Not understanding why in overcurrent protection only a certain section will isolate (why not the whole system)

Question

I'm having an issue understanding a simple concept for faults. I understand the principles of how the relays cause the circuit breaker to open etc when a larger current goes through them to provide overcurrent protection.

But I don't quite understand why only the breakers of a certain section of a circuit will trip in some cases.

For example for the following Circuit:

If a fault occurs at the highlighted section, why don't all circuit breakers open to isolate the fault. Doesn't this fault cause a much larger current to be drawn from the generator (e.g 10x the rated current?) and so wouldn't the other breakers trip as well not just the section where the fault is located. Also how does the generator supply enough current for this fault in the first place?

Essentially I'm having trouble understanding why only this section is isolated.

Thanks In advance

Answer 1

Tl; Dr

The order of breaker trips is usually laid out in a selectivity calculation. Usually the closer to the source (generator in this case), the later the breaker will trip. The order of trip is not determined by the breaker's capacity, but by it's time current curve. This is roughly speaking a logarithmic multiplication of the overload current and the time it lasts.

However, in a ring-net system like mentioned in the question, it is almost impossible to be selective enough using "time current" to cover all configurations and possible faults. In such cases there are systems which determine the "location" (Zone selectivity) of the short circuit by measuring the direction of the current, as mentioned in Transistor's answer or in a more digital way in modern breakers. There are systems where breaker communicate this data to make a final determination which ones should open. Or other methods. This page from ABB has some nice graphic examples about advanced selectivity techniques.

Order of protection

More explained below, but looking to the diagram the following order of protections should kick in, depending on the seriousness of the fault:

1. Earth fault monitoring warning / alarm
2. Earth fault monitoring trip (optional)
3. Differential current protection trip
4. Overload protection as per time current curve and possibly zone selectivity
5. Short circuit protection according zone selectivity

More on time current curves

From this source, with more detailed explanation.

Time-current curves are used to show how fast a breaker will trip at any magnitude of current. The following illustration shows how a time-current curve works. The figures along the bottom (horizontal axis) represent current in amperes. The figures along the left side (vertical axis) represent time in seconds.

To determine how long a breaker will take to trip at a given current, find the level of current on the bottom of the graph. Draw a vertical line to the point where it intersects the curve. Then draw a horizontal line to the left side of the graph and find the time to trip. For example, in this illustration a circuit breaker will trip when current remains at 6 amps for 0.6 seconds.

It can be seen that the higher the current, the shorter the time the circuit breaker will remain closed. It can be seen from the time-current curve on the following page that actual time-current curves are drawn on log-log paper, and the horizontal line is in multiples of the breaker’s continuous current rating. From the information box in the upper right hand corner, note that the time-current curve illustrated on the following page defines the operation of a CFD6 circuit breaker.

For this example a 200 ampere trip unit is selected.

Overload vs Short circuit

An overload constitutes an excess of AC current during an amount of time. Those are thermal protected according the time current curves noted above. Short circuits however are more violent and are magnetic protected. The above mentioned sources (and google) also give you more info about this subject, which is an important background.

Diagram in question

The above part hopefully answers the text part of the question, which specifically mentions over-current trips. However, the diagram attached to the question actually has an earth fault. This means 1 or more phases is connected to earth and doesn't necessarily creates an overload or short circuit current.

Measurement of earth faults

The insulation value of the installation's insulation level is usually monitored and alarmed by an earth fault protection system. Depending on the installation requirements an earth fault can lead to a trip, or not. I found a data sheet for such a device on the market, which gives some more background on this subject.

Single phase earth fault

This should not give any overload or short circuit currents. However, it may affect the insulation of the installation cables. When 1 phase is connected to earth, it's phase-earth potential becomes 0V. The other 2 phases will have an potential voltage to earth multiplied by √3. If the installation is not capable of handling this increase of voltage, the fault should be terminated by opening breaker.

From this article:

These insulation levels are discussed as follows:

100% level:

Cables in this category may be applied where the system is provided with relay protection which normally clears ground faults within 1 minute This category is usually referred to as the grounded systems.

133% level:

Cables in this category may be applied where the system is provided with relay protection which normally clears ground faults within 1 hour This category is usually referred to as the low resistance grounded, or ungrounded systems.

173% level:

Cables in this category may be applied where the time needed to de-energize the ground fault is indefinite This level is recommended for ungrounded and for resonant grounded systems.

If the installation uses cable in the first, 100% category, the earth fault measurement should cause a breaker trip. In the other 2 cases it will just raise an alarm to the plant's operators and they should intervene within the time limitations noted above. Usually earth faults are logged to not exceed the cable's service life.

Differential protection

If there are any currents to earth or mismatch in phase currents, which doesn't necessarily case an overload or short circuit, there is a differential protection which should trip breakers. The question already includes a differential measurement around the fault, which means that this should act and not any over-current relay, only if there is in fact a current to earth. A nice read on this subject can be found here.

More reading

It can be a nice start to read about the various levels of protection in this wikipedia article and use google on the terms used.

Answer 2

Why only the breakers of a certain section of a circuit will trip.

It is simple logic. By most anyone's regulations the most downstream breaker should be the one to trip OFF if the overload is part of its load(s).

There is an intentional cascading topology where the main breakers have the highest rating and are the least likely to trip. A 1200 amp main service entrance panel could be full of mostly double/triple pole breakers in the 50 to 250 amp range. The master breaker would be 1200 amp and it is large and tough to manually switch on/off.

A 250 amp breaker to a sub-feed panel (there could be many sub-feed panels in a manufacturing plant) basically protects the sub-feed panel, which has rows of 15 to 30 amp breakers to protect lights and wall outlets, in the sense that a short or arc fault has occurred and the breaker trips to prevent a small problem from turning into a large one.

This cascade effect insures that only the lowest rating breaker closest to the shorted device will trip.

Ultimately breakers and fuses mostly exist to prevent wires from burning and causing a fire. They seldom can protect semiconductors. There are expensive platinum strip fuses designed to protect large SCR's and MOSFET modules.

Answer 3

I can't speak with any great authority on this but the following thoughts occur to me.

Figure 1. Numbered diagram.

• The layout of your diagram has two paths between the 220 kV and 38 kV busbars. Ideally a single line fault should trip out only the faulty section.
• In your case the fault is between the transformer XFMR2 and the circuit breaker X4.
• By monitoring power flow at X4 we can determine whether the fault is to the left or right. In normal operation power should flow from left to right. Only in a fault condition will power flow from right to left.

This poses a little design challenge. How can we determine the power flow direction on an AC current when it is alternating by nature? The answer is to use a directional relay.

Figure 2. Induction disk overcurrent relay. Source: Open Electrical.

I can't find a good image of a directional fault current relay but the illustration in Figure 2 may suffice to explain the principle of operation. (The illustration is for an over-current relay and torque works against a spring. For a directional relay the spring isn't required as the relay should just swing fully open or fully closed depending on power flow direction.)

A rotating disc is driven by eddy currents induced by a pair of coils - one fed from line voltage and the other by a CT from the line current. The torque on the disc will be proportional to the vector product of the voltage and current and the direction of rotation will be determined by the relative phases. In the illustration the contacts will close when the disc moves clockwise so all that is required is to arrange the voltage and current phasing to open the contacts in normal power flow.

The contact of the directional relay can then be used to suppress or enable features such as earth fault tripping on a circuit breaker. In your example the contact would be open on X3 but closed on X4 to ensure that the correct breaker trips.