Question

我正在学习Modelica，一切正常。直到我想用一个电阻器向我的同事证明其原因。我们问自己一个问题：当电阻的热功率为1W时，1 Ohm电阻的压降和电流是多少（显然答案应该是1V，1A）。我没有得到0 V，0A以外的任何其他结果。从物理上讲，我对结果感到满意，因为我不希望一旦加热电阻器就可以将其转换为电源，但是我不理解这种因果关系会在电阻器模型中内置。我通过电阻-ConditionalHeatPort-HeatPort_a-HeatPort追溯了Modelica库，但据我了解，Modelica仅存在非因果方程式。有人可以照亮它吗？

谢谢！

编辑：对Rene Just Nielsen的回答：

我正在使用下面的代码。想法是，考虑到电阻流出的热量固定为1W的事实，需要在电阻上建立电流和电压，以便求解所有方程式。如果我对此进行仿真，则组件fixedHeatFlow1上的热流= 0 W，电流和电压也均为0 V和0A。当然，这彼此一致，但与-1 W的固定边界条件不一致在fixedHeatFlow1。

model ElectricalPowerFromHeat
  Modelica.Electrical.Analog.Basic.Resistor resistor1(R = 1, alpha = 0, useHeatPort = true) annotation(
    Placement(visible = true, transformation(origin = {-28, -46}, extent = {{-10, -10}, {10, 10}}, rotation = 0)));
  Modelica.Electrical.Analog.Basic.Ground ground1 annotation(
    Placement(visible = true, transformation(origin = {12, -80}, extent = {{-10, -10}, {10, 10}}, rotation = 0)));
  Modelica.Thermal.HeatTransfer.Sources.FixedHeatFlow fixedHeatFlow1(Q_flow = -1, alpha = 1)  annotation(
    Placement(visible = true, transformation(origin = {-68, 14}, extent = {{-10, -10}, {10, 10}}, rotation = 0)));
equation
  connect(resistor1.n, ground1.p) annotation(
    Line(points = {{-18, -46}, {12, -46}, {12, -70}, {12, -70}}, color = {0, 0, 255}));
  connect(fixedHeatFlow1.port, resistor1.heatPort) annotation(
    Line(points = {{-58, 14}, {-28, 14}, {-28, -56}, {-28, -56}}, color = {191, 0, 0}));
  connect(resistor1.p, ground1.p) annotation(
    Line(points = {{-38, -46}, {-38, -60}, {12, -60}, {12, -70}}, color = {0, 0, 255}));
  annotation(
    uses(Modelica(version = "3.2.2")),
    experiment(StartTime = 0, StopTime = 1, Tolerance = 1e-06, Interval = 0.002));
end ElectricalPowerFromHeat;

根据电阻器内部的方程，我希望电阻器可以用作电源：

  R_actual = R*(1 + alpha*(T_heatPort - T_ref));
  v = R_actual*i;
  LossPower = v*i;

Answer 1

如果您想找到一个1欧姆电阻的压降，假设热流为1W，则可以按以下方式建模。首先，您使用一个模型来生成简单电路的热流（在VoltageToHeatFlow中），然后将信号求逆（在Test-model中）：

package ShowInvertPower
  model VoltageToHeatFlow
    Modelica.Electrical.Analog.Basic.Ground ground
      annotation (Placement(transformation(extent={{-22,-16},{-2,4}})));
    Modelica.Electrical.Analog.Basic.HeatingResistor resistor(R_ref=1)
      annotation (Placement(transformation(extent={{-8,48},{12,68}})));
    Modelica.Electrical.Analog.Sources.SignalVoltage signalVoltage
      annotation (Placement(transformation(extent={{-58,52},{-38,72}})));
    Modelica.Thermal.HeatTransfer.Sensors.HeatFlowSensor heatFlowSensor
      annotation (Placement(transformation(extent={{52,22},{72,42}})));
    Modelica.Thermal.HeatTransfer.Components.HeatCapacitor heatCapacitor(C=1)
      annotation (Placement(transformation(extent={{84,36},{104,56}})));
    Modelica.Blocks.Interfaces.RealOutput Q_flow1
      "Heat flow from port_a to port_b as output signal"
      annotation (Placement(transformation(extent={{96,-18},{116,2}})));
    Modelica.Blocks.Interfaces.RealInput v1
      "Voltage between pin p and n (= p.v - n.v) as input signal"
      annotation (Placement(transformation(extent={{-126,-32},{-86,8}})));
  equation 
    connect(signalVoltage.n, resistor.p) annotation (Line(points={{-38,62},{-26,
            62},{-26,58},{-8,58}}, color={0,0,255}));
    connect(resistor.n, ground.p) annotation (Line(points={{12,58},{30,58},{30,4},
            {-12,4}}, color={0,0,255}));
    connect(signalVoltage.p, ground.p) annotation (Line(points={{-58,62},{-68,62},
            {-68,4},{-12,4}}, color={0,0,255}));
    connect(resistor.heatPort, heatFlowSensor.port_a) annotation (Line(points={{
            2,48},{28,48},{28,32},{52,32}}, color={191,0,0}));
    connect(heatFlowSensor.port_b, heatCapacitor.port) annotation (Line(points={
            {72,32},{84,32},{84,36},{94,36}}, color={191,0,0}));
    connect(heatFlowSensor.Q_flow, Q_flow1) annotation (Line(points={{62,22},{66,
            22},{66,-8},{106,-8}}, color={0,0,127}));
    connect(signalVoltage.v, v1) annotation (Line(points={{-48,74},{-106,74},{-106,
            -12}}, color={0,0,127}));
  end VoltageToHeatFlow;

  model Test
    ShowInvertPower.VoltageToHeatFlow voltageToHeatFlow annotation (Placement(
          transformation(
          extent={{-10,-10},{10,10}},
          rotation=180,
          origin={-2,58})));
    Modelica.Blocks.Math.InverseBlockConstraints inverseBlockConstraints
      annotation (Placement(transformation(extent={{-24,46},{16,70}})));
    Modelica.Blocks.Sources.Constant const(k=2)
      annotation (Placement(transformation(extent={{-86,46},{-66,66}})));
  equation 
    connect(voltageToHeatFlow.v1, inverseBlockConstraints.y2) annotation (Line(
          points={{8.6,59.2},{13.1,59.2},{13.1,58},{13,58}}, color={0,0,127}));
    connect(inverseBlockConstraints.u2, voltageToHeatFlow.Q_flow1) annotation (
        Line(points={{-20,58},{-12,58},{-12,58.8},{-12.6,58.8}}, color={0,0,127}));
    connect(const.y, inverseBlockConstraints.u1) annotation (Line(points={{-65,56},
            {-46,56},{-46,58},{-26,58}}, color={0,0,127}));
    annotation (Icon(coordinateSystem(preserveAspectRatio=false)), Diagram(
          coordinateSystem(preserveAspectRatio=false)));
  end Test;
  annotation (uses(Modelica(version="3.2.3")));
end ShowInvertPower;

结果是需要1 V（和1 A）。显然，可以用更简单的方法对其进行建模，但是以这种方式使用逆模型是Modelica中的标准方法。

在modelica中，是否将Electrical.Analog.Basic.Resistor的Heatport定义为仅输出？如果是，怎么办？

1 个答案: