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I've read some papers on voltage collapse, and it seems to be described as the loss of a steady state solution for the system equations. Where the Jacobian matrix is singular, and voltage is monotonically decreasing.

My understanding is that if the voltage is monotonically decreasing, and bounded below by 0, then this must necessarily converge to a single solution greater than or equal to 0.

Wouldn't this limit then have to be a steady state? (A stable equilibrium, even?)

It seems to me that these statements are contradicting each other:

  • Loss of steady state
  • Monotonically decreasing
  • Bounded below by zero

This figure seems to indicate that there are no steady-state solutions for $$\lambda=1$$

The figure is taken from this paper:
https://www.researchgate.net/publication/3345120_Voltage_Collapse_in_Power_Systems

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  • \$\begingroup\$ This is a bit beyond me. Don't know what a Jacobian is. But in essence, once a regulator is overloaded it becomes a totally different system. Different pathways are on and tons of assumptions that made it a linear system with a stable output become invalid. And it is not guaranteed that it converges to a steady state. It may be unstable as long as the overload is applied, with oscillating output current or even voltage. Thermal cycling may even be involved. \$\endgroup\$ – mkeith Jun 4 '20 at 4:04
  • \$\begingroup\$ A Jacobian matrix essentially just contains all the derivatives of the system. If you are familiar with newtons method, this is used when you have a system of equations (rather than just one equation): en.wikipedia.org/wiki/… \$\endgroup\$ – Tobias Bergkvist Jun 4 '20 at 8:32
  • \$\begingroup\$ Yeah, this is about what happens when a power grid is overloaded (or when reactive power is not balanced). From what I understand, the voltage will drop towards 0 (decreasing dynamically). There have been several large-scale blackouts attributed to this. \$\endgroup\$ – Tobias Bergkvist Jun 4 '20 at 8:38
  • \$\begingroup\$ Yeah that is pretty beyond me. \$\endgroup\$ – mkeith Jun 4 '20 at 9:09
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Given the Jacobian, what do you know about the total system energy trajectory?

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I am seeking to understand the Jacobian uses in a system model.

With the energy heading toward zero (assumed), then there no longer is energy being supplied to the system, or an enormous dampener has appeared (possibly a SHORT circuit or an OPEN circuit). Are these thoughts accurate?

If a finite dampener has suddenly been inserted, then the circulating energy will be dampened, and current and voltage amplitudes will be altered, until the new dampener losses are accommodated and a new steady state vector is reached.

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From what I've read, the power organizations seek to avoid total voltage collapse, and various industrial users are designated as "discretionary". Thus these users will the FIRST energy consumers to be SHED, to be DROPPED, to be DIS_CONNECTED from the grid.

In dropping certain users, the residual energy in the various rotating masses will be adequate to serve the remaining loads.

Realize you have energy producer rotating masses, and energy consuming rotating masses (motors), thus the enormous mass ( many thousands of horsepower, with spinning armatures) of motors may assist in stabilizing the grid.

In load shedding, the grid is forced to react to energy transients.

This is exactly what an adaptive_communication link does, with echo cancellation or using high-pass filters to reverse the losses of low-pass wiring and PCB_dissipative copper surfaces and insulation loss tangents.

Thus we can examine how the grid system becomes adaptive, with pre-selected loads_to_be_shed, with a time constant of seconds.

I was out driving yesterday, and spotted 7 new towers on the horizon. These new structures have the height and aspect ratio of the exhaust_cleansing systems attached to the exhausts of natural_gas_powered jet_turbine power generators.

These new natural_gas generators have replaced the previous dual oil-fired boilers/turbines/generators of that square kilometer power station.

Key point is the reaction time of a jet_turbine natural_gas rotating mass, under computer control. Compared to the huge thermal mass of boiler/water.

The natural_gas system NEED NOT CHANGE ANYTHING ROTATIONAL. There is no inertia to alter.

A 0.5 second change in the natural_gas throttle is the only adaptation to a huge and sudden demand upon the grid.

Thus natural_gas power plants, and fields of solar panels with their DC_AC converters needing only 1/2 power cycle to synchronize to the grid, have a wonderful potential in stabilizing our power grids.

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  • \$\begingroup\$ Well, the Jacobian tells us how a system is changing, right? And when it is singular, this is a problem if you try to use newton-rhapson to solve for/find a steady state solution. The question is not really directed against any specific circuit/power grid, but is about trying to better understand the concept of voltage collapse. What does a system look like "after" a voltage collapse has occured? \$\endgroup\$ – Tobias Bergkvist Jun 3 '20 at 15:09
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    \$\begingroup\$ @analogsystemsrf Your answer has been flagged as low quality, and is making the rounds in the delete-a-thon. I'd assume with your points that it's obvious why. Wanna tidy it up? \$\endgroup\$ – Chris Knudsen Jun 3 '20 at 20:14
  • \$\begingroup\$ The idea is that weather conditions, or changing loads can bring a system that is operating close to its limits into a voltage collapse. Several large scale blackouts have been attributed to this. Question is: how does the grid recover? what happens if you try to measure your wall output after such a blackout (assuming fuses/switches are not affected)? What will you see? \$\endgroup\$ – Tobias Bergkvist Jun 4 '20 at 8:59
  • \$\begingroup\$ I would imagine that all kinds of switches get thrown in the process of the voltage collapse. It is not like your power outlet is still connected to the generator after the failure. The generator will spin back up with no load, then loads will be added one at a time until the whole system is back up. I think that is how it works. \$\endgroup\$ – mkeith Jun 4 '20 at 9:13
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Regarding Fig 3 and your questions

One thing that might be the issue is the meaning of \$x(t)\$ and \$\lambda\$, while \$x(t)\$ is the state of the system (some signal), \$\lambda\$ is some parameter of the system, and for nonlinear systems, that change in parameters can lead to systems that have no equilibria. And in order to be stabilized you would have to come up with some control law that can (re)create the equilibria of the system.

My understanding is that if the voltage is monotonically decreasing, and bounded below by 0, then this must necessarily converge to a single solution greater than or equal to 0.

No, for some monotonic function \$f\$ , for \$x\leq y\$ then \$f(x)\leq f(y)\$. I assume that by being bounded at 0 you are referring to when \$\lambda = 0\$ and you have that equilibrium at 0, which is only bounding trajectories of \$x(t)>0\$.

Loss of steady state

It should be clear that for \$\lambda > 0\$ there will be no equilibrium. One example of such system would be

$$\dot x(t) = -x(t)^2-\lambda,$$

Which will lead to the same plot as the one you have in your question. For \$\lambda>0\$ no \$x(t)\$ can make \$\dot x(t) =0.\$

Monotonically

This only means that the trajectories go downwards until they land in an equilibrium (if there is one, otherwise they grow unbounded).

Bounded below by zero

From your drawing, this seems to only happen at \$\lambda = 0\$, in that case, solutions from \$x(t)>0\$ will go to \$0\$, but any \$x(t)<0\$ will go to \$-\infty\$. Therefore this is an unstable equilibrium, in which any error or perturbation that moves the state from the equilibrium \$x_{eq}=0\$ to any slightly negative number \$x(t)<0\$ will make \$x(t) \xrightarrow{t \xrightarrow{} \infty} -\infty\$.

Notice that this is a mathematical model, by \$x(t) \xrightarrow{t \xrightarrow{} \infty} -\infty\$ you should read trouble, regardless if it's a positive or negative infinity. The real system will not go to infinity (taking infinity energy), but it will probably crash/break/hit its limit hard.

Regarding the paper

All that said, this model is not really a model of a power system, at the paper they write that

Let’s look at a prototypical example to introduce the essential ideas. Consider a nonlinear system with a single state variable, x, and a slowly varying parameter, h. The dynamics when h = -2 are shown by the corresponding vertical line in Fig. 3.

So they are not talking about a model that directly explains a power system, the purpose is to present a bifurcation diagram and show how equilibria move as you change parameters. It also shows that those equilibria can disappear or change in number as those parameters change.

The two following figures refer to actual power system, even though they don't present the model/equations. Looking at fig 5 you will see the problem that once \$\lambda \geq \lambda^* \$ either the system will have a unstable equilibrium or no equilibrium. At \$\lambda = \lambda^* \$ any perturbation will make \$V \xrightarrow{} -\infty\$ (and yes, the real thing will stop at some point, probably zero, but the authors did not add an equilibrium at zero, so mathematically it won't).

bifurcation diagrams and phase portrait of the system

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  • \$\begingroup\$ Since the voltage phasor represents an absolute value (with an angle), it doesn't really make sense to consider this being negative. Setting the voltage phasor to be negative would just correspond to shifting its phase by 180 degrees. I guess the assumption of a sinusoidal signal might be a problem. If voltage is going towards negative infinity, doesn't this then mean that the energy of the system is going towards infinity? \$\endgroup\$ – Tobias Bergkvist Jun 7 '20 at 11:45
  • \$\begingroup\$ You are right, that first model is not a model of a power system, it is just explaining a concept related to equilibria and bifurcation. In control/dynamical system many times things will 'go to infinity', that doesn't mean the real system will though, it is just a mathematical model. But it does represent trouble (the system breaking, crashing or doing dangerous things). \$\endgroup\$ – jDAQ Jun 7 '20 at 16:28

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