# Motor Slip

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Dear All!

I can't find explanation for Motor slip.

I work for one foreign company in Russia and now I need to translate a document and for me it is not clear what is "Motor slip".

Part of text

"Speed accuracy is 10% of the motor slip, which with an 11kW motor, equals 0.3% static speed accuracy.

With a 110kW motor, speed accuracy is 0.1% without encoder (open-loop)."

Please help me. It will be better if some body puts a picture of mechanical curve of induction motor together with explanation of Motor slip.

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Dear All!

I can't find explanation for Motor slip.

I work for one foreign company in Russia and now I need to translate a document and for me it is not clear what is "Motor slip".

Part of text

"Speed accuracy is 10% of the motor slip, which with an 11kW motor, equals 0.3% static speed accuracy.

With a 110kW motor, speed accuracy is 0.1% without encoder (open-loop)."

Please help me. It will be better if some body puts a picture of mechanical curve of induction motor together with explanation of Motor slip.

Hello.

I have always known only one translation of the term "motor slip" into Russian: "skolzhenie motora". It is the lag between the rotating el magnetic field in the stator and the rotation of the rotor being carried away by that field. Both are said to be "asyncronous" to each other, so these motors are called "asynchronous" (in the English language the adjective "induction" is oftener used for these kind of el motors). Slip is minimal at no load, load increasing, slip increases ( current also of course, they are interdependent).

Slip at the rated motor's load is deducible from a motor's speed at the motor's nameplate: a one pair poles' motor may have speed (at rated load) 2850 rpm, so its slip (at rated load) will be 3000 - 2850 = 150. A two pair poles' one may be 1420 rpm , so slip 1500 - 1420 = 80.

What is precisely meant by "static speed" and the "encoder", it seems I do not know at this moment.

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Hello Yuri!

Unfortunately I also know what is "motor slip" in Russian, but here I must translate the phrase

"Speed accuracy is 10% of the motor slip, which with an 11kW motor, equals 0.3% static speed accuracy.

With a 110kW motor, speed accuracy is 0.1% without encoder (open-loop)."

and here as you can see there is nothing the same with our Russian meaning and I need somebody who can explain in English what does it mean in English version.

I also have some imagination, but I need somebody who can confirm my thinking.

If you can translate this text, please, do it here in Russian.

Best regards!

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Hello Tua

For a motor to develop torque, there must be an interaction between magnetic fields, one in the stator and one in the rotor.

The stator is connected to a three phase electric supply and this results in a magnetic field that is rotating at a speed that is determined by the supply frequency and the number of poles.

If we now consider a stationary rotor with a shorted winding within the stator, the rotating magnetic field will cause current to flow in the rotor which behaves like a short circuited secondary winding on a transformer. While the rotor is stationary (Locked Rotor) the frequency of the rotor current is equal to the frequency of the stator current.

The rotor current flowing causes a rotor magnetic field which is phase displaced from the stator field but is at the same speed relative to the stator. The rotor field and the stator field interact to create a rotational force on the rotor (torque) which is dependent on the current amplitudes and displacement angle.

Provided that the motor and driven load are not jammed, the torque will cause the motor to accelerate in the direction of the rotating fields.

The difference in speed between the stator rotating field and the rotor is called the slip. When the rotor is not rotating, the slip is 100%. If the rotor was spinning at the same speed as the field, it would be rotating at zero slip, but at zero slip, (synchronous speed) there would be no magnetic flux cutting through the rotor winding and so zero current and zero torque.

As the motor accelerates, the effective frequency of the flux as seen by the rotor reduces. This reduces the frequency of the current flowing in the rotor so that at zero rotor speed, the rotor current is at 100% of the line frequency. At 25% speed, the rotor current frequency is 75% of the line frequency. If you add the rotor speed and the slip speed, you get the line speed.

When the slip is very low, the torque is dependent on the slip. As you increase the torque, the slip increases. Once the torque has reached the maximum torque, the motor will stall, the slip rapidly increases and the torque falls.

The design of the rotor has an impact on the slip at which the maximum torque occurs and there is a tendency for larger motors to have a maximum torque at a lower slip.

The variation in slip with the variation in torque results in a speed droop with torque and impacts on the speed accuracy with a constant frequency. This can be quickly determined by looking at the rated full load speed of the motor and comparing this with the synchronous speed. The closer the full load speed is to the synchronous speed, the better the speed accuracy with a constant line frequency.

When a vector controlled VFD is used with an induction motor, there is either a mathematical model of the motor in software, or there is an encoder to provide feedback on speed. The vector controlled drive will alter the frequency applied to the drive in a manner to keep the speed constant irrespective of the load torque. The open loop system calculates the torque and slip and increases or reduces the frequency to compensate for the motor slip. If a shaft encoder is used, the actual speed is measured and so the frequency can be easily controlled to keep the speed constant.

The spped accuracy will be much better with true feedback (encoder feedback) than with a mathematical model. If the model was able to very accurately describe the motor behaviour and characteristics, the accuracy could also be very high, but the model has compromises and there is a reduction in the accuracy, especially at low speeds.

"Speed accuracy is 10% of the motor slip, which with an 11kW motor, equals 0.3% static speed accuracy.

With a 110kW motor, speed accuracy is 0.1% without encoder (open-loop)."

This is a reflection of the accuracy of the mathematical model and suggests that the model is more accurate on the larger motor than the smaller motors.

If the motor has a full load slip of 5% (150RPM for a 2 pole 50Hz machine) the speed accuracy would be 15RPM for the 11KW motor and 5RPM for the 110KW motor. (based on figures quoted. This is dependent on the ability of the model to accurately reflect the characteristics of the motor).

Best regards,

Mark.

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Hello Tua.

I am afraid your citation is looking torn away from a preceeding text.

If we abide to the usual for induction 50Hz motors slip of 5% then the 10% (from your text) of this slip = 0.5% speed accuracy. The sentence with "10%" seems to be referring to a less than 11kW motor, let's say 1kW. So (0.5% speed accuracy for 1kW motor with an encoder), 0.3% for 11kW (with an encoder), 0.1% for 110kW without encoder.

Best regards

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