Modelica.Thermal.HeatTransfer.Examples

Information

Package Content

Name	Description
TwoMasses	Simple conduction demo
ControlledTemperature	Control temperature of a resistor
Motor	Second order thermal model of a motor

Modelica.Thermal.HeatTransfer.Examples.TwoMasses

Information

This example demonstrates the thermal response of two masses connected by a conducting element. The two masses have the same heat capacity but different initial temperatures (T1=100 [degC], T2= 0 [degC]). The mass with the higher temperature will cool off while the mass with the lower temperature heats up. They will each asymptotically approach the calculated temperature T_final_K (T_final_degC) that results from dividing the total initial energy in the system by the sum of the heat capacities of each element.

Simulate for 5 s and plot the variables
mass1.T, mass2.T, T_final_K or
Tsensor1.T, Tsensor2.T, T_final_degC

Parameters

Name	Description
T_final_K	Projected final temperature [K]

Modelica.Thermal.HeatTransfer.Examples.ControlledTemperature

Information

A constant voltage of 10 V is applied to a temperature dependent resistor of 10*(1+(T-20C)/(235+20C)) Ohms, whose losses v**2/r are dissipated via a thermal conductance of 0.1 W/K to ambient temperature 20 degree C. The resistor is assumed to have a thermal capacity of 1 J/K, having ambient temperature at the beginning of the experiment. The temperature of this heating resistor is held by an OnOff-controller at reference temperature within a given bandwidth +/- 1 K by switching on and off the voltage source. The reference temperature starts at 25 degree C and rises between t = 2 and 8 seconds linear to 50 degree C. An appropriate simulating time would be 10 seconds.

Parameters

Name	Description
TAmb	Ambient Temperature [K]
TDif	Error in Temperature [K]

Modelica.Thermal.HeatTransfer.Examples.Motor

Information

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

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 convective 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.

Parameters

Name	Description
TAmb	Ambient temperature [K]