ht motor

[sapark]

I am having a 6.6 kV motor of 315 kw. This motor is having six leads brought out and normally connected in star connection.On switchgear front I am having 3.3kV supply.Can I use this motor on 3.3kV supply by connecting it in delta?

 

[edison123]

No, you cannot. Any reason you want to do that ?

 

[electricpete]

At 6kv wye connection each leg sees 6.6/sqrt(3) =~3.8kv.
You would be applying 3.3kv to each leg.

I am not sure IN THEORY why it couldn't be done. Just thinking out loud...

Supply circuit would have to be sized for sqrt(3) times FLA in order to carry full load.
Derate motor [3.3/3.8] to account for voltage below normal
Derate motor some additional amount to account for possible heating from circulating 3rd harmonics in the delta.

HT motors typically have no insulation grading like EHV transformers, so there is no concern for applying voltage toward the ends of the windings which in wye connection were called "neutral".

The one thing I am certain of is that I trust edison's judgement on these things and I wouldn't proceed until I understood his objection.

kumar/edison - Can you explain further?

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[electricpete]

... forgot to mention protective relaying also set sqrt(3) times as high if it is sensing phase current. But if you happen to have ct's in term box then protect motor as normal.

I'm not sure if allowing that 3rd harmonic current might have some vibration effect in addition to slight heating.

once again just thinking out loud and don't do anything without appropriate thorough review by a competent engineer.

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[edison123]

sapark/pete,

From 6.6 KV wye to 3.3 KV delta, the phase voltage is down by 15% and the starting torque is down by square of 3.3/3.8. We do not the motor application where such a reduced torque could result in stalling. Also, if the motor runs at rated load (315 KW), the running current will be higher by 15% which may heat up the winding. Due to these reasons, my post. From insulation point, no problem.

Kumar

 

[edison123]

correction

"we do not know the motor application .....

 

[aolalde]

I agree with Edison, unless your load is only 74% of the motor capacity (233 kW) droping the voltage to 86% of the nominal design voltage will lead overheating and short life expectancy for your motor.

 

[electricpete]

The 86% derating for steady state is the same as I mentioned (3.3/3.8)

The 74%=86%^2 applies to torque characteristic - which would need to be evaluated for starting duty. For unusual loads we might also have to consider breakdwon torque in relation to load momentary peak torque. However this number has no direct bearing on continuous load capability to my knowledge.

There remains some derating for possible delta circulating current. It is an unknown in my mind.

I am curious if anyone has ever seen a large motor reconnected in this fashion?

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[justinn4]

Epete;

If power P is the result of torque times speed, and the torque reduces with the second power of the voltage reduction. Why you state that the power reduces only proportional to the voltage reduction?

P= k *T*rpm k= 1/5250 for torque in Ft-Lb, and P in HP.
k=1/9541.6 for torque in N-m and P in kW

 

[electricpete]

justin - the load determines the torque and the motor matches it, so reduction in torque capability has nothing to do with motor rating as long as starting conditions are evaluated (and for loads which can impose severe temporary overload - evaluate breakdown torque).

What generally determines the steady state power rating of a motor is the heat generation.

If power factor remains the same, reducing voltage to 86% would increase full load current to 1/86%~114%. If we derate to limit the load to 86%, that limits the current to 100%, and current heating to approximately the motor design 100%.

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