Motor query

I am making an assumption in this question. Please correct me if I am wrong.
a) Slip of a synchronous motor varies with load
If the above statement is true then what would be the approximate variation on say a nominal 1440 rpm 7.5 kw motor running at a) no load
and b) full load ie drawing its nameplate current.
I know there are a lot of variables involved here but I am only after a very rough guide ie 3-5% or less than half a percent etc.
Regards Alan

%SLIP=Ns-n/Nsx100
Ns=sychronous speed in RPM
n=motor speed in RPM

When dealing with motors this site may help you with the calculations etc RELIANCE MOTOR

Look at all 5 sections but Applications is where the formulas are to find answers to specific values

Thanks Ron. A useful site. Alan

If I remember correctly, a syschronous motor rotates at a speed proportion to the line freqency. No slip. A squirrel cage motor has slip.

Are you sure it is a synchronous motor? 1440 rpm is not integrally divided by 50 or 60 Hz. At 50Hz synchronous speed would be 1500 rpm.

For an induction motor I think that the rated rpm is at full load. You would get closer to synchronous speed as load lightens.

My mistake, I meant a squirrel cage motor. What I am after is a rough idea of how close to synchronous does the motor get at no load.

The no load ie synchronous speed of a motor is calculated by
Ns=synchrouous speed
f=frequency in cycles per second
P=number of poles

Ns=120xf/P

Since you are down under I will assume your f=50HZ so the equation should look like Ns=120x50/4=1500

You stated the motor was a 1440 rpm..using the formula
%SLIP=Ns-n/Nsx100
1500-1440/1550=.04 x 100 =4%SLIP

Ron. What I am looking for is:
Does the slip stay constant for all loads or does a lightly loaded 1440 rpm motor run at closer to synchronous ie 1450 rpm
If the above is true then what is a very rough approximation for the variance?
Regards Alan

I think I get what you are looking for now. You want to know what the variance is from no load to full load in RPM...ie if its a 1500 rpm motor at no load...1440 at full load, this is a 7.5KW (7.5x1.34=10.05 horsepower) @230v (just picked a voltage) will run 10Amps at full load.

What I am about to do I have no idea if it is relevant or not because I am not sure of the linearity of current to load with motors. Using simple math tho I would see the above case a loss of 60rpm from no load to full load I would divide the 60rpm by the 10A and get 6rpm of slip per 1Amp, you can determine more resolution if needed.

All the above is probably ****(but made sense to me), what you probably need to do is look at it with an ampprobe and RPM gauge if its that important, if its not that important an issue then use 1440 as the constant speed when under load (any load). I have done the ampprobe and guage thing before, the motor will vary depending on load. Thats why I stated the above made sense to me, what I am not sure of is with motors that linearity applies..ie if the slip is proportional to the current at smaller load.

Slip is dependent on load. An
increase in load will cause the rotor to slow down or increase
slip. A decrease in load will cause the rotor to speed up or
decrease slip.

I think I said that, but wasnt sure if it was truly linear but have looked at the formulas etc and have determined it would have to be in direct proportion to the load...and current is a representation of the load.

That stated I see my math is wrong
7.5KW = 10HP @230v 3PH would draw 25amps(where did I get 10?)
60RPM/25A = 2.4 rpm per 1A of load OR approximately 6rpm per HP.
I leave it to you to carry it further if you need too.

Motor slip

Alan, I know I'm jumping in here a little late but I think I can help. In a squirrel cage induction motor, the nameplate rpm and amps is at full rated load. As the load is reduced to zero, the slip also reduces quite linearly to zero. The no-load speed at 50HZ will be 6000 divided by the number of poles. At 60HZ, it will be 7200 divided by the number of poles.

An induction motor running free shafted will usually run within 1 rpm of sync speed. You must remember that there is always some load, mostly fan windage and bearing/seal drag.

Slip is probably the best way to estimate motor loading in the field. A simple strobe light will work nicely. Do not be seduced into using motor amps. Torque producing amps and magnetizing amps are vector quantities and require solving the vector equation to identify the % of torque producing amps. The error is quite small very near full load but increases quickly as load reduces until, at no load all you have is magnetizing (field producing) amps. This current usually runs around 20-25% of full load amps and clearly produces no shaft torque.

Hope this helps a bit.

Thanks to Dick and Ron. Would either of you know where I could find any examples of the slip vs load curves of a motor. Regards Alan

curves

In the US, the Electrical Engineering Handbook has torque-speed curves for the various designs of NEMA motors. I don't know where you would find curves for IEC motors.

However, why use a curve? The load-slip characteristic is a straight line! And the nameplate gives you the full load point while the sync speed-zero slip gives you the no-load point. Between those two points, its a simple proportion---half rated torque would be half nameplate slip, 90% rated torque would be 90% nameplate slip, etc.

Thanks DickDV. I was under the impression the that it would be a curve not a straight line. That makes it a lot easier to calc.
What I was looking at was if 2 identical motors are doing an identical job of lifting 2 objects (but 1 object weighed twice as much as the other) then what way could I use to calculate the % difference in lift between the 2 motors. If I can calc slip difference as a % then the rest is easy to calc.
Regards Alan

motors

Right, Alan. As long as neither motor is being overloaded and they are identical motors, then the motor with twice the load will slip twice as much. Simple proportion or %.