Do we have to select a transformer (transformer loaded only by an asynchronous motor) based on the apparent power of the asynchronous motor during startup or based on the apparent power of the asynchronous motor during normal operation?
In other words, does the apparent power of the secondary windings of the transformer need to be larger than or equal to the apparent power of the asynchronous motor during startup or does it need to be larger than or equal to the apparent power of the asynchronous motor during normal operation ?
For example: asynchronous motor with the following data:
- rated shaft power = 450 kW
- rated supply Voltage = 6.600 V
- rated current = 49 A
- service factor = 1,10
- S.F.A. = 54 A
- locked-rotor current at the rated supply Voltage = 530 %
- power factor = 0,84
- efficiency at rated shaft power = 95,5 %
- motor engineered according to IEC/EN 60034-1
If this motor does not use its service factor (normal operation), this motor needs to be fed an apparent power of (square root(3)) x 6.600 V x 49 A ≈ 560.145 VA.
If this motor does use its service factor (normal operation), this motor needs to be fed an apparent power of (square root(3)) x 6.600 V x 54 A ≈ 617.303 VA.
During a DOL start on a strong grid, this motor needs an apparent power of (square root(3)) x 6.600 V x 49 A x 5,3 ≈ 2.968.770 VA.
According to EN 60034-1, the tolerance on the locked rotor current of cage induction motors with any specified starting apparatus = +20 %.
This might result in a maximum apparent power (during a DOL start on a strong grid) of (square root(3)) x 6.600 V x 49 A x 5,3 x 1,20 ≈ 3.562.524 VA.
So do the secondary windings of the transformer need to be engineered for a rated apparent power that is larger than or equal to
- 617.303 VA (if the motor does use its service factor),
- 2.968.770 VA or
- 3.562.524 VA ?
And what apparent power would the secondary windings of the transformer need to deliver if the motor would be started/driven by a frequency converter? The starting current (on the grid side) will be much smaller (order of magnitude 1 to 1,5 x rated current) then but the harmonic content will generate additional losses in the transformer. And of course the starting current of a motor being supplied by a frequency converter also depends upon the settings of the frequency converter. Is it possible to determine the starting current (on the grid side) in the selection phase of the frequency converter? If yes, then how? If no, then how do you determine the starting current in advance? I assume that I need to add a spare margin to the apparent power that the frequency convertor gets out of the grid in order to determine the apparent power that needs to be delivered by the secondary windings of the transformer? Does the apparent power that needs to be delivered by the secondary windings of the transformer also depend upon the
- THDi and/or the
If yes, is there a relation then between the THDI and/or the THDU and the spare margin? If yes, what relation?
A: In general, the transformer is sized for maximum loading current and not motor starting locked-rotor current. The economic selection of the transformer should include all expected loads but not necessarily the nameplate rated loads because normal operations of many loads will include a diversity factor as well as operating at lower loading levels. If harmonics are involved they need to be accounted for by either selecting a larger power rating or specifying a K-factor rating. Other considerations include voltage regulation, fault current, grounding, protection, etc.
You have to indicate the other factors that will be needed in transformer sizing, e.g. ambient air temperature surrounding the transformer location, driven by the asynchronous motor i.e. pump or fan as the latter has a quite a heavy start due to the inertia by the fan. Worst case fan, bearing and belt starting torque will add 10% to the load.
Also, suggest you identify the allowable minimum voltage you can accept at the motor terminals for a successful starts, then calculate the maximum acceptable transformer impedance which will allow you to start. Calculate based on low supply side of the transformer.
For the raw (unconfirmed) transformer sizing due to incomplete data or information, you may assume like this: kVA = VLine x LRA (because the OP indicates DOL starter) x sqrt of 3/1000. Substitute the values, thus the transformer apparent power in kVA to meet the motor demand is 6600V x 259.7A x 1.732/1000 = 2968.68 kVA ~ 3 MVA.
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