.Buildings.Electrical.AC.OnePhase.Loads.Resistive .Buildings.Electrical.AC.OnePhase.Loads.ResistiveModel of a resistive load. It may be used to model a load that has a power factor of one.
The model computes the complex power vector as
S = P + jQ = V ⋅ i*
where V is the voltage phasor and i* is the complex conjugate of the current phasor.
The load model takes as input the power consumed by the inductive load and
the power factor pf=cos(φ). The power
can be either fixed using the parameter P_nominal, or
it is possible to specify a variable power using the inputs y or
Pow. The different modes can be selected with the parameter
mode, see
Buildings.Electrical.Interfaces.Load for more information.
The equations of the model can be rewritten as
i1 = (P V1 + Q V2)/(V12 + V22),
i2 = (P V2 - Q V1)/(V12 + V22),
where i1, i2, V1, and V2 are the real and imaginary parts of the current and voltage phasors.
Since the model represents a load with a power factor of one, the complex power is Q = 0. This leads to the following equations where there are nonlinear equations that relate the current to the voltage
i1 = P V1/(V12 + V22)
i2 = P V2/(V12 + V22)
The non-linearity is due to the fact that the load consumes the power specified by the variable P, irrespectively of the voltage of the load. The figure below shows the relationship between the real part of the current phasor and the real and imaginary voltages of the load.
When multiple loads are connected in a grid through cables that cause voltage drops, the dimension of the system of nonlinear equations increases linearly with the number of loads. This nonlinear system of equations introduces challenges during the initialization, as Newton solvers may diverge if initialized far from a solution, as well during the simulation. In this situation, the model can be parameterized to use a linear approximation as discussed in the next section.
Given the constraints and the two-dimensional nature of the problem, it is difficult to find a linearized version of the AC load model. A solution could be to divide the voltage domain into sectors, and for each sector compute the best linear approximation. However, the selection of the proper approximation depending on the value of the voltage can generate events that increase the simulation time. For these reasons, the linearized model assumes a voltage that is equal to the nominal value
i1 = P V1/VRMS2
i2 = P V2/VRMS2
where VRMS is the Root Mean Square voltage of the AC system. Even though this linearized version of the load model introduces an approximation error in the current, it satisfies the constraints related to the ratio of the active and reactive powers.
The image below show the linearized function
The initialization problem can be simplified using the homotopy operator. The homotopy operator uses two different types of equations to compute the value of a variable: the actual one and a simplified one. The actual equation is the one used during the normal operation. During initialization, the simplified equation is first solved and then slowly replaced with the actual equation to compute the initial values for the nonlinear system of equations. The load model uses the homotopy operator, with the linearized model being used as the simplified equation. This numerical expedient has proven useful when simulating models with more than ten connected loads.
The load model has a parameter initMode that can be used to select
the assumption to use during the initialization phase by the homotopy operator.
The choices are between a null current or the linearized model.
replaceable for terminal.
initMode that can be used to
select the assumption to be used during initialization phase
by the homotopy operator.