Direct Current Transmission Heating Problem

Direct Current Transmission Heating Problem
State dimension:	$|A|$
Differential states:	$2|A|+|P|$
Discrete control functions:	0
Interior point equalities:	3

The direct current transmission heating problem models a simplified flow of direct electrical current within an electrical network with defined producers and consumers of electrical power. The model includes heating and cooling that occurs in electrical conductors due to specific resistance and heat exchange with the surrounding air. The goal is finding an optimal strategy for regulating the power production of individual producers as well as voltages and currents on specific conductors while minimizing the net power loss due to heating on all power lines over a fixed time horizon while fully supplying all consumers.

The model is largely assembled from well-known basic descriptions of phenomena easily accessible to the audience of high-school level physics courses. Mainly, these are Ohm's law, electrical resistivity, Joule heating and thermal conductivity. Note that changes of resistivity with respect to conductor temperature are taken into account.

Physical background

The aim of the controller in this model is to provide all consumers with their required amount of electric power. Electric power is given by the product of voltage and current $P=I\cdot U$. The electric power is brought to the consumers from producers via a network of power lines. Producers are assumed to have the capability of accurately regulating their power output within a certain range. Power lines are assumed to be solid blocks of an electrically conductive material surrounded by a layer of insulating material. Their cross-section is assumed to be constant along the entire length of the power line. Since both heating and power loss increase with current but not with voltage, it is always preferable to transfer direct current at the highest possible voltage which is allowed by the design of the power line. Therefore, only the electrical current is considered as a control.

The power loss in a power line is given by $W=I^2\cdot R$ where $R$ is the resistance of the power line which is derived from the resistivity of the conductor using the formula

$R=\rho \cdot \frac{l}{A}$

where $\rho$ is the resistivity of the conductor, $l$ is the length of the conductor and $A$ is the the cross-sectional area of the conducting material within the power line. Resistivity is assumed to vary exponentially with the temperature of the conducting material, meaning that $\rho ={\rho }_0\cdot e^{\alpha T}$ where ${\rho }_0$ is the resistivity of the conducting material at room temperature (which is defined as $T=0$) and $\alpha$ is a material-specific dimensionless constant. This causes a steady power loss of

$P_1={\rho }_0\cdot e^{\alpha T}\cdot \frac{l}{A}\cdot I^2.$

The lost energy is transformed into heat which raises the temperature of the conductor. As the conductor is now hotter than the air surrounding its layer of insulating material, heat is steadily lost at a rate determined by the temperature difference, the conductors surface area and the insulating material's heat conductivity. The heat loss is given by

$P_2=q\cdot S\cdot \mathrm{\Delta }T$

where $q$ is the ratio between the isolating material's thermal conductivity and its thickness, $S$ is the surface area of the power line and $\mathrm{\Delta }T$ is the difference between the conductor's temperature and the surrounding air's temperature (which we will assume to be the same as room temperature, i.e. $T=0$). In summary, the rate at which the heat stored in a power line changes is given by

$\dot{W}=P_1-P_2={\rho }_0\cdot e^{\alpha T}\cdot \frac{l}{A}\cdot I^2-q\cdot l\cdot C\cdot T$

where $C$ is the circumference of the cross-section of the power line. Assuming constant density, we can use the conducting material's volume-specific heat capacity, which we will refer to as $c$, to calculate the change in temperature from the change in heat:

$\dot{T}=\frac{\dot{W}}{c\cdot A\cdot l}=\frac{{\rho }_0}{c\cdot A^2}\cdot e^{\alpha T}\cdot I^2-\frac{q\cdot C}{c\cdot A}\cdot T=:{\beta }_1\cdot e^{\alpha T}\cdot I^2-{\beta }_2\cdot T.$

For the examples in this article, we will assume the conductor to be solid copper surrounded by a layer of cross-linked polyethylene (PEX) with a thermal conductivity of approximately $0.51W/(m\cdot K)$.

Mathematical formulation

Let $G=(V,A)$ a directed graph with vertex set $V$ and arc set $A$. Let $P,C\subset V$ be disjoint subsets of the vertex set. $P$ serves as the set of producer nodes while $C$ serves as the set of consumers. For $c\in C$ let $d_c$ denote the power demand for $c$. For $v\in V\setminus C$, let $d_v=0$.

The resulting optimal control problem is given by

$\begin{array}{ll}\underset{T,I,u}{\min }& \underset{t_0}{\overset{t_f}{\int }}\sum _{a\in A}{P}_{1,a}(t)\mathrm{d}t\\ \text{s.t.}& {\dot{T}}_a& =& {\beta }_{1,a}\cdot e^{{\alpha }_a{T}_a}\cdot I_a^2-{\beta }_{2,a}\cdot T_a& \mathrm{\forall }a\in A,\\ & d_v& =& \sum _{a\in {\delta }^{in}(v)}(U_a\cdot I_a-P_1^a)-\sum _{a\in {\delta }^{out}(v)}(U_a\cdot I_a)& \mathrm{\forall }v\in V\setminus P,\\ & -u_v& =& \sum _{a\in {\delta }^{in}(v)}(U_a\cdot I_a-P_1^a)-\sum _{a\in {\delta }^{out}(v)}(U_a\cdot I_a)& \mathrm{\forall }v\in P,\\ & T_a(t_0)& =& 0.0& \mathrm{\forall }a\in A,\\ & I_a(t)& \in & [0,I_{max,a}]& \mathrm{\forall }a\in A,t\in [t_0,t_f],\\ & T_a(t)& \in & [0,T_{max,a}]& \mathrm{\forall }a\in A,t\in [t_0,t_f],\\ & u_v(t)& \in & [0,u_{max,v}]& \mathrm{\forall }v\in P,t\in [t_0,t_f].\end{array}$

The differential state $T_a$ describes the temperature of power line $a$ while the controls $I_a$ and $u_v$ represent the current in a power line $a\in A$ and the power produced in a production node $v\in P$ respectively.

Parameters

There are multiple test scenarios. For all scenarios, we allow three types of conductors, all of which are copper ($\alpha =0.003862$, ${\rho }_0=1.68\cdot 10^{-8}\mathrm{\Omega }m$, $c=3.446\cdot 10^{9}J/(K\cdot m^3)$) cylinders surrounded by an insulating layer of PEX:

Type	Voltage	Conductor diameter	Insulator thickness	$A$	$C$	$q$
A	$330kV$	$7cm$	$2cm$	$0.003848m^2$	$0.2199m$	$25.5W/(m^2\cdot K)$
B	$220kV$	$5cm$	$1.5cm$	$0.001963m^2$	$0.1571m$	$34W/(m^2\cdot K)$
C	$110kV$	$3cm$	$0.75cm$	$0.0007069m^2$	$0.09425m$	$68W/(m^2\cdot K)$

PEX looses structural integrity at operating temperatures roughly in excess of $120^{\circ }C$ which means that $T_{max,a}=100$ for all three types of power lines.

Delayed heating

In this scenario, we consider a simple network consisting of two producers producing up to $1MW$ each and a single consumer with a demand of $1.25MW$. Both producers have direct power links of type C and length $1km$ to the consumer. The time horizon is defined as $(t_0,t_f)=(0,300)$. A JSON description of the network can be found under Direct Current Transmission Heating Problem: Delayed heating (Network).