frequencymotorrpm
I have a 3-phase motor but my knowledge on the topic is almost null.
But, you know, I have also curiosity to satisfy.
The motor's nameplate reads:
50 Hz, 7.5kW,
400V, 13.8A / 690V, 8.0A
cos phi 0.89
2910/min
60 Hz, 8.6kW,
460V, 13.5A
cos phi 0.89
3492/min
It looks like the RPM is proportional to power line frequency. Is this true? Applying i.e. 340V 50Hz, may I assume that the RPM remains 2910/min?
For applications with only moderate requirements you can implement closed loop control without a sensor by measuring the back-EMF of the motor.
You can even have a purely analog solution where the motor is driven by a power supply with negative output resistance to compensate for the motor resistance - this was used in Philips cassettes recorders 40 years ago.
This is a similar circuit to that used. The resistor Rs should be equal to the motor resistance measured when the motor is not rotating. The voltage across Rs increases with motor current and is fed back to increase the applied voltage to compensate for the voltage dropped by the internal motor resistance. The equivalent circuit in section B of the diagram helps explain the operation.
(Image from Precisionmicrodrives)
This can be generalised as RPM = 120 x f / poles.
Therefore # of poles = 120 x f / RPM.
Since induction motors slip the full-load RPM will be somewhat less than the calculated speed. From the rating plate we can see that the min -1 (revolutions per minute) is 930 to 950 at 50 Hz. This is just below 1,000 RPM so it must be a six-pole motor.
Best Answer
You have a squirrel cage induction motor. The synchronous speed \$n_S\$ (speed of the magnetic field pulling the rotor) is:
$$n_S = \frac {120 f}{P}$$ where f is frequency and P is the number of poles.
You have a 2 pole machine, which gives a \$n_S\$ of 3000rpm at 50Hz and 3600rpm at 60Hz.
It's an induction machine, because rotor speed n is 97% of \$n_S\$. Slip is 3% at full-load.
To answer the question.
At 340V, 50Hz, the motor will spin at 2910rpm dependent on the load torque.
$$ T \ \alpha \ V^2$$
Torque is proportional to the Voltage squared. 340V is 85% of 400V, so load torque would be approximately 72.3% [\$(85\%)^2\$] of full-load torque at 400V.
If the actual load torque was greater than this, the motor speed would slow down to drive the load. Slip would increase.
Edit...
Torque Reference
Red line is load torque. Motor will slow down until it is capable of driving load.