Optimal Dispatch of Power Grid Load

Many power plants export electricity to the public grid, delivering it to factories and homes (Fig. 18.1). We should optimize the distribution and scheduling of the load rate of each line to minimize the power loss of the transmission system. The amount of energy wasted in the power grid transmission process is very considerable, and many scientists, scholars, students and engineers are engaged in research in this field.

Fig. 18.1

Power grid

Many optimization methods require the establishment of a power grid model. However, since accurate models of actual power grids are difficult to establish, these optimization methods will quickly become difficult to apply in actual power grids given the complexity of the models and algorithms, the curse of dimensionality, and the computational time.

An approach that avoids these complications would be highly advantageous, especially for business and industry.

18.1 Energy Consumption in a Power Grid

There are n transmission lines to supply power in an area, the power lines in parallel. The total apparent power required is S_t, VA, the line voltage U of each transmission line is identical. The total apparent current is I₀, the resistance and current of i-th line are R_i and I_i, respectively, the total energy consumption W_t of n transmission lines is expressed as

$$ W_{t} = 3\mathop \sum \limits_{i = 1}^{n} I_{i}^{2} R_{i} = 3I_{0}^{2} \mathop \sum \limits_{i = 1}^{n} \left( {\frac{{I_{i} }}{{I_{0} }}} \right)^{2} R_{i} = 3I_{0}^{2} \mathop \sum \limits_{i = 1}^{n} \theta_{i}^{2} R_{i} $$

(18.1)

$$ I_{0} = \mathop \sum \limits_{i = 1}^{n} I_{i} $$

$$ S_{t} = \sqrt 3 UI_{0} $$

where $\theta_{i}$ represents the load rate of the i-th transmission line, expressed as

$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

(18.2)

$$ \theta_{i} = \frac{{I_{i} }}{{I_{0} }} $$

If the resistance of each line is the same as R₀, W_t of the n transmission lines is expressed as

$$ W_{t} = 3I_{0}^{2} R_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i}^{2} $$

(18.3)

18.2 Optimal Dispatching of Power Grids Composed of Lines with the Identical Resistance

Theorem:

If the resistance of each line is i R₀, for the optimization problem of W_t, the minimization of the total energy consumption of the power grid.

$$ \mathop {\min }\limits_{{ \begin{array}{*{20}c} {s.t. I_{i} > 0} \\ { \mathop \sum \limits_{i = 1}^{n} I_{i} = I_{0} } \\ { } \\ \end{array} }} W_{t} $$

(18.4)

is given by

$$ \min W_{t} = 3I_{0}^{2} R_{0} \frac{1}{n} $$

(18.5)

The optimal point is

$$ I_{1} = I_{2} = \ldots = I_{k} = \frac{{I_{0} }}{k} $$

(18.6)

That is

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

(18.7)

Proof:

We begin our inductive proof by considering the case where n = 2.

The constraint condition then becomes

$$ I_{1} + I_{2} = I_{0} $$

(18.8)

where

$$ I_{1} > 0 $$

(18.9)

$$ I_{2} > 0 $$

The objective function W_t is expressed as

$$ W_{t} = 3R_{0} \left( {I_{1}^{2} + I_{2}^{2} } \right) $$

(18.10)

The optimal condition is given for

$$ W_{t}{\prime} \left( {I_{1} } \right) = 0 $$

(18.11)

We have

$$ 4I_{1} - 2I_{0} = 0 $$

(18.12)

It is then easily verified that

$$ I_{1} = I_{2} = \frac{{I_{0} }}{2} $$

(18.13)

is an optimal point, that is

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

(18.14)

We then check the second derivative,

$$ W_{t}^{^{\prime\prime}} > 0 $$

(18.15)

Therefore, the optimal point is the unique minimum.

The minimal value of the total energy consumption of the power grid is

$$ \min W_{t} = 3I_{0}^{2} R_{0} \frac{1}{2} $$

(18.16)

We then assume that this holds for n = k. The above conclusion is extended to the case of n = k, and the optimal point is then

$$ I_{1} = I_{2} = \ldots = I_{k} = \frac{{I_{0} }}{k} $$

(18.17)

That is

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

(18.18)

The minimal value of W_t is

$$ \max W_{t} = 3I_{0}^{2} R_{0} \frac{1}{k} $$

(18.19)

Our inductive case is then given by n = k + 1. For the total energy consumption of the power grid we have

$$ W_{t} = 3R_{0} (\mathop \sum \limits_{i = 1}^{k} I_{i}^{2} + \left. {I_{k + 1}^{2} } \right) $$

(18.20)

and the minimum of the first item is

$$ \max 3R_{0} \mathop \sum \limits_{i = 1}^{k} I_{i}^{2} = 3R_{0} \left( {I_{0} - I_{k + 1} } \right)^{2} \frac{1}{k} $$

(18.21)

where

$$ I_{1} = I_{2} = \ldots = I_{k} $$

(18.22)

The expression for W_t becomes

$$ W_{t} = 3R_{0} (\frac{{(I_{0} - I_{k + 1} )^{2} }}{k} + \left. {I_{k + 1}^{2} } \right) $$

(18.23)

The optimal condition is given for

$$ W_{t}{\prime} \left( {I_{k + 1} } \right) = 0 $$

(18.24)

The solution is

$$ I_{k + 1} = \frac{{I_{0} }}{k + 1} $$

(18.25)

and

$$ I_{1} = I_{2} = \ldots = I_{k} = \frac{{I_{0} - I_{k + 1} }}{k} = \frac{{I_{0} }}{k + 1} $$

(18.26)

Therefore, the optimal point is then

$$ I_{1} = I_{2} = \ldots = I_{k} = I_{k + 1} = \frac{{I_{0} }}{k + 1} $$

(18.27)

That is

$$ \theta_{1} = \theta_{2} = \ldots = \theta_{k} = \theta_{k + 1} = \frac{1}{k + 1} $$

(18.28)

The minimal value of total energy consumption of the power grid is

$$ \min W_{t} = 3R_{0} I_{0}^{2} \frac{1}{k + 1} $$

(18.29)

Load distribution theorem: In a power grid which consists of n main lines that have the identical resistance, the optimal control method is to keep each line to have the identical current I₀/n.

18.3 Optimal Scheduling of a Power Grid with the Different Power Line

Theorem:

If the resistances of all power lines are different, respectively R1, R2, …, Rn, for the optimization problem of W_t, the minimization of the total energy consumption of the power grid.

$$ \mathop {\min }\limits_{{ \begin{array}{*{20}c} {s.t. I_{i} > 0} \\ { \mathop \sum \limits_{i = 1}^{n} I_{i} = I_{0} } \\ { } \\ \end{array} }} W_{t} $$

(18.30)

is given by

$$ \min W_{t} = 3I_{0}^{2} \left( {R_{1} \parallel R_{2} \parallel \ldots \parallel R_{n} } \right) = 3I_{0}^{2} \frac{{\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{n}}} {\mathbf{R}}_{{\mathbf{j}}} }}{{\mathop \sum \nolimits_{{{\mathbf{m}} = 1}}^{{\mathbf{n}}} (\left( {\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{n}}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{\mathbf{m}}} } \right)}} $$

(18.31)

The optimal point is

$$ {\text{I}}_{i} = \frac{{(\mathop \prod \nolimits_{{{\text{j}} = 1}}^{{\text{n}}} {\text{R}}_{{\text{j}}} )/{\text{R}}_{{\text{i}}} }}{{\mathop \sum \nolimits_{{{\text{m}} = 1}}^{{\text{n}}} (\left( {\mathop \prod \nolimits_{{{\text{j}} = 1}}^{{\text{n}}} {\text{R}}_{{\text{j}}} )/{\text{R}}_{{\text{m}}} } \right)}}{\text{I}}_{0} $$

(18.32)

Proof:

We begin our inductive proof by considering the case where n = 2.

The constraint condition then becomes

$$ I_{1} + I_{2} = I_{0} $$

(18.33)

where

$$ I_{1} > 0 $$

(18.34)

$$ I_{2} > 0 $$

The objective function W_t is expressed as

$$ W_{t} = 3\left( {I_{1}^{2} R_{1} + I_{2}^{2} R_{2} } \right) $$

(18.35)

The optimal condition is given for

$$ W_{t}{\prime} \left( {I_{1} } \right) = 0 $$

(18.36)

The solution is

$$ I_{1} = \frac{{R_{2} }}{{R_{1} + R_{2} }}I_{0} $$

(18.37)

$$ I_{2} = \frac{{R_{1} }}{{R_{1} + R_{2} }}I_{0} $$

Checking the second derivative, we see

$$ W_{t}^{^{\prime\prime}} > 0 $$

(18.38)

Thus, the optimal point is the unique minimum.

The minimal value of the total energy consumption of the power grid is

$$ \max W_{t} = 3I_{0}^{2} (R_{1} \parallel R_{2} ) = 3I_{0}^{2} \frac{{R_{1} R_{2} }}{{R_{1} + R_{2} }} $$

(18.39)

We then assume that this holds for n = k. The above conclusion is readily extended to the case of n = k, and the optimal point is then

$$ {\text{I}}_{i} = \frac{{(\mathop \prod \nolimits_{{{\text{j}} = 1}}^{{\text{k}}} {\text{R}}_{{\text{j}}} )/{\text{R}}_{{\text{i}}} }}{{\mathop \sum \nolimits_{{{\text{m}} = 1}}^{{\text{k}}} (\left( {\mathop \prod \nolimits_{{{\text{j}} = 1}}^{{\text{k}}} {\text{R}}_{{\text{j}}} )/{\text{R}}_{{\text{m}}} } \right)}}{\text{I}}_{0} $$

(18.40)

The minimal value of W_t is

$$ \max W_{t} = 3I_{0}^{2} \frac{{\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{k}}} {\mathbf{R}}_{{\mathbf{j}}} }}{{\mathop \sum \nolimits_{{{\mathbf{m}} = 1}}^{{\mathbf{k}}} (\left( {\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{k}}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{\mathbf{m}}} } \right)}} $$

(18.41)

Our inductive case is then given by n = k + 1. For the total energy consumption of the power grid we have

$$ W_{t} = 3(\mathop \sum \limits_{i = 1}^{k} I_{i}^{2} R_{i} + \left. {I_{k + 1}^{2} R_{k + 1} } \right) $$

(18.42)

and the minimum of the first item is

$$ 3\mathop \sum \limits_{i = 1}^{k} I_{i}^{2} R_{i} = 3\left( {I_{0} - I_{k + 1} } \right)^{2} \frac{{\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{k}}} {\mathbf{R}}_{{\mathbf{j}}} }}{{\mathop \sum \nolimits_{{{\mathbf{m}} = 1}}^{{\mathbf{k}}} (\left( {\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{k}}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{\mathbf{m}}} } \right)}} $$

(18.43)

The expression for W_t becomes

$$ W_{t} = 3\left( {I_{0} - I_{k + 1} } \right)^{2} \frac{{\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{k}}} {\mathbf{R}}_{{\mathbf{j}}} }}{{\mathop \sum \nolimits_{{{\mathbf{m}} = 1}}^{{\mathbf{k}}} (\left( {\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{k}}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{\mathbf{m}}} } \right)}} + 3I_{k + 1}^{2} R_{k + 1} $$

(18.44)

The optimal condition is given for

$$ W_{t}{\prime} \left( {I_{k + 1} } \right) = 0 $$

(18.45)

Based on above conclusion of n = 2, the solution is

$$ I_{k + 1} = \frac{{(\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{{\mathbf{k}} + 1}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{{\mathbf{k}} + 1}} }}{{\mathop \sum \nolimits_{{{\mathbf{m}} = 1}}^{{{\mathbf{k}} + 1}} (\left( {\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{{\mathbf{k}} + 1}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{\mathbf{m}}} } \right)}}{\mathbf{I}}_{0} $$

(18.46)

Therefore, the optimal point is then

$$ {\mathbf{I}}_{{\varvec{i}}} = \frac{{(\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{{\mathbf{k}} + 1}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{\mathbf{i}}} }}{{\mathop \sum \nolimits_{{{\mathbf{m}} = 1}}^{{{\mathbf{k}} + 1}} (\left( {\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{{\mathbf{k}} + 1}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{\mathbf{m}}} } \right)}}{\mathbf{I}}_{0} $$

(18.47)

and the minimal value of total energy consumption of the power grid is

$$ \min W_{t} = 3I_{0}^{2} \frac{{\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{{\mathbf{k}} + 1}} {\mathbf{R}}_{{\mathbf{j}}} }}{{\mathop \sum \nolimits_{{{\mathbf{m}} = 1}}^{{{\mathbf{k}} + 1}} (\left( {\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{{\mathbf{k}} + 1}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{\mathbf{m}}} } \right)}} $$

(18.48)

18.4 Optimal Scheduling of a Direct Current Power Grid

Theorem:

If not all resistances of direct current power lines are identical, that is, R₁, R₂, …, R_n, respectively, are different, for the optimization problem of W_t, the minimization of the total energy consumption of the power grid.

$$ \mathop {\min }\limits_{{ \begin{array}{*{20}c} {s.t. I_{i} > 0} \\ { \mathop \sum \limits_{i = 1}^{n} I_{i} = I_{0} } \\ { } \\ \end{array} }} W_{t} $$

(18.49)

is given by

$$ \min W_{t} = 2I_{0}^{2} \left( {R_{1} \parallel R_{2} \parallel \ldots \parallel R_{n} } \right) = 2I_{0}^{2} \frac{{\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{n}}} {\mathbf{R}}_{{\mathbf{j}}} }}{{\mathop \sum \nolimits_{{{\mathbf{m}} = 1}}^{{\mathbf{n}}} (\left( {\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{n}}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{\mathbf{m}}} } \right)}} $$

(18.50)

The optimal point is

$$ {\mathbf{I}}_{{\varvec{i}}} = \frac{{(\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{n}}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{\mathbf{i}}} }}{{\mathop \sum \nolimits_{{{\mathbf{m}} = 1}}^{{\mathbf{n}}} (\left( {\mathop \prod \nolimits_{{{\mathbf{j}} = 1}}^{{\mathbf{n}}} {\mathbf{R}}_{{\mathbf{j}}} )/{\mathbf{R}}_{{\mathbf{m}}} } \right)}}{\mathbf{I}}_{0} $$

(18.51)

The Proof is Omitted.

18.5 Conclusion

By assuming a fixed value of the total apparent current, we propose an optimal control method that does not rely on an accurate model of the grid.

The proofs of optimal control and switching methods given in this chapter are mainly based on the minimum energy consumption rule. Therefore, this optimal method has the following characteristics:

• (1) Including lined and unlined systems,

• (2) No systematic mathematical model is required,

• (3) High versatility.