Generator Voltage Drop
I am working on a project with generator, 650kW, 480/277V.
I have questions regarding generator voltage drop. I have read some articles in Internet and this forum and and one of them I copied below.
Originally Posted by bob
I most cases you could say that, but since your distance is so long you need to take into account the generator VD. A typical Xd' = 0.15.
125 amps x .15 x 1.73 = 33 volts. Check the genny and get the correct value.
It may be a smaller value.
In my case, from generator cut sheet Xd’ = 0.1798 is per unit
It means -> 1000 amps x 0.1789 x 1.73 = 309.5 volts.
Is this voltage drop correct or I made a mistake?
Moreover, the generator is about 400 ft away from the building.
I am definitely lost how to handle this issue.
Please advise, any thoughts will be great appreciated.
Thank you in advance.
I've never been good with synchronous machine theory, but I'm pretty sure that you are making a couple of small errors.
The synchronous reactance of a generator relates the open circuit voltage produced by the changing magnetic field to the terminal voltage given the actual current flow through the coils. So if you had a machine with _constant_ magnetic field strength then you could use it to calculate how the terminal voltage changes with different loads.
But with almost any generator used for electric power production, the magnetic field strength is not constant. Instead the magnetic field strength gets regulated to maintain the desired output voltage (or, in the case of paralleled generators, the field strength is regulated to maintain desired KVARs and power factor). Since the magnetic field strength is not constant, you can't use the reactance to calculate 'generator voltage drop'.
The generator output voltage regulation will be set by things such as the design of the voltage regulation system. Synchronous reactance will show up in things such as transient response and short circuit current.
I'm wondering exactly what you are trying to calculate? A 650KW generator doesn't have 1000 amps of capacity, which leads me to think you are trying to calculate the voltage drop for motor starting loads. Is that correct?
I think most manufacturers provide either curves or programs to calculate the generator voltage drop for starting large motors. You might try contacting the sales rep.
Originally Posted by Urko
When I read this I was shocked because I have no recollection of this post. I did some investigating and found it was actually written by IWIRE:grin:. It seems that he hacked the system and retrieved several of the passwords and has been posting under other names. So just disregard anything that he has posted. You do have an internal generator impedance, Xd but your regulator should compensate for any drop in the generator.
Thank you for clarification.
The generator voltage regulator should compensate the generator impedance and I should consider only the cable for voltage drop. Is this correct?
I use Caterpillar software for generator sizing but there is no consideration for cable voltage drop at all. In my case the generator is 400 ft away from the building. How can I be sure that generator will run properly?
Thank you in advance!
Originally Posted by steve66
The generator is 650KW, 0.8 power factor, 480V
Amps = (650 x 1.25)/ (1.73 * 0.480) = 978 Amps
I see. THe generator should be able to maintain 480 Volts out at the 1000 amps.
Originally Posted by Urko
I would just calculate the voltage drop along feeder. 400' usually isn't a long distance for 480V if you size the wire for the ampacity.