output of alternator - depending on cos phi of its load? 1

[kingnero]

I am currently in the middle of buying multipe generators for welding purposes. Power wise, I am talking of 20-25 gross HP fuel engines, combined with 14-16 kVA alternators.
Perhaps an alternator is what is called a generator in english? I'm not too sure about the translation.

We are currently having trouble at understanding why a generator that is rated (and tested) at 14 kVA at cos phi = 0.8, cannot deliver 14 kVA with a resistive (cos phi = 1) load (which would equal 14 kW).

So, some questions that will probably help me understand this better:

1) Does the current an alternator can deliver, depends of the type of load (resistive or capacitive)?
2) Is there a change of efficiency of the alternator depending on the cosine phi of the load?
3) Is there a physical difference between current that is out of phase with the voltage and current in phase with voltage ? Does the alternator has to "work harder" in the former or latter case?
4) If the alternator can deliver, for instance, 14 kVA with a load of cos phi = 0.8 ; (and assuming the fuel engine is large enough for this) does that mean the alternator can deliver 14 kVA at either value of cos phi between 0.5 and 1 ?

thanks in advance...

 

[waross]

KVA describes the ability of the generator or alternator to service the load without overheating.
kW describes the power of the prime mover to produce real power (kilo-Watts).
The alternator will deliver 14 KVA within its capability curve which will probably include PF 50% lagging to 100%.
However a 14 KVA @ PF 0.8 may have a prime mover capable of only 14 x 0.8 = 11.2 kW at PF 1.
The prime mover may stall if you try to produce more than 11.2 kW
Yes, if you have an oversized engine the set will produce 14 kW or 14 KVA @ PF 100%

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[kingnero]

However a 14 KVA @ PF 0.8 may have a prime mover capable of only 14 x 0.8 = 11.2 kW at PF 1.

This part, I do not understand.
If the set is capable of delivering 14 kVA at PF=0.8, then why not at PF=1 ? This is what I am struggling with, and this question is also the summary of my opening post.
The current that runs at 14 kVA, remains the same, no matter what PF ? or am I mistaken here?

 

[Skogsgurra]

1) The current that the generator is designed for is the current that the generator can deliver without overheating. It is the rated current and it is the same regardless of the type of load. Capacitive, resistive, inductive or mixed doesn't matter.

2) The efficiency is not the important thing here. The main thing is how much work the genset can deliver. If it can deliver 14 kVA at cos(phi)=0.8, it produces 0.9*14=11.2 kW of real work. So, you cannot get more work out of it - mainly because your diesel was designed to produce 11.2 kW at its output shaft (plus some extra power to cover the losses in the generator).

3) The current looks the same. And the generator does not need to work harder when the cos(phi) changes. It is the diesel that needs to work harder when cos(phi) gets close to unity.

4) Yes, but then the diesel engine needs to be "strong enough" to deliver the necessary power.

There are a few subtle details that may influence the outcome. It can be read off the capability curve of the generator. But I don't Think that will be important in your case.



Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[kingnero]

Thank you very much for the detailed answer.

So, if I understand your remarks (and most of all, remark #3) correctly,
it is not the kVA rating of the set that is important, but rather the kW rating that is a constant value ?

With a diesel engine rated at 11 kW (+ some extra to cover the losses):
If I keep my load close to, lets say 11 kW "on the real power axis", this means I can hook up inductive loads that are equivalent to 11 kW (18.3 kVA with a PF = 0.6 ; or 13.8 kVA with a PF = 0.8 ) ?

I will already be thankful to understand the general idea, I doubt the subtle details are meaningful to me at this moment.

 

[LionelHutz]

Why do you think the alternator can't produce 14kW @ 1.0 power factor? Is this from the manufacturer?

 

[kingnero]

Spec sheets indicate 14 kVA at PF = 0.8 and a max current for resistive loads (16 amps at 3phase 400V => sqrt(3) . 400 . 16 . cos phi = 11.1 kW)
Field tests have proven the 16 amps with a resistive load. This was the machine's practical maximum load.
We do not have the facilities to test a 14 kVA / PF = 0.8 claim. But I wouldn't know why or how the set would be able to produce the 20 amps in order to achieve this.

But thank you for the question, this is indeed very important to know.

 

[LionelHutz]

If it's in the specifications then asking the manufacturer why they specified it should get the right answer.

 

[kingnero]

Manufacturer can't explain this. When they attended the tests at our plant, they were surprised as well.
When we asked the producer of the fuel engines, they did not know the first thing about electricity, so again no answer from that side .
What I need is a 14 kVA / PF = 0.8 load in order to check the specifications. Don't have anything like that nor sufficient engines + load to put them in parallel to simulate this total load.

 

[Skogsgurra]

You cannot look at power alone. You also need to look at rated current.

The kW say how much "real work" you can get from the set. And that may either be limited by the diesel engine OR the generator's rated current. The first limit reached limits the output. There's also an excitation current limit.

You need to calculate (or read the nameplate) to find out what the rated current is. Then you can decide which limit you hit first in all different load situations. That is actually what you do when you draw a capability diagram. Have a look at the second frame from the end in this presentation:

http://www.egr.unlv.edu/~eebag/SYNCHRONOUS GENERATORS.pdf

It says what you need to consider when dealing with this kind of problems.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[itsmoked]

kingnero, I think you are missing a key point. A subtle detail.

KVA is describing two kinds of power. One requires engine horsepower(kW) and the other requires magnetic capability (kVAR) and no horsepower at all because it does no work, it is only there to metaphorically grease-the-rails so work can be done.

So if these two "powers" can add up to 14kVA it should be understandable that the kW one is not likely to equal 14k anything by itself. Well, that 14kW, if it could happen, would require the horsepower needed to provide it. (plus losses).

So the maker can kind-of 'specsmanship you' by trumpeting 14kVA which is a bigger number than '11kW' to the potential buyer.


In the manufacture's defence they are telling you the magnetic capability of their machine by stating the kVA and power factor. This helps a savvy buyer know more details about the machine. For instance, if the machine was capable of 14kW but had zero magnetic capability, above servicing a pf = 1 load, you could burn up the machine as soon as you tried running a 14kW load that actually wasn't pf = 1. (and thereby needed a bigger magnetic circuit including the wiring)


Keith Cress
kcress -

http://www.flaminsystems.com

 

[LionelHutz]

The manufacturer of the alternator can't explain why they put the ratings on the nameplate? Seriously?

You've got lots of power in the prime mover so my guess would be that the excitation field is being limited to limit the output. But then, not knowing how you tested the set to get 16A as the limit makes it a pretty simple guess. Did you actually increase the load until you reached the minimum acceptable output voltage or was that load just the most load you had available?

 

[waross]

I've done generators a few times. I may be restating Skogg's answer somewhat. The generator output is lmited by current.
This is determined by dividing the VA by the rated voltage and by root 3 for three phase.
That is 14000VA/400V/1.73 = 20 Amps. The generator will deliver 20 Amps without overheating.
If this generator (the same physical machine) was rated for 480 Volts, the KVA rating would be 16.6 KVA, but still 20 Amps maximum.
Again if rated for 380 Volts or 416 Volts the KVA rating would change but the maximum current of 20 Amps would stay the same.
Now power factor describes the action of the circuit when the current is not in phase with the voltage.
When the current is not in phase with the voltage the actual power in Watts is less than the apparent power calculated by VxA.
The cosine of the phase angle difference between the voltage and the current describes the ratio between real power (kW) and apparent power (VA).
Small sets are generally rated at 100% PF.
At around 14 or 15 KVA, the load is generally partly inductive, with a power factor of less than 100% or unity.
Set manufacturers have found that so few 14 KVA sets are called upon to produce 14 kW, even at 14 KVA that it is almost universal practice to rate sets at 80% power factor and so the 14 KVA set will be supplied with an engine capable of producing 80% of 14 KVA, or 11.2 kW.
If you need more than 11.2 kW then either buy the next larger set or special order a set with the next larger engine installed.
That said many small sets have a slightly oversized engine.
Connect a resistive load and increase the load up to the point that the engine starts to stall. That will tell you what real power the set is actually capable of.
The main reason to test at 80% PF is to check for overheating.
You can check for overheating by setting the voltage down to 80% of rated voltage or 320 Volts. Now you should be able to use a resistive load to run the current up to 20 Amps to check for overheating.
Note: At 320 volts the field current is less and the field heating will be a little less. The difference in the field losses may be in the order of 10 W or 20 W so the effect of the field heating is negligible.

Bill
--------------------
"Why not the best?"
Jimmy Carter