en:cs:modelling_multi-phased_electrical_system_interconnexion

Modelling multi-phased AC electrical system interconnexion

We suppose that all the multi-phased electrical systems are properly balanced: $ \forall n \in \{1,2\}; Iph_{n} = Iph_{n+1} $.

Let's call $ A = (p_{a}, V_{a}, I_{a}, Iph_{a}) $ and $ B = (p_{b}, V_{b}, I_{b}, Iph_{b}) $ two multi-phased electrical systems where $ p_{a},p_{b} \in \mathbb{N} $ are respectively the number of phases, $ V_{a}, V_{b} \in \mathbb{N} $ are the voltage per phase, $ I_{a}, I_{b} \in \mathbb{N} $ are the total intensity and $ Iph_{a}, Iph_{b} \in \mathbb{N} $ the intensity per phase.
We always have: $ I_{a} = p_{a} \times Iph_{a} $ and $ I_{b} = p_{b} \times Iph_{b} $. So the intensities properties can be deduced one from the other.

  • $ A $ and $ B $ are interconnected serially : $ A \triangleleft \triangleright B $.

How to evaluate the properties of the resulting system $ A \triangleleft \triangleright B $?
$ A \triangleleft \triangleright B = (p = min(p_{a},p_{b}), V_{A \triangleleft \triangleright B} = min(V_{a}, V_{b}), I_{A \triangleleft \triangleright B} = min(I_{a}, I_{b}), Iph_{A \triangleleft \triangleright B} = min(Iph_{a},Iph_{b})) $

Let's say you only have $ I_{a}, I_{b} $ in $ A $ and $ B $. We can deduce $ Iph_{a}, Iph_{b} $ from them.
How can we deduce $ Iph_{A \triangleleft \triangleright B} $?
$ Iph_{A \triangleleft \triangleright B} = min(I_{a}, I_{b}) \div min(p_{a},p_{b}) $
Division evaluation:

  • $ I_{a} < I_{b} \land p_{a} < p_{b} \Rightarrow I_{a} \ mod \ p_{a} = 0 \land Iph_{A \triangleleft \triangleright B} \in \mathbb{N} $ ;
  • $ I_{a} > I_{b} \land p_{a} < p_{b} \land p_{a}, p_{b} \in \{2,3\} \land I_{b} \ mod \ p_{a} = 0 \Rightarrow Iph_{A \triangleleft \triangleright B} \in \mathbb{N} $ ;
  • $ I_{a} > I_{b} \land p_{a} > p_{b} \Rightarrow I_{b} \ mod \ p_{b} = 0 \land Iph_{A \triangleleft \triangleright B} \in \mathbb{N} $;
  • $ I_{a} > I_{b} \land p_{a} < p_{b} \land p_{a}, p_{b} \in \{2,3\} \land I_{b} \ mod \ p_{a} = 0 \Rightarrow Iph_{A \triangleleft \triangleright B} \in \mathbb{N} $ ;
  • $ I_{a} > I_{b} \land p_{a} < p_{b} \land p_{a}, p_{b} \in \{2,3\} \land I_{b} \ mod \ p_{a} \neq 0 \Rightarrow Iph_{A \triangleleft \triangleright B} \notin \mathbb{N} $ ;
  • $ I_{a} < I_{b} \land p_{a} > p_{b} \land p_{a}, p_{b} \in \{2,3\} \land I_{a} \ mod \ p_{b} \ne 0 \Rightarrow Iph_{A \triangleleft \triangleright B} \notin \mathbb{N} $ ;

We can expose at leat two cases where $ Iph_{A \triangleleft \triangleright B} \notin \mathbb{N} $ if $ p_{a}, p_{b} \in \{2,3\} $

Let's say you only have $ Iph_{a}, Iph_{b} $ in $ A $ and $ B $. We can deduce $ I_{a}, I_{b} $ from them.
How can we deduce $ I_{A \triangleleft \triangleright B} $?
$ I_{A \triangleleft \triangleright B} = min(p_{a},p_{b}) \times min(I_{a}, I_{b}) \in \mathbb{N} $

So if you model an AC electrical system with the properties $ (p, V, Iph) $, you can deduce the properties of any interconnexion of them within the natural integer set.

$ I $ is the line current.
$ V $ is the line-to-neutral voltage.
$ U $ is the line-to-line voltage: $ U = \sqrt{3} \times V $.

$$ I = p \times Iph \\ Pph = Vph \times Iph \times \cos(\varphi) \\ P = \sum_{n=1}^{p} Pph_{n} = \sum_{n=1}^{p} Vph_{n} \times Iph_{n} \times \cos(\varphi) = p \times Vph \times Iph \times \cos(\varphi) = Vph \times I \times \cos(\varphi) \\ Qph = Vph \times Iph \times \sin(\varphi) \\ Q = \sum_{n=1}^{p} Qph_{n} = \sum_{n=1}^{p} Vph_{n} \times Iph_{n} \times \sin(\varphi) = p \times Vph \times Iph \times \sin(\varphi) = Vph \times I \times \sin(\varphi) \\ Sph = Vph \times Iph \\ S = \sum_{n=1}^{p} Sph_{n} = \sum_{n=1}^{p} Vph_{n} \times Iph_{n} = p \times Vph \times Iph = Vph \times I $$

$ J $ is the phase current.
$ I $ is the line current: $ I = \sqrt{3} \times J $.
$ U $ is the line-to-line voltage.

$$ J = p \times Jph \\ Pph = Uph \times Jph \times \cos(\varphi) \\ P = \sum_{n=1}^{p} Pph_{n} = \sum_{n=1}^{p} Uph_{n} \times Jph_{n} \times \cos(\varphi) = p \times Uph \times Jph \times \cos(\varphi) = Uph \times J \times \cos(\varphi) \\ Qph = Uph \times Jph \times \sin(\varphi) \\ Q = \sum_{n=1}^{p} Qph_{n} = \sum_{n=1}^{p} Uph_{n} \times Jph_{n} \times \sin(\varphi) = p \times Uph \times Jph \times \sin(\varphi) = Uph \times J \times \sin(\varphi) \\ Sph = Uph \times Jph \\ S = \sum_{n=1}^{p} Sph_{n} = \sum_{n=1}^{p} Uph_{n} \times Jph_{n} = p \times Uph \times Jph = Uph \times J $$

Low level class of electrical helper methods implementation:

export class ElectricUtils {
  public static calculateAmpTotal(nbOfPhases: number, Iph: number): number {
    return nbOfPhases * Iph;
  }
 
  public static calculatePowerPerPhase(V: number, Iph: number, cosPhi = 1): number {
    const powerPerPhase = V * Iph * cosPhi;
    if (cosPhi === 1) {
      return powerPerPhase;
    }
    return Math.round(powerPerPhase + 1);
  }
 
  public static calculatePowerTotal(nbOfPhases: number, V: number, Iph: number, cosPhi = 1): number {
    return nbOfPhases * ElectricUtils.calculatePowerPerPhase(V, Iph, cosPhi);
  }
 
  public static calculateAmpPerPhaseFromPower(nbOfPhases: number, P: number, V: number, cosPhi = 1): number {
    const power = ElectricUtils.calculateAmpTotalFromPower(P, V, cosPhi);
    const powerPerPhase = power / nbOfPhases;
    if (power % nbOfPhases === 0) {
      return powerPerPhase;
    }
    return Math.round(powerPerPhase + 1);
  }
 
  public static calculateAmpTotalFromPower(P: number, V: number, cosPhi = 1): number {
    const power = P / (V * cosPhi);
    if (cosPhi === 1 && P % V === 0) {
      return power;
    }
    return Math.round(power + 1);
  }
}

AC electrical component interface:

interface ElectricACComponent {
  nPhases: number;
  Iph: number;
  V: number;
  I?: number;
  P?: number;
}

Electrical properties calculation of an electric car connected to a AC charging station implementation (the electrical properties of the connexion cable are always beyond):

const Car = {
  nPhases: 2,
  Iph: 18,
  V: 220,
};
 
const ChargingStation = {
  nPhases: 3,
  Iph: 32,
  V: 230,
};
 
function getCombinedElectricObject(Car, ChargingStation): ElectricACComponent {
  return {
    nPhases: Math.min(Car.nPhases, ChargingStation.nPhases),
    Iph: Math.min(Car.Iph, ChargingStation.Iph),
    V: Math.min(Car.V, ChargingStation.V),
    I: ElectricityUtils.calculateAmpTotal(this.nPhases, this.Iph),
    P: ElectricityUtils.calculatePowerTotal(this.nPhases, this.Iph, this.V),
  };
}

And now for example you can define a fine grained OCPP charging profile with the following schedule for the electric car connected to the charging station that will include the number of phases used and the max intensity per phase:

const ChargingSchedulePeriod = {
  startPeriod: 60,
  limit: 15, // Sanity check: ensure the limit is below getCombinedElectricObject(Car, ChargingStation).Iph
  numberPhases: getCombinedElectricObject(Car, ChargingStation).nPhases,
};

Modelling multi-phased AC/DC electrical system

See first section.

FIXME

FIXME

Modelling multi-phased DC/AC electrical system

FIXME

FIXME

See first section.

References

  • en/cs/modelling_multi-phased_electrical_system_interconnexion.txt
  • Dernière modification: 2020/11/02 20:40
  • de fraggle