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Dol Starting Of Motors Affects Transformer Sizing


rvim002

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Hi All,

 

The quality of the content I get from this forum is just fantastic. Thanks for all the info.

 

I am in a process of specifying the transformer capacity for starting three 1250HP motors. I looked into a few websites for the transformer sizing guidelines and this is what I have come up with.

 

I assume that it is going to be a 5.7% impedance transformer. I would require the two motors running and the third motor to be started, with a maximum of 10 % voltage drop at the motor terminals.

 

Motor nameplate details:

1250 HP

3300V

Locked rotor current: 1254 A

Full load current: 198 A

Full load power factor: 86%

Code Letter: G

 

 

Calculation of motor starting kVA:

 

For Code G motor, starting kVA = 6.29 times Horsepower

 

So,

 

Motor starting kVA = 6.29 x 1250

= 7.862 MVA

 

Transformer sizing:

 

Total transformer kVA size required = ( Running kVA rating of two 1250 HP motors) + ( Transformer rating required to start the 3rd motor with 10 % Volt drop)

 

Running kVA Rating of 1250 HP motors = 1.732 * ( Rated Voltage) * ( Rated FL Current) / Assumed average power factor of 85%

=1.732 * 3300 * 198 / 0.85

= 1332 kVA

So, Running kVA of two 1250 HP Motors = 2 * 1332 kVA

= 2.664 MVA

 

Generally, Volt drop due to motor starting in % = Motor starting kVA / ( Motor starting kVA + Transformer short circuit kVA)

 

So, 10% = (7.862MVA) / ( 7.862 MVA + Transformer short circuit capacity)

So, calcualted transformer short circuit capacity = 70.76 MVA

 

Transformer Short circuit capacity = Transformer kVA rating / (Transformer impedance)

So, transformer kVA Rating = Transformer short circuit capacity * Transformer impedance

So, the transformer rating required to start the third motor with 10 % volt drop = 70.76 MVA * .057

transformer rating required to start the third motor with 10 % volt drop = 4.033 MVA

 

Recollect

Total transformer kVA size required = ( Running kVA rating of two 1250 HP motors) + ( Transformer rating required to start the 3rd motor with 10 % Volt drop)

= 2.664 MVA + 4.033 MVA

= 6.697 MVA

 

Does this approach seem right? Or am I going wrong somewhere?

 

I would actually be using softstarters in this case, but I need to prove that 5MVA transformer is not sufficient for DOL starting of third 1250 HP motor, with two other 1250 HP motors running.

 

Thanks in advance

 

Cheers

 

 

 

 

 

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  • 3 weeks later...
  • 2 weeks later...

Hello rvim002

 

Essentially, you are on the right lines.

Iwould probably use a different calculation method but the concept is similar.

 

The calculated running current is probably a little high.

The power rating is actually the shaft power, so the electrical power is the shaft power divided by the efficiency. A motor of this size could reasonable have an efficiency of 96% and a power factor of 0.95 (your figure of 0.85 power factor is probably on the low side but does include a safety margin.

 

By my calculations:

1250HP = 933KW = 933/0.96/0.95 = 1023KVA full load per motor.

Start KVA would be in the order of 7 - 8MVA. Allow for 8MVA to be sure.

 

The load with two motors runing and one starting would be in the order of 10MVA.

 

If we assume a 5.7% transfomer impedance and a voltage drop of 10%, then the transformer rating would be 5.7/10 x 10MVA so would be in the order of 5.7MVA.

 

My calculations yeild a reduced transformer size, but this is primarily due to the reduced runing KVA compared to your estimation.

You would need to get a good indication of the full load power factor and efficiency to firm up on these figures.

 

Best regards,

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Thanks Marke

 

 

The manufacturer datasheets suggests that the full load power factor is 86% and the efficiency is 96%.

 

Is there an industry standard governing the volt drop during the motor startup?

 

I see that the Nema Fire pump standards NFPA 20 suggests 15% volt drop during startup is OK. But it is only a guideline and not a standard. Have you come across any industry standard on volt drop during startup?

 

Thanks again.

 

 

Hello rvim002

 

Essentially, you are on the right lines.

Iwould probably use a different calculation method but the concept is similar.

 

The calculated running current is probably a little high.

The power rating is actually the shaft power, so the electrical power is the shaft power divided by the efficiency. A motor of this size could reasonable have an efficiency of 96% and a power factor of 0.95 (your figure of 0.85 power factor is probably on the low side but does include a safety margin.

 

By my calculations:

1250HP = 933KW = 933/0.96/0.95 = 1023KVA full load per motor.

Start KVA would be in the order of 7 - 8MVA. Allow for 8MVA to be sure.

 

The load with two motors runing and one starting would be in the order of 10MVA.

 

If we assume a 5.7% transfomer impedance and a voltage drop of 10%, then the transformer rating would be 5.7/10 x 10MVA so would be in the order of 5.7MVA.

 

My calculations yeild a reduced transformer size, but this is primarily due to the reduced runing KVA compared to your estimation.

You would need to get a good indication of the full load power factor and efficiency to firm up on these figures.

 

Best regards,

 

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