Ze of Generator

Cookie · May 17, 2021

How do you find the the Ze of a diesel generator when power a building, for example a 400kw unit?

In so far I have FLC of 1,110 amps. 7.5 (L-E) x 1,110 = 8327. 120/8327= 0.015 ohms

Somehow this seems just to easy.

timhoward · May 17, 2021

Since no one else is biting...I don't think I'm quite understanding the question...and don't think I'm completely alone in this.
Are we talking a generator wired as a TN-S setup, with one end of windings (neutral) earthed at the generator? Or is it some form of centre-tapped earth?
What is making this different from measuring Ze on a normal supply?

Is the 7.5 you mention the Zs@Db in ohms? Is this an 120v supply?
Or is the focus of the question that you are unable to disconnect loads to determine it?
Sorry for loads of questions, I'm hoping answering a couple of them will facilitate more responses for you!

Lucien Nunes · May 17, 2021

As mentioned in the thread about ADS, there is not just a single figure, since the expected variation in load from zero to full to fault conditions significantly change the behaviour of the generator. In the event of a fault, the behaviour then changes over time in a way that the public supply network does not.

Manufacturers typically give separate impedance figures for sub-transient, transient and synchronous conditions, respectively valid for a few cycles, a few seconds, and continuously. I have a link somewhere to some good reading on the subject with examples, maybe I can find it...

timhoward · May 17, 2021

This is probably a terminology thing - in my mind the term Ze meant the impedance of any external earth electrode and earthing conductor up to the point the distribution starts in the building. Is that not a fixed resistance whatever the generator is doing?
Or maybe I'm still missing the point of the question!

Cookie · May 17, 2021

timhoward said:
Since no one else is biting...I don't think I'm quite understanding the question...and don't think I'm completely alone in this.
Are we talking a generator wired as a TN-S setup, with one end of windings (neutral) earthed at the generator? Or is it some form of centre-tapped earth?
What is making this different from measuring Ze on a normal supply?

Is the 7.5 you mention the Zs@Db in ohms? Is this an 120v supply?
Or is the focus of the question that you are unable to disconnect loads to determine it?
Sorry for loads of questions, I'm hoping answering a couple of them will facilitate more responses for you!

TN-S, wye configuration, 120/208 volt supply.

Goal is to achieve required disconnection time. And to help with selective discrimination.

Cookie · May 17, 2021

timhoward said:
This is probably a terminology thing - in my mind the term Ze meant the impedance of any external earth electrode and earthing conductor up to the point the distribution starts in the building. Is that not a fixed resistance whatever the generator is doing?
Or maybe I'm still missing the point of the question!

View attachment 85842

Technically Ze is the external loop impedance. A simple multi function tester will confirm it. Add to R1+R2 and you know that fault current you are dealing with in relation to opening an MCB.

However, with a generator, the fault current is coming from a single stator and not the National Grid, so things will be a bit different but unsure as to how.

timhoward · May 17, 2021

I understand now. So it comes down to whether the generator can generate sufficient fault current to trip the MCB in required time without stalling / frying, and as Lucien pointed out there are three states to consider.
This web page is a maybe a helpful starting point: Calculating the short-circuit current across the terminals of a synchronous generator - https://tinyurl.com/cbtsxvn9

Rockingit · May 17, 2021

As I think you've now realised, in simple terms you'll never achieve final circuit ADS times on a generator, hence reliance on RCD's for protection instead. The reason is simply down to the mechanical reaction time of the generator itself and the larger the set then the worse the problem, unless you happen to be running a chain of sets so high already that Ipfc = In. You need upstream electronic protection devices to interrupt the current flow, not simply the supposed Zs on a downstream MCB.

Cookie · May 17, 2021

timhoward said:
I understand now. So it comes down to whether the generator can generate sufficient fault current to trip the MCB in required time without stalling / frying, and as Lucien pointed out there are three states to consider.
This web page is a maybe a helpful starting point: Calculating the short-circuit current across the terminals of a synchronous generator - https://tinyurl.com/cbtsxvn9

Correct

Cookie · May 17, 2021

Rockingit said:
As I think you've now realised, in simple terms you'll never achieve final circuit ADS times on a generator, hence reliance on RCD's for protection instead. The reason is simply down to the mechanical reaction time of the generator itself and the larger the set then the worse the problem, unless you happen to be running a chain of sets so high already that Ipfc = In. You need upstream electronic protection devices to interrupt the current flow, not simply the supposed Zs on a downstream MCB.

At some point a large enough generator has to trip an MCB. The more down stream circuits I can get to trip the better even if its not possible to clear every MCB upstream. As I'm calculating it a generator can produce 7 times the rated FLC for several seconds, which is enough if played correctly. RCDs on a generator are a very bad idea, leakage current or a high impedance fault should not be able to clear emergency circuits.

Cookie · May 18, 2021

Rockingit said:
As I think you've now realised, in simple terms you'll never achieve final circuit ADS times on a generator, hence reliance on RCD's for protection instead. The reason is simply down to the mechanical reaction time of the generator itself and the larger the set then the worse the problem, unless you happen to be running a chain of sets so high already that Ipfc = In. You need upstream electronic protection devices to interrupt the current flow, not simply the supposed Zs on a downstream MCB.

Picture it like this:

A 400kw unit has an output current of 1,110 amps. If the generator can export at least 300% current for 10 seconds then 1,110 amps x 3 = 3330 amps. It takes about 8 times the handle rating to trip a breaker in 5 seconds- so 3,330 / 8 = 416 amps. Thus we know that the maximum breaker that we can have the in emergency (life safety) switchboard is 400 amps.

Am I correct to say that a 400 amp breaker will trip for a fault within 5 seconds before a separately excited permanent magnet synchronous generator drops its excitation or shuts down?

Maybe an example drawing will help clear things up (370kw is the prime rating, 400kw standby rating):

Cookie · May 18, 2021

From the site specs:

400REOZJC, 60 Hz | Industrial Diesel Generators | KOHLER

kohlerpower.com

Lucien Nunes · May 18, 2021

Am I correct to say that a 400 amp breaker will trip for a fault within 5 seconds before a separately excited permanent magnet synchronous generator drops its excitation or shuts down?

It's not just about the generator actively changing the excitation, but also about the demagnetising effect of the fault current which in turn depends on the fault current pf. If low, the output winding reaction MMF more strongly demagnetises the machine with rapid reduction of terminal voltage during the transient period and also reduction of engine load. So it can maintain speed and deliver low real power in a kind of 'folded-back' state, which might occur while you are still waiting for the breaker to trip. Obviously the pf depends on the winding L and R and tends to decrease with increasing machine rating,

I don't have practical experience of speccing this size of unit for utility replacement duty and wouldn't like to call your particular situation. I would go from the actual curves of the specific machine and breaker rather than using any generalised multipliers.

Cookie · May 18, 2021

Lucien Nunes said:
It's not just about the generator actively changing the excitation, but also about the demagnetising effect of the fault current which in turn depends on the fault current pf. If low, the output winding reaction MMF more strongly demagnetises the machine with rapid reduction of terminal voltage during the transient period and also reduction of engine load. So it can maintain speed and deliver low real power in a kind of 'folded-back' state, which might occur while you are still waiting for the breaker to trip. Obviously the pf depends on the winding L and R and tends to decrease with increasing machine rating,

I don't have practical experience of speccing this size of unit for utility replacement duty and wouldn't like to call your particular situation. I would go from the actual curves of the specific machine and breaker rather than using any generalised multipliers.

You are big help none the less

You know more about this stuff than I do.

Regarding PF, does the jX of the external conductor or fault loop make a difference?

pc1966 · May 19, 2021

Cookie said:
Regarding PF, does the jX of the external conductor or fault loop make a difference?

It will, but whether it makes a significant difference is less clear.

Your example 400 kcmill supply cable is a little bigger than our 185mm (for which I have a table handy) where the cable Z is about 20% higher than R due to inductance, where as the 4/0 AWG cables are a little bigger than our 95mm size where the difference is around 5%.

With large cables, transformers, etc, it is always best to run the calculations using the full information of X & R values and then see what comes out, even if in reality something like a 5% difference is likely lost in the uncertainty of fault magnitude, etc.

Lucien Nunes · May 19, 2021

The machine's own resistance and reactance would normally dominate for a bolted fault at the main panel. The stator reaction depends on the magnitude and phase of the current w.r.t the EMF, not the terminal voltage. The difference between the two is no longer nominal nor hidden behind the AVR.

Cookie · May 19, 2021

pc1966 said:
It will, but whether it makes a significant difference is less clear.

Your example 400 kcmill supply cable is a little bigger than our 185mm (for which I have a table handy) where the cable Z is about 20% higher than R due to inductance, where as the 4/0 AWG cables are a little bigger than our 95mm size where the difference is around 5%.

With large cables, transformers, etc, it is always best to run the calculations using the full information of X & R values and then see what comes out, even if in reality something like a 5% difference is likely lost in the uncertainty of fault magnitude, etc.

Alright and agree.

However, what R & X to I assume for the generator windings when doing the equation to get the total Zs? And what voltage? I know this is a dynamic 4-D scenario, but I want to keep it simple (if practicable) to a 2-D set of equations.

Cookie · May 19, 2021

Lucien Nunes said:
The machine's own resistance and reactance would normally dominate for a bolted fault at the main panel. The stator reaction depends on the magnitude and phase of the current w.r.t the EMF, not the terminal voltage. The difference between the two is no longer nominal nor hidden behind the AVR.

As I am reading up on it, and I could be wrong, the kw on the prime mover actually drops due to the R going down and the X going up. The current is reactive, and as the voltage goes down the field is "forced" to its max excitation to keep voltage up.

Cookie · May 23, 2021

Cookie said:
As I am reading up on it, and I could be wrong, the kw on the prime mover actually drops due to the R going down and the X going up. The current is reactive, and as the voltage goes down the field is "forced" to its max excitation to keep voltage up.

Anyone know? R goes up as shirt circuits move away from the generator.

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