What is a theretical way to write i++ in Dymola? And have hybrid continous and discrete model?

Question

I am just started to learn modelica and I have one (newbie) question. The problem for me is to change the way of thinking from convential programming thinking to modelica way of thinking.

I want to do simple program. I have input array with PV output values in 5 minutes resolution. I have input array with heat load values in 60 minutes resolution. I have a energy storage that stores excess energy or takes energy fro meeting the heat demand in real time.

I wrote this in openmodelica:

`class Add
 Real PV[:] = 100:10:1000;
 Real Heat[:] = 200:300:6000;
 Real Storage;
 Real p;
 Integer j;
 Integer i;
 Boolean power,heat;
 equation
 power=sample(0,5);
 heat=sample(0,60);
 when power then
 j=j+1;
end when;

when heat then
i=i+1;

end when;

Storage= PV[j] * 2.375-Heat[i];

p=Storage+ pre(p);

end Add;`

But when I c/p to dymola it gets an error on this " p=Storage+ pre(p); " part because it says pre() cannot be used for continuous model. When I delete pre() then it says it cannot devide by 0.

Can you explain me what I am doing wrong?

Thanks!

Answer 1

I hope I understand your problem correctly. And I used Dymola to solve a simple example - I hope this works in OpenModelica, too.

If you are trying to use a time series of input data I would suggest using the model Modelica.Blocks.Sources.TimeTable. In your case the table's first column would denote hourly timesteps, i.e. 0, 3600, 7200, ...; the second column could give values for the heat demand in kW, if it is constant at 300 kW like in your example this could mean 300, 300, 300, ...;

You can reference the output of the TimeTable model in equations using its RealOutput as TimeTable.y.

A very simple example for your test case could thus look like this:

model heatStorage

  Modelica.SIunits.Conversions.NonSIunits.Energy_kWh storage "Energy content of storage in kWh";

  Modelica.Blocks.Sources.TimeTable solarThermal(table=[0,50; 3600,70; 7200,40; 10800,73]);
  Modelica.Blocks.Sources.TimeTable heatDemand(table=[0,300; 3600,300; 7200,300; 10800,
    300]);

equation
  der(storage) = (solarThermal.y - heatDemand.y)/3600;

end heatStorage;

I assumed time-varying output of a solar thermal collector. If you use PV to heat water you could include another variable and conversion equation. For the variable storage I used the definition of energy in kWh, therefore I divide the given equation by 3600. As Modelica is equation-based, writing der(storage) is the same as having the right side of the equation integrated. Thus, the calculated value for storage is the integral of the difference between input and output.

I hope this helps.