Heat Transfer · Chapter 9
0%
Heat Transfer · Chapter Nine

Natural Convection

Natural convection is a feedback process: temperature changes density, gravity turns density differences into motion, and motion changes heat transfer.

Because geometry and orientation determine whether buoyancy aids or suppresses circulation, correlation selection requires more than calculating a Rayleigh number.

Learning objectives

  • Explain how density variation and gravity generate buoyant flow.
  • Compute Grashof and Rayleigh numbers using correct characteristic lengths and film properties.
  • Select correlations for vertical, horizontal, cylindrical, finned, and enclosed configurations.
  • Assess whether natural, forced, or mixed convection controls.
  • Audit orientation, sign, units, validity ranges, and physical limits.
9–1

Physical Mechanism of Natural Convection

Near a hot surface, most fluids become less dense and rise; cooler fluid replaces them. A cold surface reverses the density change and plume direction. Gravity must have a component that can rearrange density layers: heating from below is unstable, while heating from above can remain nearly conduction dominated.

Boussinesq model

For moderate temperature differences, density is treated as constant everywhere except in the buoyancy term, where \(\rho\approx\rho_f[1-\beta(T-T_f)]\). This preserves incompressible flow while retaining its driving force. Large property variation requires variable-property analysis.

Thermal expansion coefficient
\[\beta=-\frac{1}{\rho}\left(\frac{\partial\rho}{\partial T}\right)_p,\qquad \beta=\frac{1}{T_f}\quad\text{for an ideal gas}\]
For gases, \(T_f\) must be in kelvin. Evaluate the remaining properties at \(T_f=(T_s+T_\infty)/2\) unless the correlation specifies another reference.

Physical meaning

\(\beta\Delta T\) is the fractional density-change scale. Without gravity, expansion, or a temperature difference, buoyancy vanishes and so does the natural-convection circulation.

9–2

Equation of Motion and the Grashof Number

The Grashof number compares buoyancy with viscous resistance. Multiplying by \(Pr=\nu/\alpha\) produces the Rayleigh number, which also accounts for thermal diffusion. The sign of \(T_s-T_\infty\) sets plume direction; correlation magnitudes normally use \(|T_s-T_\infty|\).

Natural-convection control parameters
\[Gr_{L_c}=\frac{g\beta|T_s-T_\infty|L_c^3}{\nu^2},\qquad Ra_{L_c}=Gr_{L_c}Pr=\frac{g\beta|T_s-T_\infty|L_c^3}{\nu\alpha}\]
The cubic length makes errors in \(L_c\) especially damaging. Use height for a vertical plate, diameter for a horizontal cylinder, and \(A_s/P\) for many horizontal-plate correlations.

Problem-solving strategy

  1. Confirm there is no dominant imposed flow and identify stable or unstable orientation.
  2. Choose geometry-specific \(L_c\), area, and sign convention.
  3. Evaluate \(\beta,\nu,k,Pr\) at the film temperature and compute \(Ra\).
  4. Select a correlation whose orientation and \(Ra,Pr\) limits match; obtain \(h=Nu\,k/L_c\).
  5. Audit direction, magnitude, radiation contribution, property variation, and the correlation range.
9–3

Natural Convection over Surfaces

Vertical plates develop rising or descending boundary layers along their height. Horizontal surfaces instead depend critically on whether buoyancy carries fluid away from the surface or presses the light fluid against it. Cylinders and spheres require curvature-specific correlations.

Representative average correlations
\[\overline{Nu}_L=0.59Ra_L^{1/4}\quad(10^4
Do not infer a universal laminar/turbulent boundary from these ranges. Use the exact range and characteristic length belonging to the selected geometry.
EXAMPLE 1Passive loss from a vertical control cabinet
Problem. One 0.60 m high by 0.45 m wide vertical side of a control cabinet is at 55 °C in still 15 °C air. At the 35 °C film temperature use \(\beta=1/308\) K\(^{-1}\), \(\nu=1.68\times10^{-5}\) m²/s, \(Pr=0.71\), and \(k=0.0268\) W/mK. Find average \(h\) and convection loss from that side.

Strategy

No fan or wind is specified, and the surface is vertical; use height to calculate \(Ra_L\), verify the vertical-plate range, then apply Newton's law.

Givens

  • \(L=0.60\) m, width \(=0.45\) m, \(\Delta T=40\) K.
  • Film-temperature properties supplied; one face only, radiation excluded.

Physical/model interpretation

Steady, isothermal plate in quiescent air with negligible edge effects. Warm air rises; positive heat is outward.

Numbered derivation steps
1. Rayleigh number. \(Ra_L=g\beta\Delta TL^3Pr/\nu^2=6.92\times10^8\).
2. Validate and correlate. \(10^4
3. Coefficient. \(\bar h=\overline{Nu}_Lk/L=(95.7)(0.0268)/0.60=4.27\ {\rm W/m^2K}\).
4. Heat rate. \(A=0.60(0.45)=0.270\ {\rm m^2}\); \(\dot Q=\bar hA\Delta T=(4.27)(0.270)(40)=46.2\) W.

Unit/sign/limit checks

Units: \(Ra\) and \(Nu\) are dimensionless; \(hA\Delta T\) gives W.

Signs: Heat leaves the hotter cabinet and the buoyant plume rises.

Limit: \(h\) of a few W/m²K is typical of still-air natural convection; the selected range is satisfied but near its upper end.

Final answer: \(\boxed{\bar h=4.27\ {\rm W/m^2K},\quad\dot Q_{\rm conv}=46.2\ {\rm W\ outward}}\)

Engineering insight

Radiation can be comparable to natural convection at this temperature and must be added separately in an enclosure thermal budget.

9–4

Natural Convection from Finned Surfaces

Passive fins add area but also obstruct the buoyant flow that cools them. Closely spaced vertical fins develop merged boundary layers and a high hydraulic resistance; excessively wide spacing wastes envelope volume. Fin efficiency must also be included because the fin is not isothermal.

One optimum-spacing model for vertical parallel fins
\[S_{\rm opt}\approx2.714\frac{L}{Ra_L^{1/4}}\]
Use only with the geometry and property definitions associated with the source correlation. Manufacturing clearance, orientation, shrouding, radiation, and chimney effects can shift the practical optimum.
9–5

Natural Convection inside Enclosures

A sealed gap can conduct heat, circulate by buoyancy, or do both. Enclosure correlations fold circulation into an effective \(Nu\). The gap width is commonly the characteristic length, but aspect ratio and tilt remain essential. If the configuration is stable or \(Ra\) is below onset, \(Nu\to1\), the pure-conduction limit.

Effective gap conductance
\[\dot Q=Nu_L\,kA\frac{T_h-T_c}{L},\qquad k_{\rm eff}=Nu_Lk\]
This relation describes gas-gap conduction plus natural circulation, not radiation between the bounding surfaces.
EXAMPLE 2Heat leak across a sealed instrument-air gap
Problem. A sealed vertical air gap in an instrument enclosure is 20 mm thick and 1.80 m² in area. Its bounding surfaces are at 38 °C and 20 °C. An enclosure correlation, evaluated for the actual aspect ratio and tilt, gives \(Nu_L=1.25\); use \(k=0.0260\) W/mK. Determine gap heat transfer and compare it with pure conduction.

Strategy

The supplied enclosure \(Nu\) already represents conduction plus buoyant circulation. Use gap thickness, then set \(Nu=1\) for the comparison baseline.

Givens

  • \(L=0.020\) m, \(A=1.80\) m², \(\Delta T=18\) K.
  • \(Nu_L=1.25\), \(k=0.0260\) W/mK; radiation excluded.

Physical/model interpretation

Steady sealed gap with correlation already matched to orientation. Positive heat flows from 38 °C toward 20 °C.

Numbered derivation steps
1. Effective conductivity. \(k_{\rm eff}=Nu\,k=(1.25)(0.0260)=0.0325\) W/mK.
2. Gap conductance. \(UA_{\rm gap}=k_{\rm eff}A/L=(0.0325)(1.80)/0.020=2.925\) W/K.
3. Actual transfer. \(\dot Q=(2.925)(18)=52.7\) W, hot to cold.
4. Conduction baseline. \(\dot Q_{\rm cond}=kA\Delta T/L=42.1\) W; circulation adds \(10.5\) W, or 25%.

Unit/sign/limit checks

Units: \(Nu\) is dimensionless, while \(kA\Delta T/L\) reduces as [W/(m·K)](m²)(K)/m = W. Thus both the 52.7 W gap result and the 42.1 W conduction baseline are dimensionally consistent heat rates.

Signs: The stated positive rate is from the 38 °C surface to the 20 °C surface.

Limit: Setting \(Nu=1\) recovers 42.1 W exactly, confirming the conduction limit.

Final answer: \(\boxed{\dot Q_{\rm gap}=52.7\ {\rm W\ hot\ to\ cold};\ 25\%\ above\ conduction}\)

Engineering insight

Increasing a gas gap does not guarantee lower heat leak because it can trigger stronger circulation; radiation and spacer bridges must also be modeled in real enclosures.

9–6

Combined Natural and Forced Convection

An imposed flow does not automatically eliminate buoyancy. The ratio \(Gr/Re^2\) compares buoyancy and inertia: values far below one suggest forced dominance, values far above one suggest natural dominance, and order-one values require a mixed-convection model. Flow direction determines whether buoyancy assists or opposes the imposed stream.

Mixed-convection screening and combination
\[\frac{Gr}{Re^2}\ll1:\ {\rm forced};\quad \frac{Gr}{Re^2}\sim1:\ {\rm mixed};\quad \frac{Gr}{Re^2}\gg1:\ {\rm natural}\]\[Nu_{\rm mix}^n=Nu_F^n\pm Nu_N^n\]
A cubic combination (\(n=3\)) is common, but the exponent and sign must come from guidance for the specific geometry. Opposing flow can be unstable; simple subtraction is not universal.
EXAMPLE 3Mixed convection on a ventilated electronics panel
Problem. For a 0.40 m heated vertical electronics panel, separate valid correlations give \(Nu_F=48\) for upward forced air and \(Nu_N=32\) for the upward buoyant plume. A geometry-specific recommendation uses \(Nu_{\rm mix}^3=Nu_F^3+Nu_N^3\). With \(k=0.0280\) W/mK, determine mixed \(Nu\), \(h\), and the increase over forced convection alone.

Strategy

Both motions are upward, so they assist. Use the supplied cubic combination rather than adding coefficients directly.

Givens

  • \(Nu_F=48\), \(Nu_N=32\), assisting flow, \(n=3\).
  • \(L=0.40\) m, \(k=0.0280\) W/mK.

Physical/model interpretation

The two component correlations already use consistent geometry, length, and properties. The result is an average coefficient; heat direction still depends on wall-to-air temperature.

Numbered derivation steps
1. Combine. \(Nu_{\rm mix}=(48^3+32^3)^{1/3}=52.2\).
2. Convert to \(h\). \(h_{\rm mix}=Nu_{\rm mix}k/L=(52.2)(0.0280)/0.40=3.65\ {\rm W/m^2K}\).
3. Forced baseline. \(h_F=48(0.0280)/0.40=3.36\ {\rm W/m^2K}\).
4. Quantify effect. \((h_{\rm mix}/h_F-1)100=(3.65/3.36-1)100=8.7\%\).

Unit/sign/limit checks

Units: Nusselt numbers are dimensionless and \(Nuk/L\) gives W/m²K.

Signs: The plus sign is justified because imposed and buoyant flows assist; \(h\) itself remains positive.

Limit: \(Nu_{\rm mix}=52.2\) exceeds each component but is less than their arithmetic sum, as a cubic combination requires.

Final answer: \(\boxed{Nu_{\rm mix}=52.2,\quad h_{\rm mix}=3.65\ {\rm W/m^2K},\quad 8.7\%\ increase}\)

Engineering insight

Even when buoyancy alone is substantial, nonlinear combination may produce only a modest increment. Opposing flow needs separate stability-aware validation rather than an automatic minus sign.

Chapter recap

  • Natural convection requires density variation, gravity, and a thermally unstable path for motion.
  • Geometry and orientation define the characteristic length and admissible correlation.
  • Enclosure \(Nu\to1\) is the conduction limit; mixed convection requires direction-aware combination.
  • Before moving on, be able to select film properties, calculate \(Ra\), verify a range, compute \(h\), and explain heat and plume direction.
← Chapter 8 · Internal Convection Chapter 10 · Boiling & Condensation →