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rockman7892 (Electrical) (OP)
27 May 09 22:55
I have read some other threads pertaining to these ratings and from what I gather "Normal Duty" and "Heavy Duty" are two different ways of describing the same vfd.

A normal duty drive is the same as a variable toruqe drive and usually has specifications of 110% overload for 1 min.  A Heavy duty drive is the same as a constant torque drive and usually has an overload rating of 150% for 1 min.  I would assume also that the normal duty drive has a higher contiuous current rating than the heavy duty one, and only the overload rating of the heavy duty drive makes the current similar to that of the normal duty  

Lets say I need a drive rated for 30hp.  A normal duty 30hp drive has a heavy duty rating of 25hp.  If a salesman or someone else tries to convince me that I need a 30hp heavy duty drive, then this would be the same thing as using a 35hp or 40hp normal duty drive.  In other words a heavy duty rating is just another way of describing a higher rated normal duty drive, and the relationship between normal duty and heavy duty will always increase lineraly.  Is my understanding of these ratings correct?   
jraef (Electrical)
28 May 09 2:04
Essentially, yes. The use of "Normal" came about because of a lot of misunderstanding of "Variable Torque" vs "Constant Torque" terminology. Since over 60% of all AC motors are used on pumps and another 10% on fans, most of which are centrifugal loads, it made sense to label VT as :Normal" and CT as the exception, or "Heavy".

It really is kind of a shell game of sorts however. VFD mfrs acknowledge that certain types of loads will require exponentially less power as the load drops off, i.e. centrifugal pumps or fans. That means that under the worst loading conditions, i.e. 1/2 speed or less and under acceleration, one can expect that a lot less will be asked of the drive. So there is no reason then to size the components and heat sinks based on 150% current for 1 minute; it will probably never be asked to do that. But if you have a CT load like a conveyor moving slow, or worse yet starting from a stop, and trying to accelerate, you will be asking MORE of the drive from an output current standpoint that if it were already running full speed continuously. Hence the 150% for 1 min. factor.

Regardless of what they call them however, the reality remains the same; if you buy a VT / "Normal Duty" drive for your pump or fan and the next guy to use it down the road has no clue as to what that means, he will try to put it on a CT load and it will fail. Murphy rules...


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ScottyUK (Electrical)
28 May 09 4:23
Murphy was an optimist...
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

Helpful Member!  DickDV (Electrical)
28 May 09 8:20
One minor clarification on jraef's post.  Using a normal duty drive on a heavy duty application will probably not cause the drive to fail but rather to fault to the point where it will not be suitable for use.

A drive that is constantly faulting on Overcurrent is almost surely a drive that was undersized for the load.

To expand a bit further on the ND and HD differences, when sizing a drive for a particular application, you really have to determine what the continuous and the short-term (typically 1 minute or less) torque demands will be.  Using that data, you then determine what continuous and short-term current the motor is going to require to provide that torque.  Using that data, you then size the drive.  Note particularly that the drive must be sized to provide both the continuous and the short-term torque.  Skimping on either will cause an unhappy result with the drive constantly faulting either on overcurrent or overtemp.
rockman7892 (Electrical) (OP)
29 May 09 8:41

So in essence if I have a 30hp conveyor or pan conveyor I will need to supply this with a 30hp Heavy Duty / Constant torque drive because of the nature of this constant torque load.  Essentially however this 30hp drive is the same as a larger (40 hp or so) normal duty drive.  So I could theoretically put a larger normal duty drive on this conveyor.  

If I have a 30hp fan or pump I can supply this with a 30hp Normal Duty / Variable Torque drive because these are variable torque loads.  If I try to put this same 30hp Normal Duty drive on my 30hp conveyor, there is a good possibility that the drive will trip.  

Do a 30hp Normal Duty and Heavy Duty drive have the same continuous current ratings?  I would imagine they would.

DickDV

You mentioned finding the continuous and short-term torque demands for a motor in order to size the drive.  What is the best approach for going about this.  If I have the motor information can these torque demands be determined from motor ratings / curves?   
DickDV (Electrical)
1 Jun 09 7:54
The continuous and short-term torque levels are determined by the load, not the motor.

So, if your conveyor never requires more than 10% short-term overtorque for its operation, then you can use a normal duty (variable torque rated) drive even tho the load torque characteristic is constant.

The key piece of information with these drive ratings is that the continuous ampacity is the same.  It's the short-term rating that is different, and that's all.

So, regardless of the torque-speed characteristic of your load, if all you require is 10% or less shortterm overtorque, use a Normal Duty drive.  If your load requires more shortterm overtorque than 10%, you will need to provide a Heavy Duty drive.

Just to make things more interesting:  If you need more than 50% shortterm overload capacity, a Heavy Duty drive will not be enough either.  You will need to go up one more hardware size to go above 50% overtorque.  Typical NEMA motors can go as far as about 220% overtorque before saturating, so that sets a limit on how far you can push the motor with larger drives.
Helpful Member!  LionelHutz (Electrical)
1 Jun 09 8:44
Lets consider 480VAC drives in North America. A 40hp drive would be rated 52A continuous and 110% or 57A for 1 minute. The manufacturer now wants a drive with the 150% for 1 minute rating so someone just calculates 57A / 1.5 = 38A. So, they call the drive 38A continuous rated and 150% or 57A for 1 minute rated. The drive is still capable of 52A continuous.

Further to the above, the manufacturer would realize that 40A is the standard FLA rating for a 30hp motor and do what is required to bump the 38A rating up to 40A and get 60A for 1 minute out of the VFD. Maybe a little extra heatsink or a slightly higher CFM fan. This could mean the drive is now capable of say 54A continuous but the manufacturer would just keep the 52A rating to match the NEMA recognized FLA of a 40hp motor. So, the manufacturer now has a drive that is rated 54A continuous and rated 60A for 1 minute. It is labeled 40hp normal duty or 30hp heavy duty.

The above is the short version of the logic used to create the dual ratings for drives. There's really nothing more complicated than that. If you wanted to rate the above drive 200% for 1 minute then it would be labeled 30A continuous or listed for use on a 25hp motor = 27A which is the closest to 30A NEMA motor rating.
 
rockman7892 (Electrical) (OP)
1 Jun 09 9:17

Thanks for the explanations guy, it makes things much clearer.

Without knowing much about the typical load, (design phase) how can you determine the torque levels that will be required by the load?  Do you have to get the load/torque curves of the load from the manufacturer?  Do these overtorque requirements include the starting of the motor as well?

I thought maybe that you could use the speed-torque curve of the motor for determining the necessary torque values, but it sounds like you need the actual load infomration.
LionelHutz (Electrical)
1 Jun 09 9:38
You find the maximum load torque required - starting or running. Now, go to the motor curve, near 100% speed (above the breakdown torque peak) find the above load torque on the motor curve. Now, for that same speed find the matching current on the motor curve. That's your required short term current rating.

As for continuous rating - well you should not be going over the motor rated torque continuously so if the drive is sized to match or be greater than the motor it will be OK.
 
rockman7892 (Electrical) (OP)
2 Jun 09 8:05

LionelHutz

I was looking at two different motor performance curves as sort of an exercise to determine the max torque required for two different motors as you explained.  On the first curve (bottom of sheet 1) I'm thinking I go right after the breakdown peak at about 1600rmp, and find the max torque required to be about 85 lb*ft.  The corrosponding current I find at this value would be about 42A.  Does this seem right.  Would this 42A be my short term current rating?

In the second motor curve (sheet 2) the infomration is given in a bit of a differnt format.  It appears that the curve essentially defines the max torque and max current for the motor.  From looking at this, it seems as if the max torque will be about 118 lb*ft and the max current will be about 47A.  Does this see correct?

Now going back to the first example, assuming I ran this motor at 1600rpm, would I always develop this max torque at this speed?  In other words just because I run this motor at a particular speed does it mean that the torque for that speed will be the value shown on the curve?  In other words if the motor is run at 1600rmp for more than a minute will this cause the drive to trip?

Even know at lower speeds say 400rmp for example, the torque is lower the current is shown as higher.  Why wouldn't we use these higher current values as the max current values?
LionelHutz (Electrical)
2 Jun 09 8:32
1st paragraph. Yes, that is basically correct.

2nd paragraph. Yes, that looks correct.

3rd paragraph. No, not all all. A VFD shifts the curve so the running rpm is reduced by the ratio of operating/rated frequency. The VFD will alway operates the motor in the part of the curve above the breakdown torque. The vast majority of the time it's operating somewhere between the 100% torque speed and the unloaded speed right at the very right end of the curve.

 Also, a motor will only draw the current required to drive the load. Well, it will attempt to draw enough current to drive the load until it reaches the breakdown torque and then it stalls. At that point, the current will go very high like you have noted and this will trip the VFD.
 
rockman7892 (Electrical) (OP)
3 Jun 09 8:42
LionelHutze

O.k. I see what you are saying regarding the speed torque curve shifting.  Basically the whole curve will shift left and will end at whatever synchronous speed corrosponds with the frequency setpoint of the VFD?

If I understand correctly, in your last paragraph above you are saying that the motor will operate at the far end of the curve (near synchronous speed) when the motor is unloaded.  When the motor starts to get loaded at a given frequency the slip will increase and the required torque will move back left on the curve until it reaches the breakdown torque and thus can no longer turn the shaft and will stall tripping the drive.  As the required torque moves left, it will reach the maximum required torque (42A in example 1 above)right before it hits the breakdown torque and stalls.  Is this correct?

What if the motor was loaded or operated at this 42A maximum torque value for more than a minute?  Would you have to have a larger drive since the drive ratings are only given for a 60sec time period?

What about the starting of the motor where the current will be much larger than 110% or 150% (usually 600%).  Does this starting current count towards the 60sec rating of the drive, or is this starting current duration usually not lasting for enough time to factor against these ratings?
LionelHutz (Electrical)
3 Jun 09 16:04
Yes.

Yes, unless the drive reaches a current limit setting. It would then keep dropping the output frequency until it found a stable operating point.

Yes.

Any current >rated counts towards the overload rating. The current could never reach 600%, the VFD could not support that high a current. The VFD starts with the slip = 0 rpm and ramps it up from there. Often, the motor will start with < FLA on a VFD.
rockman7892 (Electrical) (OP)
4 Jun 09 13:33

LionelHutz

Can you please explain the 2nd and 4th paragraphs in your last response.  I understand the general concept of them but am unsure exactly what you are saying.

When you refer to a current limit setting, are you referring to an manually entered setting, or a setting based on the rating of the drive?  How does the drive drop the frequency unitl it finds a stable operating point?

If the VFD is starting the motor with a slip=0 then does that mean that the motor is somehow being started at synchronous speed?  I do not undersand this comment.
rockman7892 (Electrical) (OP)
9 Jun 09 8:20
Can anyone please explain how when starting a motor with a VFD, the motor can be started without drawing much more than full load current?  I am having a hard time seeing this related to the speed torque curves.
CJCPE (Electrical)
9 Jun 09 9:07
The VFD is started at less than 2 Hz. At that frequency, the synchronous speed is only about 50 RPM (4 pole motor). If the normal slip is 50 RPM, the motor is at normal slip at zero speed. Since the motor is at normal slip, it can produce rated torque at the normal full load current. The frequency is increased at a controlled rate so that the slip doesn't increase beyond 150% of the normal slip, and the motor produces up to 150% of rated torque while drawing no more than 150% of rated current. If the VFD is only rated for 110% overload, the acceleration rate is limited to limit the current accordingly.
DickDV (Electrical)
9 Jun 09 12:14
rockman, the speed-torque curve you are likely looking at is for the motor at 60Hz.  You can know that by looking at the synchronous or no-load speed.  For example, if the motor is wound 4 pole, then the no-load speed would be 1800rpm and the full-load speed would be slightly less than that, the difference being the slip.

When you power a motor from an inverter, the motor sees variable frequency power.  That torque-speed curve shifts horizontally to the left as the frequency goes down.  That's the key difference.

So, when a motor is standing still, the drive starts it with 0, then 1, then 2, then 3hz.  This is equivalent to shifting the torque-speed curve so far to the left that only the sync speed and a little of the curve is visible.

The effect is to always keep the motor operating on the "front face" of the torque-speed curve.  In that range, the motor efficiently converts amps to torque so there is no need for huge inrushes.  The motor is always operated between the no-load and slightly above the full-load point on the curve.  The curve simply shifts left or right with frequency to get the various speeds (including zero speed).

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