# Induction Motors

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### #1 shermaine

shermaine

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Posted 25 March 2004 - 01:48 AM

Any one can advise me on that.
(1) Why is the no-load input current and power as a percentage of their full(rated)values recorded are considerably
higher than those of the transformer?

(2) Estimate the starting current of the motor(as a percentage of its rated value) when it is started at rated voltage.
How can the starting current be reduced to an acceptable value in practice.

### #2 marke

marke

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Posted 25 March 2004 - 10:10 AM

Hello Shermaine

Although there are similarities between the equivilent circuit and a motor, the two are not the same.
1) The induction motor behaves a little bit like a transformer, but the secondary is short circuited. The frequency of the current flowing in the rotor is proportional to the rotor slip. This greatly influences tha current and power flowing into the motor under all conditions. An induction motor, even under no load conditions, has rotational losses, frictional and windage as well as rotor losses. These losses do not occur in a transformer.

2) When an induction motor is started at full voltage, it will initially draw locked rotor current. This high current will slowly reduce as the motor accelerates, but will only reduce significantly when the motor is approaching full speed. The actual current and the speed/current curve of the motor are a function of the motor design.
The Locked rotor current is typically between 550% and 900% of the rated cuirrent of the motor.

Best regards,

### #3 shermaine

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Posted 29 March 2004 - 04:38 PM

Any one knows why the no-load current and power of the motor expressed as percentage of their full-load values is considerably higher than those of a transformer?

For Blocked Rotor Test

Can the starting current be reduced to an acceptable value in practice?

Are there many methods?

Thanks

### #4 marke

marke

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Posted 29 March 2004 - 07:30 PM

Hello shermaine
As referred to previously, the induction motor is like a transformer with a shorted circuited secondary. I also has additional losses that do not occur with a transformer, hence the current is higher.

2 The Locked rotor current can be reduced by reducing the applied voltage. This is done in industry and is referred to as reduced voltage starting. If you look on the web pages for this site you will find examples of reduced voltage starters and their characteristics.

Best regards,

### #5 shermaine

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Posted 30 March 2004 - 04:02 PM

Hello,

Any great website to recommend in reducing the starting current of the motor? I can't seem to find more detail information like how it is been done to reduce the starting current?
Thanks.

### #6 marke

marke

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Posted 30 March 2004 - 07:15 PM

Hello Shermaine

You could try this website for a start. There is a page on motor starters showin how they operate and reduce start current. http://www.lmphotonics.com/m_start.htmThere is also a page on soft starters.

Best regards,

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