When you have a standby genset with an alternator that is larger than what the generator is rated for, do you use the generator rating or the actual alternator rating?

Example. A genset is rated for 500kw. The actual alternator used is rated for 660kw/825kva. The diesel engine driving the alternator is not large enough to produce 660kw. It is only capable of 500kw. In your model do you model as 660kw or 500kw (along with all the decrement curve and time constant data from the alternator data sheet) for the purposes of fault contribution calcs, not load flow?

Hey Brent...

I would think you would use your engine rating for kW and use your other values, e.g. Xd'', etc. of your generator. I've never run into your issue before...

Mike

PS I haven't found any reference for my suspicion...

Thanks for the suggestion, Mike. Not sure that is appropriate or not. I've run into this twice in one week. Seen it a few times before. Here's why I'm skeptical of this. If the alternator is spinning at 1800 rpm when the fault occurs and the fault contribution is so brief due to the fast decay, does the HP of the engine really matter for fault contribution purposes?

I just did a quick test, and the KW you input into SKM definitely affects the contribution with all the decrement curve and time constant data being identical between the two. What am I missing here?

This is what the generator rep told me when this came up a year ago. I don't recall what I did for that model. Like I said, I've had this two more times within the last week. Once more with a Kohler generator and once with a Cummins.

After thinking about this and looking at the data sheet for one of the generators, I think I may have answered my own question. I think you have to use the alternator rating, not the entire generator set rating. My reasoning for this is, the alternator will have a "base" KVA and the decrement curve data is often given in per unit.

The alternator manufacturer doesn't necessarily know what engine the alternator will be paired with and what the final genset output rating will be. They only know what their KVA rating is at a particular voltage. Data sheets are published for the alternator. I think therefore, you must use the alternator kva rating in the SKM model for the per unit decrement curve information to be accurate.

At least for fault current modeling, not for load flow.

Make sense anyone? Agree / disagree?

Here is the nameplate and the data sheet for a 480V, 500 kw genset. As you can see, the alternator is 660kw rated at 480V. The engine-alternator combo is only 500 kw.

Brent,

But doesn't your application engineer's email confirm my first post? If the engine is the limiting factor, it won't be able to supply more than what the engine can output?

Mike

I don't know. Can it? We're talking fractions of a second. Just a few cycles or so, or until a breaker tripped. Does the engine HP make any difference? I think the inertia of the alternator and its kva rating would be the determining factor.

Brent...

I put this question into ChatGPT and here's what it spit out:

In a **short circuit current study**, you should use the **alternator (generator) rating and its impedance characteristics**, not the engine rating.

---

### Here’s Why:

1. **Short Circuit Studies Focus on Electrical Output:**

* The **engine (prime mover)** provides mechanical power, but it doesn’t directly impact the **electrical fault current**.
* During a short circuit, the **alternator’s electrical characteristics** (particularly its **subtransient reactance**) determine the **initial fault current magnitude**.

2. **Generator Constants Matter:**
The key constants for an alternator in a short circuit study are:

* **X''d (Subtransient Reactance)** – Determines the **initial fault current** (first few cycles).
* **Xd' (Transient Reactance)** – Affects the current after a few cycles.
* **Xd (Synchronous Reactance)** – Governs the longer-term behavior (seconds after the fault).

3. **Alternator Rating Is the Starting Point:**

* Use the alternator’s rated **kVA** (or kW if PF is known) and its **subtransient reactance (X''d)** in per unit or percentage based on its own base (usually its own rating).
* Fault current (initial) can be estimated as:

$$
I_{sc} = \frac{I_{\text{rated}}}{X''_d}
$$

Where $I_{\text{rated}}$ is the rated current of the alternator.

---

### Note About the Engine:

While the engine is undersized relative to the alternator (500 kW vs 660 kW), during **short-circuit conditions**, the engine’s mechanical limitation isn’t the bottleneck because:

* The engine’s inertia and governor response are too slow to limit the **initial fault current**.
* The alternator delivers high fault current using its stored **rotational energy (inertia)** and magnetic field collapse, independent of immediate engine torque.

In prolonged fault conditions, **engine capability** might matter in terms of **sustained current**, but for most **initial fault analysis**, it’s not limiting.

---

### Summary:

* **Use the alternator’s rating** (660 kW or its kVA equivalent) and impedance (X''d, etc.).
* The **engine rating is not used** in short circuit current calculation.
* Alternator constants (especially X''d) are **crucial** for calculating fault currents.

If you have the actual impedance data (e.g., subtransient reactance in % or per unit), I can help you calculate expected fault currents.


It's from ChatGPT...so it must be right!

Mike

That's great!

Never thought to use AI for an answer. In fact, I have yet to use ChatGPT or any other AI. Might have to start though.

Thanks for sharing that!