# Power Flows with Flat Voltage Profiles: an Exact Approach

Anthony B. Morton

(July 2022)

##  1 Background: the DC Power Flow conditions

Analysis of alternating–current electrical networks with prescribed power
flows is complicated by the fact that network quantities are defined by two
degrees of freedom, amplitude and phase displacement. This greatly complicates
the mathematics compared with that of DC circuits and of AC circuits with
prescribed currents.

When the focus for analysis is on the flows of active power and on the phase
angle relationships between network voltages, it is common to adopt an
idealised approach often described as _DC power flow_. Despite appearances,
this does not mean solving a DC network with the same power flows. Rather, the
‘DC’ is a fossilised reference to DC network analysers, a type of mid-20th
century analogue computer that embodied this approach to network solution in
discrete electronics.

The ‘DC power flow’ idealisations for an AC power system are as follows:

  * •

Network branches are assumed lossless or practically lossless, with series
reactances $X_{\alpha}$.

  * •

Voltage magnitudes are assumed equal or nearly equal to nominal values, thus
equal to 1 in a per-unit system based on nominal voltage $V_{\mathrm{nom}}$
and an arbitrarily chosen power base $S_{\mathrm{base}}$.

With these idealisations, the full AC power flow equations can be replaced
with a simplifed set based on the familiar power-angle formula for a lossless
branch

$P_{\alpha}=\frac{|V_{j}||V_{k}|}{X_{\alpha}}\sin\left(\delta_{j}-\delta_{k}\right)$
(1)

where $P_{\alpha}$ is the active power flow on branch $\alpha$ (nominally
directed from bus $j$ to bus $k$), $X_{\alpha}$ is the branch reactance,
$|V_{j}|$ and $|V_{k}|$ the terminal bus voltage magnitudes, and $\delta_{j}$,
$\delta_{k}$ the corresponding voltage phase angles. Since by assumption
$|V_{j}|=|V_{k}|=1$, this further simpifies to

$P_{\alpha}=B_{\alpha}\sin\left(\delta_{j}-\delta_{k}\right)$ (2)

where $B_{\alpha}=1/X_{\alpha}$ is the branch susceptance. Often the
approximation is carried one step further, by supposing the phase difference
small enough that the sine nonlinearity can be disregarded, thus

$P_{\alpha}=B_{\alpha}\left(\delta_{j}-\delta_{k}\right).$ (3)

Whether equations (2) or (3) are used to represent the network, the state of
the idealised system is described by the phase angle $\delta_{k}$ at each
busbar, and the power flow $P_{\alpha}$ on each branch. In the approximate
form (3), the network equations can be physically modelled by DC circuit
branches with conductances $B_{\alpha}$, the DC node voltages serving as
analogues of the angles $\delta_{k}$ and DC branch current (not power) as the
analogue of $P_{\alpha}$. This was the principle behind the DC network
analyser.

Neither of the above idealising assumptions is truly valid in practical AC
systems, especially that of lossless branches. All network branches have
positive resistance and this can often be significant relative to the branch
reactance. The complication, when nonzero resistances are considered, is that
active power losses are determined by both the active and reactive power flows
in the branch.

The assumption of uniform nominal or ‘flat’ voltages is also contradicted in
practical scenarios, but is arguably more defensible than the zero-resistance
assumption, at least for steady-state analysis of transmission networks where
relatively abundant voltage controls ensure magnitudes remain within a few
percent of nominal values.

This note provides methods for analysing AC power flows when the lossless
assumption is discarded but the voltage profile is still assumed ‘flat’ in the
sense that all voltages have 1pu magnitude. By moderating the level of
idealisation, these methods can yield results that are more practically
relevant in cases where network losses are non-negligible. They can also
assist in demonstrating that theoretical results, obtained for lossless
networks, are robust to nonzero resistance values. One such application is to
the existence of circulating flows in wide-area networks [2], discussed toward
the end of this note.

##  2 Definitions

Figure 1 depicts a generic branch $\alpha$ within an AC network, having series
impedance $R_{\alpha}+jX_{\alpha}$ with $X_{\alpha}>0$ and $R_{\alpha}\geq 0$.

Bus $j$$V_{j}=1\angle\delta_{j}$$P_{j},Q_{j}$$R_{\alpha}$$jX_{\alpha}$Bus
$k$$V_{k}=1\angle\delta_{k}$$P_{k},Q_{k}$ Figure 1: A single branch $\alpha$
in an AC network, with active power flow from bus $j$ to bus $k$.

The relative size of the resistance can be measured by the ratio

$\rho_{\alpha}=\frac{R_{\alpha}}{X_{\alpha}}\geq 0.$ (4)

The terminal buses (nodes) are labelled $j$ and $k$, and for present purposes
the voltage magnitude is assumed to be nominal (1pu) at both buses. The
voltage phase angles are denoted $\delta_{j}$ and $\delta_{k}$ respectively.

The complex power flow into the branch at bus $j$ (‘source end’) is defined as
$P_{j}+jQ_{j}$, with $P$ denoting active power and $Q$ denoting reactive
power. Similarly, the complex power flow out of the branch at bus $k$
(‘receiving end’) is denoted $P_{k}+jQ_{k}$. For definiteness it will be
assumed that the orientation of the branch from $j$ to $k$ matches the
direction of active power flow, and so

$P_{j}\geq P_{k}\geq 0$ (5)

with $P_{j}=P_{k}$ when the branch is lossless ($R_{\alpha}=0$).

The current flow in the branch will be denoted $I_{\alpha}$, with magnitude
$|I_{\alpha}|$, in terms of which the active and reactive power consumption
satisfy the identities

$P_{j}-P_{k}=R_{\alpha}|I_{\alpha}|^{2}\geq 0,\qquad
Q_{j}-Q_{k}=X_{\alpha}|I_{\alpha}|^{2}\geq 0,$ (6)

from which also follows the identity

$\mu_{\alpha}=X_{\alpha}P_{j}-R_{\alpha}Q_{j}=X_{\alpha}P_{k}-R_{\alpha}Q_{k}.$
(7)

The _flow coefficient_ $\mu_{\alpha}$ will play an important role in the power
flow solution below.

Lastly, the notation $\sigma_{\alpha}$ will denote the (dimensionless) ratio
between the reactive power consumption on the branch and the receiving end
active power:

$\sigma_{\alpha}=\frac{X_{\alpha}|I_{\alpha}|^{2}}{P_{k}}\qquad\text{(when
$P_{k}>0$, else $\sigma_{\alpha}=0$).}$ (8)

This is denoted the branch _coefficient of support_ and will also prove useful
below.

##  3 Exact solution for a branch

From Ohm’s law together with the fundamental complex power and voltage
identities

$V_{j}I_{\alpha}^{*}=P_{j}+jQ_{j},\qquad
V_{k}I_{\alpha}^{*}=P_{k}+jQ_{k},\qquad V_{j}V_{k}^{*}=|V_{j}||V_{k}|\exp
j\\!\left(\delta_{j}-\delta_{k}\right),$ (9)

can be derived two key formulae that provide the exact power flow solution for
the branch $\alpha$ of Figure 1. The first is the identity relating the flow
coefficient to the phase angle shift

$\mu_{\alpha}=X_{\alpha}P_{j}-R_{\alpha}Q_{j}=X_{\alpha}P_{k}-R_{\alpha}Q_{k}=|V_{j}||V_{k}|\sin\left(\delta_{j}-\delta_{k}\right)$
(10)

which generalises formula (1) to branches with arbitrary series resistances.
The second is a quadratic equation relating the (squared) magnitudes of the
endpoint voltages, written here for the receiving end $k$:

$|V_{k}|^{4}-\left[|V_{j}|^{2}-2\left(R_{\alpha}P_{k}+X_{\alpha}Q_{k}\right)\right]|V_{k}|^{2}+\left(R_{\alpha}^{2}+X_{\alpha}^{2}\right)\left(P_{k}^{2}+Q_{k}^{2}\right)=0.$
(11)

Equation (11) is a useful tool for determining $|V_{k}|$ in situations where
$|V_{j}|$, $P_{k}$ and $Q_{k}$ are known. Together with (10) it provides the
general branch flow solution when any three of the five quantities $|V_{j}|$,
$|V_{k}|$, $P_{k}$, $Q_{k}$ and $\delta_{j}-\delta_{k}$ are known.

Under the ‘flat voltage’ assumption $|V_{j}|=|V_{k}|=1$, formula (11) leads
directly to the condition

$\left(R_{\alpha}^{2}+X_{\alpha}^{2}\right)\left(P_{k}^{2}+Q_{k}^{2}\right)=-2\left(R_{\alpha}P_{k}+X_{\alpha}Q_{k}\right).$
(12)

For a given $P_{k}$, this yields a quadratic equation for $Q_{k}$

$\left(R_{\alpha}^{2}+X_{\alpha}^{2}\right)Q_{k}^{2}+2X_{\alpha}Q_{k}+\left[2R_{\alpha}P_{k}+\left(R_{\alpha}^{2}+X_{\alpha}^{2}\right)P_{k}^{2}\right]=0$
(13)

with solutions

$Q_{k}=\frac{1}{R_{\alpha}^{2}+X_{\alpha}^{2}}\left[-X_{\alpha}\pm\sqrt{X_{\alpha}^{2}-2R_{\alpha}\left(R_{\alpha}^{2}+X_{\alpha}^{2}\right)P_{k}-\left(R_{\alpha}^{2}+X_{\alpha}^{2}\right)^{2}P_{k}^{2}}\right].$
(14)

Observe that with $X_{\alpha}>0$ and $R_{\alpha},P_{k}\geq 0$ by assumption,
both solutions for $Q_{k}$ are negative or zero (the latter only when
$P_{k}=0$). The solution with the positive sign is closer to zero, while the
other is of order $1/X_{\alpha}$ (and lies on an ‘inverted branch’ of the
network $Q$–$V$ curve). Taking the former, therefore, as the ‘practical’
solution for $Q_{k}$ and invoking the ratio (4) gives

$Q_{k}=-\frac{1/X_{\alpha}}{1+\rho_{\alpha}^{2}}\left[1-\sqrt{1-2\rho_{\alpha}\left(1+\rho_{\alpha}^{2}\right)\left(X_{\alpha}P_{k}\right)-\left(1+\rho_{\alpha}^{2}\right)^{2}\left(X_{\alpha}P_{k}\right)^{2}}\right].$
(15)

Except in the quiescent condition $P_{k}=0$, the value $Q_{k}$ given by (15)
is strictly negative. That is to say, maintaining a flat voltage profile
necessitates that there be an injection of reactive power $Q_{k}$ at the
receiving end of the branch, in the contrary direction to the active power
$P_{k}$.

Aside from confirming $Q_{k}$ is negative, (15) does not give immediate
insight into the quantity required, although it points to a breakdown in the
solution (voltage collapse) when $P_{k}$ is of order $1/X_{\alpha}$. In most
practical situations the magnitude of $P_{k}$ is a good deal smaller, and it
is helpful to approximate the square root using the Taylor expansion

$\sqrt{1-x}=1-\frac{x}{2}-\frac{x^{2}}{8}-\frac{x^{3}}{16}-O\\!\left(x^{4}\right)$
(16)

which leads to

$\sqrt{1-2\rho_{\alpha}\left(1+\rho_{\alpha}^{2}\right)\left(X_{\alpha}P_{k}\right)-\left(1+\rho_{\alpha}^{2}\right)^{2}\left(X_{\alpha}P_{k}\right)^{2}}\\\
=1-\rho_{\alpha}\left(1+\rho_{\alpha}^{2}\right)\left(X_{\alpha}P_{k}\right)-\frac{\left(1+\rho_{\alpha}^{2}\right)^{3}}{2}\left(X_{\alpha}P_{k}\right)^{2}-\frac{\rho_{\alpha}\left(1+\rho_{\alpha}^{2}\right)^{4}}{2}\left(X_{\alpha}P_{k}\right)^{3}-O\\!\left(\left(X_{\alpha}P_{k}\right)^{4}\right).$
(17)

Substituting in (15) and simplifying yields a more practical formula for
$Q_{k}$ when $X_{\alpha}P_{k}\ll 1$:

$Q_{k}=-P_{k}\left[\rho_{\alpha}+\frac{\left(1+\rho_{\alpha}^{2}\right)^{2}}{2}\left(X_{\alpha}P_{k}\right)+\frac{\rho_{\alpha}\left(1+\rho_{\alpha}^{2}\right)^{3}}{2}\left(X_{\alpha}P_{k}\right)^{2}+O\\!\left(\left(X_{\alpha}P_{k}\right)^{3}\right)\right].$
(18)

This formula reveals a qualitative difference between the reactive support
requirements in a near-lossless branch relative to one with substantial
resistance. In the lossless or near-lossless case, $\rho_{\alpha}$ is
practically zero and so the leading term vanishes; the required reactive power
is $Q_{k}\approx-(X_{\alpha}/2)P_{k}^{2}$ to leading order. On the other hand,
when $\rho_{\alpha}$ is larger the leading term dominates, so that
$Q_{k}\approx-\rho_{\alpha}P_{k}$ to leading order, and scales linearly with
active power flow rather than quadratically.

Having solved for $Q_{k}$ it is straightforward to determine the current
$I_{\alpha}$, sending-end reactive power $Q_{j}$, and the phase displacement
$\delta_{j}-\delta_{k}$. To obtain practically useful formulae for these
quantities, however, it is useful to introduce a dimensionless parameter that
can temporarily set aside purely mathematical complications.

##  4 The Coefficient of Support

The _coefficient of support_ $\sigma_{\alpha}$ is defined by (8) above:

$\sigma_{\alpha}=\frac{X_{\alpha}|I_{\alpha}|^{2}}{P_{k}}\qquad\text{(when
$P_{k}>0$, else zero)}.$

To recap, if $P_{k}$ is the power flow at the ‘receiving’ end of the branch
$\alpha$, then by definition $\Delta Q=\sigma_{\alpha}P_{k}$ is the total
reactive power consumed in the branch reactance. Reactive power is a conserved
quantity, so there must be a compensating net injection of reactive power at
the terminal buses to ‘support’ the power flow $P_{k}$.

Assuming $X_{\alpha}>0$, the reactive power requirement is always positive
when $P_{k}>0$. If $P_{k}=0$, then there is clearly no current flow on the
branch and so the reactive power consumption is also zero: it will be seen
below that in fact $\sigma_{\alpha}\rightarrow 0$ as $P_{k}\rightarrow 0$,
which motivates extending the definition (8) to set $\sigma_{\alpha}=0$ when
$P_{k}=0$.

The coefficient $\sigma_{\alpha}$ can be expressed explicitly in terms of
$P_{k}$ and the branch impedance alone, using the solution in the previous
section. The identity $V_{k}I_{\alpha}^{*}=P_{k}+jQ_{k}$ immediately gives

$|I_{\alpha}|^{2}=\frac{1}{|V_{k}|^{2}}\left(P_{k}^{2}+Q_{k}^{2}\right)=P_{k}^{2}+Q_{k}^{2}$
(19)

and combining with the condition (12) for a flat voltage profile gives

$|I_{\alpha}|^{2}=P_{k}^{2}+Q_{k}^{2}=-\frac{2}{R_{\alpha}^{2}+X_{\alpha}^{2}}\left(R_{\alpha}P_{k}+X_{\alpha}Q_{k}\right).$
(20)

(Note this implies that $Q_{k}$ is both negative and large enough that
$R_{\alpha}P_{k}+X_{\alpha}Q_{k}$ is also negative.) It follows that

$\sigma_{\alpha}=-2\frac{R_{\alpha}X_{\alpha}}{R_{\alpha}^{2}+X_{\alpha}^{2}}-2\frac{X_{\alpha}^{2}}{R_{\alpha}^{2}+X_{\alpha}^{2}}\frac{Q_{k}}{P_{k}}=\frac{2}{1+\rho_{\alpha}^{2}}\left(\frac{(-Q_{k})}{P_{k}}-\rho_{\alpha}\right),$
(21)

where $Q_{k}$ is given in terms of $P_{k}$ and the branch impedance by
equation (15). The exact formula for $\sigma_{\alpha}$ that results is
cumbersome, but when the asymptotic formula (18) for $Q_{k}$ is used instead,
a much nicer expression results:

$\sigma_{\alpha}=\left(1+\rho_{\alpha}^{2}\right)\left(X_{\alpha}P_{k}\right)+\rho_{\alpha}\left(1+\rho_{\alpha}^{2}\right)^{2}\left(X_{\alpha}P_{k}\right)^{2}+O\\!\left(\left(X_{\alpha}P_{k}\right)^{3}\right).$
(22)

It is evident that this expression for $\sigma_{\alpha}$ can in principle be
continued as a power series in the quantity $X_{\alpha}P_{k}$, the
coefficients being polynomials in $\rho_{\alpha}$. To leading order
$\sigma_{\alpha}$ is just $(X_{\alpha}+R_{\alpha}^{2}/X_{\alpha})P_{k}$, and
if the branch is lossless than $\sigma_{\alpha}$ is just $X_{\alpha}P_{k}$
with residual error of order $(X_{\alpha}P_{k})^{3}$.

The power series (22) also hints at $\sigma_{\alpha}$ being a strictly
increasing function of $P_{k}$, as intuition might suggest. This property can
be confirmed by direct calculation using (21) and (15), which reveals that

$\frac{\partial\sigma_{\alpha}}{\partial
P_{k}}=\frac{\sigma_{\alpha}}{P_{k}\sqrt{\Delta}}>0\qquad\left(P_{k}>0\right)$
(23)

where $\Delta$ is the discriminant under the square root in (15). When
$P_{k}\rightarrow 0$ one may use L’Hôpital’s rule to find

$\lim_{P_{k}\rightarrow 0}\frac{\partial\sigma_{\alpha}}{\partial
P_{k}}=\frac{2}{1+\rho_{\alpha}^{2}}\lim_{P_{k}\rightarrow
0}\left(\frac{1}{2}\frac{\partial^{2}(-Q_{k})}{\partial
P_{k}^{2}}\right)=\left(1+\rho_{\alpha}^{2}\right)X_{\alpha}>0,$ (24)

agreeing with the power series (22) and showing that $\sigma_{\alpha}$ is
strictly increasing with $P_{k}\geq 0$.

With a little more effort it may also be shown that provided $P_{k}>0$,
$\sigma_{\alpha}$ is strictly increasing with the ratio $\rho_{\alpha}$, all
else being equal (in other words, strictly increasing with $R_{\alpha}$ for
fixed $X_{\alpha}$). Starting again from (15) and some algebraic
manipulations, one may show that

$\frac{\partial(-Q_{k})}{\partial\rho_{\alpha}}=\frac{2\rho_{\alpha}}{\left(1+\rho_{\alpha}^{2}\right)\sqrt{\Delta}}\left(-Q_{k}\right)+\frac{1-\rho_{\alpha}^{2}}{1+\rho_{\alpha}^{2}}\frac{P_{k}}{\sqrt{\Delta}}=\frac{P}{\sqrt{\Delta}}\left(1+\rho_{\alpha}\sigma_{\alpha}\right),$
(25)

from which it can be shown using (21) that

$\frac{\partial\sigma_{\alpha}}{\partial\rho_{\alpha}}=\frac{2X_{\alpha}}{\sqrt{\Delta}}\left(1+\rho_{\alpha}\sigma_{\alpha}\right)\left(-Q_{k}\right)>0\qquad\text{when
}P_{k}>0.$ (26)

Of course when $P_{k}=0$, then $\sigma_{\alpha}=0$ has no dependence on
$\rho_{\alpha}$.

It should likewise come as little surprise that $\sigma_{\alpha}$ is strictly
increasing with $X_{\alpha}$, when all other parameters are held constant
(including the ratio $\rho_{\alpha}$). Direct calculation using (21) and (15)
confirms that indeed

$\frac{\partial\sigma_{\alpha}}{\partial
X_{\alpha}}=\frac{\sigma_{\alpha}}{X_{\alpha}\sqrt{\Delta}}>0\qquad\text{when
}P_{k}>0.$ (27)

As a consequence of all the above, it may be seen that if two of the three
parameters $P_{k}$, $X_{\alpha}$ and $\rho_{\alpha}$ are fixed, there is a
one-to-one mapping between $\sigma_{\alpha}$ and the third parameter in the
set, such that the two variables increase or decrease together.

The coefficient of support yields simple exact formulae for all quantities
associated with the branch power flow solution, given $P_{k}$ and the branch
impedance, as follows.

Branch current: $\displaystyle|I_{\alpha}|$
$\displaystyle=\sqrt{\frac{\sigma_{\alpha}P_{k}}{X_{\alpha}}}$ (28) Power
losses: $\displaystyle P_{j}-P_{k}=R_{\alpha}|I_{\alpha}|^{2}$
$\displaystyle=\rho_{\alpha}\sigma_{\alpha}P_{k}$ (29) Receiving end Q:
$\displaystyle Q_{k}$
$\displaystyle=\left(-\frac{1+\rho_{\alpha}^{2}}{2}\sigma_{\alpha}-\rho_{\alpha}\right)P_{k}$
(30) Sending end Q: $\displaystyle Q_{j}=Q_{k}+\sigma_{\alpha}P_{k}$
$\displaystyle=\left(\phantom{-}\frac{1-\rho_{\alpha}^{2}}{2}\sigma_{\alpha}-\rho_{\alpha}\right)P_{k}$
(31) Sending end P: $\displaystyle P_{j}$
$\displaystyle=\left(1+\rho_{\alpha}\sigma_{\alpha}\right)P_{k}$ (32) Flow
coefficient: $\displaystyle\mu_{\alpha}=X_{\alpha}P_{k}-R_{\alpha}Q_{k}$
$\displaystyle=X_{\alpha}P_{k}\left(1+\rho_{\alpha}^{2}\right)\left(1+\frac{\rho_{\alpha}\sigma_{\alpha}}{2}\right)$
(33) Phase shift: $\displaystyle\delta_{j}-\delta_{k}$
$\displaystyle=\arcsin\mu_{\alpha}.$ (34)

Formulae (30) and (31), specifically, indicate how reactive power support must
be configured at the two ends of the branch in order to maintain the active
power flow $P_{k}$ under a flat voltage profile. For a lossless branch, this
amounts to a symmetric injection $Q=(\sigma_{\alpha}/2)P$ at both ends. With
nonzero resistance, there must be an additional $Q$ flow superimposed on this
symmetric injection, contrary to the direction of $P$ flow and with magnitude
$(\rho_{\alpha}+\rho_{\alpha}^{2}\sigma_{\alpha}/2)P_{k}$.

Based on the fact observed above that $\sigma_{\alpha}$ is strictly increasing
with $P_{k}\geq 0$, one may also confirm that the current $|I_{\alpha}|$, the
power losses, the flow coefficient and the phase shift are also all strictly
increasing functions of $P_{k}\geq 0$.

There is of course a limit to the power flow $P_{k}$ that can be sustained in
this manner: as $P_{k}$ increases, the discriminant $\Delta$ under the square
root in (15) decreases until eventually it reaches zero, beyond which point
there is no solution for $Q_{k}$. Equating $\Delta$ to zero reveals that the
limiting values for $P_{k}$ and $Q_{k}$ are

$P_{k}=\frac{1}{X_{\alpha}}\cdot\frac{\sqrt{1+\rho_{\alpha}^{2}}-\rho_{\alpha}}{1+\rho_{\alpha}^{2}},\qquad\text{and}\qquad
Q_{k}=-\frac{1}{X_{\alpha}}\cdot\frac{1}{1+\rho_{\alpha}^{2}}.$ (35)

Formulae (21) and (33) and a little algebra than show that under the same
conditions

$\sigma_{\alpha}=\frac{2}{\sqrt{1+\rho_{\alpha}^{2}}}\qquad\text{and}\qquad\mu_{\alpha}=\frac{1}{\sqrt{1+\rho_{\alpha}^{2}}}.$
(36)

These identities have notable practical implications for the limiting power
flow condition: first, that at the limit the reactive power consumption on the
branch can be up to twice the active power flow; and second, that the phase
shift in bus voltages at the limit is equal to the ‘impedance angle’

$\phi=\arcsin\frac{X_{\alpha}}{\sqrt{R_{\alpha}^{2}+X_{\alpha}^{2}}}.$ (37)

In particular, if $R_{\alpha}>0$ then the maximum phase shift is strictly less
than $\pi/2$. When $R_{\alpha}=0$ on the other hand one has
$P_{k}=-Q_{k}=1/X_{\alpha}$, $\sigma_{\alpha}=2$, $\mu_{\alpha}=1$ and the
phase shift is $\pi/2$ at the limit.

##  5 The power–angle relation with flat voltages

As noted at the outset, the ‘DC power flow’ approach to network solution
hinges on the relation (1) between phase displacement and active power flow in
a lossless branch:

$\sin\left(\delta_{j}-\delta_{k}\right)=X_{\alpha}P_{\alpha}.\qquad\left(|V_{j}|=|V_{k}|=1,R_{\alpha}=0\right)$

It was seen above that after factoring in resistance, this must be modified to
equation (33):

$\sin\left(\delta_{j}-\delta_{k}\right)=\mu_{\alpha}=X_{\alpha}P_{k}\left(1+\rho_{\alpha}^{2}\right)\left(1+\frac{\rho_{\alpha}\sigma_{\alpha}}{2}\right),$

where $P_{k}\leq P_{j}$ is the receiving end power, $\rho_{\alpha}$ is the
$R$-to-$X$ ratio, and the coefficient of support $\sigma_{\alpha}$ increases
monotonically with any one of $(P_{k},X_{\alpha},\rho_{\alpha})$ when the
others are held fixed.

The monotonicity properties of $\sigma_{\alpha}$ assist greatly in
understanding the _qualitative_ behaviour of the relation (33) between power
flow and angle displacement. To begin with, it asserts that all else being
equal, an increase (reduction) in any of the parameters $P_{k}$, $X_{\alpha}$
or $R_{\alpha}$ will also increase (reduce) $\mu_{\alpha}$ and, consequently,
the angle displacement. Importantly, it also asserts that if the value of
$\mu_{\alpha}$ is fixed, then an increase (reduction) in one of these same
parameters requires a corresponding reduction (increase) in one of the others
to keep $\mu_{\alpha}$ constant.

Equation (33) is however of limited assistance in obtaining _quantitative_
solutions for $\mu_{\alpha}$ or (especially) $P_{k}$. This requires working
with an explicit formula for $\sigma_{\alpha}$ in terms of
$(P_{k},X_{\alpha},\rho_{\alpha})$, which can be obtained using (21) from the
previous section and substituting (15) for $Q_{k}$. When the resulting formula
for $\sigma_{\alpha}$ is used in (33), the result is

$\mu_{\alpha}=X_{\alpha}P_{k}+\frac{\rho_{\alpha}}{1+\rho_{\alpha}^{2}}\left(1-\sqrt{1-2\rho_{\alpha}\left(1+\rho_{\alpha}^{2}\right)\left(X_{\alpha}P_{k}\right)-\left(1+\rho_{\alpha}^{2}\right)^{2}\left(X_{\alpha}P_{k}\right)^{2}}\right).$
(38)

This formula remains cumbersome, but simplifies substantially when rearranged
to clear the square root. A subtle point emerges when doing so however:
observe that the nonnegativity of the square root in (38) implies that
$\mu_{\alpha}$ and $P_{k}$ must satisfy the inequality

$\mu_{\alpha}-X_{\alpha}P_{k}\leq\frac{\rho_{\alpha}}{1+\rho_{\alpha}^{2}}.$
(39)

(This inequality ultimately stems from having excluded at the outset the
‘impractical’ solution for the reactive power $Q_{k}$.) Since (33) already
implies that $\mu_{\alpha}\geq X_{\alpha}P_{k}$, one now has two-sided bounds
for the difference between $\mu_{\alpha}$ and $X_{\alpha}P_{k}$ that arises
when $\rho_{\alpha}>0$. Furthermore, given that the quantity under the square
root in (38) is the same discriminant $\Delta$ from (15), one can readily
characterise the conditions leading to these bounds: the difference
$\mu_{\alpha}-X_{\alpha}P_{k}$ is a minimum (zero) for a lossless branch, and
is a maximum (39) in the limiting flow condition given by (35) and (36).

Isolating the square root in (38), squaring and simplifying now yields a
quadratic equation linking $P_{k}$ and $\mu_{\alpha}$:

$\left(1+\rho_{\alpha}^{2}\right)\mu_{\alpha}^{2}-2\rho_{\alpha}\mu_{\alpha}+2\left(\rho_{\alpha}-\mu_{\alpha}\right)\left(1+\rho_{\alpha}^{2}\right)\left(X_{\alpha}P_{k}\right)+\left(1+\rho_{\alpha}^{2}\right)^{2}\left(X_{\alpha}P_{k}\right)^{2}=0.$
(40)

When this is rewritten to complete the square in $P_{k}$ as

$\left[\left(1+\rho_{\alpha}^{2}\right)\left(X_{\alpha}P_{k}\right)+\left(\rho_{\alpha}-\mu_{\alpha}\right)\right]^{2}+\left(\mu_{\alpha}^{2}-1\right)\rho_{\alpha}^{2}=0,$
(41)

it can be solved directly for $P_{k}$ given $\mu_{\alpha}$. This would
ordinarily yield two solutions for $P_{k}$, but one may verify that only one
of these actually satisfies the inequality (39)—the other solution enters when
squaring both sides to clear the square root in (38), and does not actually
satisfy (38) itself. Accordingly, one has the unique solution for $P_{k}$
given $\mu_{\alpha}$

$P_{k}=\frac{1/X_{\alpha}}{1+\rho_{\alpha}^{2}}\left[\mu_{\alpha}-\rho_{\alpha}\left(1-\sqrt{1-\mu_{\alpha}^{2}}\right)\right].$
(42)

Note first that for $\rho_{\alpha}=0$ this reproduces the lossless solution
(1), in the form $X_{\alpha}P_{k}=\mu_{\alpha}$. One may also verify that when
$\mu_{\alpha}$ takes its limiting value $1/\sqrt{1+\rho_{\alpha}^{2}}$,
formula (42) reduces to formula (35) for $P_{k}$ as it should. Finally, it may
be verified that

$\frac{\partial
P_{k}}{\partial\mu_{\alpha}}=\frac{1/X_{\alpha}}{1+\rho_{\alpha}^{2}}\left(1-\frac{\rho_{\alpha}\mu_{\alpha}}{\sqrt{1-\mu_{\alpha}^{2}}}\right),$
(43)

and this is strictly positive for $\mu_{\alpha}<1/\sqrt{1+\rho_{\alpha}^{2}}$,
decreasing to zero when $\mu_{\alpha}$ reaches this limiting value. This
confirms that $P_{k}$ and $\mu_{\alpha}$ always increase together when the
branch impedance is held constant, hence that the $P_{k}$ value (35) at the
flow limit is also the largest active power transfer that can be achieved on a
branch of fixed impedance, subject to the flat voltage profile.

##  6 Circulating flows and winding numbers in large networks

The preceding analysis for a single network branch can be applied to each of
the branches making up a wider network, with varying power flows $P_{k}$ on
each branch and injections of active and reactive power at the network buses.
Equation (33) and its converse (42) provide the necessary linkage between each
branch power flow, the impedance, and the displacement between the terminal
bus voltage phase angles. In particular, they show that when branch resistance
is taken into account, the angle displacement is always larger than suggested
by the formula $\sin(\delta_{j}-\delta_{k})=X_{\alpha}P_{k}$ for lossless
branches. The difference between $\mu_{\alpha}=\sin(\delta_{j}-\delta_{k})$
and $X_{\alpha}P_{k}$ ranges from zero (when the branch is lossless) up to the
limit given by equation (39) in terms of the $R/X$ ratio $\rho_{\alpha}$.

The most immediate application is to a network with ‘string’ topology, where
successive branches are connected end-to-end, and the sending power $P_{j}$ on
one branch is the sum of receiving power $P_{k}$ for the previous branch and
any additional power injected at the intermediate bus. Assuming each
$P_{k}>0$, the voltage phase angles $\delta_{k}$ at successive busbars form a
strictly decreasing sequence, the lag being determined at each stage by the
power flow $P_{k}$ and branch impedance. Reactive power injection is also
required at each intermediate busbar in order to meet the support requirement
$\Delta Q=\sigma_{\alpha}P_{k}>0$ on each branch and maintain the flat voltage
profile.

It turns out to be possible to ‘close the loop’ on such a network, identifying
an upstream bus with a remote downstream bus to form a ring network with a
positive circulating power flow. Let such a ring network comprise $n$ buses
(numbered $0$ through $n-1$) and $n$ branches (numbered $1$ through $n$), with
branch $k$ linking buses $k-1$ and $k$, and bus indices $n$ and $0$
identified. For $1\leq k\leq n$ it is assumed that $P_{k}$, the receiving-end
power on the $k$th branch, is strictly positive. Then one may identify the
_circulating power_ $P_{\circ}$ as the minimum of $P_{k}$ over all indices
$k$, since the measured active power flow is uniform in direction and no less
than $P_{\circ}$ at any point in the ring.

Of course, since $\delta_{k-1}-\delta_{k}>0$ for all $k$ when $P_{\circ}>0$ it
is not possible for $\delta_{0}$ and $\delta_{n}$ to be identical as real
numbers, but this is not necessary: as they represent angle variables it is
only necessary that they differ by an integer multiple of $2\pi$ [2]. That is
to say,

$\sum_{k=1}^{n}\left(\delta_{k-1}-\delta_{k}\right)=\sum_{\alpha=1}^{n}\arcsin\mu_{\alpha}=2m\pi,\qquad
m\in\mathbb{Z}_{+}.$ (44)

The integer $m$ is termed the _winding number_ in the literature, and can be
defined for any cycle of branches in networks with node states characterised
by angles, on the understanding that the differences
$(\delta_{k-1}-\delta_{k})$ in (44) are always restricted to the interval
$(-\pi,\pi]$ [1]. As seen above, a nonzero winding number is necessary for
circulating power to exist, and so a winding number of zero for all cycles in
the network certifies the absence of circulating power flow.

Circulating power flows can in principle exist in any ring network of
sufficient size, with (in general) non-uniform impedances
$R_{\alpha}+jX_{\alpha}$ and nonuniform power flows, subject only to there
being net power injections at each busbar sufficient to maintain the reactive
power consumption $\sigma_{\alpha}P_{k}$ and active power losses
$\rho_{\alpha}\sigma_{\alpha}P_{k}$ on every branch, for some $P_{k}\geq
P_{\circ}$. However, the ‘homogenous’ case with uniform impedances and power
flows is worth considering in further detail as a pointer to the behaviour in
the general case.

Suppose then that in a ring network of $n$ branches, every branch has the same
impedance $R+jX$ and the same power flow $P=P_{\circ}$ at the receiving end,
directed uniformly. (Subscripts are now dropped given all branches are
identical.) To relate this to real-world applications it is useful to recall
that these quantities are expressed in a per-unit system, so that if
$R_{\mathrm{branch}}$, $X_{\mathrm{branch}}$ and $P_{\mathrm{branch}}$ are the
equivalent branch quantities in SI units and $V_{\mathrm{nom}}$ is the nominal
voltage at each bus, then

$R=\frac{R_{\mathrm{branch}}}{Z_{\mathrm{base}}},\quad
X=\frac{X_{\mathrm{branch}}}{Z_{\mathrm{base}}},\quad
P=\frac{P_{\mathrm{branch}}}{S_{\mathrm{base}}}\qquad\text{where}\qquad
Z_{\mathrm{base}}=\frac{V_{\mathrm{nom}}^{2}}{S_{\mathrm{base}}},$ (45)

with $S_{\mathrm{base}}$ chosen arbitrarily as a convenient base scale for
power flows (often taken as 100MVA in large transmission systems).

Since each branch has the same impedance and power flow, each also has the
same coefficient of support $\sigma$ and the same flow coefficient $\mu$. The
winding number condition (44) therefore becomes

$\mu=\sin\left(\frac{2m}{n}\pi\right),\qquad m\in\mathbb{Z}_{+},\qquad 1\leq
m\leq\left\lfloor\frac{n}{4}\right\rfloor,$ (46)

where the restriction on $m$ is to ensure $\delta_{j}-\delta_{k}\leq\pi/2$ for
each individual branch, and implies in particular that there are at least 4
branches making up the ring. (In the context of steady state power flows,
angle differences greater than $\pi/2$ across a single branch are not
practical, as they invert the increasing relationship between active power and
angle difference that underpins system stability, and also, much like the
second solution to (14), lead to impractical reactive power flows.)

Formula (42) then implies a unique circulating power

$P_{\circ}=\frac{1/X}{1+\rho^{2}}\left[\sin\left(\frac{2m}{n}\pi\right)-\rho\left(1-\cos\left(\frac{2m}{n}\pi\right)\right)\right],$
(47)

for each admissible winding number $m$, given the impedance of a single branch
and the number $n$ of branches. This is clearest in the lossless case $R=0$,
where the solutions are

$P_{\circ}=\frac{1}{X}\sin\left(\frac{2m}{n}\pi\right)\qquad\text{for}\qquad
1\leq m\leq\left\lfloor\frac{n}{4}\right\rfloor.$ (48)

The practice is adopted here and below (as in both (35) and (42) above) of
expressing power quantities as a multiple of $1/X$, which provides a
convenient scale for circulating flows. Note from above that using a scale of
$1/X$ for per-unit power is equivalent to using a scale of
$V_{\mathrm{nom}}^{2}/X_{\mathrm{branch}}$ for power in engineering units
(watts or VA).

For a more quantitative analysis, consider first the limiting values of
$\rho=R/X$ and $P_{\circ}$ for fixed values of $n$ and $X$. Recall that from
(36) the maximum feasible value of $\mu$ is $1/\sqrt{1+\rho^{2}}$. Combined
with taking $m=1$ in (46) to get the _smallest_ feasible value of $\mu$ for
fixed $n$, this prescribes an _upper_ bound on $\rho$ to achieve any
circulating flow in a ring of $n$ branches. This upper bound $\rho_{\max}$ is
tabulated in Table 1 along with the $P_{\circ}$ value at $\rho_{\max}$ from
either (35) or (47), the reactive power consumption $\Delta Q=\sigma
P_{\circ}$ (where $\sigma=2\mu$ by (36)), and the active power losses on each
branch.

Ring | Largest | Limiting flow |  $Q$ consumption |  $P$ losses
---|---|---|---|---
size | feasible | $P_{\circ}(\rho_{\max})$ | per branch $\sigma P_{\circ}$ | per branch $\rho\sigma P_{\circ}$
$n$ |  $R/X$ ratio | (all in units of $V_{\mathrm{nom}}^{2}/X_{\mathrm{branch}}$)
4 | 0 | 1 | 2 | 0
5 | 0.3249 | 0.6572 | 1.25 | 0.4061
6 | 0.5774 | 0.4330 | 0.75 | 0.4330
7 | 0.7975 | 0.2944 | 0.4603 | 0.3671
8 | 1 | 0.2071 | 0.2929 | 0.2929
9 | 1.1918 | 0.1504 | 0.1933 | 0.2304
10 | 1.3764 | 0.1123 | 0.1320 | 0.1817
Table 1: Limits on $R/X$ and circulating power for ring networks with small
numbers of branches.

It should be stressed the values in Table 1 only pertain to the (normalised)
limiting flow at this largest feasible $R/X$ ratio $\rho_{\max}$. So it should
not be too surprising that for this extreme limiting flow, the $Q$ consumption
on each branch is often larger than the actual active power flow
$P_{\circ}$—indeed by (36) the coefficient of support $\sigma$ at the limit
exceeds 1 whenever $\rho<\sqrt{3}$. Likewise, for networks of 7 branches or
more having the largest possible resistance in accordance with $\rho_{\max}$,
the $P$ losses on each branch actually exceed $P_{\circ}$. (The threshold case
is that of 6 branches, where $\mu=\sqrt{3}/2$, $\rho_{\max}=1/\sqrt{3}$ and
$\sigma=\sqrt{3}$, so $\rho\sigma=1$ and the losses are the same as
$P_{\circ}$, both $\sqrt{3}/4$ in per-unit.)

If the actual $R/X$ ratio is smaller than $\rho_{\max}$, flows with larger
circulating power $P_{\circ}$ are possible. So for example when $n=7$, a flow
with winding number $m=1$ must have $\mu=\sin(2\pi/7)=0.7818$ irrespective of
the ratio $\rho=R/X$, but any value $\rho\leq 0.7975$ yields feasible
solutions $P_{\circ}$ given by equation (42), right down to $\rho=0$ where the
solution is $P_{\circ}=\mu/X=0.7818/X$. It is readily seen from (42) that
$XP_{\circ}$ increases monotonically with decreasing $\rho$ when $\mu$ is
fixed, from $0.2944$ at $\rho=\rho_{\max}$ to $0.7818$ at $\rho=0$. This also
follows from equation (33) due to the monotonicity properties of $\sigma$
found earlier.

It follows that for a fixed reactance $X$ per branch and a fixed number of
branches $n$, the values $P_{\circ}(\rho_{\max})$ in Table 1 are in fact the
_minimum_ feasible circulating flows, corresponding to the _maximum_ feasible
network resistances. Conversely, the _maximum_ feasible circulating flows with
a fixed $X$ and $n$ (and winding number 1) are the values given by (48) with
$m=1$ and $R=\rho=0$. For values intermediate between these extremes, there
will be a ratio $\rho=R/X\in(0,\rho_{\max})$ that achieves this value of
circulating power with a winding number of 1.

Larger winding numbers are achievable when $n$ is sufficiently large, as
suggested by (46). Each winding number generates a fixed value of $\mu$ by
(46), which when equated to $1/\sqrt{1+\rho^{2}}$ yields an equivalent
$\rho_{\max}$ value. A continuum of feasible circulating flows $P_{\circ}$
again results, corresponding to values of $\rho$ between zero and
$\rho_{\max}$. The endpoints of the feasible continuum for $P_{\circ}$ are
given by (48) for $\rho=0$ and by (35) or (47) for $\rho=\rho_{\max}$.

One may further note, as already found above, that whenever $R>0$ the
circulating power $P_{\circ}$ is always accompanied by a circulating $Q$ flow
in the opposite direction, of magnitude $(\rho+\rho^{2}\sigma/2)P_{\circ}$.
This circulating $Q$ flow is additional to the consumption $\Delta Q=\sigma
P_{\circ}$ on each branch, which must be supplied as a $Q$ injection at each
intervening busbar.

##  7 Conclusion

Working through the exact AC power flow solution with a ‘flat’ voltage profile
provides important insights into the relationship between active and reactive
power flows, reactive compensation requirements and phase angle differences
when the resistance of network branches is taken into account. Through use of
the ‘coefficient of support’ associated with the power flow on each network
branch, the inevitable mathematical complications can be de-emphasised without
sacrificing precision.

The simple relation $\sin(\delta_{j}-\delta_{k})=XP$ for a lossless network
branch is generalised by recognising that with a flat voltage profile, the
correct value on the right hand side of this relation is the flow coefficient
$\mu=XP-RQ$. Solving for $Q$ from first principles, it is found that a
positive injection of reactive power is always required to sustain a flat
voltage profile with $|V|=1$pu at both ends of a network branch with positive
reactance $X$. The required injection is readily quantifiable in terms of the
coefficient of support $\sigma=X|I|^{2}/P$. For branches with positive
resistance one always has $\sin(\delta_{j}-\delta_{k})>XP$, and bounds for the
difference $\sin(\delta_{j}-\delta_{k})-XP$ are readily derived.

The generalised analysis has a ready application to the study of circulating
power flows that can arise in large ring systems. For the case of a uniform
ring with a circulating flow, the analysis provides bounds on the maximum
allowable network resistance and the minimum attainable power flow in terms of
the other configuration parameters. It also reveals that when resistance is
taken into account, a real circulating power flow is always accompanied by a
contrary flow of reactive power.

## References

  * [1] S. Jafarpour, E.Y. Huang, K.D. Smith, and F. Bullo.  Flow and elastic networks on the $n$-torus: Geometry, analysis and computation.  SIAM Review, 64(1):59–104, February 2022.
  * [2] N. Janssens and A. Kamagate.  Loop flows in a ring AC power system.  Electrical Power and Energy Systems, 25:591–597, 2003.