Generator KW

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delectric123

Senior Member
Location
South Dakota
I recently tried out a 120/2403 ph. delta 100 kW standby generator for a hog facility. On the generator's digital readout the load was at 50 kW at a 0.93 PF. Does that mean the generator could actually safely put out more than 100 kW? Just curious......
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
130525-1511 EDT

A generator is going to be limited by current rather than power. So for your generator if the continuous rating is 100 kW at 240, then maximum continuous load current should not be above 416 A. An 0.5 PF load would have to be limited to 50 kW. This is a thermal problem related to heating of the generator windings.

.
 

delectric123

Senior Member
Location
South Dakota
that makes sense. The generator is rated at 100 kw with worst case scenario 0.8 PF. At 240 v, that would be 300 amps. So if the PF is less than 0.8, then the generator will not be capable of 300 amps without overheating.
 

eHunter

Senior Member
130525-1511 EDT

A generator is going to be limited by current rather than power. So for your generator if the continuous rating is 100 kW at 240, then maximum continuous load current should not be above 416 A. An 0.5 PF load would have to be limited to 50 kW. This is a thermal problem related to heating of the generator windings.

.

If I am not mistaken, the 416 A would be for single phase at unity PF and the OP stated he has a 3 phase genset which is good for approximately 240.84 A.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
130525-1541 EDT

100 kW at 0.8 PF and 240 V is a current of 100,000/(0.8*240) = 520 A.

This means the generator can supply 100 kw to a load with an 0.8 PF. Thus, the current available is greater than for a resistive load. It means the engine is the same as for a 100 kW resistive load, but the generator is larger than what could be used for a resistive load.

.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
I recently tried out a 120/2403 ph. delta 100 kW standby generator for a hog facility. On the generator's digital readout the load was at 50 kW at a 0.93 PF. Does that mean the generator could actually safely put out more than 100 kW? Just curious......

If the 100kW rating is continuous, then the generator can safely put out a lot more power for short periods of time.
Long term the two limits are generator heating and maximum power output of the engine. The latter will depend on the condition of the engine and your willingness to use up its rated useful life.
Very short term, the limit is the rotational inertial of the engine and generator and any overcurrent devices present.
Medium term, there may be a limit to the engine's ability to keep the generator within your required frequency and voltage limits.
 

delectric123

Senior Member
Location
South Dakota
130525-1541 EDT100 kW at 0.8 PF and 240 V is a current of 100,000/(0.8*240) = 520 A.This means the generator can supply 100 kw to a load with an 0.8 PF. Thus, the current available is greater than for a resistive load. It means the engine is the same as for a 100 kW resistive load, but the generator is larger than what could be used for a resistive load..
Its 3 ph. you forgot to multiply by the factor of 1.732
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
130525-1554 EDT:

eHunter:

I totally lost the 3 phase part because of the 120/240.

Corrections: If it is 240 leg to leg, then leg to theoretical midpoint is 240/1.732 = 138.6 . Thus, line current for 100 kw resistive is 100,000 / (3*138.6) = 240.5 . At 0.8 PF this is 300 A for 100 kW.

delectric123:

If the power factor was less than 0.8, then the maximum power would be less than 100 kW. You would need to stay within the 300 A per leg rating. Effectively the engine becomes over-sized.

.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
delectric123:

If the power factor was less than 0.8, then the maximum power would be less than 100 kW. You would need to stay within the 300 A per leg rating. Effectively the engine becomes over-sized.

.

Depends in part on whether there was a power factor rating in the specs of the generator.
If the standard is that a generator shall deliver its rated power output into a load with PF of .8 or higher, then you are exactly right. But can we be sure that this generator was rated that way?
The .8 power factor limitation can also be interpreted as a current limitation of 1.25 times the load current for full power into a resistive load. But an OCPD cannot tell the difference.
 

delectric123

Senior Member
Location
South Dakota
130525-1554 EDT:eHunter:I totally lost the 3 phase part because of the 120/240.Corrections: If it is 240 leg to leg, then leg to theoretical midpoint is 240/1.732 = 138.6 . Thus, line current for 100 kw resistive is 100,000 / (3*138.6) = 240.5 . At 0.8 PF this is 300 A for 100 kW..
its 120/240 DELTA 3 phase
 

jumper

Senior Member
i always get mixed up which voltage should go first

There is no mandatory rule, but convention/custom is as I stated. I believe IEEE has a standard that the higher voltage is listed first and the NEC has this:

220.5 Calculations.
(A) Voltages. Unless other voltages are specified, for purposes
of calculating branch-circuit and feeder loads, nominal
system voltages of 120, 120/240, 208Y/120, 240, 347,
480Y/277, 480, 600Y/347, and 600 volts shall be used.

But nothing is written in stone.:)

Gar is an EE and simply missed the "3 phase" part when you wrote 120/240. Ie: if I see 120/208, I assume a 1 phase system derived from a 3 phase 208Y/120 system.
 

kwired

Electron manager
Location
NE Nebraska
There is no mandatory rule, but convention/custom is as I stated. I believe IEEE has a standard that the higher voltage is listed first and the NEC has this:



But nothing is written in stone.:)

Gar is an EE and simply missed the "3 phase" part when you wrote 120/240. Ie: if I see 120/208, I assume a 1 phase system derived from a 3 phase 208Y/120 system.

I can see that having a standard of using the higher voltage first means it is three phase and if the lower number is first it is single phase would possibly have some merit. But I think there is currently too many people not following any such standard to just assume what is written with no additional information to verify just what is there. I remember when I was in school our instructors used the lower voltage first for everything, and that became my habit also simply because that is how I was trained. I usually am looking for additional adjectives to determine just exactly what is there because of the loose use of simply stating voltages in no particular order.
 

jumper

Senior Member
I can see that having a standard of using the higher voltage first means it is three phase and if the lower number is first it is single phase would possibly have some merit. But I think there is currently too many people not following any such standard to just assume what is written with no additional information to verify just what is there. I remember when I was in school our instructors used the lower voltage first for everything, and that became my habit also simply because that is how I was trained. I usually am looking for additional adjectives to determine just exactly what is there because of the loose use of simply stating voltages in no particular order.

No argument, there is no defined standard for all.:(

I simply meant to explain why Gar may have thought 120/240 was 1 phase at a glance, I would.
 

JoeStillman

Senior Member
Location
West Chester, PA
In answer to the original post, the maximum kW output of the generator is determined by the engine. The kVA is determined by the alternator. The engine is limited by its ability to drive a physical load. The alternator is limited by the amount of current it can carry without overheating.

The 100 kW generator has a rated power factor of 80%, which gives the genset a rating of 125 kVA. But if you place a load of 125 kVA at 95% power factor on the engine, it will stall because 125 kVA @ 95% power factor is 119 kW and exceeds the rating of the engine. The alternator can handle it, but the engine can't.
 

Sahib

Senior Member
Location
India
In answer to the original post, the maximum kW output of the generator is determined by the engine. The kVA is determined by the alternator. The engine is limited by its ability to drive a physical load. The alternator is limited by the amount of current it can carry without overheating.

The 100 kW generator has a rated power factor of 80%, which gives the genset a rating of 125 kVA. But if you place a load of 125 kVA at 95% power factor on the engine, it will stall because 125 kVA @ 95% power factor is 119 kW and exceeds the rating of the engine. The alternator can handle it, but the engine can't.
The alternator design power factor is 0.8. So when the alternator is operating at a higher power factor than 0.8, its exciter circuit is likely to be overloaded, even when the gen set is delivering part load. Need to check up the capability curve of the generator.......
 
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