Introduction & Context

The convective heat-transfer coefficient h quantifies how effectively heat is carried away from a surface by an adjacent moving fluid. In process engineering it is the key parameter that links the driving temperature difference to the rate of heat removal, so it appears in every heat-exchanger, reactor, furnace, or cooler design calculation. A quick estimate of h is often obtained from a conductive-layer model: the fluid very close to the wall is assumed to behave as a stagnant film whose thickness δ is fixed by the flow regime; once δ is known, h follows directly from the fluid thermal conductivity k. The sheet below shows how to compute h and the resulting heat duty q for a single flat plate of length L and unit width.

Methodology & Formulas

Convert the film thickness from millimetres to metres: \[ \delta = \frac{\delta_{\text{mm}}}{1000} \]
Compute the convective heat-transfer coefficient: \[ h = \frac{k}{\delta} \] where k is the thermal conductivity of the fluid evaluated at the film temperature.
Determine the temperature driving force: \[ \Delta T = T_{\text{plate}} - T_{\text{air}} \]
Calculate the heat-transfer area (unit width): \[ A = L \cdot \text{width} \]
Obtain the total heat-flow rate: \[ q = h \cdot A \cdot \Delta T \]

Flow regime	Typical δ range (mm)	Remarks
Laminar boundary layer	0.5 – 2	Natural or low-speed forced convection
Turbulent boundary layer	0.1 – 0.5	High-speed forced convection

The convective heat-transfer coefficient, usually written as h, is the proportionality constant that links the heat flux (q) to the temperature difference between a surface and the moving fluid: q = h A ΔT. A larger h means more heat can be moved for the same area and ΔT, so it directly sets the required heat-transfer area for exchangers, heaters, and coolers. In short, it tells you how “hard” the fluid is working to carry heat away from or toward the wall.

Velocity: doubling velocity in turbulent regime can raise h 60–80 %.
Thermal conductivity: appears to the 0.6–0.8 power in most correlations.
Viscosity: higher viscosity thickens the boundary layer and drops h.
Density & specific heat: grouped as ρ·Cp, they set the fluid’s heat-carrying capacity.
Pipe diameter: smaller tubes give higher velocities and thinner boundary layers, increasing h.

Pick a correlation matched to your geometry (Dittus-Boelert for turbulent pipes, Sieder-Tate if large viscosity changes, etc.).
Evaluate all physical properties at the film temperature—the average of wall and bulk temperatures.
Check the Reynolds and Prandtl numbers to be inside the correlation’s stated range.
Add a fouling resistance on the side where deposits are expected; this lowers the effective h.
Apply a 20–30 % safety factor on the final area if the exchanger is critical and cannot be cleaned easily.

Temperature, velocity, and even phase condition evolve along the length, so local h varies. Simulators split the exchanger into small segments, recalculate properties and h in each segment, then integrate the overall duty. This segmented approach captures effects such as viscosity changes in heavy oils or partial condensation in vapor streams.

Worked Example – Estimating the Convective Heat-Transfer Coefficient for a Thin Steel Plate

A small process heater uses a 1 mm-thick steel plate to transfer heat to an air stream. To verify that the plate can dissipate 280 W under laminar flow conditions, the convection coefficient must be checked against the allowable temperature difference.

Knowns

Air thermal conductivity, \(k\) = 0.028 W m^-1 K^-1
Boundary-layer thickness, \(\delta\) = 1.2 mm = 0.0012 m
Plate temperature, \(T_{\text{plate}}\) = 80 °C
Air free-stream temperature, \(T_{\text{air}}\) = 20 °C
Characteristic length (flow length), \(L\) = 0.2 m
Plate width, \(w\) = 1.0 m (unit width for area calculation)
Required heat duty, \(q\) = 280 W

Step-by-Step Calculation

Compute the temperature difference driving convection: \[ \Delta T = T_{\text{plate}} - T_{\text{air}} = 80 - 20 = 60\ \text{K} \]
Determine the convective area: \[ A = L \times w = 0.2 \times 1.0 = 0.2\ \text{m}^2 \]
Estimate the convection coefficient from the boundary-layer approximation (laminar flow over a flat plate): \[ h = \frac{k}{\delta} = \frac{0.028}{0.0012} = 23.333\ \text{W m}^{-2}\ \text{K}^{-1} \]
Calculate the actual heat-transfer rate: \[ q_{\text{calc}} = h\,A\,\Delta T = 23.333 \times 0.2 \times 60 = 280\ \text{W} \]

Final Answer

The estimated convective heat-transfer coefficient is 23.3 W m^-2 K^-1, yielding a heat flux of 280 W for the specified 60 K temperature difference—confirming the plate meets the thermal requirement.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle