calculate RPM from VFD

[smcmanus]

Hi. I have a AB PowerFlex VFD. I need to calculate in the PLC(SLC 500)the speed the wheel in RPM's. All I know in the PLC is the Frequency of my drive in HZ. My gear ratio is 30. The number of tooth on the drive sprocket is 22. The number of teeth on the driven sprocket is 100. Appreciate your response.

 

[Steve Bailey]

The number of tooth on the drive sprocket is 22. The number of teeth on the driven sprocket is 100.

Given that, the speed of the driven sprocket is 22/100 times the speed of the drive sprocket.

My gear ratio is 30

Given that, and assuming it's a gear reduction, the output speed of the gearbox is 1/30 times the input speed.
You still haven't provided enough information. You haven't told us the motor speed at 60Hz.

 

[manmeetvirdi]

Hi there
If at 50Hz motor rpm is 1500, then
rpm at gear shaft is 1500/30=50

now one full turn (360 degree) of gear shaft will result in
22/100 rpm of the driven shaft
so at 50hz driven rpm will be
0.22*50=11rpm

 

[smcmanus]

Information on the name plate of the motor is 480VAC, 3Phase, 60HZ, RPM 1725, F.L.A = 0.08.

 

[Steve Bailey]

Your motor will turn at 1725 RPM when the drive's output is 60 Hz. Now you have enough information to work it out on your own. Please do it and then show your results here.

 

[TConnolly]

The synhronous speed of three phase motors is computed using the equation

Freq*120/p

Where Freq is the input frequency of the motor and p is the number of poles in the motor.

A 60hz 3600 RPM motor has two poles, an 1800 RPM motor has four poles, a 1200 RPM motor has six poles.

The equation computes the synchronous speed of the motor, however no three phase induction motor can run at synchronous speed - there must be some slip otherwise no induction can occur. A motor will typically run about 30 to 50 RPM lower than the synchronous speed. The more the load the more the slip. According to your nameplate data, your motor slips 75RPM.

You can sometimes determine the RPM by using a proximity sensor and high speed input to count gear teeth. If you need exact RPM then consider using an encoder or resolver.

 

[smcmanus]

I am a little confused. This system has no feedback. It just has a VFD which the PLC send's it a command speed. Instead of the operator inputing the speed in HZ in the HMI, I want them to be able enter the speed in RPM's. The PLC calculation will then convert the RPM's to HZ which will be sent to the drive as the command speed.

 

[manmeetvirdi]

Ok
1725RPM=60Hz
Now its linear all the way down

 

[Steve Bailey]

We've given you enough information to do the calculation. Use a motor speed of 1800 RPM at 60 HZ. Don't worry about slip. There's a good chance that you can enter that as a drive parameter.

Wheel Speed(RPM) * Total Ratio = Motor Speed(RPM)

Motor Speed(RPM) * 60 Hz/1800 RPM = Motor Frequency(Hz)

 

[TConnolly]

This system has no feedback.

Exactly. Thats the problem. If you can approximate speed and don't need exact speed then we can make a few calculations to put you in the ball park, probably better than +/- 100 RPM on the motor. Multiply that through your reducer to get an idea of what the final precision will be (+/-3-1/2 RPM). If you need more precise speeds then you must add feedback.

The equation for the motor synchronous RPM was given in post #6.

RPM = freq*120/p.

A three phase induction motor CANNOT run at synchronous speed because at synchronous speed the rotating magnetic field matches the rotor and then it cannot induce any current into the rotor windings, causing a loss of magnetic field in the rotor. The exact amount of slip between the magnetic fields will vary with the amount of load on the motor - so your motor speed will ALWAYS be less than what you calculate it to be based on frequency alone.

http://en.wikipedia.org/wiki/Induction_motor

Now you have a gear reducer that you say is a 30:1 ratio. That means that the input shaft speed is 30 times the output shaft speed. BTW, your gear tooth count doesn't jive with what your say the reduction speed is. But running with what you have given us so far, lets do a little math.

Motor_RPM = 30 * Reducer_RPM

Now a little algebra to rearrange the equation from post 6.
step by step:

Motor_RPM = freq*120/p ->

Motor_RPM*p = freq*120/p ->

Motor_RPM*P/120 = freq*120 - >

Freq = Motor_RPM*p/120


Keep in mind that this DOES not account for motor slip. What we know from the nameplate is that slip is 75RPM at the rated horsepower and frequency, but slip under other conditions is unknown. If your drive is a vector drive it may be albe to give you slip feedback, otherwise for exact RPM you will have to add instrumentation.

Not sure which powerflex you are using, but I think that some powerflex models allow you to input the comandspeed in RPM instead of frequency. If your drive supports that then use that option and then you don't have to worry about the conversions.

 

[Lancie1]

SMC,
If you know what speed range your operator needs (or PREFERS) to input (0 to 100, 0 to 1725, 1 to 10, whatever) then simply scale those numbers from X-Y RPM to 0-60 HZ and you are done. I usually just use 0-to-100% for speed inputs.

Input = 0 to 1725 RPM
Output (Scaled) = 0 to 60 HZ

Scaled value = input value X slope + offset
Slope = scaled max - scaled min / input max - input min
Offset = scaled min - (input min X slope)

Slope = (60-0)/ 1725-0 = 60/1725 = 0.03478
Offset = 0 - (0 X 0.03478) = 0

So, Output (HZ) = 0.03478 RPM, or RPM/28.75

 

[DickDV]

As has already been mentioned, frequency can only be used to estimate speed. That is due to the slip in the motor. With the given motor nameplate data, the difference between the no-load speed and the full-load speed is 75rpm (1800-1725). At 1800rpm, that is only about 4%.

Unfortunately, as you slow down, the difference between the no-load and full-load speed holds constant at 75rpm so, at 900rpm, the slip represents over 8%. At 150rpm, the slip is 50% and at 75rpm, it is 100%.

This speed error from slip passes thru to whatever the output speed calculates to so, for example, if the power train ratio is 100/1 and the motor speed at no-load is 150rpm, then the output speed would be 150/100 = 1.5rpm. Unfortunately, at full-load, since the error is still 50%, the output speed will have fallen to .75rpm.

This a perfect example of how sensorless vector or flux vector control of motors gives better slow speed control---by reducing the load-induced slip error. As shown above, what was only a small error at full speed, becomes huge error at very slow speeds unless something like vector control acts to reduce the slip speed error.

Hope this helps to make this a bit more understandable.

 

[Peter Nachtwey]

Why can't the PLC just read the RPM or frequency from the PLC?

Isn't there a way to get the PLC to communicate with the drive?

 

[manmeetvirdi]

If the load remains more or less constant then run the motor at maximum speed then note down the speed of finally driven object(for example speed of the conveyor in m/min) this value is say at 60Hz, then again assume this relationship is linear.
Do this and check whether the margin of error is within acceptable specified limit or not.

 

[DickDV]

Peter, sure there are ways to get the PLC to communicate with the drive but the drive doesn't know the rpm either unless it is a vector drive. Open loop scalar drives only know frequency and that, as I described above, is useful only as an estimate of rpm with the error getting larger fast at slower speeds.