Power Flow Tracing in Complex Networks

Abstract

The increasing share of decentralized renewable power generation represents a challenge to the current and future energy system. Providing a geographical smoothing effect, long-range power transmission plays a key role for the system integration of these fluctuating resources. However, the build-up and operation of the necessary network infrastructure incur costs which have to be allocated to the users of the system. Flow tracing techniques, which attribute the power flow on a transmission line to the geographical location of its generation and consumption, represent a valuable tool set to design fair usage and thus cost allocation schemes for transmission investments. In this article, we introduce a general formulation of the flow tracing method and apply it to a simplified model of a highly renewable European electricity system. We review a statistical usage measure which allows to integrate network usage information for longer time series, and illustrate this measure using an analytical test case.

Appendix

Consider a network consisting of three connected nodes with

$$\begin{aligned} P_{1}&= P_{1}^{+}> P_{2} = P_{2}^{+} > 0~,\nonumber \\ P_{3}&= -P_{3}^{-} < 0~, \end{aligned}$$

(40)

and unit reactance for all links. It holds $P_{1}^{+}+P_{2}^{+}=P_{3}^{-}$. The power flows are given by

$$\begin{aligned} F_{1\rightarrow 2}&= \frac{1}{3}\left( P_{1}^{+} - P_{2}^{+}\right) ~,\nonumber \\ F_{1\rightarrow 3}&= \frac{1}{3}\left( P_{1}^{+} + P_{3}^{-}\right) ~,\nonumber \\ F_{2\rightarrow 3}&= \frac{1}{3}\left( P_{2}^{+} + P_{3}^{-}\right) ~.\nonumber \\ \end{aligned}$$

(41)

For some source colouring $\{q_{1\alpha }^{+},q_{2\alpha }^{+}\}$ we obtain via the flow tracing algorithm

$$\begin{aligned} F_{1\rightarrow 2}^{\alpha }&= q_{1\alpha }^{+}\frac{1}{3}\left( P_{1}^{+}-P_{2}^{+}\right) ~,\nonumber \\ F_{1\rightarrow 3}^{\alpha }&=q_{1\alpha }^{+}\frac{1}{3}\left( P_{1}^{+} + P_{3}^{-}\right) ~,\nonumber \\ F_{2\rightarrow 3}^{\alpha }&=q_{2\alpha }F_{2\rightarrow 3} = q_{2\alpha }^{+}P_{2}^{+} + q_{1\alpha }^{+}F_{1\rightarrow 2} = q_{2\alpha }^{+}P_{2}^{+} + q_{1\alpha }^{+}\frac{1}{3}\left( P_{1}^{+}-P_{2}^{+}\right) \nonumber \\&= \frac{1}{3}\left( \left( 3q_{2\alpha }^{+}-2q_{1\alpha }^{+}\right) P_{2}^{+}+q_{1\alpha }^{+}P_{3}^{-}\right) ~. \end{aligned}$$

(42)

Here we have used $P_{1}^{+}+P_{2}^{+}=P_{3}^{-}$. For the sink it holds

$$\begin{aligned} q_{3\alpha }P_{3}^{-}&= F_{1\rightarrow 3}^{\alpha } + F_{2\rightarrow 3}^{\alpha }\nonumber \\&= q_{1\alpha }^{+}\frac{1}{3}\left( P_{1}^{+} + P_{3}^{-}\right) +q_{2\alpha }^{+}P_{2}^{+} + q_{\alpha 1}^{+}\frac{1}{3}\left( P_{1}^{+}-P_{2}^{+}\right) \nonumber \\&= q_{1\alpha }^{+}P_{1}^{+} + q_{2\alpha }^{+}P_{2}^{+}~. \end{aligned}$$

(43)

Since there is only one sink in this network this is a necessary result.

As an analytical test case for the link ownership measure we consider this three-node example with the choice

$$\begin{aligned} P_{1}^{+}(x)&= x~,\nonumber \\ P_{2}^{+}(x)&=1-x~,\nonumber \\ P_{3}^{-}(x)&=P_{1}^{+}(x) + P_{2}^{+}(x) =1~. \end{aligned}$$

(44)

Here x is a random variable drawn from the uniform distribution on $[0,\sigma ]$, with $\sigma \ge 0.5$ as a parameter. The flow on the link between node 1 and 2 is given by

$$\begin{aligned} F_{1\rightarrow 2} = \left\{ \begin{array}{lrccl}0 \quad &{} 0 &{}\le &{} x &{} \le 0.5\\ \frac{1}{3}\left( P_{1}^{+}-P_{2}^{+} \right) = \frac{1}{3}\left( 2x-1 \right) \quad &{} 0.5 &{}< &{}x &{}\le \sigma \end{array}\right. ~, \end{aligned}$$

(45)

and

$$\begin{aligned} F_{2\rightarrow 1} = \left\{ \begin{array}{lrccl} \frac{1}{3}\left( P_{2}^{+}-P_{1}^{+} \right) = \frac{1}{3}\left( 1-2x \right) \quad &{} 0 &{}\le &{} x &{} \le 0.5\\ 0 \quad &{} 0.5 &{}< &{}x &{}\le \sigma \end{array}\right. ~. \end{aligned}$$

(46)

The probability distribution of x is given by

$$\begin{aligned} p(x) = \left\{ \begin{array}{lrccl} 0 \quad &{} x &{} < &{} 0\\ \frac{1}{\sigma } \quad &{} 0 &{}\le &{} x &{} \le \sigma \\ 0 \quad &{} x &{} > &{} \sigma \end{array}\right. ~. \end{aligned}$$

(47)

The undirected flow $f=F_{1\rightarrow 2}+F_{2\rightarrow 1}$ is a random variable $f=g(x)$, with the function g(x) determined by Eqs. (46) and (47). The distribution function P(f) of f is given by

$$\begin{aligned} P(f) = \int _{0}^{f}\mathrm{{d}}f' p(f') = \int _{g^{-1}[0,f]}p(x)\mathrm{{d}}x~. \end{aligned}$$

(48)

Note that as long the random variable x is in the interval [0, 0.5), there is only a flow $F_{2\rightarrow 1}$, with the maximum value $F_{2\rightarrow 1}=1/3$ for $x=0$. If x is larger than 0.5, we obtain a flow $F_{1\rightarrow 2}$ with a maximum value

$$\begin{aligned} f_{\sigma }=\frac{1}{3}\left( 2\sigma -1\right) ~. \end{aligned}$$

(49)

It is easy to see that

$$\begin{aligned} g^{-1}[0,f] = \left\{ \begin{array}{lrcl} \left[ x(f),1-x(f)\right] \quad &{} f &{} < &{} f_{\sigma }\\ \left[ x(f),\sigma \right] \quad &{} f &{}\ge &{} f_{\sigma } \end{array}\right. ~, \end{aligned}$$

(50)

where

$$\begin{aligned} x(f)=\frac{1}{2}\left( 1-3f\right) \end{aligned}$$

(51)

is the inverse of g(x) defined on the interval [0, 0.5). We thus obtain

$$\begin{aligned} P(f) = \frac{1}{\sigma }\left\{ \begin{array}{lrcl} 1-2x(f) \quad &{} f &{} \le &{} f_{\sigma }\\ \sigma -x(f) \quad &{} f &{} > &{} f_{\sigma } \end{array}\right. ~, \end{aligned}$$

(52)

and with $p(f)=P'(f)$

$$\begin{aligned} p(f) = \left\{ \begin{array}{lrcl} \frac{3}{\sigma } \quad &{} f &{} \le &{} f_{\sigma }\\ \frac{3}{2\sigma } \quad &{} f &{}> &{} f_{\sigma } \end{array}\right. ~, \end{aligned}$$

(53)

where we have used the specific form of x(f). Another way to calculate p(f) without the detour using the distribution function is via

$$\begin{aligned} p(f)=\int dx \,\delta \left[ g(x)-f\right] p(x)~, \end{aligned}$$

(54)

which yields the same result as above.

For the flow tracing we use nodal colouring, that is $\alpha \in \{1,2\}$ and $q_{i\alpha }=\delta _{i\alpha }$ . For the determination of the link ownership measure we observe that for $f\le f_{\sigma }$ the flow over the link between node 1 and 2 goes in either direction with equal probability, whereas for $f> f_{\sigma }$ the flow always goes from node 2 to node 1. We thus have

$$\begin{aligned} h_{1}(f) = \left\{ \begin{array}{lrcl} \frac{1}{2} \quad &{} f &{} \le &{} f_{\sigma }\\ 0 \quad &{} f &{}> &{} f_{\sigma } \end{array}\right. \end{aligned}$$

(55)

and

$$\begin{aligned} h_{2}(f) = \left\{ \begin{array}{lrcl} \frac{1}{2} \quad &{} f &{} \le &{} f_{\sigma }\\ 1 \quad &{} f &{}> &{} f_{\sigma } \end{array}\right. ~. \end{aligned}$$

(56)

The weight $w_{1}(K)$ on this link is then given by

$$\begin{aligned} w_{1}(K)&=\frac{1}{1-P(K)}\int _{K}^{F_{\max }}\mathrm{{d}}f\,p(f)h_{1}(f)\nonumber \\&= \frac{\sigma }{2x(K)-(1-\sigma )}\int _{K}^{F_{\sigma }}\mathrm{{d}}f\,\frac{3}{\sigma }\cdot \frac{1}{2}\nonumber \\&= \frac{3\left( F_{\sigma }-K\right) }{2\left( 2x(K)-(1-\sigma )\right) }~. \end{aligned}$$

(57)

Here we used $h_{1}(f>f_{q})=0$. Substituting $f_{\sigma }$ and some rearranging yields

$$\begin{aligned} w_{1}(K)=1-\frac{x(K)}{2x(K)-(1-\sigma )}~. \end{aligned}$$

(58)

The part $K_{1}$ of the capacity $F_{\max }=1/3$ used by node 1 then finally is obtained via the integral

$$\begin{aligned} K_{1}=\int _{0}^{F_{\max }}\mathrm{{d}}K\,w_{1}(K)=\int _{0}^{f_{\sigma }}\mathrm{{d}}K\left( 1-\frac{x(K)}{2x(K)-(1-\sigma )}\right) ~. \end{aligned}$$

(59)

This can be rewritten as

$$\begin{aligned} K_{1}=f_{\sigma }-\frac{2}{3}\int _{1-\sigma }^{1/2}\mathrm{{d}}x\,\frac{x}{2x-(1-\sigma )}~, \end{aligned}$$

(60)

which can be solved as

$$\begin{aligned} K_{1}=\frac{1}{6}\left[ 2\sigma -1+\left( 1-\sigma \right) \ln \left[ \frac{1-\sigma }{\sigma }\right] \right) ~. \end{aligned}$$

(61)

Note that this gives $K_{1}=0$ for $\sigma =0$, and $K_{1}=1/6=F_{\max }/2$ for $\sigma =1$.