Calculate Speed

Tim Ganz

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T
Join Date
Dec 2010
Location
Dallas, Texas
Posts
689
Hello

First this ia really great site. Glad i found it.

Here is my question.

I need to calculate the full speed of a driven auger that is controlled by a vfd. The motor nameplate has 1705 rpm at 60 hz but the full speed of the vfd is 80hz.

I calculated this to be 2400 rpm synchronus and 2306 rpm with the slip compensated. Is this correct? This is the first time i have had to do this. Also there is a gearbox with a ratio of 14.38 stamped on the side. How would i go about calculating the final shaft speed at 80 hz with this ratio? Do i just divide the 2306 by 14.38?
 
From what I can tell your numbers look correct, as does your determination of final shaft speed by dividing by gear ratio, assuming you are accounting for all the gearing (no additional chains or belts).

However, your drive may have something to say about the actual final speed. The nameplate RPM is listed at full load as if the motor were powered across the line. If that motor were not loaded to full load then the RPM will be higher. A VFD can do some things with the output waveform to account for loading. If your VFD is doing this then you can't simply assume that the shaft speed will be nameplate speed. In fact, unless you know the loading, you can't assume the shaft speed even if the VFD isn't doing anything but putting out a given voltage and frequency.

Keith
 
I think that it should be.
RPM = (Frequency 60 Hz * 120) / # of poles 4 in the motor
RPM = (60Hz * 120) / 4
RPM = 1800
But the Nameplate says 1705, so the difference is your slip rating.

The VFD's that I have used have all been pretty linear, so that you can do a straight ratio.
A motor turning 1705 RPM @ 60hz will turn 2273 RPM @ 80 Hz.
Of course I am by no means an expert, so you'll have to take this with a grain of salt.
​
 
I think the base speed of your motor is 1800 RPM assuming it is designed for 60 Hz and has 4 poles. So using this it does work out to 2400 RPM at 80 Hz.

"The motor nameplate has 1705 rpm at 60 hz"
Can you verify the above statement.

Hope this helps
 
%Slip = ((SynSpeed-RatedSpeed)/SynSpeed)*100
= ((1800-1705)/1800)*100
= 5.27%

So 2400rpm (80hz 4pole)

= 2400 * (5.27/100)
= 126.5 rpm (loss due to slip @ FLT)
= 2400-126.5
= 2273.5 rpm

Hope it helps
 
For an induction motor:
n = f x 60 / p

Where f = supply frequency in Hz, p = number of pairs of poles, and n = synchronus rotor speed in r.p.m

For a 2 pole motor at 60 Hz: n = 60 x 60 / 1 = 3600 rpm.
For a 4 pole motor at 60 Hz: n = 06 x 60 / 2 = 1800 rpm.

So your motor is obviously a 4 pole motor with a synchronus speed of 1800 rpm

At 80 Hz and 4 poles: n = 80 x 60 / 2 = 2400 rpm.

The slip of the motor will vary with the mechanical load on the motor. So the slip will increase with an increase in load and decrease with a decrease in load.

Whether the slip changes linearly with load and frequency changes is questionable. This may be an acceptable assumption to make but should then be tested on the machine for acceptable operation with your application
 

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