Thermal conductor

Linear heat conduction: Q = G·ΔT.

  • thermal
  • Modelica Standard Library
  • 2 ports
  • 1 parameter
BlockThermal conductorKindthermalConductorLibraryThermal

Description

Linear heat conduction between two ports with no heat storage: the heat flow is proportional to the temperature difference.

Example

Cooling a hot part A 10 W part cools through a thermal resistance into a heatsink that convects and radiates to 20 °C.

10 WPart 100 J/KHeat flowR 2 K/WHeatsink 500 J/Kh·A 5 W/KConvectionAir 20 °CRadiationRoom 20 °CPart temperatureAcross R350 KHeld at 350 KG 1 W/KBody 100 J/K20 W at 1500 sHeaterBody temperature
In Gradara, select a Thermal conductor block and press F1, then choose Open example. It opens as a new model in My models, ready to run.

Also in this example:ConstantStepHeat capacitorThermal resistorConvectionRadiationFixed temperatureTemperature sourceFixed heat flowHeat flow sourceTemperature sensorHeat flow sensorTemperature difference sensor

Ports

Conserving terminals 2

  • aport_athermal

    Heat port a. Heat flow Q from a to b enters here.

  • bport_bthermal

    Heat port b. Q leaves here.

Parameters

  • ConductanceG1 W/K

    ≥ 0

    Thermal conductance, in W/K. For a slab, G = k · A / L.

Equations

Modelica
ΔT = port_a.T − port_b.T
Q = G · ΔT
port_a.Q_flow = Q, port_b.Q_flow = −Q

Implementation

Modelica Standard Library 4.1.0Modelica.Thermal.HeatTransfer.Components.ThermalConductor
MSL documentation

Assumptions and limitations

  • Constant conductance, independent of temperature. No heat storage.

Used in

These larger examples use it too. Open them from Examples in the app.

  • Data center cooling control

See also

Select a block in Gradara and press F1 to open its page offline.