DickDV ... I shure hope you will get in this one.

Sir,

Has you know, being a programmer is only one of the steps to be a fully recognised integrator. Some will stay at this level and some will evolve.

The pryor steps are trough various paths. Seldom have seen programmers which origine flows trough a high level motor related path.

BUT... without motors, we would not exist. They are the end results of our profession... mostly we control forces... mostly motors.

For many years I've asked many, many, many peolpes/experts/EE/my dad on this particular motor basic knowledge... but nobody knows ...

So here it is:

Calculating the work needed to lift a load from A to B is simple.

Calculating the Force neede to do it in a certain time is also not that difficult.

Going the other way around is also not that bad, I mean breaking this load to a stop in X time.

So heres my problem.

A weight of 100 pounds goes down 10 feet. What can we calculate to bring it down in 10 seconds, 100 seconds, infinity... (holding it in mid air).

With the new VFDs mentionning Full tork, 100 Duty, 0 Hz ... They should be able to hold my weight into mid-air for infinity ...

Then how many Amps would they pull ( Say 440 Vac / 3 Ph. )

It's not really PLC but how can I become a good programmer if I don't know what I'm trying to control ?

Thanks Dick

PS: I'm having a ball with your earlyer posts...

PPS: What is fantastic about you is that you give up those "Magic Numbers"... In any feild we have rough estimate numbers... and you are telling us some of those in the motor field...

Drive-Motors on Hanging Loads

Golly, Pierre! Thanks for the comments. The only trouble is now I can't get my hat to fit anymore. Some of my shirts won't button around the neck either!

Enough of that!

Rather than give you specific numbers on your hanging load application, a better approach, if I understand you correctly is to exercise a little engineering judgement on this.

First, if you have a drive-motor system that is capable of developing enough torque to start the load upward and lift it, then it can be reasonably assumed that the motor will have enough torque to lower it and stop it in midair.

Second, lowering a load and stopping it is essentially a braking function, not a motoring function. Therefore, the drive must be capable of dealing with the braking energy. Most will not have this provision unless specified that way. Then the drive will be designed as a four-quadrant drive (this simply means that it is capable of motoring and braking up to its rated capacity in both directions) or it will be a conventional two-quadrant drive with additional hardware added to it to dissipate the braking energy. This hardware is often called dynamic braking which is a throwback to DC motors. A more precise term is snubber braking since the DC buss in the center section of the AC drive has its voltage "snubbed" at a certain level by the brake hardware.

Third, while the drive and motor may be able to lower and hold the load for a few seconds, it is mandatory that the thermal capacity of the motor to develop full torque as a holding brake at zero speed be evaluated. This is particularly brutal treatment of a motor and most will experience thermal runaway very quickly. When that happens, the drive's motor model becomes invalid and the load will fall. For this reason, every hoist manufacturer that I have contact with will ALWAYS equip the hoist motor with a holding brake which engages as soon as the motor brings the load to a stop. My view is that, regardless of the drive-motor holding torque capacity or the sophistication of the drive, a holding brake is absolutely required for responsible levels of safety on a hoist.

Pierre, you can see that I assumed that your question related to applications where an AC motor-drive system was in use. If a DC system was what you had in mind, or if I misunderstood your question, please reply and I will try again.

I'll stick my 2¢ in here.

First, for a vertical lift hp = (lbs x ft/min)/33,000

And given the power and speed Torque lb-ft = (hp x 5,250)/rpm

To get torque at "stall" you need to look at the mechanical linkage between the load and the motor shaft, but essentially T = lbs x ft for a simple lever or pulley, and divide by the gear ratio.

Now, Dick can correct me, but most VFDs aren't much good below 6 Hz, and the motors tend to "cog" or move with a little jerking. That may be a problem.

To do what you want you need dynamic braking and regenerative capabilities in your VFD, which not all have. And to get the motor to act like the brake when motion is stopped close to stopping you need to use DC injection braking, and the specs for a typical VFD I looked up indicate the braking torque is 30% of nominal torque. I don't know what the current is, but it obviously isn't going to exceed the VFD's AC output current rating. If you have a gearbox in your system you may be able to use a worm gear, and at ratios greater than around 30:1 they are self locking unless you have a lot of vibration.

And, to emphasize what Dick said, IF THERE IS ANY SAFETY ISSUE WHATSOEVER USE A MECHANICAL BRAKE.

Hoist drive questions

Tom, you are correct concerning 6Hz and below for scalar or V/Hz drives. But, today there are flux vector drives(use an encoder on the motor shaft) or sensorless vector drives (no encoder) that can develop full motor torque at stall. You have to be careful about sensorless vector. Many manufacturers are using this term (more snake oil!) and not delivering much better torque management especially at slow speed than if the drive were V/Hz. I am very familiar with ABB's Direct Torque Control system and have often used it to develop full torque at zero speed. There may be others but I haven't had any contact with them.

If Pierre's question was about holding torque and the resulting motor amps required, then you are also correct that the mechanical system between the hoist cable and the motor shaft must be considered. If the needed motor shaft torque can be calculated, then I can come up with motor amps, given motor nameplate data.

I wasn't real sure just what Pierre was looking for, so maybe this will encourage him to reply and clarify.

Hoist Drive Appl

One other thing, Tom. DC injection is not suitable for this kind of work simply because you cannot use ramp-to-stop drive control. DC injection braking only works when the drive is set to coast-to-stop which would be an out-of-control condition on a hoist. And, as you point out, DC injection doesn't give you much in torque either.

Much better to ramp to stop and use regen or snubber braking to enforce the ramp, then when stop occurs, hold the load with the motor only long enough to engage the mechanical holding brake.

Of course, the tricky part is restarting a hanging load either up or down! To do that properly, you have to energize the motor and develop the necessary torque BEFORE you release the mechanical brake. Getting this just right so you don't get jerky operation is often harder than getting the stop correct.

Sorry guys, I had to work this past week ... but here is what I meant.

3 years ago, I was called to "fix" a dynamic braking problem in a tobaco factory. They where lifting tobaco boxes (4'x4'x4') 3 floors up. They tought that bringing them empty boxes back down was a piece of cake... so they did not fit there drives with braking resistors... Big mistake.

The thing was that the weight of the elevator was about 75% of what they where lifting ...

So after a few runs, they destroyed an elevator cage. The elevator came down 3 floors, accelerating ... Voila!

The thing to do was very simple. First listen to the "expert". So they where called...

The mechanical eng. which design the system did'nt say a word, he looked at the motor supplyer's own Eng. who then looked at the Drive supplyer Eng. who ...was a great acrobat

What happened was that when they started this system, they ran the elevator down very fast... they slow the elevator in a short distance and a few time a day the drive failed (overvoltage on the buss of course...).

They had a brillant idea, "Lets slow it down, this way it will need less force to decelerate at the end of its travel".

That was there second big mistake.

When you brake a load against gravity, it's not the same thing has braking one in movement in an horizontal plane. The one in an horizontal plane does not get force from the "outside" !!! You just have to "release" the energy it has.

When Mother-Earth is pulling that bady down, you've got to do a little more...

The solution to this past problem was simple. First they added a machanical brake. Then they added external resistors. Done!

But for me it was only the beginning of a reflexion which I did not solve yet.

I calculated the energie involved in stopping the load in X feet, going at a certain speed.

I calculated the energie involved in stopping the load in X seconds going at a certain speed.

To proof the maths, its always nice to take a look at what is the results of making a variable infinite. So Stopping the load in an infinite amount of time ... Gosh this one really bugged me. No movement, no work, hence no force needed ...

So this system is like watching football on TV, spending Energie but doing not measurable Work ...

If its possible to use Energie but not do any Work, then it's possible to do Work without spending Energie... OK I'm only kidding on this but Work IS related to Force, Distance and Time...

From what I've read, I guess I hunderstood where you where leading me.

To solve this I have to get the Tork involved with the mechanical links to this elevator. Then the ratio of the gear box... then to the tork on the motor shaft... then this resulting Tork IS what my Drive will give...

That quest for knowledge puped-up again last week. My Danfoss rep. was telling me there new AC Drives could give me 100% Tork at 0 Hz, 100% Duty...

I asked them... So it could hold a weight in mid-air at 0 Hz to infinity... They said "Yap!" So I asked... how many amps will they pull.

The supplyer Eng looked at the Drives Eng. who look at... well you already know this story

Hanging Loads-Pierre

Pierre, before I respond to the technical issues in your last post, I want to say how delightful it is to read your posts. I don't mean any offense here, to be sure, but I suspect that English is not your native language. None the less, your beautiful choice of words and sense of humor are a joy to read even when the subject matter isn't pretty! I can only imagine how your prose must read in French but then I wouldn't be able to read it--pity!

Another issue I should mention. In the USA, people that design and build hoists and hoist systems must be licensed. I don't know how it is in Canada, but the highly hazardous conditions you mention serve as an excellent illustration as to the wisdom of doing so. I am not licensed for hoist work but I have done numerous hoist systems while partnering with a person that was licensed.

Let's say, for example, that you have a cable hoist with a load on the hook such that the total cable tension is 1000 lbs. And, let's say that the cable drum is two feet in diameter or 1 foot radius. Clearly, the drum shaft will see 1000 ft-lb torque under these conditions. Let's also say that the drum shaft is connected thru a gearbox with a ratio of 50/1 to an AC induction motor. Ignoring friction losses, etc. that would result in a motor shaft torque of 20 ft-lbs.

Let's also assume that the motor is four-pole which is 1800 rpm at 60Hz. You can solve the hp formula HP = T x R/5252 for one hp and you will find that each hp results in 3 ft-lbs torque. This is a handy number to remember for four pole motors. So, at 20 ft-lbs, a 7.5hp motor could be used but there would be no spare capacity for anything over 1000 lbs. It is likely that at least 10hp would be chosen which would result in a maximum available motor torque of 30 ft-lbs. A very important point here is that the amps required to develop a certain amount of torque at any speed is dependent on the motor involved. This is primarily due to the different magnetizing amps for the various hp ratings--the amps needed to produce the same torque will be quite similar over the different size motors.

If we assume a 10 hp motor, then the nameplate full load amps will be right around 15 amps. Lacking precise data from the motor mfgr, we can estimate magnetizing amps at 20% of FLA which would be 3 amps. Now we know two sides of the right triangle so, solving the vector equation twice, once for full load and once for 20 ft-lbs, we find that total amps would be about 10.3 amps.

Considering that this is at zero speed and that the motor formulas tend to fall apart somewhat at such low frequencies, I would take the 10.3 amp figure as the low limit. My experience is that actual amps might be a little higher. But, this gets you close and, if you need to estimate it, will probably be good enough.

But, I go back to my earlier comment of several posts back. You will always need more torque for lifting so the motor will always be large enough to hold the load. The drive sizing will also have to be large enough to do the lifting, as well. I conclude, therefore, that if you always specify the drive for identical braking and motoring amps, you should have the application covered with some to spare. This rule only applies to hoists, if you please. Other applications may need more or less braking. (I did one job some years ago where the braking hp was 10 times larger than the motoring hp!)

Questions? or is this clear?

It's cristal clear.

Merci beaucoup !