transformers: 330 volt secondary 2

[rangesparky]

I teach in an electrical apprenticeship program, and focus on motor controls. Recently, we received 4 donated 330/575V (8.5/4.9A) three phase, six lead (Y or delta), 3.7Kw (European) motors for our lab. In the lab, we have a three phase 120/208 panel, and a step down 480 to 208Y/120 transformer. We have three motor control stations, and we also have three transformer stations. We are using this transformer to step up to approximately 460 volts (with appropriate taps) to supply other motors at each lab station. Our secondary is an ungrounded delta system, and we have installed phase monitoring lights. My goal is to utilize these new donated motors to teach Y-delta starting. So, I need to produce 330 volt three phase power for this application. My problem is that having searched the common manufacturers for standard in-stock step-down and buck/boost transformers, I have been unable to find anything that will satisfy this requirement. I have a budget for lab equipment and supplies, but I am beginning to think that I will have to have these transformers custom made. I don't think that my budget will allow for this. It could be either one transformer or transformer bank large enough to run three motors (with no load), or three separate installations at each of our transformer stations. I'm hoping that someone knows of a way to do this or knows of a product that I have overlooked. Please help.

 

[waross]

First if these are 50 Hz motors you will need 330 x 6/5 = 396 Volts
For that, two 120/240:240/480 Volt dry type transformers wired for 240:240. Connect them as 240:480 auto-transformers in open delta.
This will give you 416 Volts from 208 Volts. 396 + 10% =435 Volts so you are within the 10% allowance.
If the motors are 60 Hz, try the same transformers but connected as auto-transformers for 240:120 Volts. on 208 Volts that will give you 312 Volts. That will be within your 10% allowance.
KVA. The step up winding will take full load current. So for the 240:480 Volt connection multiply full load current by 240 Volts / 1000
For an 8.5 Amp motor that will be 2.04 KVA You can probablly use 2 KVA transformers.
To calculate KVA use the motor full load current and the voltage increase for the voltage but use the transformer rated voltage not the actual voltage. For instance we are using a 240 Volt rated winding to increase the voltage by 208 Volts. Use 240 Volts, not 208 Volts to calculate needed KVA.
If you need 330 Volts the voltage increase is 104 Volts derived from a 120 Volt winding. Again 8.5 Amps x 120 Volts / 1000 = 1.02 KVA.
Use a 1 KVA transformer.
Why use rated voltage to calculate KVA instead of actual voltage?? Because we are not calculating KVA as such, we are calculating the size of transformer needed when the transformer may be derated.
Example: A 1 KVA transformer will safely carry 8.33 Amps through the 120 Volt winding. For a given current, the KVA times the rated voltage will give the KVA needed. If the transformer is used at less than rated voltage the KVA rating must be derated. The current is the limiting factor. Using actual current and rated voltage gives the right size transformer for the current carrying capacity. If it is used on a lower voltage the allowable KVA must be reduced.
You will be able to teach auto-transformers as well as motor starting.
You mention an ungrounded delta. Is your transformer a 480V delta : 120/208V star?
That is a connection with some unique issues and problems. If that is what you have let me know and I can suggest some interesting experiments for your gifted students.

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

 

[LionelHutz]

You don't need rated voltage to run the motors unloaded. A 120/208V 3-phase source would work for spinning the motors.

 

[rangesparky]

@waross. Thank you. I was so hung up on trying to use the 460v that we have in the lab,that I hadn't given much consideration to the 120/208 that is also available to me. Question though: Isn't it true that open delta only gives you 57.7% of the power that a true three phase transformer would give. The way you are calculating the transformer Kva seems to be for each individual phase, and since one phase will be "derived" from the other two, would I have to increase your calculated power by 1.732 for an open delta. I realize that we will have no load on the motors, and for the short amount of time that we will be running the motors, these transformers will probably suffice, but to do things properly as we would in the field, is that correct? I don't particularly like to take too many shortcuts as I try to always train these apprentices to do things the correct way whether it be code issues or AC theory issues. In response to our large transformer for the lab, it is a 480delta to 208Y120 step-down, which we are using to step-up the panel 208 voltage. The previous instructor for this particular part of the curriculum installed it and used the primary taps to get close to a 460 volt output, but he failed to catch the fact that he had created an ungrounded delta. When I began teaching this class, and caught it, I had my students install the phase monitoring lights, and each year I have the intentionally ground one corner to demonstrate how the lights will respond. I have yet to have a student that is willing to stay in the room with me when I close the safety switch for this demonstration. Lol.
@LionelHutz, hadn't considered your solution either, but as I stated in the previous paragraph, I'd prefer to do it the most accurate way, . . .but thank you for the input.

 

[ScottyUK]

Most small transformer shops will do a non-standard winding at little variance to the price of the standard windings. Most of the costs are in the core and copper, and within a reasonable range the voltage ratings don't have a significant influence on cost. As Lionel has suggested, you can probably get away with a much smaller transformer than strictly needed if these motors were working under load.

Just to satisfy my own curiousity, where in Europe uses that voltage?

 

[waross]

If you look at the vector diagram of the open delta auto-transformer boost circuit you will see that the boost winding carries the motor current, no more no less. As long as the motor current does not exceed the current rating of that winding you are safe.
I will derive the value from the motor KVA for you so you may be able to field questions from any gifted students.
Motor current 8.5 Amps
Motor Voltage 312 Volts
KVA 8.5A x 312V x 1.73 = 4.59 KVA

Required three phase KVA for derating to open delta 1/.58 or 1.73 x 5.9 = 8 KVA
Voltage De-rating factor 104V/120V = 1.15
1.15 x 4.59KVA = 9.15 KVA
Each transformer = 3.06 KVA (We are calculating a three phase bank and have to divide by 3 instead of 2. This is de-rated to allow for open delta use.)
Auto-transformer factor 1/3 = 1.02 KVA
Final voltage = 312 Volts
Boost voltage = 104 Volts
Auto-transformer factor 104V/312V = 1/3
Easier... The boost winding must carry the full motor current, 8.5 Amps. The boost winding is rated at 120 Volts. 8.5A x 120 V / 1000 = 1.02 KVA.
Only your best students may be able to follow the rigourous calculations but all should be able to remember "Motor current times winding rated voltage equals KVA."

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

 

[waross]

Take a look at my Dec 23 post in this thread.
24940Y/600Y 3Phase Padmount Transformer Wiring
thread238-335356
Seeing as you have a four-wire-wye:delta transformer I can suggest some interesting experiments if you wish. If you have observed any strange behavior this may explain it.
Yours
Bill


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