Every alternator is different. Only the manufacturers can answer that question for you, as davidbeach implied.
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An induction motor has form 5 to 7 time power at starting than in nominal running.
you have a motor: P=110kW -> S=138 KVA at nominal running.
when starting : S=(5to7)x 138 = 690 to 966 KVA needed !it but yoy have just a generator about 480 KW-> 600 KVA.
i thing you need to change the starting mode or buying a larger generator.
If you are able to accept severe voltage and frequency dips when the motor is starting, you may use a ratio as little as 2:1 This would be an instance where there are no other loads on the generator other than the motor, the load is easy to start and the motor has a fairly low locked rotor current.
If the generator has other loads the minimum sizing is generally 3:1
However, each instance should be evaluated by someone familiar with generator applications.
respectfully
An induction motor does draw a starting current that can be in the order of 5 to 7 times running current but that does not equate to 5 to 7 times power. The power factor of a starting motor is very low, lots of reactive power, but not a great increase in real power above running power. Depending on the load type the starting (real) power could be well below the running (real) power even though the starting current is well above the running current.
Hi David;
I agree with you.
I have never started one on a generator, but a high slip motor such as are used for shears may use less power starting than full load power.
My experience with motors and generators has been mostly with air conditioners, which unfortunately do draw more power (real power) starting than running.
When we try to start a motor on an undersized generator, it usually follows the following sequence:
1> The motor load overloads the engine and pulls the frequency down.
2> As the frequency drops below about 58Hz. the Under Frequency Roll Off feature of the voltage regulator starts to lower the voltage setpoint proportionally.
3> The voltage and frequency drop to a level that allows the contactor in the automatic transfer switch to drop out. It does this with considerable current flowing and creates a substantial arc.
Note: As long as the frequency is below 59 or 60 Hz. the governor on the engine will be wide open. The engine will be producing maximum power throughout this sequence.
4> With no load, the engine accelerates and the voltage and frequency build up.
5> When the voltage is sufficient the contactor closes. (Note, it may be the contactor or the contactor control relay cycling on low voltage. It doesn't matter, the effect is the same.)
6> The cycle repeats until the contactor is destroyed.
Note: This sequence applies to a small set with a relay controlled transfer switch with no time delays. With a more sophisticated transfer switch, there are some time delays added to the sequence. Also the engine alternator typically reaches 60Hz. and the governor closes to the high idle position before the contactor closes.
The end result is that it takes a little longer to destroy the contactor with a more sophisticated transfer switch.
I have found by trial and error that the minimum reserve capacity to start an A/C unit is about 2:1 There will be excessive voltage drop and some frequency drop but it will start without causing damage.
At 3:1 there is some voltage drop and little frequency drop.
Some years ago there were a number of gensets sold in Central America that were too small for the number of connected A/C units.
When word spread that I knew something about gen-sets, I got to see most of them. It was a process of:
1> Experiment to see how many A/Cs the set would start before it overloaded.
2> Timer circuits and electrically latched circuits to prevent the A/Cs from starting simultaneously, and in some cases to prevent automatic restarting. (This was a challenge, because the customers often wanted the sets in the sleeping areas to restart automatically, but were willing to manually restart A/Cs in other areas.)(I think you would have enjoyed this one Keith.)
3> Customer education.
It is on this adventure with about 12 sets that I base my recommendation:
Less than 2:1 capacity and you are in for problems.
More than 3:1 capacity and you are safe.
Between 2:1 and 3:1, it depends.
I have used these guidelines for sizing many more sets with no problems whatsoever.
respectfully
>>The power factor of a starting motor is very low, lots of reactive power, but not a great increase in real power above running power. Depending on the load type the starting (real) power could be well below the running (real) power even though the starting current is well above the running current.<<
What could be a typical PF of 4-ton central A/C at startup?
Another question is: could a genset's circuit breaker trip by instanteneous startup surge current even if average real power is not exceeding its rating?
No idea on the central A/C. The circuit breaker is looking at amps and has no knowledge whether the amps contribute to watts or vars, so the breaker is only protecting the current rating of the generator (related to the kVA rating). Yes, it could trip for surge current that has very little real power content.
If you are afraid that reactive power could damage generator you can lover his active power making him to produce more reactive power. Now if that is not good for further use, maybe you can try to run it over capacitors in parallel, they should suffice the difference in reactive power you have. Of course you should calculate appropriate amount by power factor you want to achieve. Generally If you are afraid of motor burning down generator you should use frequency converter. You have maximum current at 100% of motor power, no starting overload currents, that are usually 6 times larger than working one. It is much cheaper way than buying new generator, and having more power than you need after startup sequence.
