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Anonymous Poster #1

Water Pump

05/16/2012 6:19 AM

shall i connect 60Hz frequency to water pump? if no , why?

shall i run with AC Drive with 60Hz frequency?

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#1

Re: water pump

05/16/2012 6:45 AM

A1) if the pump is suitable for 60Hz supply, why not <rhetorical question>?

A2) if the drive is suitable for 60Hz supply, why not <rhetorical question>?

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#2
In reply to #1

Re: water pump

05/16/2012 6:50 AM

a1)motor pump designed for 50HZ

a2) we can get more than 60HZ frequency from AC Drive

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#3
In reply to #2

Re: water pump

05/16/2012 9:36 AM

a1) If the motor is designed for operating at 50Hz, it cannot be operated at 60Hz, continuously, though momentary increase is permitted.

a2) you can very well get even 400Hz from the VFD.

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#4
In reply to #2

Re: water pump

05/16/2012 12:46 PM

"shall i connect 60Hz frequency to water pump? if no , why?

shall i run with AC Drive with 60Hz frequency?

...

a1)motor pump designed for 50HZ

a2) we can get more than 60HZ frequency from AC Drive"

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balrajbontha,

Assuming you are the original poster and have now registered (thank you), here are the issues:

1) If the motor was designed for 50Hz and you run it at 60Hz without a corresponding increase in voltage, your motor will LOSE torque at a rate equivalent to the square of the voltage difference if the same V/Hz ratio had been maintained. So for example if you have a motor designed for 380V 50Hz and you want to run it at 60Hz, you would need to maintain the same V/Hz ratio of 7.6:1 (380/50). So to get the SAME torque from the motor, the voltage would need to be 60 x 7.6 = 456V (+-10%, i.e. 460V). So if you do NOT supply 456V and run it at 60Hz, then you have an effective voltage reduction of 380/456 = 83.33%, which means the motor torque will drop by the square of that, so 69.44% torque.

2) AT THE SAME TIME, you are increasing the SPEED of the motor shaft to 120% (60/50 = 1.20). If you have a CENTRIFUGAL pump, (i.e. a "quadtratic" load), then it will follow the Affinity Laws and the LOAD on the motor will change by the CUBE of the change in speed. So 1.203 = 1.728, meaning your LOAD on the motor will become 173% of what it was at 50Hz.

Now combine those two: Motor capacity is reduced to 69%, load is increased to 173%. What do you think the result will be?

By the way, EVEN IF you were to boost the motor voltage to 460V so that you have the SAME torque at 60Hz, you now still have a situation where motor capacity = 100%, LOAD increases to 173%. Result?

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