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Vfd Retrofit On 2 Speed Motor


anu_rags

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Dear All,

 

My customer has a 2 speed motor and he wants to retrofit a VFD on the motor. The application is mixer in polyvinyl industry. The motor details are as follows:

 

Make: Siemens

Model No: 1LA02230YK41

Frame: 225

Frequency Voltage Motor Connection Power Rating KW Current Rating A Speed RPM

50 415V Delta 38 68 1475

50 415V Star 45 80 2955

 

Im attaching the motor connection diagram. I have used a 45KW VFD with 90A rated current. I connected my VFD so that the motor remains permanently connected in Fast mode. When I try to start the motor, the speed does not rise above 1200 rpm and the current rises upto 103A. The VFD is not allowing the motor to further increase the speed as it is going in Maximum current. When I attach the motor in slow mode, the motor accelerates upto 1500rpm but again the current rises upto 103A.

 

Customer wants slow speed during the start for 2 minutes and then he wants to switch over to fast speed. Any ideas how can I succeed in retrofiting VFD in this application?

 

Regards,

Anurag

Siemens_Motor_connection.pdf

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Hello.

Did you put bridge over R1B1Y1 and connect R2B2Y2 directly to the VFD?

103A is displayed on the VFD ? (I supose it is not a measurement with an ammeter)

If so I'm a bit wondering how 90A rated VFD tolerates that biggish (permanent) overcurrent.

If the overcurrent takes indeed place try to disunite the mixer from the motor's shaft and measure the current then.

Best regards.

 

 

 

 

 

 

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Hello.

Did you put bridge over R1B1Y1 and connect R2B2Y2 directly to the VFD?

103A is displayed on the VFD ? (I supose it is not a measurement with an ammeter)

If so I'm a bit wondering how 90A rated VFD tolerates that biggish (permanent) overcurrent.

If the overcurrent takes indeed place try to disunite the mixer from the motor's shaft and measure the current then.

Best regards.

 

The current was not that high when we were running the mixer without load in fast mode. But during normal operation, the customer fills the mixer with raw material and then he starts the mixer in slow speed. After 2 minutes or so when the mixer gets into motion at slow speed, it is put in fast mode. Now I tried with both fast and slow modes, with the mixer full of raw material, but the current was rising upto 103A. I had to stop the motor then otherwise it would have damaged the motor itself.

 

Regards,

Anurag

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What was the current drawn before the VFD was fitted?

 

What is the supply voltage before the start and after the motor starts and is running at full load?

 

Are all parameters concerning the motor entered correctly on the VFD (50Hz, 415V, 45kW, 80A, 2955rpm)?

 

What stands for low speed and for high speed in the relevant parameters?

 

Are the contactors now disconnected, there is put the bridge across R1B1Y1, and the R2B2Y2 are connected to the VFD's output?

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Having said "what stands for the speeds" is it, for instance, "30Hz" for the low speed and "50Hz" for the high ?

 

Also, are all other param. entered correctly (f/V relation, number of poles, etc) ?

 

I may be missing some important point in my suggestions, and maybe other add to my comments or correct me in some.

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Hello anu_rags

 

The problem that you have is that when you vary the speed of a motor using a VFD, the maximum torque is limited by the flux in the iron and as motors are designed to operate close to saturation flux, the torque can not be increased beyond rated torque at rated speed. As you reduce the speed of the motor by reducing the frequency, you must also reduce the voltage to prevent the motor from becoming saturated. The maximum torque is limited to the maximum torque at rated speed.

If you reduce the motor speed by changing poles, windings or mechanical means, the torque can increase beyond the full speed torque.

 

Power is torque times speed. P = T x N

When a VFD is used to control the speed, as the torque is limited by the maximum flux in the iron, the power of the motor reduces with speed.

If a mechanical speed changer is used, the motor power stays constant and so the torque increases as the speed reduces. If the speed is halved and the power is constant, then the torque is doubled.

If the speed is altered by switching poles, there are a number of different winding configurations that can be used and the selection of motor design determines the power of the motor at the lower speed. If the motor is designed to give a constant power, then the torque at the low speed will be increased.

With the motor used, the power rating at full speed is 45 KW and the power rating at half speed is 38 KW.

Using a VFD with the full speed winding, the rating at full speed is 45KW and the rating at half speed is 22.5KW which is too low.

 

One solution to your problem is to increase the motor size so that there is sufficient torque capacity at the lower speed.

Another solution is to change the mechanical coupling ratio so that at full speed of the mixer, the motor is operating faster. If you change the ratio so that at full speed, the motor operates at 75Hz, then and half speed, the motor will be operating at 37.5 Hz and the torque will be increased by 50%. You would need to ensure that there was sufficient power available at 75Hz to operate the driven load.

 

VFDs can not be used to replace mechanical speed changers and pole switching speed control without the engineering calculations being done to ensure that there is sufficient torque available to operate the machine correctly.

 

Best regards,

Mark.

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Hello Mark.

I still am in doubt as to what the VFD installed might have done to the motor.

If, for instance, in the 2 pole mode the motor probably worked normally (the poster does not say it however, I am only assuming this): current rated, speed rated - why, now with the VFD, in the same mode there are 103A and the speed more than two times lower than one which should be?

Best regards

 

However, behind this may lie such a thing: the motor being undersized for some reason, the client had the hope a VFD would solve the problem. (the poster only said "The current was not that high when we were running the mixer without load ")

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Hello Yuri

 

When the motor is connected in 4 pole, it has a rating of 38KW.

When the motor is connected in 2 pole, it has a rating of 45KW.

If we connect the motor in 2 pole and reduce the speed to half speed with a VFD, then at 4 pole speed, it will have a rating of 45KW/2 = 22.5KW. If the load at half speed is higher than 22.5KW, the motor will be overloaded when operating at half speed with a VFD and 2 pole connection, but will not be overloaded in 4 pole connection.

 

Best regards,

Mark.

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Hello Yuri

 

When the motor is connected in 4 pole, it has a rating of 38KW.

When the motor is connected in 2 pole, it has a rating of 45KW.

If we connect the motor in 2 pole and reduce the speed to half speed with a VFD, then at 4 pole speed, it will have a rating of 45KW/2 = 22.5KW. If the load at half speed is higher than 22.5KW, the motor will be overloaded when operating at half speed with a VFD and 2 pole connection, but will not be overloaded in 4 pole connection.

 

Best regards,

Mark.

 

Dear Mark,

 

What you say is correct, the motor is getting overloaded at lower speed. Im not able to understand how the poles in motor get changed with only 6 terminals. Does the motor have 2 separate windings for 2 speed or it is tapping on the same winding? If I connect the motor in slow speed for 38kw and then run the motor upto 3000rpm using VFD, will it be a problem for the motor and whether I can keep the current within limits of the motor. Will the motor winding of slow speed be damaged if they are been run upto 3000rpm. Or if it is the same winding for both the speed then it should not be a problem?

 

Regards,

Anurag

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