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Starting Ac Motor With Low Voltage


AB2005

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Dear All,

 

I want to clear some confusion which is related with the theory of AC motors.

For example, a motor is designed for 400V at delta connection and 690V at star connection. If we connect the motor in star configuration but the voltage is 400V, then, the winding would draw the current 58% of full load-no sufficient torque would be produced-its slip would be increased and first, the rooter would heat up then all motor. Is it true or I am mistaking?

 

"Don't assume any thing, always check/ask and clear yourself".

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Well, you are partially correct except that the 58% only applies to voltage plus the rules will apply to everything, not just FLA. But let's step back a second.

 

By connecting in Star, you have effectively reduced the motor terminal voltage by the square root of 3, (1.732) or 58% as you said. But the torque, and therefore the current is reduced by the SQUARE of the voltage reduction. So 58% x 58% = 33% or 1/3. So your current and torque is now 1/3 of what it would have been if the motor were connected in Delta (or you had applied 690V). That 1/3 value however applies to everything as I said earlier, including the STARTING current, which is typically 600% of FLA. So on startup, your new starting current is 200% of FLA (1/3 of 600%) and in some cases that may be enough to accelerate the motor. That is the basic principal behind a Star-Delta motor starter. Torque follows the same rule. Normal starting torque is usually 160% of FLT but in Star it will be 54% of FLT (1/3 of 160%). Again, it may still be enough to accelerate the motor if it is more than the load torque requirement without full coupling, as in a variable torque load like a pump or fan. Once it is running and the load is fully coupled however, that is when you get into trouble if you leave the motor connection in Star. Your effective voltage is still at 58% therefore torque is still at 1/3 of the motor's rated output. So if that is not enough to keep your load spinning at normal slip speed, the slip increases and the motor pulls more current to try to return to slip speed, but it can't because of the loss of torque, so it will overload.

"He's not dead, he's just pinin' for the fjords!"
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Thank you very much Mr. Jaraef

 

Well explanation. I have faced a problem that a motor is designed for the agitator in our new glue kitchen. When it started, it drew 115% current (with load) and heated up after some time. There was pressure on me for starting the motor because of making the glue for new machine. I changed the connection and run the motor in star configuration. Now current reduced at 87% (with load) and speed reduced at 85%. After some time, it also heated up.

That was the reason for asking the question in my first post.

 

"Don't assume any thing, always check/ask and clear yourself".

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