__Choosing how to benefit from Triple Harmonic Injection (THI____)__

Power flow in a transmission line is approximately given by:

P = 3 V^{2} Sinδ / X

where V is the working line-to-ground voltage,

δ is the phase angle difference between the ends of the line and

X is the reactance of each phase of the line.

A phasor diagram illustrating the operation (and origin) of this equation is given above. The sending end phasor, E, is advanced by angle δ with respect to the receiving end phasor V. This creates a voltage difference between the sending end voltage waveform and the receiving end voltage waveform. This voltage when applied to the reactance of the line causes a current I to flow. The current is approximately in phase with V and so power is transmitted. To obtain exactly in-phase current E should be a little longer than V. The maths employed here is simplified to make a general point.

THI enables the amplitude of the fundamental of the line-to-ground voltage waveform to be increased by 15.6% with no accompanying increase in the amplitude of the overall (combined) waveform.

There are a number of ways this can be employed to advantage.

We assume that X remains unchanged but δ is variable (and V is increased in all cases by 15.6%).

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**(1) Keep δ unchanged. **

The maximum power capability of the line, given approximately by P = 3V^{2} Sinδ / X, is increased by 33%.

The power equation, P = 3V^{2} Sinδ / X, can be rewritten as P = 3 V ( V Sinδ / X ).

( V Sinδ / X ) corresponds closely to the current in each phase.

So increasing the line-ground voltage by 15.6% also increases the phase current by 15.6%.

Power loss in the wires due to resistance is given by Pd = I^{2}R.

Hence Pd is increased by 33% if the wire size is unaltered.

To bring the I^{2}R losses back to their original value the conductor cross-sectional area will need to be increased by 33% (i.e. the diameter of the wires will need to be increased by 15.6%).

The VA rating of the transformers at each end of the power line will need to be increased to match the new voltage and current requirements.

**(2) Reduce δ (reduce Sin****δ by 15.6%).**

The maximum power capability of the line, given approximately by P = 3V^{2} Sinδ / X, is increased by 15.6%.

The power equation, P = 3V^{2} Sinδ / X, can be rewritten as P = 3 V ( V Sinδ / X ).

( V Sinδ / X ) corresponds closely to the current in each phase.

So increasing the line-ground voltage by 15.6% increases the phase current by 15.6% but reducing Sinδ by 15.6% negates this.

Power loss in the wires due to resistance is given by Pd = I^{2}R and is unchanged since I is unchanged.

The VA rating of the transformers at each end of the power line will need to be increased to match the new voltage but not the current (unchanged). Hence in the retrofit situation the transformer voltage may be boosted on the THI-line side by booster transformers. The reduction in δ may also be welcome as stability (in response to a fault condition) is improved.

**(3) Reduce δ (reduce Sin****δ by 33%).**

The maximum power capability of the line, given approximately by P = 3V^{2} Sinδ / X, is altered by 0% (unchanged)

The power equation, P = 3V^{2} Sinδ / X, can be rewritten as P = 3 V ( V Sinδ / X ).

( V Sinδ / X ) corresponds closely to the current in each phase.

So increasing the line-ground voltage by 15.6% but reducing Sinδ by 33% results in a 15.6% reduction in line current (but the power rating of the system remains the same).

Power loss in the wires due to resistance is given by Pd = I^{2}R and hence the I^{2}R losses in the conductors (and the transformers) are reduced by 33%. The reduction in δ may be also welcome as stability (in response to a fault condition) is improved.

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**Similarities with quadrature voltage injection and related benefits.**

In the power equation, P = 3 V ( V Sinδ / X ), the voltage component, V Sinδ, is largely in quadrature (leading) with the receiving-end voltage waveform. It is applied across the line reactance X to produce a current in the lines which is largely (ideally) in phase with the receiving-end voltage waveforms, and hence conveys power.