Modelica.Thermal.FluidHeatFlow.Examples

Examples that demonstrate the usage of the FluidHeatFlow components

Information

This package contains test examples:

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

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

Package Content

Name	Description
SimpleCooling	Simple cooling circuit
ParallelCooling	Cooling circuit with parallel branches
IndirectCooling	Indirect cooling circuit
PumpAndValve	Cooling circuit with pump and valve
PumpDropOut	Cooling circuit with drop out of pump
ParallelPumpDropOut	Cooling circuit with parallel branches and drop out of pump
OneMass	Cooling of one hot mass
TwoMass	Cooling of two hot masses
WaterPump	Water pumping station
TestOpenTank	Test the OpenTank model
TwoTanks	Two connected open tanks
TestCylinder	Two cylinder system
Utilities	Utility models for examples

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

Name	Description
medium	Cooling medium
TAmb	Ambient temperature [K]

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

Name	Description
medium	Cooling medium
TAmb	Ambient temperature [K]

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

Name	Description
outerMedium	Outer medium
innerMedium	Inner medium
TAmb	Ambient temperature [K]

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:

drive the pump with variable speed and let the valve full open to regulate the volume flow rate of coolant
drive the pump with constant speed and throttle the valve to regulate the volume flow rate of coolant

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

Parameters

Name	Description
medium	Cooling medium
TAmb	Ambient temperature [K]

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

Name	Description
medium	Cooling medium
TAmb	Ambient temperature [K]

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

Name	Description
medium	Cooling medium
TAmb	Ambient temperature [K]

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

Name	Description
medium	Cooling medium
TAmb	Ambient temperature [K]
TMass	Initial temperature of mass [K]

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

Name	Description
medium	Cooling medium
TAmb	Ambient temperature [K]
TMass1	Initial temperature of mass1 [K]
TMass2	Initial temperature of mass2 [K]

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).

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):

The level of the medium in the tank decreases.
The temperature of the medium in the tank remains unchanged.

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

The level of the medium in the tank increases again.
The temperature of the medium in the tank decreases (mixing temperature).

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

Modelica.Thermal.FluidHeatFlow.Examples.TwoTanks

Two connected open tanks

Information

Two tanks are connected with a pipe:

Tank 1: initial level = 0.9 m, T = 40°C
Tank 2: initial level = 0.1 m, T = 20°C

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).

Modelica.Thermal.FluidHeatFlow.Examples.TestCylinder

Two cylinder system

Information

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

cylinder1: A = 0.1 m2, L=10. m, initial position of piston at s=L/2
cylinder2: A = 1.0 m2, L=1.0 m, initial position of piston at s=L/2

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

the force at piston2 is ten times the force at piston1
movement of piston1 is ten times the movement of piston2

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).

Automatically generated Thu Dec 19 17:20:24 2019.