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Transformer sizing for small motor loads

Transformer sizing for small motor loads

Transformer sizing for small motor loads

(OP)
Hi, this is question about sizing step-up transformer for a bunch of kitchen equipment loads that have various sizes compressors for refrigerators.  These are small size compressors like 1/3 hp, 1/4hp, for ice machine, refrigerators, collers, etc.  I have discussed the transformers issues under a seperate post before. Here I have a question about sizing of these transformers for these small refrigerator loads.  Becasue of the initial start-up and in-rush current (600%) for motors in general, I suggested we need to oversize the transformers such that the in-rush will not cause severe undervoltage condition (limit to within 10% of the motor voltage), such that the motor might fail during start up.  I know this is true for big motors, but I am not certain if they are of concern for small motors like them.  Each kitchen equipemnt load has a different voltage (European models) so a lot of step-up transformers are used. The contractor felt oversizng transformers isnot necessary (cost) and state there is no real concers for such small motors.  But I wonder if that is true or not and like to ask others if anyone has any experience in this areas.  Thanks.

RE: Transformer sizing for small motor loads

First a comment about voltage drop and transformer impedance.
The normal voltage drop of a transformer is dependent on the regulation, not the impedance, with normal loads.
Motor starting, however, is highly reactive and voltage drop calculations based on impedance are a good approximation.
If the transformer has a %imp voltage of 3%, then expect 6 X 3% / 1.25 or 14.4% or maximum voltage drop under motor starting if the transformer is sized exactly at 1.25% of the motor rating. Generally it is not possible to purchase a small transformer rated at exactly 1.25% of motor KVA and the transformers are invariably over sized resulting in less voltage drop. The other factor is that the voltage drop tends to reduce the current which tends to reduce the voltage drop, so the actual voltage drop is less than the simple calculations would indicate. If your contractor has taken 125% of the motor KVA and selected the next larger transformer, it should work well.
If you have a marginally sized transformer feeding lighting and utility circuits as well as a lot of small motors that will all start together following a power outage, then you need to do a system survey or analysis. That is beyond the scope of "Rules of Thumb" or quick tips.
For the ratios of starting to running KVAs for various sized motors see the Cowern Papers.
http://baldor.com/support/literature_load.asp?LitNumber=PR2525
See:
CODE LETTER The code letter is an indication of the amount of inrush or locked rotor current that is
required by a motor when it is started. (See "Locked Rotor Code Letters" for more details.)
Types of Motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5  (Page 9)
and
Locked Rotor Code Letters and Reduced Voltage Starting Methods . . . . . . . . . . . . . . . . . . . 27

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Transformer sizing for small motor loads

(OP)
Hi Bill, Thanks for the excellent tips.  It is very helpful.  I am still a little confused why the current will reduce with voltage drops.  I lkike to confirm that what you have discussed above is that the actual voltage drop during motor startup is actually less than than calculation based upon the LRA (or I typically use 6x the FLA), due to voltage regulation by the transformer under loaded condition?  Is this statement correct?  ALso, is there a direct relationship between % voltage regulation and % impedance?

I really appreciet the reference you pointed out!

RE: Transformer sizing for small motor loads

A starting motor can be represented as a constant impedance, so lower voltage results in lower current by Ohm's Law.  This is why reduced voltage starters work.

Reg = R·cosØ + X·sinØ + 0.5·(X·cosØ - R·sinØ)²

Z² = R² + X²
cosØ = power factor of load
 

RE: Transformer sizing for small motor loads

A transformer impedance consists of resistance and inductive reactance.
A load may consist of resistance, inductive reactance and/or capacitive reactance. Under normal loading with a resistive load, the equivalent circuit is a series circuit of transformer resistance, transformer inductive reactance and load resistance. When you plot the voltage drops, the transformer terminal voltage drop will be at an angle to the load voltage drop and so the resistive component will be the predominant effect. This describes transformer regulation.
Under short circuit conditions, the the current phase angle is determined by the transformer impedance and X:R ratio and the impedance accurately describes the voltage drop.
Under normal or light loading with a resistive load, the voltage drop will be described by the %regulation. As the load becomes inductive, the phase angle lags, the % voltage drop will increase until it will be maximum and equal to the %impedance when the load phase angle equals the phase angle described by the X:R ratio of the transformer.
 

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Transformer sizing for small motor loads

Reg, Z, X, and R are in per unit in my previous post.  Divide Z% by 100 to get per unit.
Multiply Reg in per unit by 100 to get percent regulation.

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