Maximum Active Power Transmission Efficiency and Apparent Power Definition with Arbitrary Waveforms

2.1 Maximum transmission efficiency with respect to the power source

This is the approach adopted to define the system apparent power used in standards like the IEEE1459 and DIN 40110.

In the circuit of Figure 1, the waveforms are not assumed sinusoidal, or even periodic (i.e., arbitrary waveforms).

It is assumed that both the resistances and the parasitic inductances of the lines can be different one from each other, although as will be seen in the demonstration that follows, it will not be necessary to know the value of the parasitic inductances to determine the maximum possible transmission efficiency (since these do not dissipate energy).

It is assumed that the functions of the waveforms of the voltages and currents are continuous, bounded, and satisfy the necessary conditions to permute the derivation and integration operations (practical waveforms have continuous derivatives bounded to values less than infinity).

The impedances of the load can be non-linear and eventually active (that is, they can contain voltage or current generators with arbitrary waveforms).

To calculate the energy transferred to the load, the instantaneous power is integrated during a time τ corresponding to the operating time of the system, or to the chosen analysis interval (which in the particular case of periodic waves can be one or multiple periods of the network).

Let m be the number of phases, the instantaneous power delivered by the source is:

$p=\sum_{x=1}^m v_x i_x+v_N i_N$     (1)

where, v_N and i_N  are the neutral voltage and current.

Therefore, the energy consumed is:

$W=\int_0^\tau p d t=\int_0^\tau\left(\sum_{x=1}^m v_x i_x+v_N i_N\right) d t$     (2)

and the average active power is defined as:

$P_{a v}=W / \tau$     (3)

An instantaneous power loss will dissipate in power lines, expressed by:

$p_S=\sum_{x=1}^m R_{S_x} i_x^2+R_N i_N^2$    (4)

During the time τ the energy lost in the lines will be:

$W_S=\int_0^\tau p_S d t=\int_0^\tau\left(\sum_{x=1}^m R_{S_x} i_x^2+R_N i_N^2\right) d t$    (5)

The root mean square value of each current is defined as:

$I_x^2=\frac{1}{\tau} \int_0^\tau i_x^2 d t ; \forall x=1,2,3, \ldots, m, N$    (6)

and that of each voltage as:

$V_x^2=\frac{1}{\tau} \int_0^\tau v_x^2 d t ; \forall x=1,2,3, \ldots, m, N$    (7)

Defining the transmission efficiency as (W-W_S)⁄(W) in order to obtain its maximum, the minimum of W_S will be found for a constant total transmitted energy W. For this, the Lagrange multipliers method will be used.

1.png

Figure 1. Circuit for the proof of the maximum efficiency theorem (m: number of phases)

It will be assumed that the functions corresponding to the waveforms of voltages and currents admit permutation between the operations of summation, integration and derivation (which happens for the waveforms used in practice, which are continuous and have continuous derivatives with bounded values less than infinity).

The minimum of W_S will be found with the restrictions that the total energy (W) transmitted from the source is constant:

$f\left(i_x\right)=\int_0^\tau p d t-W=0 ; \forall x=1,2,3, \ldots, m, N$      (8)

and the restriction imposed by the node equation:

$g\left(i_x, i_N\right)=\sum_{x=1}^m i_x+i_N=0$     (9)

which is equivalent to:

$\int_0^\tau g d t=\int_0^\tau\left(\sum_{x=1}^m i_x+i_N\right) d t=0$     (10)

With which, from the Lagrange method, the system of equations in partial derivatives to solve is:

$\frac{\partial W_S}{\partial i_x}-\lambda_1 \frac{\partial f}{\partial i_x}-\lambda_2 \frac{\partial \int_0^\tau g d t}{\partial i_x}=0$    (11)

By exchanging derivation and integration, the system to solve becomes:

$\frac{\partial p_s}{\partial i_x}-\lambda_1 \frac{\partial p}{\partial i_x}-\lambda_2 \frac{\partial g}{\partial i_x}=0 ; \forall x=1,2,3, \ldots, m, N$     （12）

Since p_S is always greater than zero, after finding its minimum instantaneous value, it will be integrated to obtain the minimum total energy lost.

Notice that if a minimum instantaneous function p_Sis found for each instant, its integral during time τ (which is W_S) must also be minimum.

Deriving (1), (4) and (10), and substituting in (12) one obtains:

$2 i_x-\lambda_1\left(v_x / R_{S_x}\right)-\left(\lambda_2 / R_{S_x}\right)=0 ; \forall x$$=1,2 \ldots m$      (13a)

$2 i_N-\lambda_1\left({ }^{v_N} / R_N\right)-\left(\lambda_2 / R_N\right)=0$    （13b）

Adding member to member (13a) and (13b) gives:

$2\left(\sum_{x=1}^m i_x+i_N\right)-\lambda_1\left[\sum_{x=1}^m\left(v_x / R_{S_x}\right)+\right.$

$\left.+\left({ }^{v_N} / R_N\right)\right]-\left(\lambda_2 / R_{S / /}\right)=0$     (14)

where,

$R_{S / /}=1 /\left[\sum_{x=1}^m\left(1 / R_{S_x}\right)+\left(1 / R_N\right)\right]$    (15)

is the result of associating in parallel all the parasitic resistances of the lines.

Applying the law of nodes (9), one gets:

$\lambda_2=-\lambda_1 v_O$     (16)

where,

$v_O=R_{S / /}\left[\sum_{x=1}^m\left(v_x / R_{S_x}\right)+\left(v_N / R_N\right)\right]$     (17)

is a reference voltage that will be zero when the phase voltages have no homopolar component and the neutral voltage is negligible.

Substituting (16) and (17) in (13a) and (13b), it results:

$i_x=\left(\lambda_1 / 2\right)\left[\left(v_x-v_0\right) / R_{S_x}\right]$    (18a)

$i_N=\left(\lambda_1 / 2\right)\left[\left(v_N-v_O\right) / R_N\right]$      (18b)

Substituting in the equation of energy losses (W_S) (5), yields:

$W_S=\left(\lambda_1 / 2\right)^2 \int_0^\tau\left\{\left[\sum_{x=1}^m\left(v_x-v_O\right)^2 / R_{S_x}\right]+\cdots+\right.$

${\left.\left[\left(v_N-v_O\right)^2 / R_N\right]\right\} d t }$    (19)

Defining:

$V_{x o^2}=\frac{1}{\tau} \int_0^\tau\left(v_x-v_O\right)^2 d t ; \forall x=1,2,3, \ldots m, N$    (20)

and substituting in (19), exchanging sum and integral, one obtains:

$W_S=\left(\lambda_1 / 2\right)^2 \tau\left[\left(\sum_{x=1}^m V_{x O^2}{ }^2 / R_{S_x}\right)+\left(V_{N O}{ }^2 / R_N\right)\right]$    (21)

from where it follows:

$2 / \lambda_1=\sqrt{\tau / W_S} \sqrt{\left(\sum_{x=1}^m V_{x O}{ }^2 / R_{S_x}\right)+\left(V_{N O}{ }^2 / R_N\right)}$    (22)

Using definitions (5) and (6):

$W_S=\tau\left[\left(\sum_{x=1}^m R_{S_x} I_x^2\right)+R_N I_N^2\right]$     (23)

Substituting (23) in (22) it results:

$2 / \lambda_1$$=\sqrt{\left(\sum_{x=1}^m V_{x o}{ }^2 / R_{S_x}\right)+\left(V_{N O}{ }^2 / R_N\right)} / \sqrt{\left(\sum_{x=1}^m R_{S_x} I_x^2\right)+R_N I_N^2}$     (24)

On the other hand, from (18a) and (18b):

$v_x=\left(2 / \lambda_1\right) R_{S_x} i_x+v_0 ; \forall x=1,2,3, \ldots, m$     (25a)

$v_N=\left(2 / \lambda_1\right) R_N i_N+v_O$     (25b)

Substituting (25a) and (25b) in (2) yields:

$W=\tau P_{a v}=\int_0^\tau\left\{\left(2 / \lambda_1\right)\left[\left(\sum_{x=1}^m R_{S_x} i_x^2\right)+\right.\right.$

$\left.\left.R_N i_N^2\right]+v_O\left(\sum_{x=1}^m i_x++i_N\right)\right\} d t$     (26)

Applying the law of nodes $\sum_{x=1}^m i_x+i_N=0$, it results:

$W=\tau P_{a v}=\left(2 / \lambda_1\right) \int_0^\tau\left[\left(\sum_{x=1}^m R_{S_x} i_x^2\right)+\right.$

$\left.R_N i_N{ }^2\right] d t$      (27)

With the definitions given by (6), exchanging summation and integration, one obtains:

$W=\tau P_{a v}=\tau\left(2 / \lambda_1\right)\left[\left(\sum_{x=1}^m R_{S_x} I_x^2\right)+R_N I_N^2\right]$     (28)

Substituting (24) in (28), gives:

$P_{a v}=\sqrt{\left(\sum_{x=1}^m V_{x O}{ }^2 / R_{S_x}\right)+\left(V_{N O}{ }^2 / R_N\right)} \sqrt{\left(\sum_{x=1}^m R_{S_x} I_x{ }^2\right)+R_N I_N{ }^2}$     (29)

A total equivalent system voltage (or DC equivalent voltage) can be defined as:

$V_{e q-t o t}=\sqrt{R_{S / /}} \sqrt{\left(\sum_{x=1}^m V_{x O}^2 / R_{S_x}\right)+\left(V_{N O}{ }^2 / R_N\right)}$    (30)

and a total equivalent system current (or DC equivalent current):

$I_{e q-t o t}=\sqrt{\left(\sum_{x=1}^m R_{S_x} I_x^2\right)+R_N I_N^2} \sqrt{R_{S / /}}$    (31)

Using both definitions, one gets:

$P_{a v}=V_{e q-t o t} \cdot I_{e q-t o t}$     (32)

To show that the found losses W_S correspond to a minimum, the second order derivatives of the auxiliary function of the Lagrange method must be considered:

$h_{\left(i_x\right)}=p_S-\lambda_1 f_{\left(i_x\right)}-\lambda_2 g_{\left(i_x\right)} ; \forall x=1,2,3 \ldots m, N$     (33)

where, the constants λ₁ and λ₂ can be calculated using (16), (17) and (24).

Using (16), the auxiliary function (33) can be written as:

$h_{\left(i_x\right)}=\left(\sum_{x=1}^m R_{S_x} i_x^2\right)+R_N i_N^2-\lambda_1\left[\sum_{x=1}^m\left(v_x-\right.\right.$

$\left.\left.-v_O\right) i_x+\left(v_N-v_O\right) i_N\right]$     (34)

and the second order derivatives are:

$\partial^2 h / \partial i_x^2=2 R_{S_X} ; \forall x=1,2,3, \ldots, m$     (35a)

$\partial^2 h / \partial i_N^2=2 R_N$     (35b)

$\partial^2 h / \partial i_x \partial i_y=0 ; \forall x \neq y$     (36)

The Hessian matrix that can be formed with the second-order partial derivatives of the Lagrange's method auxiliary function, is a diagonal matrix in which all the determinants of its principal submatrices (called principal minors) are greater than zero (positive definite matrix according to the Sylvester criterion). Therefore, the singular point found is a minimum.

For example, for m = 3 it is:

${\left[\begin{array}{cccc}\partial^2 h / \partial i_1{ }^2 & \partial^2 h / \partial i_1 \partial i_2 & \partial^2 h / \partial i_1 \partial i_3 & \partial^2 h / \partial i_1 \partial i_N \\ \partial^2 h / \partial i_2 \partial i_1 & \partial^2 h / \partial i_2{ }^2 & \partial^2 h / \partial i_2 \partial i_3 & \partial^2 h / \partial i_2 \partial i_N \\ \partial^2 h / \partial i_3 \partial i_1 & \partial^2 h / \partial i_3 \partial i_2 & \partial^2 h / \partial i_3{ }^2 & \partial^2 h / \partial i_3 \partial i_N \\ \partial^2 h / \partial i_N \partial i_1 & \partial^2 h / \partial i_N \partial i_2 & \partial^2 h / \partial i_N \partial i_3 & \partial^2 h / \partial i_N{ }^2\end{array}\right] }$

$=2\left[\begin{array}{cccc}R_{S_1} & 0 & 0 & 0 \\ 0 & R_{S_2} & 0 & 0 \\ 0 & 0 & R_{S_3} & 0 \\ 0 & 0 & 0 & R_N\end{array}\right]$

In this matrix it is evident that all the principal minors, that is, all the determinants of the principal submatrices, are greater than zero, so both the instantaneous power p_S and its integral, which is the lost energy, are minimal.

For the constant energy W, the losses found are minimal and the transmission efficiency is maximum, so the opposite is also valid, that is, for constant losses, the transferred power calculated through (29) will be the maximum possible for these effective values of current and voltage (rms values with periodical waveforms).

Consequently, this maximum possible power is the one that should be adopted as the denominator to calculate the power factor, and (29) can be adopted as the definition of apparent power for the more general case of a polyphase system discussed here.

Particular Cases of Special Interest.

There are two situations of particular interest:

(1) First, when the parasitic resistances of the lines are equal to each other, but different from the resistance of the neutral conductor. That is:

$R_{S_x}=R_S \neq R_N ; \forall x=1,2,3, \ldots, m$     (37)

With this condition one gets:

$R_{S / /}=1 /\left[\left(m / R_S\right)+\left(1 / R_N\right)\right]$     (38)

and defining:

$\rho=R_N / R_S$     (39)

it results:

$R_{S / /}=R_S /[m+(1 / \rho)]$     (40)

Substituting (40) in (17):

$v_O=\left(\sum_{x=1}^m v_x+v_N\right) /[m+(1 / \rho)]$    (41)

and then (29) yields:

$P_{a v}=\sqrt{\sum_{x=1}^m V_{x O^2}{ }^2+\left(V_{N O}{ }^2 / \rho\right)} \sqrt{\sum_{x=1}^m I_x^2+\rho I_N^2}$    (42)

(2) Case in which all the parasitic resistances of the lines are equal to each other. That is:

$R_S=R_N \Rightarrow \rho=1$     (43a)

and

$R_{S / /}=R_S /(m+1)$   (43b)

Therefore:

$v_O=\left(\sum_{x=1}^m v_x+v_N\right) /(m+1)$     (44)

turns out to be the homopolar, or zero sequence, voltage of the polyphase system formed by v₁, v₂, v₃, ..., v_m, v_N.

Thus, it results:

$P_{a v}=\sqrt{\sum_{x=1}^m V_{x O^2}{ }^2+V_{N O}{ }^2} \sqrt{\sum_{x=1}^m I_x^2+I_N^2}$     (45)

An equivalent system voltage per phase can be defined as:

$V_{e q}=\sqrt{\left(\sum_{x=1}^m V_{x O}{ }^2+V_{N O}{ }^2\right) / m}$    (46)

and an equivalent system current per phase:

$I_{e q}=\sqrt{\left(\sum_{x=1}^m I_x^2+I_N^2\right) / m}$     (47)

which is the equivalent current originally proposed by Buchholz. With these definitions it results:

$P_{a v}=m V_{e q} I_{e q}$     (48)

which is the maximum active power obtainable in a balanced and symmetric polyphase system with m phases operating with the given currents and effective voltages.

2.2 Maximum transmission efficiency with respect to the power delivered to the load

Observing Figure 1, it can be concluded that it would be more interesting and appropriate to search the maximum efficiency by minimizing transmission losses for a constant power consumed in the load, instead of imposing that the power delivered by the source be kept constant. That is, finding the minimum of W_S given by (5), keeping constant the total energy consumed by the load:

$W^{\prime}=\tau P_{a v}{ }^{\prime}=\int_0^\tau p^{\prime} d t=\int_0^\tau\left(\sum_{x=1}^m v_x{ }^{\prime} i_x\right) d t$    (49)

where,

$p^{\prime}=\sum_{x=1}^m v_x^{\prime} i_x$     (50)

is the instantaneous power consumed by the load and

$P_{a v}{ }^{\prime}=W^{\prime} / \tau$     (51)

is the average active power consumed by the load.

The instantaneous power losses p_S continues to be given by (4), and therefore the energy lost in transmission is also W_S given by (5).

Now, using the method of Lagrange multipliers, the minimum of W_S will be found keeping W' constant, with which the first restriction is:

$f\left(i_x\right)=\int_0^\tau p^{\prime} d t-W^{\prime}=0 ; \forall x=1,2,3, \ldots, m$     (52)

and the second restriction is the law of nodes, expressed by (9) or (10).

In an entirely similar way to what was done before, one must solve the system formed by:

$\frac{\partial p_S}{\partial i_x}-\lambda_1 \frac{\partial p^{\prime}}{\partial i_x}-\lambda_2 \frac{\partial g}{\partial i_x}=0 ; \forall x=1,2,3, \ldots, m, N$     (53)

Differentiating (4), (9) and (50), one obtains:

$2 i_x-\lambda_1\left(v_x^{\prime} / R_{S_x}\right)-\left(\lambda_2 / R_{S_x}\right)=0 ; \forall x$$=1,2, \ldots m$     (54)

$2 i_N-\left(\lambda_2 / R_N\right)=0$     (55)

Adding member to member (54) and (55) yields:

$2\left(\sum_{x=1}^m i_x+i_N\right)-\lambda_1\left[\sum_{x=1}^m\left(v_x{ }^{\prime} / R_{S_x}\right)\right]-\left(\lambda_2 / R_{S / /}\right)=0$     (56)

where, $R_S / /$ is given by (15) and applying the law of nodes (9) it results:

$\lambda_2=-\lambda_1 v_O^{\prime}$    (57)

where,

$v_O{ }^{\prime}=R_S\left[/ \sum_{x=1}^m\left(v_x{ }^{\prime} / R_{S_x}\right)\right]$   (58)

Substituting (57) in (54) gives:

$i_x=\left(\lambda_1 / 2\right)\left[\left(v_x{ }^{\prime}-v_O{ }^{\prime}\right) / R_{S_x}\right]$    (59)

and from (55) and (57) it results:

$i_N=-\left(\lambda_1 / 2\right) v_O^{\prime} / R_N$     (60)

Substituting in the expression of energy losses (W_S), (5):

$W_S=\left(\lambda_1 / 2\right)^2 \int_0^\tau\left\{\left[\sum_{x=1}^m\left(v_x^{\prime}-v_O^{\prime}\right)^2 / R_{S_x}\right]\left(v_O^{\prime 2} / R_N\right)\right\} d t$     (61)

Defining:

$V_{x O}^{\prime}{ }^2=\frac{1}{\tau} \int_0^\tau\left(v_x{ }^{\prime}-v_O{ }^{\prime}\right)^2 d t ; \forall x=1,2,3, \ldots m$    (62)

$V_O^{\prime 2}=\frac{1}{\tau} \int_0^\tau\left(v_O^{\prime}\right)^2 d t$    (63)

and substituting in (61), exchanging summation and integral, gives:

$W_S=\left(\lambda_1 / 2\right)^2 \tau\left[\left(\sum_{x=1}^m V_{x O}^{\prime 2} / R_{S_x}\right)+\left(V_O^{\prime 2} / R_N\right)\right]$    (64)

from which it yields:

$2 / \lambda_1=\sqrt{\tau / W_S} \sqrt{\left(\sum_{x=1}^m V_{x O}^{\prime 2} / R_{S_x}\right)+\left(V_o^{\prime 2} / R_N\right)}$   (65)

Substituting (23) in (65), yields:

$2 / \lambda_1=\sqrt{\left(\sum_{x=1}^m V_{x o}^{\prime 2} / R_{S_x}\right)+\left(V_o^{\prime 2} / R_N\right)} / \sqrt{\left(\sum_{x=1}^m R_{S_x} I_x^2\right)+R_N I_N^2}$    (66)

Substituting (59) into (49):

$\begin{aligned} W^{\prime}=P_{a v}^{\prime} \tau=& \int_0^\tau\left[\left(2 / \lambda_1\right)\left(\sum_{x=1}^m R_{S_x} i_x^2\right)+\right.\\ &\left.+v_O^{\prime}\left(\sum_{x=1}^m i_x\right)\right] d t \end{aligned}$     (67)

Solving from the node equation, it is $\sum_{x=1}^m i_x=-i_N$ and using (60):

$W^{\prime}=P_{a v}^{\prime} \tau=\left(2 / \lambda_1\right) \int_0^\tau\left[\left(\sum_{x=1}^m R_{S_x} i_x^2\right)+\right.\left.R_N i_N{ }^2\right] d t$    (68)

Permuting integral and summation, using the definitions given by (6), one obtains:

$W^{\prime}=P_{a v}^{\prime} \tau=\left(2 / \lambda_1\right) \tau\left[\left(\sum_{x=1}^m R_{S_x} I_x^2\right)+R_N I_N^2\right]$    (69)

Substituting (66) into (69) gives:

$P_{a v}^{\prime}= \sqrt{\left(\sum_{x=1}^m V_{x o}^{\prime}{ }^2 / R_{S_x}\right)+\left(V_o^{\prime 2} / R_N\right)} \sqrt{\left(\sum_{x=1}^m R_{S_x} I_x^2\right)+R_N I_N^2}$     (70)

As in the previous case, a total equivalent system voltage (or DC equivalent voltage) and a total equivalent system current (or DC equivalent current) can be defined such that:

$P_{a v}^{\prime}=V_{\text {eq-tot }}^{\prime} I_{\text {eq-tot }}^{\prime}$    (71)

where,

$V_{e q-t o t}^{\prime}=\sqrt{R_S / /} \sqrt{\left(\sum_{x=1}^m V_{x O}^{\prime 2} / R_{S_x}\right)+\left(V_O^{\prime 2} / R_N\right)}$    (72)

$I_{e q-t o t}^{\prime}=I_{e q-t o t}=$$\sqrt{\left(\sum_{x=1}^m R_{S_x} I_x^2\right)+R_N I_N^2} / \sqrt{R_{S / /}}$     (73)

where, this current is the same as that of the preceding case.

2.3 Comparison between both approaches

The expression of the maximum power obtained with the second approach is very similar to that obtained with the first one, which has been the preferred criterion by most international standards. The numerical results do not differ substantially between the two criteria when the neutral voltages are small and the transmission efficiencies are high. However, note that in the case of optimizing the efficiency with respect to the power transferred to the load, the reference voltage $v_O^{\prime}$, given by (58), does not depend on the neutral voltage, which, as can be deduced from observing Figure 1, is a voltage difficult to measure from the point of electrical energy consumption where the user is located.

For the case in which the parasitic resistances of the lines are equal to each other, but different from the resistance of the neutral conductor (condition 37), it is:

$v_O^{\prime}=\{m /[m+(1 / \rho)]\} v_{\Sigma}^{\prime}$     (74a)

where, $v_{\Sigma}^{\prime}$ is the homopolar component of the load voltages:

$v_{\Sigma}^{\prime}=(1 / m) \sum_{x=1}^m v_x^{\prime}$    (74b)

Thus, (72) and (73) become:

$V_{e q-t o t}^{\prime}=\sqrt{\left(\sum_{x=1}^m V_{x O}^{\prime 2}\right)+\left(V_O^{\prime 2} / \rho\right)} / \sqrt{m+(1 / \rho)}$    (75)

$I^{\prime}{ }_{\text {eq-tot }}=\sqrt{\sum_{x=1}^m I_x^2+\rho I_N^2} \sqrt{m+(1 / \rho)}$    (76)

This second approach based on (70) allows defining the apparent power without the need to know the neutral voltage drop as:

$S^{\prime}=\sqrt{\left(\sum_{x=1}^m V_{x O}^{\prime 2}\right)+\left(V_O^{\prime 2} / \rho\right)} \sqrt{\sum_{x=1}^m I_x^2+\rho I_N^2}$    (77)

which would be a simpler definition than those based on the first approach (which is used in IEEE Std. 1459).

When there is no homopolar component in the output voltages, (75) and (76) are simplified as:

$V_{\text {eq-tot }}^{\prime}=\sqrt{\left(\sum_{x=1}^m V_x^{\prime 2}\right) / m}$    (78)

$I^{\prime}{ }_{\text {eq-tot }}=\sqrt{\sum_{x=1}^m I_x{ }^2+\rho I_N{ }^2} \sqrt{m}$     (79)

and (77) becomes:

$S^{\prime}=\sqrt{\left(\sum_{x=1}^m V_{x O}^{\prime 2}\right)} \sqrt{\sum_{x=1}^m I_x^2}$    (80)

In both optimization approaches, losses can be expressed as:

$P_S=\left(\sum_{x=1}^m R_{S_x} I_x^2\right)+R_N I_N^2$    (81)

with which, (29) and (70) become:

$\left.P_{a v}=V_{e q-t o t}\right\rfloor_{c r i t \# 1} \sqrt{P_S} / \sqrt{R_S / /}$    (82a)

$P_{a v}^{\prime}=\left.V_{e q-t o t}^{\prime}\right|_{c r i t \# 2} \sqrt{P_S} / \sqrt{R_S / /}$    (82b)

where, $\left.V_{e q-t o t}\right|_{c r i t \# 1}$ and $\left.V_{e q-t o t}^{\prime}\right|_{c r i t \# 2}$ are given by $(30)$ and (72) respectively. From where the equivalent voltages per phase can be defined, according to:

$\left.V_{e q}\right|_{c r i t \# 1}=\left.V_{e q-t o t}\right|_{c r i t \# 1} / \sqrt{m}$    (83a)

$\left.\left.V_{e q}^{\prime}\right\rfloor_{\text {crit#2 }}=V^{\prime}{ }_{e q-t o t}\right\rfloor_{c r i t \# 2} / \sqrt{m}$     (83b)

Note that in the most general case it results:

$\left.\left.\left.V_{e q}^{\prime}\right\rfloor_{c r i t \# 1}=V_{e q}\right\rfloor_{c r i t \# 1}-R_{S / /} I_{e q-t o t} \neq V_{e q}^{\prime}\right\rfloor_{c r i t \# 2}$      (84)

which should be taken into account when proposing the equivalent circuit models.

With either criteria, at least three equivalent circuit topologies can be proposed for a polyphase system with arbitrary waveforms (Figure 2).

(a) Polyphase equivalent model with sinusoidal alternating current, with symmetrical source and balanced load (Figure 2a)

The current of each phase is adopted as $I_{e q}=I_{e q-t o t} / \sqrt{m}$ and the equivalent voltages per phase $V_{e q}$ and $V_{e q}^{\prime}$ will have expressions that will depend on the optimization criteria adopted. The loss equivalent series resistors will be:

$R_{S_{e q}}=R_{S / /}$, such that:

$P_S=m I_{e q}^2 R_{S_{e q}}=I_{e q-t o t}{ }^2 R_{S / /}=$$\left(\sum_{x=1}^m R_{S_x} I_x^2\right)++R_N I_N^2$    (85)

When it is:

$R_{S_x}=R_S ; \forall x=1,2,3, \ldots m$    (86a)

$R_N=\rho R_S$    (86b)

based on (42), the model can be expressed by:

$R_{S_{e q}}=m R_{S / /}=R_S /[1+(1 / m \rho)]$    (87)

$I_{e q}=\sqrt{\sum_{x=1}^m I_x^2+\rho I_N^2} \sqrt{1+(1 / m \rho)}$     (88)

with $V_{e q}$ given by (83a), for criterion # 1, or $V_{e q}^{\prime}$ given by (83b) for criterion $\# 2$.

111.png

Figure 2. Equivalent circuit models, (a) Sinusoidal polyphase equivalent system with symmetric source and balanced load, (b) Sinusoidal single-phase equivalent system, (c) Direct current equivalent system

Using (87) and (88) one may verify that P_S results the same given by (85).

It is clear that since all quantities are sinusoidal, phasors can be used.

b) Single-phase sinusoidal equivalent model (Figure 2b)

In this case, the sinusoidal voltage source will have an effective voltage equal to V_eq-tot (with the corresponding expression according to the optimization criterion adopted) and the sinusoidal current will have an rms value V_eq-tot, given by (31) and the equivalent series loss resistance will be $R_{S / /}$.

c) Bifilar equivalent model of direct current (Figure 2c)

From Figure 2c it follows that:

$R_{S_{\text {tot }}}=R_{S_{(+)}}+R_{S(-)}=R_{S / /}$    (89)

The DC voltage $V_B$ will be $V_B=V_{e q-t o t}$ (with the expression corresponding to the optimization criteria adopted) and the direct current will be $I_{D C}=I_{e q-t o t}$, given by (31).

In Figure 3, a hybrid system composed of three subsystems (numbered from # 1 to #3), with different nature and topologies of direct current and alternating current, is presented. It will be shown here how to define the apparent powers of each subsystem applying a definition based on the second optimization criterion proposed in Section 2, using (72) and (73) to define by (91) the apparent power of each subsystem.

It is assumed that the effective values of the voltages and currents indicated in the figures were experimentally measured, as well as the active powers of each load phase. With these data, the apparent powers and the power factors of each subsystem can be obtained and then through (93) or (94), the total apparent power and the total power factor.

4.1 Subsystem #1 (Figure 3a)

The subsystem shown in Figure 3a is a two-wire direct current power supply system with a number of phases m=1. Therefore, to apply definition (78), or also (91), one first gets from (58):

$V_{O \# 1}^{\prime}=\frac{1}{m+1} V_{\# 1}^{\prime}=\frac{1}{2} V_{\# 1}^{\prime}$    (E.1)

from which the total equivalent system effective voltage (75) results:

$V_{e q-t o t}^{\prime}{ }_{\# 1}=\sqrt{\left(V_{\# 1}^{\prime}-V_O^{\prime}{ }_{\# 1}\right)^2+V_O^{\prime}{ }_{\# 1}^2}=\frac{1}{\sqrt{2}} V_{\# 1}^{\prime}$     (E.2)

and the total equivalent system effective current is obtained by applying (76):

$I_{\text {eq-tot }}{ }_{\# 1}=\sqrt{I_{\# 1}{ }^2+I_N{ }^2}=\sqrt{2} I_{\# 1}$    (E.3)

Therefore:

$S_{\# 1}^{\prime}=V_{e q-t o t_{\# 1}}^{\prime} I_{e q-t o t_{\# 1}}^{\prime}=V_{\# 1}^{\prime} I_{\# 1}$    (E.4a)

Using the first optimization criterion, the following would have been obtained:

$S_{\# 1}=V_{e q-t o t} I_{\# 1} I_{e q-t o t_{\# 1}}=V_{\# 1} I_{\# 1}$    (E.4b)

which was naturally expected according to definition (93).

Assuming negligible voltage drops in the lines, it results: $V_{\# 1} \cong V_{\# 1}^{\prime}$.

As $V_{\# 1} \cong V_{\# 1}^{\prime}=500 \mathrm{~V}$ and $I_{\# 1}=50 \mathrm{~A}$, it is: $S_{\# 1} \cong S_{\# 1}^{\prime}=$ $25 \mathrm{kVA}$.

Since the load is pure resistive, it results: $F P_{\# 1}=1$.

4.2 Subsystem #2 (Figure 3b)

In Figure 3b a multi-wire direct current subsystem is shown, with 4 wires whose parasitic resistances are all the same. As in the previous case it will be assumed that voltage drops in the lines can be neglected compared to the voltages in the loads.

The measured voltages and currents are: $V_{1 \# 2}=400 \mathrm{~V}$, $\begin{aligned} V_{2 \# 2}=300 \mathrm{~V}, V_{3 \# 2}=260 \mathrm{~V}, I_{1 \# 2}=10 \mathrm{~A}, I_{2 \# 2}=20 \mathrm{~A}, I_{3 \# 2}=30 \mathrm{~A} .\end{aligned}$

In this subsystem it is m = 3 and from (58): $V_{O_{\# 2}}^{\prime}=\frac{1}{4}\left(V_{1 \# 2}+V_{2 \# 2}+V_{3 \# 2}\right)=240 \mathrm{~V}$.

According to (75) the total equivalent system voltage is:

$V^{\prime}{ }_{\text {eq-tot } \# 2}=\sqrt{\left(V_{1 \# 2}^{\prime}-V_{O_{\# 2}}^{\prime}\right)^2+\left(V_{2 \# 2}^{\prime}-V_{O_{\# 2}}^{\prime}\right)^2+\left(V_{3 \# 2}^{\prime}-V_{O_{\# 2}}^{\prime}\right)^2+V_{O_{\# 2}}^{\prime}}{ }^2=295.3 \mathrm{~V}$      (E.5)

and the total equivalent system current according to (76) is:

$I_{e q-t o t_{\# 2}}=I^{\prime}{ }_{e q-t o t_{\# 2}}=\sqrt{I_{1_{\# 2}}{ }^2+I_{2 \# 2}{ }^2+I_{3^2}{ }^2+I_{N_{\# 2}}{ }^2}$     (E.6)

where according to the law of nodes is:

$I_{N \# 2}{ }^2=\left(I_{1 \# 2}+I_{2 \# 2}+I_{3 \# 2}\right)^2$     (E.7)

222.png

Figure 3. Hybrid system, (a) Two-wire direct-current subsystem, (b) Multiwire DC subsystem, (c) Three phase symmetric subsystem with rectangular waves

Substituting (E.7) in (E.6) and replacing by the numerical values given as data, yields:

$I_{e q-t o t_{\# 2}}=I_{e q-t o t_{\# 2}}=70.7 \mathrm{~A}$    (E.8)

With the values given by (E.5) and (E.8) one obtains:

$S_{\# 2}^{\prime}=V^{\prime}{ }_{e q-t o t_{\# 2}} I_{e q-t o t_{\# 2}}=20.88 \mathrm{kVA}$    (E.9)

According to data, the total active power of subsystem #2 is:

$P_{\text {tot#2 }}=V_{1_{\# 2}} I_{1 \# 2}+V_{2 \# 2} I_{2 \# 2}+V_{3 \# 2} I_{3 \# 2}=17.8 \mathrm{~kW}$     (E.10)

and consequently, with (E.9) and (E.10), the power factor of subsystem #2 is:

$P F_{\# 2}=P_{\text {tot# } 2} / S_{\# 2}^{\prime}=0.852$    (E.11)

Note that the sum of the powers calculated as the product of each voltage and each current would also be the total apparent power considering each source and its respective load as separate subsystem from the others grouped as subsystem #2 (according to what was proposed in Section 3) which would be in disagreement with the result obtained through (E.9). What this equation shows is that more power could be transferred to the load with the same level of transmission losses, if the multi-wire system of Figure 3b were replaced by the equivalent two-wire system formed by a direct voltage source $V^{\prime} _{e q-tot_{\# 2}}$  delivering a continuous power to the load, equal to S'_#2.

Obviously, when establishing a standard, it should be decided the criteria to address this issue regarding DC multi-wire systems.

When the source voltages are different and are intended to supply loads independent of each other, it should be more appropriate to consider each source with its load as a separate subsystem (even if they share the neutral).

In contrast, in symmetrical sources (delivering opposite voltages: +V_B and -V_B), either single-pair or multi-pair (such as those used in some HVDC power transmission links) it is recommended to adopt a definition of apparent power (and consequently of power factor) based on the application of the maximum transmission efficiency theorem shown here (with either of the two optimization criteria exposed) because in this way, a figure of merit would be available to show the wasted transmission capacity of the link as a consequence of eventual load asymmetries.

4.3 Subsystem #3 (Figure 3c)

Figure $3 \mathrm{c}$ shows an inverter capable of delivering load phase voltages with rectangular waveforms, with a pulse width $D T / 2$ where $T$ is the period and $D$ is the duty cycle, between $0<\mathrm{D} \leq 2 / 3$ (see Figure 4). The rms currents are: $I_{1 \# 3}=11 \mathrm{~A}, I_{2_{\# 3}}=20 \mathrm{~A}, I_{3_{\# 3}}=14 \mathrm{~A}, I_{N_{\# 3}}=5 \mathrm{~A}$.

The active powers consumed by the non-linear impedances: $P_{Z_{-} 01 \# 3}=1100 \mathrm{~W}, P_{Z_{-} 02 \# 3}=2500 \mathrm{~W}, P_{Z_{-} O 3 \# 3}=1400 \mathrm{~W}$.

The symmetrical DC supply voltages are: $V_{(+)_{\# 3}}=V_{(-)_{\# 3}}=300 \mathrm{~V} .$

There are two possible values for D that are of particular interest:

a) For D = 2⁄3 the pulse width is 2π⁄3 and the waveform of the phase voltage obtained does not contain harmonic components of order 3 or their multiples, nor does it contain even components, so the first harmonic component that appears is of order 5 (which would facilitate filtering if it were decided to do so).

b) For D = 1⁄2 the crest factor (or peak factor) of the resulting rectangular wave is the same that would correspond to a sine wave ($F_{C R_{D=1 / 2}} \quad=F_{C R_{\sin }}=\hat{V} / V=\sqrt{2}$). This can be advantageous when powering equipment that must receive a peak voltage similar to that of a sinusoidal power line, for example, equipment powered by sources that include a rectifier with a direct capacitive filter and simultaneously, equipment containing transformers, filament lamps or heaters, which require an rms voltage similar to that of the public main.

4.png

Figure 4. Waveforms of Subsystem # 3, which includes a rectangular wave inverter with non-linear and unbalanced load

In Figure 4 the phase voltages are shown for an arbitrary value of the duty cycle comprised between $0<D \leq 2 / 3$.

The peak voltage of the rectangular wave is:

$V_P \cong V_{(+)_{\# 3}}=V_{(-)_{\# 3}}$    (E.12)

According to Figure 4 the pulse width is Dπ and ∆θ is:

$\Delta \theta=\pi\left(\frac{2}{3}-D\right)$     (E.13)

and also:

$v_{O_{\# 3}}^{\prime}=\frac{1}{4} v_{\Sigma}^{\prime}=\frac{1}{4} \sum_{x=1}^3 v_{x \# 3}^{\prime}$    (E.14)

Notice that the period of $v_{O_{\# 3}}^{\prime}$ is 3 times less than that of the phase voltages. From Figure 4 it is concluded that:

$\begin{aligned} \frac{1}{2 \pi} \int_0^{2 \pi}\left(v_{x \# 3}^{\prime}-v_{O_{\# 3}}^{\prime}\right)^2 d \theta \\ &=\frac{1}{2 \pi} \int_0^{2 \pi}\left(v_{x \# 3}^{\prime}{ }^2-2 v_{x \# 3}^{\prime} v_{O_{\# 3}}^{\prime}\right.\\ &\left.+v_{O_{\# 3}^{\prime}}^{\prime}{ }^2\right) d \theta={V_{x \# 3}^{\prime}}^2+V_{O_{\# 3}}^{\prime}{ }^2 \end{aligned}$     (E.15)

because,

$\int_0^{2 \pi} v_{x_{\# 3}}^{\prime} v_{O_{\# 3}}^{\prime} d \theta=0$     (E.16)

being:

$V_{x_{\# 3}}^{\prime}{ }^2=\frac{1}{2 \pi} \int_0^{2 \pi} v_{x \# 3}^{\prime}{ }^2 d \theta$    (E.17a)

$V_{O_{\# 3}}^{\prime}{ }^2=\frac{1}{2 \pi} \int_0^{2 \pi} v_{O_{\# 3}}^{\prime}{ }^2 d \theta$    (E.17b)

which gives:

$V^{\prime}{ }_{\text {eq-tot }}=\sqrt{3 V_{\# 3}^{\prime}{ }^2+4 V_{O_{\# 3}}^{\prime}{ }^2}$     (E.18)

From Figure 4 one obtains:

$V_{\# 3}^{\prime}{ }^2=D V_P{ }^2$    (E.19)

$V_{O_{\# 3}^{\prime}}{ }^2=\frac{\Delta \theta}{\pi / 3}\left(V_P / 4\right)^2$    (E.20)

Substituting (E.13), (E.19) and (E.20) in (E.18) gives:

$V_{\text {eq-tot }}^{\prime}{ }_{\# 3}=\frac{V_P}{2} \sqrt{9 D+2}$    (E.21)

Being $V_P=300 \mathrm{~V}$, for $D=2 / 3$ it results:

$\left.V_{\text {eq-tot }}^{\prime}{ }_{\# 3}\right|_{D=2 / 3}=\sqrt{2} V_P=424.3 \mathrm{~V}$    (E.22)

and for $D=1 / 2$ it is:

$\left.V^{\prime}{ }_{\text {eq-tot }}{ }_{\# 3}\right|_{D=1 / 2}=1.275 V_P=382.4 \mathrm{~V}$    (E.23)

The total equivalent system current (76) is:

$I_{e q-t o t_{\# 3}}=I_{\text {eq-tot }}{ }_{\# 3}=\sqrt{I_{1 \# 3}{ }^2+I_{2 \# 3}{ }^2+I_{3 \# 3}{ }^2+I_{N_{\# 3}}{ }^2}=27.24 \mathrm{~A}$    (E.24)

and the apparent power, neglecting the losses in the lines, can be expressed as:

$S_{\# 3} \cong S_{\# 3}^{\prime}=V_{e q-tot_{\# 3}}^{\prime} I_{e q-t o t_{\# 3}}$    (E.25)

If it were $D=2 / 3$ it would be:

$S_{\# 3}=11.56 \mathrm{kVA}$    （E.26a）

If instead it were $D=1 / 2$ it would result:

$S_{\# 3}=10.42 \mathrm{kVA}$    （E.26b）

The total active power of subsystem #3 is:

$P_{\text {tot#3 }}=P_{Z_{-} 01_{\# 3}}+P_{Z_{-} 02_{\# 3}}+P_{Z_{-} 03_{\# 3}}=5 \mathrm{~kW}$    (E.27)

and with these values the power factors may be easily obtained:

$\left.F P_{\# 3}\right\rfloor_{D=2 / 3}=P_{\text {tot }_{\# 3}} / S_{\# 3}=0.43$    (E.28a)

$\left.F P_{\# 3}\right\rfloor_{D=1 / 2}=P_{\text {tot#3 }} / S_{\# 3}=0.48$    (E.28b)

REMARK: In a real case, when modifying the duty cycle, the currents would be modified (depending on the converter topology and load features) and a different I_eq-totwould have to be considered.