***This Message Is Cross-Posted With A Duplicate Thread Found At The Canadian Electrical Code Area. Attached is A Pasted Copy Of That Message, Plus A Link To That Thread ***P.S.: Schematics for polyphase "Buck/Boost" Autotransformer connections are available in the Technical Reference area. Check the "Menu" located at the Tech. Reference area's main page for links to these drawings, or just search the forum's pages.

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bigpapaMaybe I can offer a bit of assistance with your Motor / Transformer questions.

1:

3 PH 40HP 460V motor FLA = 52A wire size is min #6 CU

600 to 480V autotransformer would be 13.5 KVA

(600V - 480V) x 1.73 x 52A = approx 11KVA

For the Motor vs. Transformer KVA capacity in this example, the Transformer would require a "Per-Winding" Capacity of at least 23.920 KVA for an Open Delta connection scheme, or 14.404 KVA per winding for a "3 Transformer / Closed" Wye configuration.

2:

Use three 600V-120V 5 kva transformer in a buck configuration for a 15KVA bank, you will be wiring a motor that requires a #6

copper wire to feed it, to a transformer that has # 12 GA wires coming out of it. It doesnt seem right.

I calculate the individual Transformers for a Voltage Bucking (Output Voltage Reduced) setup - using 600V X 120V Isolated Transformers - to be minimum of 7.5 KVA each.

A 5 KVA Isolated Transformer with a Secondary Voltage of 120 VAC is only good for 41.67 Amps, which falls below the FLA of the Motor.

With an Isolated Transformer of 7.5 KVA Capacity - and a Secondary Voltage of 120 VAC, the maximum load Amperes of which the Secondary Winding is able to _Safely_ handle continuously is 62.5 Amps - which exceeds the FLA rating of 52 Amps for the Motor.

The only part of the Transformer that will be carrying the Motor's Load Current (for the "Buck-Boost" Transformer arrangement), will be the Secondary Winding - or Windings (if "Split-Coil" type is used).

Therefore, the "Primary" Windings' terminal leads may be of small conductors - such as the # 12's you mentioned, without any threats of Barbecued Wires!

The "Primary" side will only draw Magnetizing Currents, which will likely not exceed a few Amperes during normal operation (may be quite high during Motor's Starting).

The rated load current for the Primary Windings of these Isolated Transformers are:

* 5 KVA - 600 VAC Primary = 8.334 Amps,

* 7.5 KVA - 600 VAC Primary = 12.5 Amps.

3:

I bought a 30 KVA autotransformer that has #12 wires for connecton. The #6 wires that feed the motor get a bit warm during normal

use, I can imagine how warm the #12 GA wiring will be.

Am I missing something here?

This Autotransformer may be too small in KVA size for this Motor's FLA rating.

My figures show:

* MINIMUM 45 KVA total capacity - for a "Full/Closed" Wye Autotransformer configuration (Motor Load should be no less than 14.404 KVA "Per Line" for a continuous load, across 3 separate 1 Phase 2 Wire Transformers),

* MINIMUM 50 KVA total capacity - for an Open Delta Autotransformer configuration (Motor load should be no less than 23.920 KVA per Winding set, with a total capacity of no less than 47.840 KVA across 2 separate 1 Phase 2 Wire Transformers).

4:

So, my question is... how is it that we can calculate what the curent carrying capability of the windings are? My guess is that the primary windings would need the ampacity of the load. Our codes do not address this issue and there is very little information out there on it.

Applying the expected Load to find the MINIMUM CAPACITY of an Isolated Transformer - when creating an Autotransformer arrangement, is done as follows:

1: Determine the Maximum Continuous Full Load Current of the Load to be driven,

2: Figure what the required OUTPUT Volatge should be to the Load, and compare this to the "Existing" System Voltage available,

3: Find a Transformer with a Secondary Voltage which will bring the Available System Voltage to the desired level of the Load's rated Voltage (by either increasing the Voltage or decreasing the Voltage),

4: Multiply the Voltage rated for the Secondary of the Transformer, by the Load's designed FLA, to find the minimal KVA rating for the Secondary Winding (and overall - the entire Transformer)

*Example:

We have a Motor with nameplate ratings of 500 VAC, and Full Load Amperes (FLA) of 10 Amps.

The existing Power System is only 400 VAC, so we need to increase the Voltage another 100 Volts.

The MINIMUM KVA Capacity for the Isolated Transformer to be used for this scenario will be 1.0 KVA - since the Secondary will be wound for 100 Volts, and we need to carry 10 Amps through it.

This is true for Single Phase Motors and 3 Phase Motors / Loads.

You need to figure the _Highest_ expected and continuous Full Load Amperes for the Load(s), then how much the Voltage needs to be adjusted - and apply the diiference in Voltage X the FLA to calculate minimum Transformer Secondary Capacity.

Again, the "Primary Windings" will not be carrying the Load Current, only Magnetizing Currents.

Good luck - and be sure to respond back if you have more questions (or if things are unclear).

Scott35

p.s. I am "Cross-Posting" this message to the other one in the "Canadian Electrical Code" area.

Sizing Autotransformers For Motors S.E.T.