Package Modelica.​Thermal.​FluidHeatFlow.​Examples
Examples that demonstrate the usage of the FluidHeatFlow components

Information

This package contains test examples:

  1. SimpleCooling: Heat is dissipated through a media flow
  2. ParallelCooling: Two heat sources dissipate through merged media flows
  3. IndirectCooling: Heat is dissipated through two cooling cycles
  4. PumpAndValve: Demonstrate the usage of an IdealPump and a Valve
  5. PumpDropOut: Demonstrate shutdown and restart of a pump
  6. ParallelPumpDropOut: Demonstrate the shutdown and restart of a pump in a parallel circuit
  7. OneMass: Cooling of a mass (thermal capacity) by a coolant flow
  8. TwoMass: Cooling of two masses (thermal capacities) by two parallel coolant flows
  9. WaterPump: Water pumping station
  10. TestOpenTank: Test the OpenTank model
  11. TwoTanks: Two connected open tanks
  12. TestCylinder: Test the Cylinder model

Extends from Modelica.​Icons.​ExamplesPackage (Icon for packages containing runnable examples).

Package Contents

NameDescription
IndirectCoolingIndirect cooling circuit
OneMassCooling of one hot mass
ParallelCoolingCooling circuit with parallel branches
ParallelPumpDropOutCooling circuit with parallel branches and drop out of pump
PumpAndValveCooling circuit with pump and valve
PumpDropOutCooling circuit with drop out of pump
SimpleCoolingSimple cooling circuit
TestCylinderTwo cylinder system
TestOpenTankTest the OpenTank model
TwoMassCooling of two hot masses
TwoTanksTwo connected open tanks
UtilitiesUtility models for examples
WaterPumpWater pumping station

Model Modelica.​Thermal.​FluidHeatFlow.​Examples.​SimpleCooling
Simple cooling circuit

Information

1st test example: SimpleCooling

A prescribed heat source dissipates its heat through a thermal conductor to a coolant flow. The coolant flow is taken from an ambient and driven by a pump with prescribed mass flow.
Results:
output explanation formula actual steady-state value
dTSource Source over Ambient dtCoolant + dtToPipe 20 K
dTtoPipe Source over Coolant Losses / ThermalConductor.G 10 K
dTCoolant Coolant's temperature increase Losses * cp * massFlow 10 K

Extends from Modelica.​Icons.​Example (Icon for runnable examples).

Parameters

TypeNameDefaultDescription
MediummediumModelica.Thermal.FluidHeatFlow.Media.Medium()Cooling medium
TemperatureTAmb293.15Ambient temperature

Model Modelica.​Thermal.​FluidHeatFlow.​Examples.​ParallelCooling
Cooling circuit with parallel branches

Information

2nd test example: ParallelCooling

Two prescribed heat sources dissipate their heat through thermal conductors to coolant flows. The coolant flow is taken from an ambient and driven by a pump with prescribed mass flow, then split into two coolant flows connected to the two heat sources, and afterwards merged. Splitting of coolant flows is determined by pressure drop characteristic of the two pipes.
Results:
output explanation formula actual steady-state value
dTSource1 Source1 over Ambient dTCoolant1 + dTtoPipe1 15 K
dTtoPipe1 Source1 over Coolant1 Losses1 / ThermalConductor1.G 5 K
dTCoolant1 Coolant's temperature increase Losses * cp * totalMassFlow/2 10 K
dTSource2 Source2 over Ambient dTCoolant2 + dTtoPipe2 30 K
dTtoPipe2 Source2 over Coolant2 Losses2 / ThermalConductor2.G 10 K
dTCoolant2 Coolant's temperature increase Losses * cp * totalMassFlow/2 20 K
dTmixedCoolant mixed Coolant's temperature increase (dTCoolant1+dTCoolant2)/2 15 K

Extends from Modelica.​Icons.​Example (Icon for runnable examples).

Parameters

TypeNameDefaultDescription
MediummediumModelica.Thermal.FluidHeatFlow.Media.Medium()Cooling medium
TemperatureTAmb293.15Ambient temperature

Model 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 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

Extends from Modelica.​Icons.​Example (Icon for runnable examples).

Parameters

TypeNameDefaultDescription
MediumouterMediumModelica.Thermal.FluidHeatFlow.Media.Medium()Outer medium
MediuminnerMediumModelica.Thermal.FluidHeatFlow.Media.Medium()Inner medium
TemperatureTAmb293.15Ambient temperature

Model Modelica.​Thermal.​FluidHeatFlow.​Examples.​PumpAndValve
Cooling circuit with pump and valve

Information

4th test example: PumpAndValve

The pump is running with half speed for 0.4 s, afterwards with full speed (using a ramp of 0.1 s).
The valve is half open for 0.9 s, afterwards full open (using a ramp of 0.1 s).
You may try to:

Extends from Modelica.​Icons.​Example (Icon for runnable examples).

Parameters

TypeNameDefaultDescription
MediummediumModelica.Thermal.FluidHeatFlow.Media.Medium()Cooling medium
TemperatureTAmb293.15Ambient temperature

Model Modelica.​Thermal.​FluidHeatFlow.​Examples.​PumpDropOut
Cooling circuit with drop out of pump

Information

5th test example: PumpDropOut

Same as 1st test example, but with a drop out of the pump:
The pump is running for 0.2 s, then shut down (using a ramp of 0.2 s) for 0.2 s, then started again (using a ramp of 0.2 s).

Extends from Modelica.​Icons.​Example (Icon for runnable examples).

Parameters

TypeNameDefaultDescription
MediummediumModelica.Thermal.FluidHeatFlow.Media.Medium()Cooling medium
TemperatureTAmb293.15Ambient temperature

Model Modelica.​Thermal.​FluidHeatFlow.​Examples.​ParallelPumpDropOut
Cooling circuit with parallel branches and drop out of pump

Information

6th test example: ParallelPumpDropOut

Same as 2nd test example, but with a drop out of the pump:
The pump is running for 0.2 s, then shut down (using a ramp of 0.2 s) for 0.2 s, then started again (using a ramp of 0.2 s).

Extends from Modelica.​Icons.​Example (Icon for runnable examples).

Parameters

TypeNameDefaultDescription
MediummediumModelica.Thermal.FluidHeatFlow.Media.Medium()Cooling medium
TemperatureTAmb293.15Ambient temperature

Model Modelica.​Thermal.​FluidHeatFlow.​Examples.​OneMass
Cooling of one hot mass

Information

7th test example: OneMass

A thermal capacity is coupled with a coolant flow. Different initial temperatures of thermal capacity and pipe's coolant get ambient's temperature, the time behaviour depending on coolant flow.

Extends from Modelica.​Icons.​Example (Icon for runnable examples).

Parameters

TypeNameDefaultDescription
MediummediumModelica.Thermal.FluidHeatFlow.Media.Medium()Cooling medium
TemperatureTAmb293.15Ambient temperature
TemperatureTMass313.15Initial temperature of mass

Model Modelica.​Thermal.​FluidHeatFlow.​Examples.​TwoMass
Cooling of two hot masses

Information

8th test example: TwoMass

Two thermal capacities are coupled with two parallel coolant flow. Different initial temperatures of thermal capacities and pipe's coolants get ambient's temperature, the time behaviour depending on coolant flow.

Extends from Modelica.​Icons.​Example (Icon for runnable examples).

Parameters

TypeNameDefaultDescription
MediummediumModelica.Thermal.FluidHeatFlow.Media.Medium()Cooling medium
TemperatureTAmb293.15Ambient temperature
TemperatureTMass1313.15Initial temperature of mass1
TemperatureTMass2333.15Initial temperature of mass2

Model Modelica.​Thermal.​FluidHeatFlow.​Examples.​WaterPump
Water pumping station

Information

There are two reservoirs at ambient pressure, the second one 25 m higher than the first one. The ideal pump is driven by a speed source, starting from zero and going up to 1.2 times nominal speed. To avoid water flowing back, the one way valve is used.

Extends from Modelica.​Icons.​Example (Icon for runnable examples).


Model Modelica.​Thermal.​FluidHeatFlow.​Examples.​TestOpenTank
Test the OpenTank model

Information

First, the medium is pumped out of the tank (initial level = 0.5 m, T = 40°C) to an (infinite) ambient (T = 20°C):

Subsequently the medium is pumped into the tank from an (infinite) ambient:

Extends from Modelica.​Icons.​Example (Icon for runnable examples).


Model Modelica.​Thermal.​FluidHeatFlow.​Examples.​TwoTanks
Two connected open tanks

Information

Two tanks are connected with a pipe:

Within 1.5 s (dependent on the flow resistance of the pipe) the level = 0.5 m in both tanks is the same, medium flows from tank 1 to tank 2. The temperature of tank 1 remains unchanged, the temperature of tank 2 is increased.

Extends from Modelica.​Icons.​Example (Icon for runnable examples).


Model Modelica.​Thermal.​FluidHeatFlow.​Examples.​TestCylinder
Two cylinder system

Information

Test of a system with 2 cylinders (with same volume):

A force is applied that presses from 0.25 s to 0.50 s with 1 Nm on piston1. Due to the ratio of areas 10:1

At piston2 a mass is mounted which is moved and presses the springDamper. When the force at piston1 is removed, the springDamper pushes back the mass and a damped oscillation occurs.

Note: Take care of the initial conditions. The unstretched spring length is cylinder2.L/2, i.e. when piston2 is the middle of its cylinder the spring applies no force to the mass (and piston2).

Extends from Modelica.​Icons.​Example (Icon for runnable examples).


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