modelIndirectCooling
Indirect cooling circuit
Extends from Modelica.Icons.Example (Icon for runnable examples).
Information
3rd test example: IndirectCooling
A 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.
Results:
| output | explanation | formula | actual steady-state value |
| dTSource | Temperature difference between heat source and ambient condition | dtouterCoolant + dtCooler + dTinnerCoolant + dtToPipe | 40 K |
| dTtoPipe | Temperature difference between heat source and coolant in pipe 1 | Losses / ThermalConductor.G | 10 K |
| dTinnerCoolant | Inner Coolant's temperature increase | Losses * cp * innerMassFlow | 10 K |
| dTCooler | Coolant temperature difference between inner pipe and outer pipe | Losses * (innerGc + outerGc) | 10 K |
| dTouterCoolant | Outer coolant's temperature increase | Losses * cp * outerMassFlow | 10 K |
Parameters
| Type | Name | Default | Description |
|---|---|---|---|
| FluidHeatFlow.Media.Medium | outerMedium | FluidHeatFlow.Media.Medium() | Outer medium |
| FluidHeatFlow.Media.Medium | innerMedium | FluidHeatFlow.Media.Medium() | Inner medium |
| SI.Temperature | TAmb | 293.15 | Ambient temperature |