Supply 3ph motor with 2ph reduced voltage 2

[davva]

We have a 3ph induction motor (350kW, 3ph, 440V, FLC=550A, 60Hz).

When the motor is switched off, a contactor closes in the mcc to energise a step down 440/35V transformer which puts 35V across 2 of the windings (U and W phase). I think the concept is to use the motor windings themselves as a form of anti-condensation heater when the motor is off.

We are experiencing motor failures and wonder if this circuit may be the cause.

I’m want to calculate the current (and subsequently work out the heat) for this arrangement.

Should I just use the stator resistance and stator reactance from the per phase equivalent circuit diagram and then double the answer (for 2 phases) to get the impedance of the two windings? (cable impedance will be included)

Any other thoughts on this arrangement would also be welcome.

 

[ScottyUK]

In your motor you have an air-gapped transformer with shorted turns in the form of the rotor rather than a simple series R-L circuit. You need to account for the losses in the rotor too.


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[edison123]

Measuring the current at 35 V would be much more easier.

Why only 2 phases ? If it is delta connected stator, two phases will see 50% of the current of the third phase. If it is wye connected, the 3rd phase wouldn't see any current at all.

You could use anti-condesation heaters to ensure uniform heating under motor shut down. Around 300 watts heater should do fine.

 

[starkopete]

It's called "trickle" heating. See attached.

 

[starkopete]

Here's another version;

 

[jraef]

As starkopete pointed out, this is done all the time. There is no inherent danger to the winding insulation if it is done right, but that is sometimes the issue; many people have seen it done and think "I can do that" but lack the knowledge to avoid the problems.

1) You MUST use a timer to wait for the motor field to collapse and to stop spinning before applying the trickle voltage. If not, you can create a regen problem that damages the transformer or worse.

2) You MUST fuse the transformer. You didn't say one way or the other, I mention it only because I see people do this with no fuses and they don't understand why they lost the motor. Without fuses if something else goes wrong, the transformer can short and fry the windings too.

3) You must size the transformer correctly. Many people think they can just put 24V (a common autotransformer used in buck-boost applications) on the motor because they saw that used on one before. The problem is, they saw it on a 120V or 240V motor and they have a 460V motor. Plus the size of the motor makes a difference too. As that paper says, the voltage ratio must be correct or it doesn't work; 10 to 20% of nameplate. In larger motors however, there is more surface area to dissipate heat so you will need to be at around 20% of the motor's nameplate rating. In your case that is 88V yet you have less than half of that, in fact you are not even applying 10% which is the minimum needed for smaller motors.

So if I had to guess (and my compulsion makes me do it), I'd say that someone recognized the need for motor winding heat in your application, but failed to implement it correctly and it is not solving the problem.


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[davva]

jreaf,

Transformer is 2.1kVA (440/35V) fused 63A on secondary, 6 amp primary.

There is a timer applied on the circuit that was originally set to 1 sec. This has been changed to 30secs recently as we thought that this was too short for the same reason that you give.

Thanks for all responses otherwise.

 

[jraef]

I still think your secondary voltage is too low for a motor of that size. The winding resistance is probably high enough that although your transforer is capable of 63A, a lot less energy than that is going through the circuit. Remember, the goal is to turn the windings into a resistive heater element, but the only way that you get resistance in this is by virtue of the length of copper wires in the winding. That means the watt density will be very low and dissipating quickly into a large surface area. So the only way to affect a decent temperature rise will be to have as much wattage as possible. Increasing the voltage will increase the wattage through the windings.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
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[davva]

jraef (and others)

Thanks for your advise, I think we have eliminated the trickle heating circuit as a possible cause of the stator failures anyway.