Direct Current Transmission Heating Problem

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Direct Current Transmission Heating Problem
State dimension: 1
Differential states: 3
Discrete control functions: 1
Interior point equalities: 3


The direct current transfer 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. An optimal strategy for regulating the power production of individual producers as well as voltages and currents on specific conductors with the goal of 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 \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 \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 materials volume-specific heat capacity which we will refer to as c, we can 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 with a thermal conductivity of approximately 0.51 W / (m \cdot K). We will assume the insulator to be 1 cm thick. Thus, q = 51 W / (m^2 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}{llcl}
 \displaystyle \min_{x, w} & \int_{t_0}^{t_f} \sum_{a \in A} P_1^a(t) \,\mathrm{d} t   \\[1.5ex]
 \mbox{s.t.} & \dot{T}^a(t) & = & x_0(t) - x_0(t) x_1(t) - \; c_0 x_0(t) \; w(t), \\
 & \dot{x}_1(t) & = & - x_1(t) + x_0(t) x_1(t) - \; c_1 x_1(t) \; w(t),  \\
 & \dot{x}_2(t) & = & (x_0(t) - 1)^2 + (x_1(t) - 1)^2,  \\[1.5ex]
 & x(0) &=& (0.5, 0.7, 0)^T, \\
 & w(t) &\in&  \{0, 1\}.
\end{array}

Here the differential states (x_0, x_1) describe the biomasses of prey and predator, respectively. The third differential state is used here to transform the objective, an integrated deviation, into the Mayer formulation \min \; x_2(t_f). The decision, whether the fishing fleet is actually fishing at time t is denoted by w(t).

Parameters

These fixed values are used within the model.


\begin{array}{rcl}
[t_0, t_f] &=& [0, 12],\\
(c_0, c_1) &=& (0.4, 0.2).
\end{array}

Reference Solutions

If the problem is relaxed, i.e., we demand that w(t) be in the continuous interval [0, 1] instead of the binary choice \{0,1\}, the optimal solution can be determined by means of Pontryagins maximum principle. The optimal solution contains a singular arc, as can be seen in the plot of the optimal control. The two differential states and corresponding adjoint variables in the indirect approach are also displayed. A different approach to solving the relaxed problem is by using a direct method such as collocation or Bock's direct multiple shooting method. Optimal solutions for different control discretizations are also plotted in the leftmost figure.

The optimal objective value of this relaxed problem is x_2(t_f) = 1.34408. As follows from MIOC theory<bibref>Sager2008</bibref> this is the best lower bound on the optimal value of the original problem with the integer restriction on the control function. In other words, this objective value can be approximated arbitrarily close, if the control only switches often enough between 0 and 1. As no optimal solution exists, two suboptimal ones are shown, one with only two switches and an objective function value of x_2(t_f) = 1.38276, and one with 56 switches and x_2(t_f) = 1.34416.


Source Code

Model descriptions are available in

Variants

There are several alternative formulations and variants of the above problem, in particular

  • a prescribed time grid for the control function <bibref>Sager2006</bibref>, see also Lotka Volterra fishing problem (AMPL),
  • a time-optimal formulation to get into a steady-state <bibref>Sager2005</bibref>,
  • the usage of a different target steady-state, as the one corresponding to  w(t) = 1 which is (1 + c_1, 1 - c_0),
  • different fishing control functions for the two species,
  • different parameters and start values.

Miscellaneous and Further Reading

The Lotka Volterra fishing problem was introduced by Sebastian Sager in a proceedings paper <bibref>Sager2006</bibref> and revisited in his PhD thesis <bibref>Sager2005</bibref>. These are also the references to look for more details.

References

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