\mathrm{Nu} = \frac{hL}{k_f} \sim \frac{L}{\delta_T}$$The Nusselt number is just the ratio of the plate length to the thermal boundary layer thickness. A thinner thermal layer means higher Nu means higher h.
Step 2: Find δ from the momentum balance. The velocity boundary layer grows because viscosity "recruits" fluid as the flow travels downstream. A scaling balance between convection ($u_\infty \partial u/\partial x \sim u_\infty^2/L$) and diffusion ($\nu\, \partial^2 u/\partial y^2 \sim \nu\, u_\infty/\delta^2$) gives:
$$\delta \sim \frac{L}{\sqrt{Re_L}}, \qquad Re_L = \frac{u_\infty L}{\nu}$$Step 3: Relate δT to δ through Pr. For Pr > 1, the thermal BL is thinner than the velocity BL. Within the thermal layer ($y \lesssim \delta_T \ll \delta$) the velocity profile is approximately linear: $u \approx u_\infty y/\delta$. A scaling balance on the thermal BL energy equation ($u\,\partial T/\partial x \sim \alpha\,\partial^2 T/\partial y^2$) with this linear profile gives:
$$\frac{u_\infty \delta_T}{\delta}\frac{\Delta T}{L} \sim \alpha \frac{\Delta T}{\delta_T^2} \quad\Rightarrow\quad \delta_T \sim \delta\,\mathrm{Pr}^{-1/3}$$Step 4: Combine.
$$\mathrm{Nu} \sim \frac{L}{\delta_T} \sim \frac{L}{\delta}\,\mathrm{Pr}^{1/3} \sim \sqrt{Re_L}\,\mathrm{Pr}^{1/3}$$Conclusion: The exact Blasius solution gives the constant as 0.664, but scaling tells us everything else: Nu grows as Re1/2 (thinner BL at higher speed) and as Pr1/3 (thinner thermal BL at higher Pr). The Pr exponent of 1/3 — not 1/2 — is the signature of the linear velocity profile inside the thermal layer.
Flat Plate Correlations (External Flow)
For laminar flow over a flat plate at uniform surface temperature, the local and average Nusselt numbers are given by the Pohlhausen solution (exact for Pr ≥ 0.6):
The laminar boundary layer equations (continuity + momentum + energy) admit a similarity solution by the substitution $\eta = y\sqrt{u_\infty / (\nu x)}$, which collapses the two-dimensional PDE system into two coupled ODEs. The momentum ODE (the Blasius equation) gives the velocity profile; the thermal ODE then gives the temperature profile for any Pr. Evaluating the temperature gradient at the wall and integrating over the plate length yields the 0.664 prefactor. The derivation appears in graduate heat transfer texts (e.g., Incropera & DeWitt, Chapter 7).
For mixed laminar-turbulent flow (plate long enough for the flow to transition):
Internal Flow: Pipes and Channels
When fluid flows inside a pipe or channel, the boundary layers from opposite walls eventually merge, forming a fully developed velocity profile. After that point, the velocity shape no longer changes along the pipe, though the temperature profile continues to evolve. For fully developed turbulent pipe flow the most widely used correlation is Dittus-Boelter:
Laminar pipe flow (Re < 2,300) has a much simpler result when fully developed: $\mathrm{Nu}_D = 3.66$ for constant wall temperature, or $\mathrm{Nu}_D = 4.36$ for constant heat flux. These are exact analytical results, independent of Re and Pr, because the velocity profile is fixed and only conduction across the fixed profile matters.
Injection mold cooling channels carry chilled water (typically 10–15°C) at high enough flow rate to stay in turbulent regime. A 6 mm diameter channel with water at 1.5 m/s gives Re ≈ 90,000, well into the turbulent range. Nu ≈ 300–400, giving h ≈ 30,000–40,000 W/m²K. This high $h$ is why the mold wall, not the water, controls the cooling time for thick polymer parts. The mold wall conduction resistance dominates over the convective resistance.
Dittus-Boelter is valid only for Re > 10,000. Applying it at Re = 5,000 gives an error of 30% or more. For transitional flow (2,300 < Re < 10,000), use the Gnielinski correlation instead: $\mathrm{Nu}_D = (f/8)(Re_D - 1000)\mathrm{Pr} / [1 + 12.7(f/8)^{1/2}(\mathrm{Pr}^{2/3} - 1)]$ where $f$ is the Darcy friction factor.
Interactive: Boundary Layer Explorer
Adjust flow conditions and see how the velocity and thermal boundary layers compare, along with the resulting Nusselt number and film coefficient.
Worked Examples
Example 9.1: Air Knife Cooling of an Extruded HDPE Sheet
An HDPE sheet exits a die and passes under an air knife that blows 3 m/s air across the top surface. The sheet is 0.5 m long in the flow direction. Air properties at film temperature: $\nu = 1.5 \times 10^{-5}$ m²/s, $k_f = 0.026$ W/mK, Pr = 0.71. The surface is at 180°C and the air is at 30°C. Find the average film coefficient and the total heat flux removed from the top face.
Solution:
$$Re_L = \frac{u_\infty L}{\nu} = \frac{3 \times 0.5}{1.5\times10^{-5}} = 100{,}000 \quad (\text{laminar} \checkmark)$$ $$\overline{\mathrm{Nu}}_L = 0.664 \times (100{,}000)^{0.5} \times (0.71)^{1/3} = 0.664 \times 316.2 \times 0.892 = 187$$ $$h = \frac{187 \times 0.026}{0.5} = 9.7 \;\text{W/m}^2\text{K}$$ $$q = h\,\Delta T = 9.7 \times (180 - 30) = 9.7 \times 150 = 1{,}460 \;\text{W/m}^2$$For a sheet 1 m wide the total power removed from one face is $1{,}460 \times 0.5 = 730$ W/m. Switching to a water mist or higher-velocity jet would increase $h$ by a factor of 5–50.
Example 9.2: Cooling Water in a Mold Channel
A mold cooling channel has diameter $D = 0.05$ m. Water at $\bar{T} = 20°$C flows at $u = 2$ m/s. Properties: $\rho = 1000$ kg/m³, $\nu = 10^{-6}$ m²/s, $k_f = 0.60$ W/mK, Pr = 4.0. The mold wall is held at 80°C. Find the Nusselt number and film coefficient (heating the fluid).
Solution:
$$Re_D = \frac{uD}{\nu} = \frac{2 \times 0.05}{10^{-6}} = 100{,}000 \quad (\text{turbulent} \checkmark)$$ $$Re_D^{0.8} = (10^5)^{0.8} = 10^4 = 10{,}000$$ $$\mathrm{Pr}^{0.4} = 4^{0.4} = 2^{0.8} \approx 1.741$$ $$\mathrm{Nu}_D = 0.023 \times 10{,}000 \times 1.741 = 400$$ $$h = \frac{400 \times 0.60}{0.05} = 4{,}800 \;\text{W/m}^2\text{K}$$This high $h$ means the water-side convective resistance is very small. The thermal bottleneck for mold cooling is almost certainly the conduction resistance through the mold steel, not the water-side film.
Concept Questions
You double the air velocity over a flat plate while keeping everything else the same. By what factor does the average film coefficient $h$ change?
For turbulent pipe flow (Dittus-Boelter), if you double the flow velocity, by what factor does $h$ increase?
Key Concept
The velocity exponent in the Nu correlation tells you how $h$ scales with flow rate. Laminar flat plate: Nu ∝ Re1/2 ∝ u1/2, so doubling $u$ increases $h$ by √2. Turbulent pipe: Nu ∝ Re4/5 ∝ u4/5, so doubling $u$ increases $h$ by 20.8 ≈ 1.74. Both are sub-linear: the benefit of higher velocity diminishes at high flow rates. Switching from air to water is far more effective than increasing air velocity.
A polymer melt has Pr ≈ 10,000. Compared to the velocity boundary layer, the thermal boundary layer is:
Which of the following best explains why the Nusselt number increases with the Prandtl number (for Pr > 1)?
Key Concept
$\delta_T / \delta = \mathrm{Pr}^{-1/3}$. When Pr is large, the thermal boundary layer is much thinner than the velocity boundary layer. Within this thin zone the temperature drops steeply from $T_s$ to $T_\infty$. A steep gradient at the wall means a large conductive flux $q = -k_f (dT/dy)|_\text{wall}$, which corresponds to a high $h$. This is why high-Pr fluids (oils, glycerin, polymer solutions) have large Nusselt numbers even at modest Reynolds numbers.
Problem Set
Part A — Conceptual Problems
Explain in your own words why the film coefficient $h$ decreases as you move further downstream along a flat plate in laminar flow. Use the concept of boundary layer growth in your answer.
The Nusselt number for fully developed laminar pipe flow is Nu = 3.66, regardless of Reynolds number or Prandtl number. Explain physically why Nu is constant in this case, even though for turbulent flow Nu depends strongly on Re.
A process engineer wants to increase the heat transfer coefficient $h$ in a polymer pipe-cooling system. She proposes two options: (a) increase the water flow rate by 50% (turbulent regime), or (b) switch from smooth to corrugated pipe walls to promote turbulent mixing. Without doing a full calculation, compare the likely impact of each option. Which do you expect to produce the larger increase in $h$? Reference the Reynolds number exponent in your reasoning.
A polymer melt (Pr ≈ 50,000) flows slowly over a heated die wall. Air (Pr = 0.71) flows at the same Reynolds number over an identical surface. (a) Which fluid has the thicker thermal boundary layer? (b) Which has the higher Nusselt number? (c) Explain why these two answers may seem contradictory at first.
(a) Scaling derivation. At the wall, the heat flux scales as $q \sim k_f\,\Delta T/\delta_T$, so $h \sim k_f/\delta_T$ and $\mathrm{Nu} = hL/k_f \sim L/\delta_T$. Use the boundary-layer results $\delta \sim L/\sqrt{Re_L}$ and $\delta_T \sim \delta\,\mathrm{Pr}^{-1/3}$ to show that $\mathrm{Nu} \sim Re_L^{1/2}\,\mathrm{Pr}^{1/3}$. State in one sentence what the Pr1/3 factor accounts for physically.
(b) Application to the FDM printer. A part-cooling fan blows air over a freshly deposited PLA bead (length ~ 5 mm). The fan can produce 1 m/s or 3 m/s. Air: $\nu = 1.5\times10^{-5}$ m²/s, $k_f = 0.026$ W/mK, Pr = 0.71. Using $\overline{\mathrm{Nu}} = 0.664\,Re^{1/2}\,\mathrm{Pr}^{1/3}$, estimate $h$ at each velocity. Show that $h \propto u^{1/2}$ from your two results.
(c) Cross-chapter link to Ch. 8. If the bead has $\rho = 1250$ kg/m³, $c_p = 1600$ J/kgK, $L_c = 0.2$ mm, use $\tau = \rho c_p L_c / h$ to find τ at each fan speed. By what factor does τ change when you triple the fan speed?
(d) Practical limit. Explain in one or two sentences why running at 3 m/s fan speed can cause print defects even though cooling is faster.
Part B — Calculation Problems
Air at $\nu = 1.5 \times 10^{-5}$ m²/s flows at $u_\infty = 3$ m/s over an HDPE sheet of length $L = 0.5$ m. Calculate the Reynolds number at the trailing edge. Is the flow laminar?
Use: $Re_L = u_\infty L / \nu$; laminar if $Re_L < 5 \times 10^5$
Re = (inertial forces) / (viscous forces) — tells you whether the flow can sustain turbulent fluctuations or whether viscosity damps them out
Using the same flow from P1 ($Re_L = 100{,}000$, Pr = 0.71, $k_\text{air} = 0.026$ W/mK, $L = 0.5$ m), find the average Nusselt number and the average film coefficient $\bar{h}$. Report $\bar{h}$ to three significant figures. Note: $(0.71)^{1/3} = 0.892$.
Use: $\overline{\mathrm{Nu}}_L = 0.664\, Re_L^{1/2}\, \mathrm{Pr}^{1/3}$; then $\bar{h} = \overline{\mathrm{Nu}}\, k_f / L$
Nu = (actual convective transfer) / (conduction across a stagnant fluid layer) — once you know Nu, multiply by $k_f/L$ to get the dimensional $h$
A polymer melt has dynamic viscosity $\mu = 100$ Pa·s, specific heat $c_p = 2{,}000$ J/kgK, and thermal conductivity $k = 0.20$ W/mK. Calculate its Prandtl number. Comment briefly on what this value implies about the relative thickness of the thermal and velocity boundary layers.
Use: $\mathrm{Pr} = \mu c_p / k$
Pr = (momentum diffusivity) / (thermal diffusivity) — large Pr means the fluid spreads momentum much faster than heat, producing a thin thermal BL and high $h$
Cooling water flows through a mold channel of diameter $D = 0.05$ m at velocity $u = 2.0$ m/s. Water properties: $\nu = 1.0 \times 10^{-6}$ m²/s, $k_f = 0.60$ W/mK, Pr = 4.0. The mold wall is hotter than the water ($n = 0.4$). Find the Nusselt number and the film coefficient $h$. Useful: $(10^5)^{0.8} = 10^4$ exactly; $4^{0.4} = 2^{0.8} \approx 1.741$.
Use: $\mathrm{Nu}_D = 0.023\, Re_D^{4/5}\, \mathrm{Pr}^{0.4}$; then $h = \mathrm{Nu}\, k_f / D$
The $Re^{0.8}$ exponent (turbulent) is stronger than $Re^{0.5}$ (laminar flat plate) — turbulence makes $h$ more responsive to velocity changes
A part-cooling fan blows air over a flat HDPE surface (100 mm × 200 mm). The film coefficient from the fan is $h = 25$ W/m²K, the air temperature is $T_\infty = 25$°C, and the surface is at $T_s = 85$°C. Calculate the total heat removal rate $\dot{Q}$ from that surface.
Use: $\dot{Q} = h A (T_s - T_\infty)$
$hA$ is the overall surface thermal conductance (W/K) — multiply by the temperature driving force to get total power removed
In Unit Ops Lab you are running a shell-and-tube heat exchanger to cool a viscous polymer solution. Your overall heat transfer coefficient $U$ is measured at 80 W/m²K, but you calculated it should be around 400 W/m²K based on tube-side and shell-side film coefficients from textbook correlations. (a) Without doing a full calculation, identify the three resistances in series between the two fluid streams and state which you suspect is dominant based on the discrepancy. (b) You increase the tube-side flow rate by 50% and remeasure $U$: it barely changes. What does that tell you about which side is limiting? (c) Propose one modification to the exchanger and one additional measurement that would help confirm your diagnosis. Frame your answer using the concept of individual film coefficients and the overall resistance.
Chapter Summary
Forced convection is conduction through a thin fluid layer whose thickness is set by the balance between viscous forces and inertia. The velocity boundary layer grows as $\delta \propto x / \sqrt{Re_x}$ along a flat plate; the thermal boundary layer is related by $\delta_T / \delta = \mathrm{Pr}^{-1/3}$, so high-Pr fluids (oils, polymer melts) concentrate all their heat transfer activity in a very thin zone near the wall.
The Reynolds number determines whether flow is laminar or turbulent, with transition at $Re \approx 5 \times 10^5$ for flat plates and $Re \approx 2{,}300$ for pipes. Turbulent flow dramatically thins the effective boundary layer, raising $h$ by factors of 5 to 50. The Nusselt number is the dimensionless film coefficient; once Nu is computed from a correlation, $h = \mathrm{Nu}\, k_f / L$ recovers the dimensional value.
The two correlations used most often in polymer processing are the Pohlhausen solution for laminar flat plates ($\overline{\mathrm{Nu}} = 0.664\, Re^{1/2}\, \mathrm{Pr}^{1/3}$) and the Dittus-Boelter equation for turbulent pipe flow ($\mathrm{Nu}_D = 0.023\, Re_D^{0.8}\, \mathrm{Pr}^n$). In both cases the film coefficient scales sub-linearly with velocity, so switching fluid (air to water) usually gives a far larger gain in $h$ than simply increasing flow rate.
By the final exam, you will be expected to write these equations from memory.
In the next chapter, we remove the fan: Chapter 10 examines natural convection, where the flow is driven entirely by buoyancy forces arising from density differences created by the heat transfer itself. This regime governs heat loss from freely suspended polymer parts, passive cooling of heated beds, and melt-pool circulation in large-volume printers.