VFD Fault alarms at low speed

Jraef said:
Well, YOU said the “NC contact changes status to NO”, so SOMETHING is programmed to do that. I was just speculation as to what that might be. You really need to call your Vacon supplier and get their help.
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Ok, will try to find out more from Vacon or Danfoss.
Thanks Jraef!
 
From post 17 and based on what it says in the parts of the manual you posted, terminals 24-25-26 are by default fault outputs to send some signal to some other component when the drive is in a fault condition. Since you don't seem to be having any fault conditions they must be configured to some the optional functions and you maybe need to check the parameters mentioned in those instructions of G2.3.1 or possibly P2.3.9 to determine what this output relay is set to do, because whatever it is it must apparently be seeing a condition it was set to respond to and is apparently associated to a run or enable input somehow that wouldn't give you any fault if it opens the circuit but rather just a stop or maybe a zero frequency command for whatever reason.
 
Jraef said:
Well, YOU said the “NC contact changes status to NO”, so SOMETHING is programmed to do that. I was just speculation as to what that might be. You really need to call your Vacon supplier and get their help.
Click to expand...
Sure, I will call Vacon to get their help.
 
kwired said:
From post 17 and based on what it says in the parts of the manual you posted, terminals 24-25-26 are by default fault outputs to send some signal to some other component when the drive is in a fault condition. Since you don't seem to be having any fault conditions they must be configured to some the optional functions and you maybe need to check the parameters mentioned in those instructions of G2.3.1 or possibly P2.3.9 to determine what this output relay is set to do, because whatever it is it must apparently be seeing a condition it was set to respond to and is apparently associated to a run or enable input somehow that wouldn't give you any fault if it opens the circuit but rather just a stop or maybe a zero frequency command for whatever reason.
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Yes, I just found P2.3.9 But I still have to find what application the drive is programmed for to select the correct Output signal
 
ptonsparky said:
What is it being used on?
A fan or pump?

An auger delivering 250# of product per hour?
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It is used to run a conveyor motor at a low speed of 3m/mt (12.50Hz). So, it could be Local/Remote application. Currently I am away, should be able to work on this equipment in the next week.
 
As I understood, Speed = 120f/P. So for a given frequency, More poles = lower speed, more torque, Less poles = higher speed, less torque
This motor is running at a low speed of 3m/mt, Frequency 12.50 Hz. So, generally speaking is the motor current high at low speed and low at Higher speed?
How does the current get affected by number of poles?
 
Rob2025 said:
As I understood, Speed = 120f/P. So for a given frequency, More poles = lower speed, more torque, Less poles = higher speed, less torque
This motor is running at a low speed of 3m/mt, Frequency 12.50 Hz. So, generally speaking is the motor current high at low speed and low at Higher speed?
How does the current get affected by number of poles?
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That is somewhat a loaded question. With a variable torque driven load, current will generally be lower at lower speed.

Constant torque load which applies to your conveyor - current can be same at low speeds as it is at high speeds, this really depends on the driven load characteristics. For simplicity lets say the motor draws 10 amps at 60 hertz when fully loaded. Now change frequency to 30 hertz and you have half speed, if current remains at 10 amps and voltage remains at same volts/hertz ratio, your output power is half what it is at full speed. That is a truly contestant torque situation there..

replacing a 4 pole motor with a 2 pole motor of same HP and driving same load - the two pole motor should draw more, but is doing about 2 times the work because it is running at 2 times the speed. Place gearbox, pulleys in the mix to get same final output speed and the current should be similar, but will vary based on efficiency and power factor differences in the motors themselves along with any losses in the speed reduction method.
 
kwired said:
That is somewhat a loaded question. With a variable torque driven load, current will generally be lower at lower speed.

Constant torque load which applies to your conveyor - current can be same at low speeds as it is at high speeds, this really depends on the driven load characteristics. For simplicity lets say the motor draws 10 amps at 60 hertz when fully loaded. Now change frequency to 30 hertz and you have half speed, if current remains at 10 amps and voltage remains at same volts/hertz ratio, your output power is half what it is at full speed. That is a truly contestant torque situation there..

replacing a 4 pole motor with a 2 pole motor of same HP and driving same load - the two pole motor should draw more, but is doing about 2 times the work because it is running at 2 times the speed. Place gearbox, pulleys in the mix to get same final output speed and the current should be similar, but will vary based on efficiency and power factor differences in the motors themselves along with any losses in the speed reduction method.
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Need to add it will do about 4 times the work if a centrifugal fan or pump as the load increase with those kind of loads is exponential with speed increase.
 
kwired said:
Need to add it will do about 4 times the work if a centrifugal fan or pump as the load increase with those kind of loads is exponential with speed increase.
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Exponential, yes. It’s only 4x if the speed is exactly double. The power requirement varies at the cube of the speed change. So at 2x speed, a centrifugal load requires 8x power, which happens to work out as 4x. But at 1.5x speed, it’s 3.375x the power requirement, not 6x. Likewise at 3x speed, it would require 27x the power not 12x!
 
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