BondLib.Thermal.HeatTransfer.Examples.Motor

# .BondLib.Thermal.HeatTransfer.Examples.Motor

Second order thermal model of a motor

## Information

This example contains a simple second order thermal model of a motor. The periodic power losses are described by table "lossTable":

time | winding losses | core losses
0 | 100 | 500
360 | 100 | 500
360 | 1000 | 500
600 | 1000 | 500

Since constant speed is assumed, the core losses keep constant whereas the winding losses are low for 6 minutes (no-load) and high for 4 minutes (over load).
The winding losses are corrected by (1 + alpha*(T - T_ref)) because the winding's resistance is temperature dependent whereas the core losses are kept constant (alpha = 0).

The power dissipation to the environment is approximated by heat flow through a thermal conductance between winding and core, partially storage of the heat in the winding's heat capacity as well as the core's heat capacity and finally by forced convection to the environment.
Since constant speed is assumed, the cinvective conductance keeps constant.
Using Modelica.Thermal.FluidHeatFlow it would be possible to model the coolant air flow, too (instead of simple dissipation to a constant ambient's temperature).

Simulate for 7200 s; plot Twinding.T and Tcore.T.

This example was copied from an equivalent example of the corresponding sub-library of the Modelica standard library. However, the simulation results obtained are slightly different.

The reason is that the conductance and convection models used by the standard library operate on real resistors. Yet, this is not a realistic assumption. Resistors heat up. They generate additional entropy. This entropy needs to be routed back into the thermal circuit.

In an electrical circuit or in a mechanical system, it may make sense to ignore the heat generated by the resistors. However here, we operate already in the thermal domain. Hence ignoring the heat generated by the resistors makes no sense whatsoever.

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