.Modelica.Thermal.FluidHeatFlow.Examples.IndirectCooling

Indirect cooling circuit

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

Generated at 2025-06-12T18:20:45Z by OpenModelicaOpenModelica 1.25.0 using GenerateDoc.mos