Generator Protection - Selection of Fuse based on Sustained Current

[Gen_61]

Hello everyone

I am designing a protection system for a local grid with one synchronous generator and consumers.

A mechanical fuse (for example, NH 100 A) is installed at every consumer, and one fuse at the generator. A PMG supplies the AVR system, and thus, the generator has short-circuit capabilities by providing a sustained short-circuit in a given period, depending upon the type of short circuit. The Short-Circuit Decremental Curve is to be found at https://www.stamford-avk.com/sites/stamfordavk/files/AGN005_D.pdf in Figure 1 as an example.

The initial maximum short-circuit current is based on the sub-transient reactance Xd'' of the generator . The sub-transient and transient periods decay rapidly due to the interactions between the damper winding and the field winding and the stator winding.
The t'd is 0.03 s, t''d is 0.008 s, and Ta is 0.007 s. I would not expect the fuse to react within t'd + t''d before entering the sustained current state (steady-state). A circuit breaker may react in the instantaneous trip region.
So, I assume that the value of the sub-transient nor the transient currents should not be used to check the breaker time. Instead, I will be using the value of the sustained current. This is a safer approach by guarantee to trip on the sustained current, and hence, covering regardless of the earlier sub-transient spike.

Should I use the value of the sustained current from the Decremental Curve in order to determine the breaker time? I think it would be wrong to use the maximum initial sub-transient current, as this is only a peak value. Also, the generator impedance can be calculated in the sustained period from the decremental curve. I will be using this impedance in order to calculate the short circuit at any buses in the grid. What do you think of this?
The Decremental Curve is only based on a short circuit at the generator terminals.

This approach is inspired by this thread https://www.eng-tips.com/threads/sustained-short-circuit-current-from-synchronous-generator.468321/

 

[wcaseyharman]

It depends. On a close in bolted 3PH fault an instantaneous breaker would likely trip. A single phase fault away from the generator might not generate enough current to trip the instantaneous element, and if it’s tripping on the breaker thermal curve then I would expect it to be in the sustained region.

If you’re looking for accurate trip times on thermal curves with changing fault current, the proper way is to integrate over time as the fault current is changing. There’s a good discussion of this in “Generator Protection” by Donald Reimert. I’ve never had to do it myself, checking the maximums in minimums in a fault program has been sufficient.

 

[dpc]

Why are you relying on fuse? You will need a circuit breaker for synchronization, I assume.

To determine the sustained fault current, you also have to consider the excitation system. Many generator decrement curves are based on a constant excitation, but many small to medium generators provide some type of current boost system (often the exciter PMG) to allow something like 300% current for 30 cycles or so. This allows for proper fault clearing. For many generators the synchronous reactance is over 100% meaning that the sustained fault current is less than full output current. This is normally handled by voltage-restrained or voltage-controlled overcurrent elements in a protective relay. Ideally, the fault is cleared long before this backup protection operates.

 

[Gen_61]

Why are you relying on fuse? You will need a circuit breaker for synchronization, I assume.

To determine the sustained fault current, you also have to consider the excitation system. Many generator decrement curves are based on a constant excitation, but many small to medium generators provide some type of current boost system (often the exciter PMG) to allow something like 300% current for 30 cycles or so. This allows for proper fault clearing. For many generators the synchronous reactance is over 100% meaning that the sustained fault current is less than full output current. This is normally handled by voltage-restrained or voltage-controlled overcurrent elements in a protective relay. Ideally, the fault is cleared long before this backup protection operates.

The sustained current shown on the decrement curve (https://www.stamford-avk.com/sites/stamfordavk/files/AGN005_D.pdf) is at the generator terminals (shorted), and thus, the current will be less at the faulted bus due to the line
impedances, right? This curve states how the maximum “clearing time”, which I assume is the boosting time of the excitation system.

I do not have access to protection software to perform the calculation—it must be done by “hand calculation”. I was thinking of calculating the generator's reactance in the sustained region at the terminals (stated in the decrement curve) and using this to estimate the true sustained current at any given bus in the grid. Would this be a proper way to proceed?

 

[waross]

The dozens of standby generators that I have installed have a factory supplied and installed circuit breaker sized at or slightly above the full load current of the generator.
This is the overload protection.
This breaker may provide short circuit protection or you may need additional short circuit protection.
If you add short circuit protection, it depends on the generator.
Is the generator:
Self excited?
PMG excited?
Use current transformer excitation boost?
The sub-transient and the transient current offer small time windows of opportunity for operation.
If for any reason you miss the time window for a trip, it is unlikely that you will later get enough current for a trip.
Voltage collapse may drop the short circuit current below the full load current level and even the overload protection will not trip.
Each installation must be evaluated on its own merits.
No one answer fits all.
But as a tip: Adding CT boost to the excitation may be part of a short circuit protection scheme for sets subject to voltage collapse.