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VFD and Locked Rotor Torque 2

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GearsNsparks

Mechanical
Jun 24, 2009
49
Howdy,

If a 20HP NEMA B AC induction motor has a locked rotor torque of 200% of full load torque, would a typical VFD be capable of developing that locked rotor torque condition at start up? The current is 600% of full load current at start up according to the speed vs current curve chart.

Is the start up torque normally limited by the VFD as opposed to the motor?

Thanks for any help.
 
 https://files.engineering.com/getfile.aspx?folder=cc36c2c9-01ad-4263-a87a-5f1408cf92c8&file=Motor_Chart.png
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That chart applies only to Across-the-Line (DOL) starting, meaning a fixed frequency applied to the motor (it shows different voltages). Current generally follows torque, but as you see, at start-up, current is much much higher than torque. That's because when you start ATL, the current goes high because the rotor is not yet moving and the difference between the rotor speed and the rotating magnetic field speed of the stator is extreme, i.e. maximum slip. Most of that current then is not "active current" creating torque, it is "reactive current", i.e the Power Factor is at something like .2 initially. As the rotor speeds up toward it's normal slip speed, the current and torque come back into alignment.

But almost nothing about that chart applies to using a VFD. With a VFD, you are varying the voltage AND the frequency together, matching what the motor is designed for. In that process then, the motor can create 100% of it's rated torque the entire time, and in that, the current and torque will follow as they would at ATL running speed. So although the ATL starting will produce more torque initially, the ratio (Amps PU* of torque) is lower with the VFD. *Per Unit

VFDs, assuming a "Heavy Duty" or "Constant Torque" sizing, are capable of making the motor deliver 150% of rated torque, but only for 60 seconds, and most can also make the motor deliver 200% for 2-3 seconds. that's based on the ability of the VFD heat sinks to safely dissipate the switching losses in the transistors. Exceeding that time frame will run a risk of damaging them. This current capacity profile is based on what you could expect, torque wise, from a Design B motor: 150% Locked Rotor Torque, 200% Break Down Torque. But at no time will a VFD allow, or NEED to allow, current to get to those higher values you see in that chart.

But I suspect from that chart that it is a Design C torque-speed curve, not a Design B as you claimed in your initial data.
NEMA_Torque_Speed_Curve_ti7ik6.jpg

If your load requires 200% torque for more than 2 seconds in order to accelerate, you will have to seriously over size an VFD.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
 
jraef; At full frequency and 10% slip the design "B" motor develops breakdown torque or 200% torque.
Assuming a suitably oversized VFD, will a design "B" motor develop breakdown torque (200%) at any speed when the slip is 10%
Thanks.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Bill,
Yes, a good "vector control" (sensorless or otherwise) drive would be able to make the motor deliver 200% torque at any time. That's why a vector control drive can maintain speed accuracy so well under a step change in load, it can briefly make the motor 2x what it is, almost instantly. In SVC, that change can be effected in as short as 300 radians. In FVC, some top performers are as little as 18 radians (less than 3 revolutions). But don't forget that 200% torque = 200% current so to do that for more than a couple of seconds would require some serious consideration of the VFD sizing, and you must also take into account that the motor will be pulling 200% current as well. On a Class 20 OL curve, that gives you about 2 minutes maximum for the motor, maybe less. After that you are flirting with the thermal damage curve. You then also have to factor in a cool-down time for that, so you can't do it over and over again either.

Then 2 minutes from the VFD might as well be an eternity, so if you DID want to deliver that much torque for that long, count on doubling the size of the VFD.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
 
Thanks Jeff.
That was my understanding.
I wanted to be sure.
That leads me to my next question:
Can we predict motor performance with a VFD by reading the torque curve backwards?
That is to say, with the curves that you have posted, if we reverse the "Percent Full-Load Speed" scale across the bottom of the graph and call it the "RPM Slip" scale starting from 100% point and reading to the left.
May we then safely use the curve at any commanded speed by considering the slip RPM?
Bearing in mind that for a 1760 RPM motor we should be using 40 RPM slip for full load current.
Said another way, 40 RPM slip at any given speed will be full load.
Have I got this right, Jeff.


Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Essentially, yes.
61347d1350737031-vfd-flying-start-function-vfd-speed-change.jpg



" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
 
Thanks Jeff.
I find it easier to concentrate on slip RPM. Same slip RPM means the same torque and the same current.
Once a VFD is involved, I just look at the motors rated speed to get the slip RPM and use that.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Does a motor starter/soft start achieve the DOL starting performance (locked rotor torque) shown in the manufacturer performance chart or is that a different behavior as well?
 
So to achieve the locked rotor torque value on start up, a VFD with the 1-2 second characteristic of 200% nameplate torque is the only method or is there another common alternative?
 
Across-the-Line (DOL) starting is where that Torque / Speed chart comes from, so a basic "Motor Starter" will make the motor behave as shown. A Soft Start is used because you DON'T want that added torque expressed on your load. You would NOT use a soft starter if you needed Locked Rotor Torque from the motor.

A VFD sized for Constant Torque applications can make a standard Design B motor deliver Locked Rotor Torque (150%) for up to 1 minute. That 200% value you showed, as I said, is not a Design B motor.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
 
A quality soft-starter can be set to connect the motor to full line voltage and produce the same torque as a full-voltage starter can.

A VFD only operates the motor to the right of the breakdown torque, so the locked rotor part of the curve has no use when considering the possible performance with a VFD.
 
A 1750 RPM design B motor with a VFD will develop 200% torque when operated with a slip of about 175 RPM. 100% - 90% on the graph.
At about half that slip or 87.5 RPM slip it will develop about 150% torque. 100% - 95% slip on the graph.
The percentages are only valid at 60 Hz applied current.
Convert to RPM slip and use that figure for other speeds.


Bill
--------------------
"Why not the best?"
Jimmy Carter
 
So if I'm sizing a motor for a conveyor with a large hopper on the end, it needs lots of starting torque until the material flows (transitions from static friction to kinetic friction) at the hopper and not so much once the belt starts moving. Should I size the motor based on 150% of name plate torque, locked rotor torque (200%), or breakdown torque (~250%) before it goes to the electrical engineer?

It seems safe to assume the EE could go with either a motor start or VFD and run 150% of rated torque during startup.

It also looks like the EE could size the VFD to provide 200% torque for a few seconds to get things moving and then ramp up to belt speed while limiting the torque to 125-150% range for up to 60 seconds.

Could the EE also select a VFD that would enable use of the breakdown torque rating for a few seconds at start up?

 
The first question to answer is how do you determine what is the 100% torque requirement. You will find that it can vary so much due to, for example, moisture, particle size, etc., that the only number you can compute is the torque required for a a best case condition with material flowing smoothly. Then pick a motor and gearbox where the overload rating is 10 times that, if you want the conveyor to start up under load.
 
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