How do you calculate the heat transfer coefficient?

Via the Nusselt number: first determine the Reynolds number Re = v · L / ν and the Prandtl number of the fluid. A suitable correlation then gives Nu, for example Nu = 0.023 · Re^0.8 · Pr^0.4 for turbulent tube flow. Finally α = Nu · λ / L. If the heat flow has been measured, α also follows directly from α = Q̇ / (A · ΔT).

What is the heat transfer coefficient of air?

For natural convection, α for air is about 2 to 25 W/(m²·K); for forced convection with a fan, about 25 to 250 W/(m²·K). In the fin channels of fan-cooled heat sinks, values of around 30 to 100 W/(m²·K) are common.

What is the heat transfer coefficient of water?

About 50 to 1,000 W/(m²·K) for natural convection and about 100 to 20,000 W/(m²·K) for forced convection. In cold plates with channels of a few millimetres and velocities around 1 m/s, several thousand W/(m²·K) are typical; microchannels can exceed 10,000 W/(m²·K).

What is the difference between the heat transfer coefficient and the overall heat transfer coefficient?

The heat transfer coefficient α describes only the transfer between one surface and a fluid. The overall heat transfer coefficient U describes the whole path from one fluid through a wall to the other fluid. For a plane wall, 1/U = 1/α₁ + s/λ + 1/α₂. Both have the unit W/(m²·K).

What is the difference between heat transfer coefficient and thermal conductivity?

Thermal conductivity λ is a material property in W/(m·K) and describes heat conduction inside a material. The heat transfer coefficient α in W/(m²·K) describes convective transfer at a surface and depends on the flow. Both appear in the Nusselt number Nu = α · L / λ, where λ is the thermal conductivity of the fluid.

What is the unit of the heat transfer coefficient?

Watts per square metre and kelvin, W/(m²·K). Because a temperature difference of 1 K equals 1 °C, W/(m²·°C) is equivalent. In US customary units, 1 BTU/(h·ft²·°F) ≈ 5.678 W/(m²·K).

How can the heat transfer coefficient be increased?

With higher flow velocity, smaller hydraulic diameters, turbulent instead of laminar flow, and structures that keep restarting the boundary layer, such as pin fins, turbulators or interrupted fins. A fluid with higher thermal conductivity also helps. Almost all of these measures increase the pressure drop.

Heat transfer coefficient: values, formulas and calculator

Q: What is the heat transfer coefficient?

The heat transfer coefficient α (or h) states the heat flow transferred per square metre of surface and per kelvin of temperature difference between a wall and a flowing fluid. It is given in W/(m²·K) and depends on the fluid, the flow velocity and the geometry.

The heat transfer coefficient α, usually written h in English literature, states how much heat is transferred per square metre of surface and per kelvin of temperature difference between a wall and a flowing fluid. Its unit is W/(m²·K). Unlike thermal conductivity, it is not a material property: it depends on the fluid, the flow velocity and the geometry, and ranges from a few W/(m²·K) in still air to more than 10,000 W/(m²·K) for water in narrow channels.

What is the heat transfer coefficient?

The convective heat transfer coefficient α (or h) is the proportionality factor between the heat flux at a surface and the temperature difference between that surface and the fluid. A value of 1,000 W/(m²·K) means that every square metre transfers 1,000 W per kelvin of temperature difference.

Symbol: α (alpha) in German and European literature, h in most English textbooks. Unit: W/(m²·K).
α is a property of the system, not of a material. It depends on the fluid (thermal conductivity, viscosity, heat capacity), the flow velocity, the flow regime (laminar or turbulent) and the surface geometry.
The reciprocal 1/α is an area-specific thermal resistance in m²·K/W. For a surface area A, the convective thermal resistance is Rth = 1/(α · A) in K/W.
In air cooling, the transfer from the heat sink to the air is often the largest single resistance in the entire heat path.

Heat transfer coefficient and heat flow calculator

You know α, for example from the table below or from a simulation, and want the transferred heat flow and the convective thermal resistance.

Heat transfer coefficient α

W/(m²·K)

Heat transfer area A

cm²

Area in cm² (1 m² = 10,000 cm²). For finned heat sinks use the total wetted surface, for cold plates the wetted channel wall.Wall-to-fluid temperature difference ΔT

Formula: Newton's law of cooling

The convective heat flow between a surface and a fluid is described by Newton's law of cooling. α combines all influences of the flow into a single number.

Q̇ = α · A · (T_s − T_f)

Q̇: transferred heat flow rate [W]
α: heat transfer coefficient [W/(m²·K)]
A: heat transfer surface area [m²]
T_s: surface temperature, T_f: fluid temperature, for channels usually the mean fluid temperature [°C or K]

Rth = 1 / (α · A)

Rearranged, this gives the convective thermal resistance in K/W. It slots directly into the series chain of thermal resistances from the semiconductor to the coolant.

Calculate thermal resistance →

Heat transfer coefficient vs. thermal conductivity vs. overall heat transfer coefficient

The three quantities are often confused, but they describe different things. λ belongs to the material, α to the interface between wall and fluid, and U to the whole component from fluid to fluid.

Quantity	Unit	Describes	Depends on
Thermal conductivity λ (k)	W/(m·K)	Heat conduction inside a material	Material and temperature
Heat transfer coefficient α (h)	W/(m²·K)	Heat transfer between a surface and a flowing fluid (convection)	Fluid, velocity, flow regime, geometry
Overall heat transfer coefficient U	W/(m²·K)	Total heat transfer through a wall from one fluid to the other	Both convective coefficients, wall thickness and wall conductivity

1/U = 1/α₁ + s/λ + 1/α₂

For a plane wall, three partial resistances add up: convection on one side, conduction through the wall of thickness s and convection on the other side. The smallest α dominates, so improving the weaker side almost always pays off first.

Heat transfer coefficient table: typical values

The values show orders of magnitude. Where a specific case falls depends on velocity, geometry and temperature. For an actual design there is no way around a calculation via the Nusselt number, a simulation or a measurement.

Application	α in W/(m²·K)	Note
Natural convection, air and gases	2 to 25	Passive heat sinks and enclosure surfaces without a fan.
Forced convection, air and gases	25 to 250	Fan-cooled heat sinks. In fin channels at a few m/s, values are usually around 30 to 100.
Natural convection, water and other liquids	50 to 1,000	Stagnant liquid, for example a tank or bath without circulation.
Forced convection, water and water-glycol in channels	100 to 20,000	Liquid cold plates. With channels of a few millimetres at around 1 m/s, typically several thousand. Water-glycol is clearly lower than water.
Microchannels with water (Dh below 1 mm)	about 5,000 to 25,000	Estimated from Nu ≈ 4 to 8 at Dh = 0.2 to 0.5 mm. High pressure drop.
Boiling and condensation (phase change)	2,500 to 100,000	Two-phase cooling as well as evaporation and condensation inside heat pipes and vapor chambers.

Ranges according to Incropera, Fundamentals of Heat and Mass Transfer (Table 1.1), and the VDI Heat Atlas. Values for fin channels and microchannels are estimated with the Nusselt correlations on this page.

Heat transfer coefficient of air and water compared

Water transfers heat at a surface orders of magnitude better than air. The reasons are its roughly 23 times higher thermal conductivity and its roughly 17 times lower kinematic viscosity, which gives a higher Reynolds number at the same velocity. Two examples for a 100 mm long flat plate in parallel flow, calculated with the tool above:

Air, 25 °C, 5 m/s: Re ≈ 32,000 (laminar), α ≈ 28 W/(m²·K).
Water, 25 °C, 0.5 m/s: Re ≈ 56,000 (laminar), α ≈ 1,740 W/(m²·K).
At a tenth of the velocity, water delivers a heat transfer coefficient about 60 times higher. That is why a cold plate gets by with a few hundred square centimetres of wetted area, while an air-cooled heat sink needs many times more fin area for the same power.
Water-glycol 50/50 lies in between but much closer to water. At the same velocity it reaches roughly one third to two thirds of the α of pure water, depending on the flow regime.

Calculating the heat transfer coefficient: Nusselt, Reynolds and Prandtl numbers

α cannot simply be looked up; it is determined via dimensionless numbers. The Nusselt number is the dimensionless heat transfer coefficient. Empirical correlations give Nu as a function of the Reynolds and Prandtl numbers, and α follows from Nu.

Nu = α · L / λ ⇒ α = Nu · λ / L

Nusselt number: ratio of convective to purely conductive heat transfer in the fluid. L is the characteristic length, i.e. the plate length or the hydraulic diameter Dh. λ is the thermal conductivity of the fluid, not of the wall.

Re = v · L / ν

Reynolds number: ratio of inertial to viscous forces. It decides whether the flow is laminar or turbulent. In tubes and channels, flow below Re ≈ 2,300 is considered laminar; on a flat plate transition occurs around Re ≈ 5 · 10⁵.

Pr = ν / a = μ · c_p / λ

Prandtl number: a pure fluid property, the ratio of momentum diffusivity to thermal diffusivity. Air ≈ 0.7, water at 25 °C ≈ 6, water-glycol 50/50 ≈ 30, oils well above 100.

Dh = 4 · A / P

Hydraulic diameter for non-circular cross sections: four times the cross-sectional area A divided by the wetted perimeter P. For a tube Dh equals the inner diameter, for a flat gap it is roughly twice the gap width.

Calculate the Reynolds number →

Nusselt correlations for plates, channels and tubes

The following equations are standard forced-convection correlations from Incropera and the VDI Heat Atlas. Fluid properties are evaluated at the film temperature for plates and at the mean fluid temperature for channels. The calculator above uses exactly these equations.

Flat plate, laminar (average Nu over length L)

Nu = 0.664 · Re^0.5 · Pr^(1/3)

Re < 5 · 10⁵, Pr ≥ 0.6, isothermal plate.

Flat plate, turbulent

Nu = 0.037 · Re^0.8 · Pr^(1/3)

5 · 10⁵ < Re < 10⁸, 0.6 < Pr < 60. Applies to a boundary layer that is turbulent from the leading edge. With a laminar leading section, 871 is subtracted: Nu = (0.037 · Re^0.8 − 871) · Pr^(1/3).

Channel or tube, laminar, fully developed

Nu = 3.66 (constant wall temperature) or Nu = 4.36 (constant heat flux)

Re < 2,300, round tube. Square duct 2.98 or 3.61, flat gap heated on both sides 7.54 or 8.23. In the thermal entry region Nu is higher.

Tube, turbulent: Dittus-Boelter

Nu = 0.023 · Re^0.8 · Pr^0.4

Re > 10,000, 0.6 < Pr < 160, L/D > 10, fluid being heated (exponent 0.3 when cooled). Deviations of up to about 25 % are possible.

Tube, transition and turbulent: Gnielinski

Nu = (f/8) · (Re − 1000) · Pr / [1 + 12.7 · (f/8)^0.5 · (Pr^(2/3) − 1)]  with  f = (0.790 · ln Re − 1.64)^(−2)

3,000 < Re < 5 · 10⁶, 0.5 < Pr < 2,000. More accurate than Dittus-Boelter and also applicable in the transition region.

Between Re ≈ 2,300 and 10,000, channel flow is in the transition region. Whether it stays laminar or becomes turbulent depends on the inlet, bends and roughness, and results are much less certain. For a robust design, avoid this range or calculate both limiting cases.

Worked example: water in the cooling tube of a cold plate

Water at 25 °C flows through a cold plate with a pressed-in tube. The tube has an 8 mm inner diameter and the mean velocity is 1.5 m/s (about 4.5 l/min). The wetted inner surface of the tube is 150 cm². We want α, the convective thermal resistance and the heat flow at a 10 K difference between tube wall and water.

Water properties at 25 °C: λ = 0.607 W/(m·K), ν = 0.893 mm²/s, Pr = 6.1
Re = v · D / ν = 1.5 · 0.008 / (0.893 · 10⁻⁶) ≈ 13,440 → turbulent
Nu = 0.023 · 13,440^0.8 · 6.1^0.4 ≈ 95.2
α = Nu · λ / D = 95.2 · 0.607 / 0.008 ≈ 7,220 W/(m²·K)
Rth = 1 / (α · A) = 1 / (7,220 · 0.015) ≈ 0.0092 K/W
Q̇ = α · A · ΔT = 7,220 · 0.015 · 10 ≈ 1,080 W

The convective resistance is about 0.009 K/W. Gnielinski gives Nu ≈ 98.7 and α ≈ 7,490 W/(m²·K) for the same case, just under 4 % more. In a real cold plate the effective area is smaller because heat does not arrive evenly around the whole tube circumference. Heat spreading in the base plate and the interface between tube and plate therefore belong in the calculation too.

What α means for heat sinks and cold plates

For a design, the product α · A is what counts. There are two levers: more surface area or a higher heat transfer coefficient. Both come at a price.

Fins increase the area A. For air, with α in the range of 10 to 100 W/(m²·K), this is the most important lever. Fin efficiency drops as fins get taller, though, and fin spacing that is too tight chokes the flow.
Higher velocity raises α: roughly with v^0.8 in turbulent flow and with v^0.5 on a laminar plate. In fully developed laminar channel flow, however, Nu is constant, so more flow barely raises α there.
Turbulators, pin fins and interrupted fins restart the boundary layer again and again. A young, thin boundary layer transfers more heat than a fully developed one.
Small hydraulic diameters raise α directly, because α = Nu · λ / Dh. Microchannels below 1 mm reach very high values even in laminar flow.
The price is pressure drop. It rises disproportionately with velocity and sharply in small channels. The pump or fan has to supply that pressure.
The coolant matters: because of its lower thermal conductivity and higher viscosity, water-glycol 50/50 reaches only about one third to two thirds of the α of pure water at the same velocity.

Related calculators and basics

Frequently asked questions about the heat transfer coefficient

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