Heat transfer rate calculator: conduction, convection and radiation

What is the heat transfer rate? Definition and unit

• Symbol: Q̇ (Q with a dot) for the heat transfer rate, q̇ for the heat flux. The dot denotes the time derivative of the heat Q in joules.
• Unit: watt (W) for Q̇, W/m² for q̇. In electronics cooling, W/cm² is also common: 1 W/cm² = 10,000 W/m².
• Driving force: heat only flows when there is a temperature difference ΔT, and always from the higher to the lower temperature.
• Three mechanisms: conduction in solids and stationary fluids, convection at a surface exposed to a flowing fluid, and thermal radiation between surfaces.
• In power electronics, the heat flow to be removed in steady-state operation equals the power dissipation of the components.

Heat transfer rate formula: conduction, convection and radiation

Each mechanism has its own formula. They share the same structure: the heat transfer rate grows with the area and with the temperature difference. For radiation, however, temperature enters to the fourth power and as an absolute value in kelvin.

• Q̇: heat transfer rate [W]
• q̇: heat flux [W/m²]
• λ: thermal conductivity [W/(m·K)]
• α: heat transfer coefficient [W/(m²·K)], often written as h
• ε: emissivity [–], between 0 and 1
• σ: Stefan-Boltzmann constant, 5.670 · 10⁻⁸ W/(m²·K⁴)
• A: area [m²], d: layer thickness [m]
• ΔT: temperature difference [K], T₁ and T₂: absolute temperatures [K]
• ṁ: mass flow rate [kg/s], c_p: specific heat capacity [J/(kg·K)]

Watch out with radiation: temperatures enter as absolute values in kelvin, T = ϑ + 273.15. Using °C gives wrong results. For conduction and convection, the difference ΔT is all that matters, and it has the same numerical value in K and °C.

Heat transfer rate and thermal resistance: Ohm's law for heat

Heat transfer rate and temperature difference behave like electric current and voltage. The thermal resistance R_th = ΔT / Q̇ in K/W corresponds to electrical resistance. This lets you calculate entire thermal paths like a circuit.

• Conduction: R_th = d / (λ · A)
• Convection: R_th = 1 / (α · A)
• Radiation, linearized: R_th = 1 / (α_rad · A) with α_rad = Q̇ / (A · ΔT)
• Series connection, e.g. base plate, thermal paste and heat sink: R_total = R₁ + R₂ + R₃. The same heat flow passes through all layers and the temperature differences add up.
• Parallel connection, e.g. convection and radiation from the same surface: 1 / R_total = 1 / R_conv + 1 / R_rad. Here the heat flows add up.

With the total resistance, the heat transfer rate follows directly from Q̇ = ΔT_total / R_total. Conversely, ΔT = Q̇ · R gives the temperature rise across each individual layer.

Worked example: heat flow through TIM, base plate and cooling channel

An IGBT module dissipates 300 W through a base area of 60 cm² (0.006 m²) into a cold plate. In between are 0.1 mm of thermal paste (λ = 3 W/(m·K)) and 10 mm of aluminium EN AW-6060 (λ = 200 W/(m·K)). In the cooling channel, α = 3,000 W/(m²·K) with a wetted area of 120 cm². We are looking for the temperature rise from the module base to the cooling water.

1. q̇ = 300 W / 0.006 m² = 50,000 W/m² = 5 W/cm²
2. R_TIM = 0.0001 m / (3 W/(m·K) · 0.006 m²) = 0.0056 K/W → ΔT = 1.7 K
3. R_plate = 0.010 m / (200 W/(m·K) · 0.006 m²) = 0.0083 K/W → ΔT = 2.5 K
4. R_conv = 1 / (3,000 W/(m²·K) · 0.012 m²) = 0.0278 K/W → ΔT = 8.3 K
5. R_total = 0.0056 + 0.0083 + 0.0278 = 0.0417 K/W → ΔT_total = 300 W · 0.0417 K/W = 12.5 K

The module base sits around 12.5 K above the cooling water temperature, and the convective transfer into the water accounts for the largest share. The TIM layer stands out: 0.1 mm of thermal paste costs 1.7 K, two thirds of what 10 mm of aluminium causes. At 5 W/cm², the application is clearly in liquid cooling territory.

Simplification: one-dimensional thermal path without heat spreading and without contact resistances. In practice, the base plate spreads the heat over a larger area, while contact resistances add up at every interface. With R_th,JC from the datasheet, you can then calculate the junction temperature.

Worked example: radiation share of a black anodized heat sink

A passively cooled finned heat sink made of black anodized aluminium (ε = 0.85) reaches 60 °C at an ambient temperature of 25 °C. Its envelope area is 0.10 m², the total fin surface 0.40 m². For natural convection, we assume α = 5 W/(m²·K).

1. T₁ = 60 + 273.15 = 333.15 K, T₂ = 25 + 273.15 = 298.15 K
2. T₁⁴ − T₂⁴ = 1.2319 · 10¹⁰ K⁴ − 0.7902 · 10¹⁰ K⁴ = 4.417 · 10⁹ K⁴
3. q̇_rad = 0.85 · 5.670 · 10⁻⁸ W/(m²·K⁴) · 4.417 · 10⁹ K⁴ = 213 W/m²
4. Q̇_rad = 213 W/m² · 0.10 m² = 21 W, α_rad = 213 W/m² / 35 K = 6.1 W/(m²·K)
5. Q̇_conv = 5 W/(m²·K) · 0.40 m² · 35 K = 70 W → radiation share 21 W / 91 W ≈ 23 %
6. With a fan (α = 50 W/(m²·K)): Q̇_conv = 700 W → radiation share ≈ 3 %
7. Bare aluminium (ε = 0.09): Q̇_rad = 2.3 W instead of 21 W

With natural convection, radiation contributes almost a quarter of the heat dissipation in this example, because α_rad at around 6 W/(m²·K) is of the same order of magnitude as α_conv. With a fan, its share drops to about 3 % and becomes negligible. Black anodizing increases the radiated power by a factor of about 9 compared with bare aluminium.

When does thermal radiation matter?

Whether you need to include radiation depends on the ratio of α_rad to α_conv. At surface temperatures up to about 100 °C and high emissivity, α_rad is around 5 to 8 W/(m²·K). That is a lot compared with natural convection and little compared with forced convection.

• Radiation matters for passively cooled devices, in closed enclosures without fans and at high surface temperatures, because q̇_rad grows with T⁴.
• A black anodized or painted surface raises ε from below 0.1 to 0.85 to 0.95. With forced convection, this brings hardly any thermal benefit.
• On finned heat sinks, the fin surfaces see each other. For radiation to the surroundings, only the envelope area counts as an approximation, not the entire fin surface.
• The formula assumes large surroundings at a uniform temperature. If nearby warm components or enclosure walls radiate back, the net heat flow decreases.

Reference values for α according to Incropera et al., Fundamentals of Heat and Mass Transfer, and the VDI Heat Atlas. α_rad calculated for a surface temperature of 60 °C and 25 °C ambient.

Interpreting heat flux: air or liquid cooling?

The heat flux q̇ is a good first indicator for the cooling concept. What counts is the flux at the mounting surface of the cooler, where the heat passes from the component into the heat sink. The ranges are guideline values, not hard limits.

A thick base plate or a copper inlay spreads the heat and lowers q̇ at the cooler surface. That is why it pays to look at both areas: that of the component and that of the cooler.

Related calculators and topics

• Calculate thermal resistance (Rth) →
• Calculate the heat transfer coefficient α →
• Emissivity: values for metals and surfaces →
• Thermal conductivity: table and formula →
• Calculate semiconductor power loss →
• Calculate junction temperature →
• Calculate cooling power and flow rate →
• Calculate the Reynolds number →
• Heat sinks for power electronics →