How to determine a frequency reference for VFD

[MikiBg]

Hi all,

Mixers for flocculation in WTP need to be rotated in desired speed. Motors are driven by vector control capable VFDs, but no speed sensors or any external control loops are applied.
Is there a way to calculate, at least approximate frequency reference for VFD which will result in desired shaft speed without any additional measurement of speed or complex model? Mixer manufacturer gave us some frequency-shaft speed diagram, but for one motor type only, and for unknown shaft load. In that diagram, within desired speed range, slip isn't constant and n times lower frequency doesn't result in n times lower shaft speed.
I'll appreciate your help.

Miki

 

[waross]

The difference between VFD frequency and the corresponding synchronous speed should not be more than 40 to 60 RPM. (2% to 4% at rated speed.)
What speed variations do you normally need?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[MikiBg]

Ok, but I need a connection between VFD frequency and speed of shaft rotation, not actual synchronous speed. I found some simplified formula which gives absolute slip for frequencies under rated, but assuming that peak motor torque is significantly larger than load torque. I'm not sure that condition is satisfied because very large torques and slips are involved. Anyway, according to diagram mentioned above, for 3kW, 50Hz, 4 pole motor: 50Hz gives 1415rpm shaft speed, hence 5,7% relative and 85rpm absolute slip; 30Hz gives 815rpm shaft speed, hence 9.4% relative and again 85rpm absolute slip. That means constant absolute slip, which complies with formula for constant load torque. Now I wonder can I use constant absolute slip for other motors and regulation ranges as well, since torque of propeller submersed in water can hardly be independent of rotation speed?

 

[LionelHutz]

Check the drive manual more carefully because vector control drives often are capable of calculating the rotor speed and compensating.

 

[jraef]

A VFD in vector mode can maintain 100% torque at any speed continuously, up to 220% briefly. If you are in vector mode and your slip is varying greatly, then that probably means one of two things:

1) That at 100% torque, the load is still requiring more than the motor can provide and that means your motor is likely too small for the task at hand.

Or...

2) You have not properly set up the vector mode. Some drives ship with it turned on but that requires you to tune it before the drive will operate. Because of that, other mfrs ship it with vector mode turned off so that the drive operates right out of the box with no tuning, figuring that if you want vector mode, you will know that it requires tuning to the motor and know how to do that. If you did not do the commissioning and don't know the status, I'd start with that issue first.

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[ScusaMe]

....from basic motor theory ....


Speed (N) in rpm is equal to the product of a constant (120) times the frequency (f) divided by the number of pole pairs (P) .... or stated mathematically ....

..... N = (120 x F) ÷ P

This renders the SYNCHRONOUS speed.

Actual shaft speed for an Asynchronous motor is dependent upon the amount of slip. . . . or shaft speed = synchronous speed - Slip

For 60 Hz motors (USA) ..... base speed (rpm) at rated frequency

# Pole-Pairs[tab]Sync. Speed
[tab][tab]2[tab][tab][tab][tab]3600
[tab][tab]4[tab][tab][tab][tab]1800
[tab][tab]6[tab][tab][tab][tab]1200
[tab][tab]8[tab][tab][tab][tab]900
[tab][tab]10[tab][tab][tab][tab]720

If your application uses IEC motors or motors that have a 50 Hz base speed, you should be able to calculate its base speed from the information above.

Hope this helps.

 

[cswilson]

For an induction motor in the standard operating range, the generated torque is close to linearly proportional to the slip frequency. (It's a good first approximation, anyway.) Typically, the specified nameplate velocity for the motor gives the speed for direct-off-line operation at a torque level the motor can maintain (thermally) indefinitely. You specify a 4-pole motor rated for 1415rpm at 50 Hz at its nominal torque level, so equivalent to 85rpm slip. If the torque is only 50% of nominal, the slip would be equivalent to 42.5rpm, for a speed of about 1457.5 rpm at 50 Hz.

An open-loop VFD, which doesn't know what the motor rotor is doing, simply puts out the commanded electrical frequency with a proportional voltage (i.e. constant V/Hz ratio). As you note, the slip in absolute terms (not percentage) is the same at different electrical frequencies. So at 30 Hz and the rated torque, the slip is 85 rpm for a speed of 815 rpm, and at half the rated torque, the slip is 42.5 rpm, for a speed of 857.5 rpm.

If the VFD is capable of "sensorless vector" control (really "shaft-sensorless vector") and is put in this mode, it automatically uses a motor model along with voltage and current measurements from inside the drive to calculate/estimate the speed of the rotor. In the normal operating range, this estimate is really very good, even without an explicit speed/position sensor on the shaft. In this mode, the command to the VFD is for the motor speed, and the drive will use the estimated speed and a built-in control loop to decide what voltage and frequency it needs to maintain that speed. For example with this motor, if the commanded motor speed is 1000 rpm, and the drive calculates that it needs half of the rated torque to reach/maintain this speed, it will calculate that it needs a slip of 42.5 rpm, so it will output a frequency equivalent to 1042.5rpm.

 

[MikiBg]

Thank you all for comments and tips. As some of you said, I also thought that solution might be "inside" the drive and contacted mfr for more details. These are Schneider Electric Altivar 61 drives.

 

[FreddyZ]

If you need EXACT speed control, you might consider the use of the VFD to operate an EC [Electrically commutated, aka Brushless DC] type motor.

 

[MikiBg]

...and here is what mfr said:

If you configured 1415rpm as nominal motor speed and speed reference is assigned to analogue input 0-10V with scaling HSP=50Hz, then voltage 10V present on analog input will cause:

a) motor speed in steady state will be 1500rpm providing that:
- Motor control law is either SVCU, SVCI or FVC and
- Slip compensation = 100%

b) motor speed in steady state will be lower than 1500rpm (for example 1450 rpm) providing that:
- Motor control law is either V/f 2 pts or V/f 5 pts or V/f quadratic or
- Motor control law is SVCU, SVCI or FVC, but slip compensation is lower than 100%

In a conclusion:
Reference is directly related to the motor speed if SVCU, SVCI or FVC is used and slip compensation is 100%
Reference is directly related to the stator frequency in V/f mode.

SVCU, SVCI or FVC with slip compensation = 100% means that motor speed does not depend on the load. If slip compensation will be lower than 100%, then motor speed will also be lowered (e.g. 10V = 1490rpm) when load increases. If slip compensation will be greater than 100%, then motor speed will also go higher (e.g. 10V = 1520rpm) with increased load. Formula is here :
Motor speed = reference (%) * synchronous speed * [ 1 + (slip compensation - 100%) * nominal slip * load (%) ]

For example slip comensation= 200%, 10V reference (=100%), nominal slip (1500-1415)/ 1500 = 5.6%, load is 67%, then
Motor speed = 100% * 1500 * [1 + (200% - 100%) * 5.6%* 67% ] = 1500 * [1 + 1*0.056*0.67] = 1557rpm

In V/f mode, parameter slip compensation is not applicable, so motor speed is always lower than adequate synchronous speed (unless breaking) and it depends on the load. For example:
- 10V as reference and load 100% will force motor to rotate at 1415rpm aprox
- 10V as reference and load 50% will force motor to rotate at 1457rpm aprox (middle point between 1415 and 1500).
- 5V as reference and load 100% will force motor to rotate at 707 rpm aprox.

Sorry for delay. Hope that this will be helpful to someone, as it was for me.