Natural Convection: Buoyancy, Rayleigh Numbers, and Churchill–Chu Correlations
In forced convection a fan or pump drives the flow. But what happens when there is no external driver at all? Hot surfaces still cool, and cold surfaces still warm, even in perfectly still air. The engine is buoyancy: warm fluid is less dense, rises, pulls cool fluid in from below, and a self-sustaining circulation forms with no motor required. This chapter quantifies that process.
Close your notes. Try each question from memory, then reveal the answer. The ones you cannot recall are exactly what this chapter reinforces.
RC-1. In forced convection over a flat plate at $Re = 10^5$ and $Pr = 0.71$, you computed $h \approx 9.7\ \text{W/m}^2\text{K}$. What two things would you change to get a higher $h$?
RC-2. What is the key difference between forced convection and natural (free) convection?
RC-3. Why does a hot surface in still air produce an upward current? What two material properties determine how strong that current is?
10.1 The Cooling Oven Door
You open a polymer-processing oven and leave the door ajar. Within minutes the metal surface that was at 200 °C has cooled enough to touch. No fan runs. The room air is still. Yet heat leaves the surface at a steady rate. Where does the energy go?
Touch the side of any hot oven: you feel a column of warm air rising. That column is natural (free) convection. The hot surface warms a thin layer of air, making it less dense. Buoyancy lifts it; cooler room air slides in to replace it. A circulation loop forms with no pump, consuming only the thermal energy stored in the surface.
For most polymer processes — quiescent melt pools, oven interiors, jacketed reactors left idle, cooling flanges in still air — natural convection is the dominant heat-transfer mode on the gas side. The heat-transfer coefficients it produces are modest (typically $h \approx 2\text{--}25\ \text{W/m}^2\text{K}$ in air), but understanding them is essential for predicting cooling rates, set-point temperatures, and energy losses.
Before reading on — you already know that $Re$ is the single number that governs whether forced convection flow is laminar or turbulent. This chapter introduces $Ra_L = Gr_L \cdot Pr$, which plays exactly the same role in natural convection. The key difference: in forced convection you set the velocity externally; in natural convection buoyancy sets it for you. Same mathematics, completely different physics — and that is precisely what makes Ra so elegant.
10.2 Buoyancy: The Driving Force
Consider a vertical heated plate at surface temperature $T_s$ in quiescent fluid at $T_\infty < T_s$. The fluid directly adjacent to the plate is heated, its density drops, and buoyancy generates an upward body force per unit volume:
$$f_b = \rho g \beta (T - T_\infty)$$
Here $\beta$ is the volumetric thermal expansion coefficient (units K$^{-1}$) and $g = 9.81\ \text{m/s}^2$. This buoyancy force drives the upward flow, while viscosity $\mu$ resists it. The competition between buoyancy and viscosity determines whether the resulting flow is laminar or turbulent.
Unlike forced convection, the flow velocity in natural convection is not an independent variable — it is determined by the buoyancy force itself. This makes natural convection inherently coupled: the temperature field drives the flow, and the flow transports heat.
10.3 The Grashof Number
The Grashof number $Gr_L$ plays the role in natural convection that $Re$ plays in forced convection. It is the ratio of buoyancy forces to viscous forces:
$$Gr_L = \frac{g\,\beta\,\Delta T\,L^3}{\nu^2}$$
where $L$ is the characteristic length (plate height for a vertical surface), $\Delta T = T_s - T_\infty$, and $\nu$ is kinematic viscosity. Notice the $L^3$ dependence: a surface twice as tall produces $2^3 = 8\times$ stronger buoyancy-driven flow.
The laminar–turbulent transition for a vertical plate occurs near $Gr_L \sim 10^9$.
All fluid properties are evaluated at the film temperature:
$$T_{\text{film}} = \frac{T_s + T_\infty}{2}$$
10.4 The Rayleigh Number
In practice, the Grashof number almost always appears multiplied by the Prandtl number. This product is the Rayleigh number:
$$Ra_L = Gr_L \cdot Pr = \frac{g\,\beta\,\Delta T\,L^3}{\nu\,\alpha}$$
where $\alpha = k/(\rho c_p)$ is the thermal diffusivity. Physically, $Ra_L$ captures how much buoyancy-driven inertia exists relative to the product of viscous and thermal diffusion that both damp the flow. All empirical correlations for natural convection are expressed in terms of $Ra_L$.
Physical interpretation of Ra
- $Ra \lesssim 10^4$: Conduction dominates; buoyancy too weak to generate significant flow.
- $10^4 \lesssim Ra \lesssim 10^9$: Laminar natural convection; most practical engineering cases.
- $Ra \gtrsim 10^9$: Turbulent natural convection; tall surfaces, large $\Delta T$.
Buoyancy driving force ÷ (viscous × thermal damping)
Boundary layer thickens as $L^{3/4}$; local $h$ decreases up the plate
10.5 The Thermal Expansion Coefficient $\beta$
For an ideal gas (which includes air and most processing gases):
$$\beta = \frac{1}{T_{\text{film}}} \quad [\text{K}^{-1}], \quad T \text{ in Kelvin}$$
This is one of the most important simplifications in natural convection. For air at a film temperature of 350 K, for example: $\beta = 1/350 = 2.86 \times 10^{-3}\ \text{K}^{-1}$.
For liquids (polymer melts, water, oil), $\beta$ must be looked up in property tables, since liquids are not ideal gases. Polymer melts typically have $\beta \approx 5\text{--}10 \times 10^{-4}\ \text{K}^{-1}$.
10.6 Churchill–Chu Correlation: Vertical Plate
The most widely used correlation for a vertical isothermal plate over the full range of $Ra$ is the Churchill–Chu (1975) equation:
$$\overline{Nu}_L = \left[0.825 + \frac{0.387\,Ra_L^{1/6}}{\left(1 + \left(\dfrac{0.492}{Pr}\right)^{9/16}\right)^{8/27}}\right]^2$$
Valid for: vertical plate or cylinder, $0 \le Ra_L \le 10^{12}$, any Pr.
The average heat-transfer coefficient follows from the definition of Nusselt number:
$$h = \frac{\overline{Nu}_L \cdot k_f}{L}$$
For air ($Pr \approx 0.71$) in the laminar range ($Ra \sim 10^5\text{--}10^9$), a simpler approximation works: $\overline{Nu}_L \approx 0.59\,Ra_L^{1/4}$ (laminar), $\overline{Nu}_L \approx 0.10\,Ra_L^{1/3}$ (turbulent). But Churchill–Chu covers all regimes in one formula.
10.7 Horizontal Surfaces
For horizontal plates, the orientation of the hot or cold surface matters because buoyancy can either assist or oppose the flow.
| Configuration | Range of $Ra$ | Correlation |
|---|---|---|
| Heated upper surface (or cooled lower surface) | $10^4 \le Ra \le 10^7$ | $Nu = 0.54\,Ra^{1/4}$ |
| Heated upper surface (or cooled lower surface) | $10^7 \le Ra \le 10^{11}$ | $Nu = 0.15\,Ra^{1/3}$ |
| Heated lower surface (or cooled upper surface) | $10^5 \le Ra \le 10^{11}$ | $Nu = 0.27\,Ra^{1/4}$ |
The characteristic length $L$ for horizontal plates is defined as $L = A_s / P$, where $A_s$ is the surface area and $P$ is the perimeter. For a $W \times D$ plate: $L = WD / [2(W+D)]$.
10.8 Interactive: Natural Convection Explorer
Adjust the surface temperature and plate height to see how the thermal plume develops and how $Ra$, $Nu$, and $h$ change. Air properties at film temperature are used automatically.
10.9 Worked Examples
Example 10.1 — Rayleigh Number for an Oven Wall
A vertical oven wall (height $L = 0.3\ \text{m}$) is at $T_s = 100\,°\text{C}$ in a room at $T_\infty = 20\,°\text{C}$. Using air properties at the film temperature ($\beta = 3.0 \times 10^{-3}\ \text{K}^{-1}$, $\nu = 1.5 \times 10^{-5}\ \text{m}^2/\text{s}$, $\alpha = 2.1 \times 10^{-5}\ \text{m}^2/\text{s}$), compute $Ra_L$.
Solution:
$\Delta T = 100 - 20 = 40\,°\text{C}$
$$Ra_L = \frac{g\,\beta\,\Delta T\,L^3}{\nu\,\alpha} = \frac{9.81 \times 3.0\times10^{-3} \times 40 \times (0.3)^3} {1.5\times10^{-5} \times 2.1\times10^{-5}}$$
Numerator: $9.81 \times 3.0\times10^{-3} \times 40 \times 0.027 = 9.81 \times 3.24\times10^{-3} = 0.03178$
Denominator: $1.5\times10^{-5} \times 2.1\times10^{-5} = 3.15\times10^{-10}$
$$Ra_L = \frac{0.03178}{3.15\times10^{-10}} \approx 1.01 \times 10^8$$
This is in the laminar range ($Ra < 10^9$). ✓
Example 10.2 — Churchill–Chu $Nu$ and $h$
For the oven wall in Example 10.1 ($Ra = 10^8$, $Pr = 0.714$, $k_{\text{air}} = 0.028\ \text{W/mK}$, $L = 0.3\ \text{m}$), find $\overline{Nu}_L$ and $h$ using Churchill–Chu.
Solution:
Step 1: $Ra^{1/6} = (10^8)^{1/6} = 10^{8/6} = 10^{1.333} \approx 21.54$
Step 2: $(0.492/0.714)^{9/16} = (0.689)^{0.5625} \approx 0.795$
Step 3: $(1 + 0.795)^{8/27} = 1.795^{0.296} \approx 1.156$
Step 4: $0.387 \times 21.54 / 1.156 \approx 7.21$
Step 5: $\overline{Nu} = (0.825 + 7.21)^2 = (8.035)^2 \approx 64.6$
(Slight rounding to $Ra = 10^8$ gives $Nu \approx 61$ with exact Ra.)
$$h = \frac{Nu \cdot k}{L} = \frac{61 \times 0.028}{0.3} = \frac{1.708}{0.3} \approx 5.7\ \text{W/m}^2\text{K}$$
This is a typical natural convection $h$ for air: modest, but significant for long cooling times.
10.10 Concept Questions
You double the height $L$ of a heated vertical plate while keeping $\Delta T$ constant. By what factor does $Ra_L$ change?
Why is $\beta = 1/T_{\text{film}}$ only valid for ideal gases, and not for water or polymer melts?
A horizontal heated plate faces upward versus downward. Which orientation gives a higher $h$, and why?
An engineer claims that adding insulation to the outside of a vertical polymer melt tank will slow natural convection cooling on the inside (melt side). Is this reasoning correct?
Derive the Rayleigh number from first principles using physical scaling.
(a) Consider a vertical plate with surface temperature $T_s$ in quiescent fluid at $T_\infty$. The buoyancy force per unit volume is $f_b = \rho g\beta\Delta T$. The viscous resistance force per unit volume scales as $f_v \sim \mu\,u / L^2$. Balance these to find the natural convection velocity scale $u_{\text{nc}}$ in terms of $g$, $\beta$, $\Delta T$, $L$, and $\nu$.
(b) Now form the thermal Péclet number $Pe = u_{\text{nc}} L / \alpha$ (ratio of convective to conductive heat transport). Substitute your $u_{\text{nc}}$ from part (a) and simplify. What familiar dimensionless group do you recover?
(c) From the scaling $Nu \sim Pe^{1/2}$ (thermal boundary-layer argument), write $Nu$ as a function of $Ra$. What exponent on $Ra$ do you expect for laminar natural convection, and does this match the Churchill–Chu simplified form $Nu \approx 0.59\,Ra^{1/4}$?
(d) A polymer melt pool has $g\beta\Delta T \approx 0.02\ \text{m/s}^2$, $L = 0.1\ \text{m}$, $\nu = 1\times10^{-4}\ \text{m}^2/\text{s}$, $\alpha = 1\times10^{-7}\ \text{m}^2/\text{s}$. Compute $Ra$ and $u_{\text{nc}}$. Is the flow laminar? Is convection significant ($Pe \gg 1$)?
10.11 Problem Set
Part B — Calculation Problems
A vertical polymer film (height $L = 0.30\ \text{m}$) exits an oven at $T_s = 100\,°\text{C}$ into still air at $T_\infty = 20\,°\text{C}$. Air properties at the film temperature: $\beta = 3.0 \times 10^{-3}\ \text{K}^{-1}$, $\nu = 1.5 \times 10^{-5}\ \text{m}^2/\text{s}$, $\alpha = 2.1 \times 10^{-5}\ \text{m}^2/\text{s}$. Compute $Ra_L$.
Use: $Ra_L = g\,\beta\,\Delta T\,L^3\,/\,(\nu\,\alpha)$
Buoyancy driving force ÷ (viscous damping × thermal diffusion) — the larger $Ra$, the stronger the natural convection plume
Air flows naturally along a hot surface. The surface temperature is $T_s = 427\,°\text{C}$ and the ambient temperature is $T_\infty = 27\,°\text{C}$. Calculate $\beta$ for air at this condition using the ideal-gas approximation. Express your answer in $\text{K}^{-1}$.
Use: $\beta = 1/T_{\text{film}}$, with $T_{\text{film}}$ in Kelvin
Ideal gas: volume grows in direct proportion to absolute temperature — so 1 K rise at 350 K expands the gas by 1/350 of its volume
For the vertical plate in P1 ($Ra_L = 10^8$, $Pr = 0.714$, $k_{\text{air}} = 0.028\ \text{W/mK}$, $L = 0.30\ \text{m}$), find $\overline{Nu}_L$ using the Churchill–Chu correlation.
Use: $\overline{Nu}_L = \!\left[0.825 + 0.387\,Ra^{1/6}\big/\!\left(1+(0.492/Pr)^{9/16}\right)^{8/27}\right]^2$
$Ra^{1/6} = (10^8)^{1/6} \approx 21.54$; $(0.492/0.714)^{9/16} \approx 0.795$; $(1.795)^{8/27} \approx 1.156$; then bracket = $0.825 + 7.21 \approx 8.04$
Now use your $Nu$ from P3a to find the film coefficient $h$ (same plate: $k_{\text{air}} = 0.028\ \text{W/mK}$, $L = 0.30\ \text{m}$).
Use: $h = \overline{Nu}\, k_f / L$
Churchill–Chu blends the laminar and turbulent regimes into one formula — $Nu$ tells you the dimensionless heat transfer; multiply by $k_f/L$ to get dimensional $h$
A jacketed polymer reactor has a vertical outer wall ($0.5\ \text{m}$ wide, $1.0\ \text{m}$ tall) at $T_s = 70\,°\text{C}$. The factory air is at $T_\infty = 20\,°\text{C}$. Using a natural convection coefficient of $h = 6\ \text{W/m}^2\text{K}$, what is the total heat loss from this wall?
Use: $\dot{Q} = h\,A\,\Delta T$
Newton's law of cooling: $h$ (W/m²K) × area (m²) × temperature gap (K) = power lost (W)
A polymer slab cools by convection from one face. In still air (natural convection) $h_{\text{nat}} = 5\ \text{W/m}^2\text{K}$. A fan provides forced convection with $h_{\text{forced}} = 25\ \text{W/m}^2\text{K}$. Using the lumped capacitance result $\tau = \rho c_p L_c / h$, by what factor does adding the fan reduce the cooling time constant?
Since $\tau \propto 1/h$: $\tau_{\text{nat}}/\tau_{\text{forced}} = h_{\text{forced}}/h_{\text{nat}}$
Cooling time constant = thermal mass ÷ heat-removal rate. Doubling $h$ halves $\tau$ — the most direct lever for faster cooling
In your unit operations lab, a 200-L jacketed stainless steel reactor (wall area $A_{\text{outer}} \approx 1.5\ \text{m}^2$, outer wall at $T_s = 80\,°\text{C}$) cools in still room air ($T_\infty = 22\,°\text{C}$). The operator expected the batch to reach 40 °C within 2 hours, but it took 6 hours.
(a) Estimate the natural convection $h$ on the outside of the reactor using Churchill–Chu with $L = 1.0\ \text{m}$ (effective height), $Ra_L \approx 1.5 \times 10^9$ (turbulent). Use $k_{\text{air}} = 0.028\ \text{W/mK}$.
(b) The heat must also flow through 3 mm of stainless steel ($k_{\text{steel}} = 16\ \text{W/mK}$) and the jacket wall. Compute the thermal resistance for conduction through the steel and compare it to the convection resistance.
(c) The engineer wants to cut cooling time by 3$\times$. Propose and justify two feasible modifications to the process setup.
10.12 Chapter Summary
Key Equations
| Quantity | Formula | Notes |
|---|---|---|
| Film temperature | $T_f = (T_s + T_\infty)/2$ | Evaluate all properties here |
| Thermal expansion (ideal gas) | $\beta = 1/T_f$ (K−1) | $T_f$ must be in Kelvin |
| Rayleigh number | $Ra_L = g\beta\Delta T L^3/(\nu\alpha)$ | $= Gr_L \cdot Pr$ |
| Churchill–Chu (vertical plate) | $\overline{Nu} = \!\left[0.825 + 0.387\,Ra^{1/6}\big/\!\left(1+(0.492/Pr)^{9/16}\right)^{8/27}\right]^2$ | All $Ra$, any $Pr$ |
| Horizontal plate (heated up) | $Nu = 0.54\,Ra^{1/4}$ ($10^4 \le Ra \le 10^7$) $Nu = 0.15\,Ra^{1/3}$ ($Ra > 10^7$) |
|
| Heat-transfer coefficient | $h = \overline{Nu} \cdot k_f / L$ | |
| Heat transfer | $\dot{Q} = h A \Delta T$ | Newton's law |
Key Ideas
- Natural convection is driven entirely by buoyancy; no fan or pump is needed.
- The Rayleigh number $Ra = Gr \cdot Pr$ replaces the Reynolds number as the governing parameter.
- For an ideal gas, $\beta = 1/T_{\text{film}}$ (Kelvin) — memorize this.
- The Churchill–Chu correlation covers laminar and turbulent regimes in one formula.
- Typical $h$ for natural convection in air: $2\text{--}25\ \text{W/m}^2\text{K}$; for forced convection in air: $25\text{--}250\ \text{W/m}^2\text{K}$; for water: $500\text{--}10{,}000\ \text{W/m}^2\text{K}$.
- Adding a fan can reduce cooling time by 5–20$\times$.