.Modelica.Thermal.FluidHeatFlow.Examples.IndirectCooling

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	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

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