Calculating start/stop time for induction motor 1

You want a simple answer to a complex question?
By far the easiest, and most accurate determination may be simply measuring the starting time of similar motors driving similar loads.
And my question; Why?
A properly configured VFD is able to start and accelerate a motor up to speed without drawing excess current.
Starting times for DOL start motors are irrelevant for VFD driven motors.



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Ohm's law
Not just a good idea;
It's the LAW!

Thanks, motor starting time mainly depends on the load torque so I wonder if there is a table or a chart that can figure the approx. starting time for each load type.
my answer to your question why ? this is to know the exact time setting for the vfd parameter " run0up time "

I strongly suggest that such a chart, if it exists, would be based on DOL starting and so would not be particularly useful for VFD operation.

From this website:

Using inertia to determine torque and acceleration

The distinction between these two inertia equations, WR2 and WK2, is important because in AC induction motor applications, inertia is used to determine the motor torque required to achieve a desired speed within a given time.

T=WR²*N/308*t

T = acceleration torque (lb-ft)

W = weight of load to be accelerated (lb)

R = radius (or K, radius of gyration) (ft)

N = change in speed (rpm)

t = time to accelerate (s)

Note: 308 is a constant that incorporates the conversions from lbf to lbm, and from rotations per minute to radians per second

The equation notwithstanding, the acceleration torque is not constant, but varies as the motor accelerates from standstill to rated speed.

But all is not lost. Using a VFD makes things very much simpler.
Torque is proportional to current. You can set a current limit with a VFD and accelerate at almost constant torque.
A motor develops full torque at rated speed.
The motor current and torque are dependant on the frequency of the induced current in the rotor, or slip frequency, not the absolute speed.
Look at a motor current graph. Current over speed.
The speed is often given in PU of synchronous speed.
If the speed in PU is converted to RPM, for an 1760 RPM motor, we will see 100% torque or rated torque at 1760 RPM.
1800 RPM - 1760 RPM = 40 RPM slip, or 40 RPM/1800 RPM * 60 Hz (3600 cycles per minute) = 1.3 Hz slip frequency.
Ignoring second order effects, the motor will develop full torque at 40 RPM slip, or at 1.3 Hz slip frequency.
Although the torque curve is non-linear, you may extrapolate between about 20 RPM slip and 100 RPM slip with fairly good accuracy.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

If you don't want to overload the VFD at any point in time then assume the motor starting torque = motor rated torque.

Starting torque would be motor rated torque - load required torque. Load required torque is the torque required to keep the load spinning.

Find the inertia of the motor, inertia of the load and the load required torque and then use the formula Bill posted.

Typical (NEMA 1 per unit) load inertias are found in NEMA MG 1 Table 20.1 (squirrel cage induction). The formula to create the table is found in Section 20.11. Synchronous machines use Table 21.6, the equation is found in Section 21.12. Machine inertias are harder to define, since each manufacturer uses slightly different geometry (and possibly materials) - the inertia can be found using waross' calculation.

Once you have the inertia OF THE DRIVE TRAIN, you can figure out how long it will take to accelerate based on the current limit you set with the drive. For the most part, the motor output torque will be proportional to applied current (e.g., 1 pu current = 1 pu torque).

Converting energy to motion for more than half a century

Thanks all for your inputs,
I found this formula and it seems simple to use and applicate :
Starting time = ( moment of inertia x rpm ) / ( 9.55 x acceleration torque )

-where acceleration torque = 1.5 motor torque – load torque

Not sure it is the same above or maybe similar.

SS Ibrahim: Be cautious. The scalar quantity (1.5) in your equation corresponds to the current limit setting for your drive - i.e., 150% of rated current is what you intend to hold the current limit to during the acceleration portion of the operating cycle. If you have a different current limit setpoint, you will need a different scalar quantity.

Converting energy to motion for more than half a century

Expecting a continuous 1.5x motor torque during starting is overly aggressive for a typical VFD start.

Correct. Even the best VFDs, sized for Constant Torque/Heavy Duty loads, will deliver 150% of the VFD’s rated current for only 60 seconds before tripping off to protect themselves.

The main thing to understand about the difference in starting with a VFD vs starting with anything else is that the VFD is capable of full rated torque (100% of motor FLA) from the very outset and consistently throughout the acceleration process, and that being at it under the FLA can be done virtually for as long as it takes with no harm to the motor; 10 seconds, 10 minutes, 10 hours, 10 days, whatever. Any other starting method runs you up against the thermal damage curve of the motor, so the load must be accelerated in less than that time.

So if you start DOL (Across the Line) on a motor that needs a Class 10 protection curve, you must get it started within 10 seconds at 600% current or the motor is potentially damaged, but hopefully the overload protection will trip it off line first. But at the same time, that 600% current is only buying you 160% torque from that motor, which means the motor is heating up faster than the relative work it is performing. This is where most of the start time formulae are coming from, but again, this doesn’t necessarily apply to starting with a VFD.

In my experience, start time considerations with a VFD generally come down to external factors in the load, such as preventing surges, mechanical strains on other drive train components, timing issues with other parts of a machine or process, things like that.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

The more I think about it, the more I'd say use a lower torque, say maybe 75% or 50% of motor rated.

With a VFD, you just don't typically accelerate with it pinned on the current limit which means it doesn't accelerate with a fixed motor acceleration torque. The motor acceleration torque used or produced by the motor will vary as it accelerates. Expect 100% torque the whole time and the VFD will cause less than 100% torque for part of the accel and more than 100% torque (also more than 100% FLA) for part of the accel.