This information is part of the Modelica Standard Library maintained by the Modelica Association.
Example: indirect cooling circuit
3rd test example: IndirectCoolingA prescribed heat sources dissipates its heat through a thermal conductor to the inner coolant cycle. It is necessary to define the pressure level of the inner coolant cycle. The inner coolant cycle is coupled to the outer coolant flow through a thermal conductor.
Inner coolant's temperature rise near the source is the same as temperature drop near the cooler.
|output||explanation||formula||actual steady-state value|
|dTSource||Source over Ambient||dtouterCoolant + dtCooler + dTinnerCoolant + dtToPipe||40 K|
|dTtoPipe||Source over inner Coolant||Losses / ThermalConductor.G||10 K|
|dTinnerColant||inner Coolant's temperature increase||Losses * cp * innerMassFlow||10 K|
|dTCooler||Cooler's temperature rise between inner and outer pipes||Losses * (innerGc + outerGc)||10 K|
|dTouterColant||outer Coolant's temperature increase||Losses * cp * outerMassFlow||10 K|