AC Drive Load Acceleration

[Modula2]

Does anyone know of a source for methodology to calculate motor acceleration by a drive. The load torque will vary from standstill to full rpm, so that seems to eliminate the formulae in the Nema application guide, for a more exact calculation.

My first thought was that with a frequency ramp, for each small step, a frequency perturbation would need to be applied to the motor to calculate motor speed change dynamics.

 

[electricpete]

The rotational equivalent of F=ma is

T = J * dw/dt
where
T = accelerating torque = Motor Torque minus Load torque
J = rotating inertai
w = radian rotational frequency = 2*pi*f
f = rotational frequency (rotations / second)

Rearrange
w(t) = (1/J) integral (T) dt

You can do a numerical integration with motor torque a function of both time (ramping changes torque speed curve over time) and speed (where are you on the torque speed curve) and pump torque a function of speed only.

I'm not sure if this is what you're looking for.

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[CJCPE]

Section 6.3.7 of the NEMA AC drive application guide can be used as follows (ft. lb. sec units):

With a linear acceleration ramp, the drive will attempt to apply a constant accelerating torque.

Attempted Accel Torque = Wk^2 X (RPM/time)/308 for any incremental speed change

Since the ramp is linear, RPM/time is (ramp time) / (total speed change)

The available accel torque is drive torque capacity minus the load's steady-speed torque requirement

If, at any speed, attempted accel torque is less than available accel torque, the drive will go into current limit or shut down

A good drive should be capable of accelerating at constant torque in current limit. While in current limit, the incremental speed change will be given by:

RPM change = (308 X time increment) / Wk^2

 

[electricpete]

Isn't torque supposed to be in the numberator of last equation?

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[CJCPE]

Your are absolutely right

RPM change = Torque X (308 X time increment) / Wk^2

 

[CJCPE]

Oops again:

RPM change = (Torque X 308 X time increment) / Wk^2

 

[Modula2]

Electricpete, a ref is "Power Electronics, Converters, Applications, and Design" by Mohan, Underland, Robbins, ch. on Induction Motor Drives - startup considerations.

In a small step of frequency and voltage (V/f = constant), these should change immediately, but the rotor speed will take time, so it would seem that the rotor slip frequency also changes immediately.

 

[aolalde]

Modula2.

Any rotating system obeys the laws of physics. It does not mater if you have a variable frequency drive or not.

As Electricpete stated the basic relation is:

dw/dt = T/j
When a VFD is used, ti (the time allowed in each step, before the frequency is increased ) must be large enough to allow acceleration to full speed of the system inertia (j). That will give the time extent of the accelerating ramp.

Note that the formula given by CJCPE is a particular case when the units are; Torque in Lb-ft, speed in rpm,constant of inertia in Lbf-Ft^2 and time in sec.

 

[Modula2]

Does that mean that if the ramp is too fast, the rotor speed would slip more and more behind the supplied frequency?

But is there a way the drive can know the rotor speed, other than direct measurement?

 

[CJCPE]

If the ramp is too fast, the rotor slips more, the motor develops more torque and the current increases. Most drives measure current in some way. A simple drive will simply shut down if the current is too high. More sophisticated drives limit the current by cutting back the voltage and/or acceleration rate. Sensorless vector drives estimate the slip mathematically and adjust the voltage to maximize torque per amp and also prevent the current from exceeding a safe value.