Optimal Control and Scheduling of Distribution Stations

Abstract

The electricity generated by the power plant is sent to the distribution station, and then transmitted to factories, mines and households through the public power grid. An enterprise power distribution station may also have many transformers, and the number of transformers used and power distribution are adjusted to transmit power to each workshop. The electricity consumed by power transmission and distribution is very considerable, and a large number of scientists, scholars, students and engineers are engaged in research in this field. This chapter will discuss this issue.

The electricity generated by the power plant is sent to the distribution station, and then transmitted to factories, mines and households through the public power grid. An enterprise power distribution station may also have many transformers, and the number of transformers used and power distribution are adjusted to transmit power to each workshop (Fig. 17.1). The electricity consumed by power transmission and distribution is very considerable, and a large number of scientists, scholars, students and engineers are engaged in research in this field. This chapter will discuss this issue.

Fig. 17.1

Power transmission and distribution

17.1 Energy Relations for Multiple Transformers in a Distribution Station

Assume that there are n transformers in a distribution station to supply power to all devices in a factory area, the total power output of the station is P₀, assuming that P₀ is a fixed value, the power output by the i-th transformer is P_i, and all operating transformers consume the total electric energy is P_t, P_t is the total input of the station, and the expression of P_t is

$${P}_{t}=\sum\limits_{i=1}^{n}\frac{{P}_{i}}{{\eta }_{i}({P}_{i})}$$

(17.1)

$${P}_{0}=\sum\limits_{i=1}^{n}{P}_{i}$$

The unit of measurement for electric energy is watts.

Assume the energy efficiency curve of the i-th transformer is shown in Fig. 17.2.

Fig. 17.2

The energy efficiency curve of the i-th transformer

In Fig. 17.2, P_im is the maximum power output of the i-th transformer, η_ie is the maximum energy efficiency of the i-th transformer, Pie is the power output when the i-th transformer has the maximum energy efficiency, and η_i(P) has the following characteristics:

$$ \begin{array}{*{20}c} {0 \le {\text{P}} \le P_{im} { }} \\ {\eta_{ie} = \eta_{i} \left( {P_{ie} } \right)} \\ {\begin{array}{*{20}c} {0 \le \eta_{i} \left( P \right) \le \eta_{ie} } \\ {\begin{array}{*{20}c} {\eta_{i} \left( 0 \right) = 0} \\ {\eta^{\prime\prime}_{i} \left( P \right) < 0} \\ \end{array} { }} \\ \end{array} } \\ \end{array} $$

(17.2)

Assume the energy efficiency functions η₁(P) and η_i(P) of the first and the i-th transformer as shown in Fig. 17.3.

Fig. 17.3

The energy efficiency curves of the first and the i-th transformer

If the following formula holds:

$${\eta }_{i}\left({P}_{i}\right)={\eta }_{1}\left(\frac{{P}_{1}}{{\beta }_{i}}\right)$$

(17.3)

where β_i is a constant, we say that the i-th transformer and the first transformer are the transformers with similar energy efficiency, referred to as “similar energy efficiency transformer”, and β₁ = 1.

The transformer with similar energy efficiency has the following characteristics:

$$\begin{array}{c}{\eta'_i}\left({P}_{i}\right)=\frac{{\eta'_1}({P}_{1})}{{\beta }_{i}}\\ \begin{array}{c}{P}_{ie}={\beta }_{i}{P}_{1e}\\ {\beta }_{i}=\frac{{P}_{ie}}{{P}_{1e}}\end{array}\end{array}$$

(17.4)

17.2 Optimal Scheduling of Multiple Transformers in a Grid

• 1. If the n transformers are identical, the energy expression of the power supply becomes

  $$ P_{t} = P_{0} \sum\limits_{i = 1}^{n} {\frac{{\theta_{i} }}{{\eta (\theta_{i} P_{0} )}}} $$

  (17.5)

where

$${\theta }_{i}=\frac{{P}_{i}}{{P}_{0}}$$

(17.6)

$${\sum }_{i=1}^{n}{\theta }_{i}=1$$

Consider the minimization problem of total power consumption

$$ \begin{gathered} \min P_{t} \hfill \\ s.t.\begin{array}{*{20}c} {} \\ \end{array} \theta_{i} > 0,i = 1,2,...n \hfill \\ \begin{array}{*{20}c} {} & {} \\ \end{array} \sum\limits_{i = 1}^{n} {\theta_{i} = 1} \hfill \\ \begin{array}{*{20}c} {} & {} \\ \end{array} P_{0} = cons\tan t \hfill \\ \end{gathered} $$

(17.7)

This problem can also be written as

$$ \begin{gathered} \min P_{0} \sum\limits_{i = 1}^{n} {\frac{{\theta_{i} }}{{\eta (\theta_{i} P_{0} )}}} \hfill \\ s.t.\begin{array}{*{20}c} {} \\ \end{array} \theta_{i} > 0,i = 1,2,...n \hfill \\ \begin{array}{*{20}c} {} & {} \\ \end{array} \sum\limits_{i = 1}^{n} {\theta_{i} = 1} \hfill \\ \begin{array}{*{20}c} {} & {} \\ \end{array} P_{0} = cons\tan t \hfill \\ \end{gathered} $$

(17.8)

We consider three cases:

• (1) n = 2

The system has two variables and has

$$ \begin{gathered} \theta_{1} + \theta_{2} = 1 \hfill \\ \theta_{1} > 0 \hfill \\ \theta_{2} > 0 \hfill \\ \end{gathered} $$

(17.9)

The objective function P_t can be expressed as

$$ P_{t} = P_{0} \left( {\frac{{\theta_{1} }}{{\eta (\theta_{1} P_{0} )}} + \frac{{\theta_{2} }}{{\eta (\theta_{2} P_{0} )}}} \right) = P_{0} \left( {\frac{{\theta_{1} }}{{\eta (\theta_{1} P_{0} )}} + \frac{{1 - \theta_{1} }}{{\eta ((1 - \theta_{1} )P_{0} )}}} \right) $$

(17.10)

The optimization condition is

$$ P'_t(\theta_{1} ) = 0 $$

(17.11)

and it is easy to see that

$$ \theta_{1} = \frac{1}{2} $$

(17.12)

is an optimization point. Then we have

$$ \theta_{2} = 1 - \theta_{1} = \theta_{1} = \frac{1}{2} = \frac{1}{n} $$

(17.13)

That is, the optimal control method is to keep

$$ P_{1} = P_{2} = \frac{{P_{0} }}{2} = \frac{{P_{0} }}{n} $$

(17.14)

The total power dissipation is

$${P}_{t}=\frac{{P}_{0}}{\eta \left( {\frac{{P}_{0}}{2}} \right)}=\frac{{P}_{0}}{\eta \left(\frac{{P}_{0}}{n}\right)}$$

(17.15)

Since the shape of the overall efficiency curve of the distribution station is the same as that of a single transformer, so the second derivative of the P_t is also greater than zero

$${{P}_{t}}^{{\prime}{\prime}}\left({\theta }_{1}\right)>0$$

(17.16)

P_t is the only minimum value.

$$\text{min}\,{P}_{t}=\frac{{P}_{0}}{\eta \left(\frac{{P}_{0}}{n}\right)}$$

(17.17)

The overall energy efficiency η_t of the distribution station is the only maximum value.

$$max\,{\eta }_{t}=\eta \left(\frac{{P}_{0}}{n}\right)$$

(17.18)

• (2) n = 3

The system has three variables, based on known conditions, we have

$$ \begin{gathered} \theta_{1} + \theta_{2} + \theta_{3} = 1 \hfill \\ \theta_{1} > 0 \hfill \\ \theta_{2} > 0 \hfill \\ \theta_{3} > 0 \hfill \\ \end{gathered} $$

(17.19)

The P_t expression becomes

$$ P_{t} = P_{0} \left(\frac{{\theta_{1} }}{{\eta (\theta_{1} P_{0} )}} + \frac{{\theta_{2} }}{{\eta (\theta_{2} P_{0} )}} + \frac{{\theta_{3} }}{{\eta ((\theta_{3} P_{0} )}}\right) $$

(17.20)

Assuming that θ₃ is fixed and an optimization point, only θ₁ and θ₂ are variables, we have

$${\theta }_{1}+{\theta }_{2}=1-{\theta }_{3}=constant$$

(17.21)

Based on the conclusion of n = 2 above, we have

$${\theta }_{1}={\theta }_{2}$$

(17.22)

to be the optimal point.

Assuming that θ₂ is fixed and is an optimization point, only θ₁ and θ₃ are variables, there are

$${\theta }_{1}+{\theta }_{3}=1-{\theta }_{2}=constant$$

(17.23)

According to the conclusion of n = 2 above, we have

$$ \theta_{1} = \theta_{3} $$

(17.24)

to be the optimal point.

Similarly, assuming thatθ₁ is fixed and is an optimization point, only θ₂ and θ₃ are variables, we have

$${\theta }_{2}+{\theta }_{3}=1-{\theta }_{1}=constant$$

(17.25)

According to the conclusion of n = 2 above, we have

$${\theta }_{2}={\theta }_{3}$$

(17.26)

to be the optimal point.

So, we have the optimal point

$$ \theta_{1} = \theta_{2} = \theta_{3} = \frac{1}{3} = \frac{1}{n} $$

(17.27)

That is, the optimal control method is to keep

$$ P_{1} = P_{2} = P_{3} = \frac{{P_{0} }}{3} = \frac{{P_{0} }}{n} $$

(17.28)

The minimum total power dissipation is

$$ \min P_{t} = \frac{{P_{0} }}{{\eta \left(\frac{{P_{0} }}{3}\right)}} = \frac{{P_{0} }}{{\eta \left(\frac{{P_{0} }}{n}\right)}} $$

(17.29)

The maximum overall energy efficiency η_t is

$${\mathit{max} \eta }_{t}=\eta \left(\frac{{P}_{0}}{3}\right)=\eta \left(\frac{{P}_{0}}{n}\right)$$

(17.30)

• (3) n = k

The system has n variables, and the above conclusion can be extended to the case of n = k, the optimal point is

$$ \theta_{1} = \theta_{2} = ... = \theta_{k} = \frac{1}{k} $$

(17.31)

That is, the optimal control method is to keep

$$ P_{1} = P_{2} = ... = P_{k} = \frac{{P_{0} }}{k} $$

(17.32)

The minimum total power dissipation is

$$ \min P_{t} = P_{0} \frac{1}{{\eta \left(\frac{{P_{0} }}{n}\right)}} $$

(17.33)

The maximum overall energy efficiency η_t is

$${\mathit{max} \eta }_{t}=\eta \left(\frac{{P}_{0}}{k}\right)=\eta \left(\frac{{P}_{0}}{n}\right)$$

(17.34)

Conclusion: When there are n sets of identical transformers supplying power to a factory area on the same power grid, and the load allocated to each transformer is the same, it is the optimal dispatching scheme.

• 2. If the n transformers are the energy efficiency similarity transformers, the energy expression of the power supply becomes

$${P}_{t}={P}_{0}{\sum }_{i=1}^{n}\frac{{\theta }_{i}}{{\eta }_{i}({\theta }_{i}{P}_{0})}$$

(17.35)

where

$${\theta }_{i}=\frac{{P}_{i}}{{P}_{0}}$$

(17.36)

$${\sum }_{i=1}^{n}{\theta }_{i}=1$$

Consider the minimization problem of total power consumption

$${\mathit{minP}}_{t}$$

(17.37)

$$s.t.\begin{array}{c}\end{array}{\theta }_{i}>0,i=\text{1,2},\dots n$$

$$\begin{array}{cc}& \end{array}\sum\limits _{i=1}^{n}{\theta }_{i}=1$$

$$\begin{array}{cc}& \end{array}{P}_{0}=cons\mathit{tan}t$$

This problem can also be written as

$${\mathit{minP}}_{0}{\sum }_{i=1}^{n}\frac{{\theta }_{i}}{{\eta }_{i}({\theta }_{i}{P}_{0})}$$

(17.38)

$$s.t.\begin{array}{c}\end{array}{\theta }_{i}>0,i=\text{1,2},\dots n$$

$$\begin{array}{cc}& \end{array}\sum\limits_{i=1}^{n}{\theta }_{i}=1$$

$$\begin{array}{cc}& \end{array}{P}_{0}=cons\mathit{tan}t$$

We consider three cases:

• (1) n = 2

The system has two variables and has

$$ \begin{gathered} \theta_{1} + \theta_{2} = 1 \hfill \\ \theta_{1} > 0 \hfill \\ \theta_{2} > 0 \hfill \\ \end{gathered} $$

(17.39)

The objective function P_t can be expressed as

$${P}_{t}={P}_{0}(\frac{{\theta }_{1}}{{\eta }_{1}({\theta }_{1}{P}_{0})}+\frac{{\theta }_{2}}{{\eta }_{2}({\theta }_{2}{P}_{0})})={P}_{0}(\frac{{\theta }_{1}}{{\eta }_{1}({\theta }_{1}{P}_{0})}+\frac{1-{\theta }_{1}}{{\eta }_{2}((1-{\theta }_{1}){P}_{0})})$$

(17.40)

The optimization condition is

$$ P'_t (\theta_{1} ) = 0 $$

(17.41)

and it is easy to see that

$${\theta }_{1}=\frac{1}{1+{\beta }_{2}}=\frac{{\beta }_{1}}{{\beta }_{1}+{\beta }_{2}}$$

(17.42)

for the optimization point. Then we have

$${\theta }_{2}=1-{\theta }_{1}=\frac{{\beta }_{2}}{1+{\beta }_{2}}=\frac{{\beta }_{2}}{{\beta }_{1}+{\beta }_{2}}$$

(17.43)

That is, the optimal control method is to keep

$$\begin{array}{c}{P}_{1}=\frac{1}{1+{\beta }_{2}}{P}_{0}\\\\{P}_{2}=\frac{{\beta }_{2}}{1+{\beta }_{2}}{P}_{0}\end{array}$$

(17.44)

The total power dissipation is

$${P}_{t}=\frac{{P}_{0}}{{\eta }_{1}(\frac{{P}_{0}}{1+{\beta }_{2}})}$$

(17.45)

Since the shape of the overall efficiency curve of the distribution station is the same as that of a single transformer, so the second derivative of the P_t is also greater than zero

$${{P}_{t}}^{{\prime}{\prime}}\left({\theta }_{1}\right)>0$$

(17.46)

P_t is the only minimum value.

$$\text{min}\,{P}_{t}=\frac{{P}_{0}}{{\eta }_{1}(\frac{{P}_{0}}{1+{\beta }_{2}})}$$

(17.47)

The overall energy efficiency η_t of the distribution station is the only maximum value.

$$max\,{\eta }_{t}={\eta }_{1}(\frac{{P}_{0}}{1+{\beta }_{2}})$$

(17.48)

• (2) n = 3

The system has three variables, based on known conditions, we have

$${\theta }_{1}+{\theta }_{2}+{\theta }_{3}=1$$

(17.49)

$${\theta }_{1}>0$$

$${\theta }_{2}>0$$

$${\theta }_{3}>0$$

The P_t expression becomes

$${P}_{t}={P}_{0}\left( {\frac{{\theta }_{1}}{{\eta }_{1}({\theta }_{1}{P}_{0})}} \right)+\left( {\frac{{\theta }_{2}}{{\eta }_{2}({\theta }_{2}{P}_{0})}} \right)+ \left( {\frac{{\theta }_{3}}{{\eta }_{3}({\theta }_{3}{P}_{0})}} \right)$$

(17.50)

Assuming that θ₃ is fixed and is an optimization point, only θ₁ and θ₂ are variables, there are

$${\theta }_{1}+{\theta }_{2}=1-{\theta }_{3}=constant$$

(17.51)

Based on the conclusion of n = 2 above, we have

$$\begin{array}{c}{\theta }_{1}=\frac{1}{1+{\beta }_{2}}\\\\ {\theta }_{2}=\frac{{\beta }_{2}}{1+{\beta }_{2}}\end{array}$$

(17.52)

to be is the optimal point.

Assuming that θ₂ is fixed and is an optimization point, only θ₁ and θ₃ are variables, there are

$${\theta }_{1}+{\theta }_{3}=1-{\theta }_{2}=constant$$

(17.53)

According to the conclusion of n = 2 above, there are

$$\begin{array}{c}{\theta }_{1}=\frac{1}{1+{\beta }_{3}}\\\\ {\theta }_{3}=\frac{{\beta }_{3}}{1+{\beta }_{3}}\end{array}$$

(17.54)

is the optimal point.

Similarly, assuming that θ₁ is fixed and is an optimization point, only θ₂ andθ₃ are variables, we have

$${\theta }_{2}+{\theta }_{3}=1-{\theta }_{3}=constant$$

(17.55)

According to the conclusion of n = 2 above, we have

$$\begin{array}{c}{\theta }_{2}=\frac{{\beta }_{2}}{{\beta }_{2}+{\beta }_{3}}\\\\ {\theta }_{3}=\frac{{\beta }_{3}}{{\beta }_{2}+{\beta }_{3}}\end{array}$$

(17.56)

to be the optimal point.

So, we have the optimal point, and the optimal control method is to keep

$$\begin{array}{c}{\theta }_{1}=\frac{{\beta }_{1}}{{{\beta }_{1}+\beta }_{2}+{\beta }_{3}}\\\\{\theta }_{2}=\frac{{\beta }_{2}}{{{\beta }_{1}+\beta }_{2}+{\beta }_{3}}\\ {\theta }_{3}=\frac{{\beta }_{3}}{{{\beta }_{1}+\beta }_{2}+{\beta }_{3}}\end{array}$$

(17.57)

The minimum total power consumption is

$${\mathit{minP}}_{t}=\frac{{P}_{0}}{{\eta }_{1}(\frac{{P}_{0}}{1+{\beta }_{2}+{\beta }_{3}})}$$

(17.58)

The maximum overall energy efficiency η_t is

$${\mathit{max} \eta }_{t}={\eta }_{1}\left( {\frac{{P}_{0}}{1+{\beta }_{2}+{\beta }_{3}}} \right)$$

(17.59)

• (3) n = k

The system has n variables, the above conclusion can be extended to the case of n = k, the optimal point and the optimal control method is to keep

$${\theta }_{i}=\frac{{\beta }_{i}}{{\sum }_{l=1}^{k}{\beta }_{l}}$$

(17.60)

The minimum total power dissipation is

$${\mathit{minP}}_{t}={P}_{0}\frac{1}{{\eta }_{1}(\frac{{P}_{0}}{{\sum }_{l=1}^{k}{\beta }_{l}})}$$

(17.61)

The maximum overall energy efficiency η_t is

$${\mathit{max} \eta }_{t}={\eta }_{1}\left(\frac{{P}_{0}}{{\sum }_{l=1}^{k}{\beta }_{l}}\right)$$

(17.62)

17.3 Optimum Number of Transformers Operating in a Distribution Station

• 1. A distribution station has m identical transformers, if the following equations are satisfied

$$\eta\left ( {\frac{{P}_{0}}{n}}\right)\ge \eta\left({\frac{{P}_{0}}{n-1}}\right)$$

(17.63)

$$\eta \left(\frac{{P}_{0}}{n}\right)\ge \eta \left(\frac{{P}_{0}}{n+1}\right)$$

$$n\le m$$

Then n is the number of transformers with optimal operation.

• 2. A distribution station has m energy efficiency similarity transformers, if the following equations are satisfied

$${\eta }_{1}\left(\frac{{P}_{0}}{{\sum }_{l=1}^{n}{\beta }_{l}}\right)\ge {\eta }_{1}\left(\frac{{P}_{0}}{{\sum }_{l=1}^{n1}{\beta }_{l}}\right)$$

(17.64)

$$n\le m$$

$${n}_{1}\le m$$

n₁ is any combination other than the optimal combination of n units this time, and also include other combinations of n units. The number n is the number of transformers with optimal operation.

17.4 Optimal Switching Rules for Multiple Transformers in Distribution Stations

• 1. A distribution station has m identical transformers, and the number n is the number of transformers currently in optimal operation. If P₀ increases to P₀₁, the following relation holds:

$$\eta \left( {\frac{{P}_{01}}{n}} \right)=\eta \left( {\frac{{P}_{01}}{n+1}} \right)$$

(17.65)

$$n\le m$$

Then P₀₁ is the optimal switching point between the operation of n transformers and the operation of n + 1 transformer. When the total required power P₀ is greater than P₀₁, the optimal number of transformers in operation is switched from n to n + 1. If P₀ increases until P₀/n = P_1m, there isn’t the point P₀₁, then P₀/n = P_1m is the switching point from n to n + 1.

If P₀ is reduced to P₀₂, the following relation is established

$$\eta \left( {\frac{{P}_{02}}{n}} \right)=\eta \left( {\frac{{P}_{02}}{n-1}} \right)$$

(17.66)

$$n\le m$$

Then P₀₂ is the optimal switching point between the operation of n transformers and the operation of n−1 transformers. When the total required power P₀ is less than P₀₂, the optimal number of transformers in operation is switched from n to n−1. If P₀ reduces until P₀/(n−1) = P_1m, there isn’t the point P₀₂, then P₀/(n−1) = P_1m is the switching point from n to n−1.

The analysis process is shown in Fig. 17.4.

Fig. 17.4

Energy efficiency comparison curve

• 2. A distribution station has m energy efficiency similarity transformers, and the number n is the number of transformers currently in optimal operation. If P₀ increases to P₀₁, the following relation holds:

$$\begin{array}{c}{\eta }_{1}\left(\frac{{P}_{0}}{{\sum }_{l=1}^{n}{\beta }_{l}}\right)={\eta }_{1}\left(\frac{{P}_{0}}{{\sum }_{l=1}^{k1}{\beta }_{l}}\right)\\ n\le m\end{array}$$

(17.67)

Then P₀₁ is the optimal switching point between the operation of n transformers and the operation of k₁ transformers. When the total required power P₀ is greater than P₀₁, the optimal number of transformers in operation is switched from n to k₁. If P₀ increases until P₀/${\sum }_{l=1}^{n}{\beta }_{l}$=P_1m, there isn’t the point P₀₁, then P₀/${\sum }_{l=1}^{n}{\beta }_{l}$=P_1m is the switching point from n to k₁.

If P₀ is reduced to P₀₂, the following relation is established

$$\begin{array}{c}{\eta }_{1}\left(\frac{{P}_{0}}{{\sum }_{l=1}^{n}{\beta }_{l}}\right)={\eta }_{1}\left(\frac{{P}_{0}}{{\sum }_{l=1}^{k2}{\beta }_{l}}\right)\\ n\le m\end{array}$$

(17.68)

Then P₀₂ is the optimal switching point between the operation of n transformers and the operation of k₂ transformers. When the total required power P₀ is less than P₀₂, the optimal number of transformers in operation is switched from n to k₂. If P₀ reduces until P₀/${\sum }_{l=1}^{k2}{\beta }_{l}$ =P_1m, there isn’t the point P₀₂, then P₀/${\sum }_{l=1}^{k2}{\beta }_{l}$ =P_1m is the switching point from n to k₂.

The analysis process is shown in Fig. 17.5.

Fig. 17.5

Energy efficiency comparison curve

k1 and k2 are any combination other than the optimal combination of n units this time, and also include other combinations of n units. Point ${P}_{0}/{\sum }_{l=1}^{k1}{\beta }_{l}$ is the point closest to ${P}_{0}/{\sum }_{l=1}^{n}{\beta }_{l}$ to the left of point ${P}_{0}/{\sum }_{l=1}^{n}{\beta }_{l}$. Point ${P}_{0}/{\sum }_{l=1}^{k2}{\beta }_{l}$ is the point closest to ${P}_{0}/{\sum }_{l=1}^{n}{\beta }_{l}$ to the right of point ${P}_{0}/{\sum }_{l=1}^{n}{\beta }_{l}$. If all transformers are identical, we have k1 = n + 1 and k2 = n−1.