The memory of the March, 1989 solar flare that produced a one –million amp surge in the Ottawa power grid remains fresh in the minds of many. As images from the overloaded circuits attested, the flare’s induced currents actually melted two-inch thick transmission cables. Service was out for at least two days,

The U.S. power grid is itself vulnerable, not only to intense solar flares but other breakdowns. Let’s face it the grid must be able to withstand fault currents which are thousands of times more frequent than 1-million amp induced flare currents. These fault currents are current surges from a short to a ground. Currently, tacit protection is afforded by circuit breakers designed to open within a few tenths of a second and some can support as much as 80,000 A for a brief time. The trouble is, as power sources have been added to expanding grids, fault currents have increased and demand for fault current limiters has grown.

Is there a better and cheaper way (long term) to address the problem of fault currents, and perhaps even those induced by moderate to large solar flares? The answer appears to be ‘yes’ with the onset of high temperature superconducting (HTS) cable technology, specifically rotating machinery of the type shown in Fig. 1. Here we have the most startling offshoot of HTS technology which can be applied directly to the grid: a dynamic synchronous condenser (capacitor). This is basically a rotating machine like a generator but minus an external mechanical energy source driving the motor. It provides out of phase currents to the power grid.

“Out of phase” may not be clear to most readers who’ve never taken university general physics, so I illustrate it using a C-R alternating circuit diagram in Fig. 2. Here a capacitor (C) is in a circuit with a resistor R and the respective voltage differences are V_C and V_R across each. The phase diagram derived from this circuit is shown in the lower right and we see that V_R is in phase with the current, I, but that V_C lags the current I by pi/2.

The resultant voltage V can be given in terms of the current I, the resistance R and what we call the capacitative reactance (X_C):

V = I(R^2 + X_C), and we know: V_R = IR, and V_C = IX_C

The phase angle for the configuration is given by: tan F = V_C/V_R = X_C/ R

Or F = arctan (X_C/R)

Now, let’s turn to the synchronous condenser in the HTS setting, to see how it addresses issues, say for a grid. In this case we use Fig. 3 which features an inductor L, along with a resistance R. The AC source has an rms (root mean square) voltage V(o) and via angular frequency w = 2pi f couples to the resistive load through a reactance X(L) which we call the “inductive reactance”. This X(L) in a model represents transmission lines, transformers or other ancillary elements in a power grid. The resulting rms current I has real (in phase) and imaginary (out of phase) components, such that:

I = V(o) / R + I X(L) = I(o)[ cos F – I sin F]

Where F is again the designated phase angle

More specifically:

I(o) = V(o)/ [R^2 + X(L)^2]^1/2 and

The U.S. power grid is itself vulnerable, not only to intense solar flares but other breakdowns. Let’s face it the grid must be able to withstand fault currents which are thousands of times more frequent than 1-million amp induced flare currents. These fault currents are current surges from a short to a ground. Currently, tacit protection is afforded by circuit breakers designed to open within a few tenths of a second and some can support as much as 80,000 A for a brief time. The trouble is, as power sources have been added to expanding grids, fault currents have increased and demand for fault current limiters has grown.

Is there a better and cheaper way (long term) to address the problem of fault currents, and perhaps even those induced by moderate to large solar flares? The answer appears to be ‘yes’ with the onset of high temperature superconducting (HTS) cable technology, specifically rotating machinery of the type shown in Fig. 1. Here we have the most startling offshoot of HTS technology which can be applied directly to the grid: a dynamic synchronous condenser (capacitor). This is basically a rotating machine like a generator but minus an external mechanical energy source driving the motor. It provides out of phase currents to the power grid.

“Out of phase” may not be clear to most readers who’ve never taken university general physics, so I illustrate it using a C-R alternating circuit diagram in Fig. 2. Here a capacitor (C) is in a circuit with a resistor R and the respective voltage differences are V_C and V_R across each. The phase diagram derived from this circuit is shown in the lower right and we see that V_R is in phase with the current, I, but that V_C lags the current I by pi/2.

The resultant voltage V can be given in terms of the current I, the resistance R and what we call the capacitative reactance (X_C):

V = I(R^2 + X_C), and we know: V_R = IR, and V_C = IX_C

The phase angle for the configuration is given by: tan F = V_C/V_R = X_C/ R

Or F = arctan (X_C/R)

Now, let’s turn to the synchronous condenser in the HTS setting, to see how it addresses issues, say for a grid. In this case we use Fig. 3 which features an inductor L, along with a resistance R. The AC source has an rms (root mean square) voltage V(o) and via angular frequency w = 2pi f couples to the resistive load through a reactance X(L) which we call the “inductive reactance”. This X(L) in a model represents transmission lines, transformers or other ancillary elements in a power grid. The resulting rms current I has real (in phase) and imaginary (out of phase) components, such that:

I = V(o) / R + I X(L) = I(o)[ cos F – I sin F]

Where F is again the designated phase angle

More specifically:

I(o) = V(o)/ [R^2 + X(L)^2]^1/2 and

cos (F) = R/ [R^2 + X(L)^2]^1/2

Note that the

P = V(o)I(o) cos F

There is also a volt-amp-reactive power:

(VAR)power = Q = V(o)I(o) sin F

Which originates with the imaginary, out of phase current. For inductive reactances, Q describes the average rate at which the magnetic energy is absorbed then returned to the source per cycle.

Now, if we compare the quantities P and Q we can quickly see what magnitude phase angle F and minimum current enables a given real power delivered to the load. Since we want the out of phase Q to approach zero, and the value of P to be maximal, this means F = 0 so sin F = 0 and Q = 0, hence vanishing imaginary power. At the same time P= I(o)V(o), since cos(F) = cos(0) = 1.

Why do we want Q = 0? Because this imaginary power plays a critical role in determining the magnitude of stress on the grid. (And if F increases,

What happens in a given power grid to cause a black out? Basically, the voltage drop across the load R decreases until the system becomes unstable and the voltage collapses. Considering the circuit diagram for the model (Fig. 3) as soon as many users simultaneously demand power – say for air conditioning – their many (parallel) resistance loads combine to reduce the load R.

Since V_L = IX(L), we see that the current and the voltage both drop across the inductive reactance. At the same time, since V_R = IR, the voltage drop across the load decreases until – when it goes below a critical threshold- the system destabilizes. We get a black out.

As we see from Fig. 3, the current lags behind the voltage for an inductive reactance, but from Fig. 2, the current leads the voltage. This provides a clue that a timely insertion or introduction of capacitative elements might prevent or mitigate the voltage collapse occurring in black outs. We will look at this more closely in Part II.

Note that the

*in-phase (real) component*determines the average power transmitted to the load:P = V(o)I(o) cos F

There is also a volt-amp-reactive power:

(VAR)power = Q = V(o)I(o) sin F

Which originates with the imaginary, out of phase current. For inductive reactances, Q describes the average rate at which the magnetic energy is absorbed then returned to the source per cycle.

Now, if we compare the quantities P and Q we can quickly see what magnitude phase angle F and minimum current enables a given real power delivered to the load. Since we want the out of phase Q to approach zero, and the value of P to be maximal, this means F = 0 so sin F = 0 and Q = 0, hence vanishing imaginary power. At the same time P= I(o)V(o), since cos(F) = cos(0) = 1.

Why do we want Q = 0? Because this imaginary power plays a critical role in determining the magnitude of stress on the grid. (And if F increases,

*additional current*is required to deliver the same power to the grid.)What happens in a given power grid to cause a black out? Basically, the voltage drop across the load R decreases until the system becomes unstable and the voltage collapses. Considering the circuit diagram for the model (Fig. 3) as soon as many users simultaneously demand power – say for air conditioning – their many (parallel) resistance loads combine to reduce the load R.

Since V_L = IX(L), we see that the current and the voltage both drop across the inductive reactance. At the same time, since V_R = IR, the voltage drop across the load decreases until – when it goes below a critical threshold- the system destabilizes. We get a black out.

As we see from Fig. 3, the current lags behind the voltage for an inductive reactance, but from Fig. 2, the current leads the voltage. This provides a clue that a timely insertion or introduction of capacitative elements might prevent or mitigate the voltage collapse occurring in black outs. We will look at this more closely in Part II.

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