Current in motor input line ? 3

[Skyba]

Hi all,

I have a motor rated @ 370 W (1/2 HP), 59 N.m and 59 RPM. I would like it to run at 40 RPM and 25 N.m (=109 W) with the help of a VFD in V/Hz mode. The torque limit is for security reasons. I'd like to know the current in the motor line in order to buy the appropriate motor starters / overload protection (and the Variator)!

I have done some calculations and I would like to share it with you, because I am not quite sure about them. To illustrate my thoughts I have done a little graph =)

http://img818.imageshack.us/img818/5006/u71r.png

Point 1->2 : To account for the speed reduction, we would have to reduce the frequency and the voltage to 68% times the initial values.

Point 2->3 : To account for the torque reduction of 25/59 = 0.43, we would have to reduce current by square_root(0.43) = 0.66. Why ? In my understanding the motor torque varies in square proportion to the input current. If we reduce current by 0.66, we´d additionaly have to reduce the voltage to 66% the value of point 2.

So at the end, we would have reduced the Frequency to 68% and the voltage/frequency ratio to 66%. Since the motor is rated for 440V and 60 Hz, this would mean that the input is :
V=0.68*0.66*440V = 197 V
Freq = 60*0.68 = 41 Hz
I suppose the current would then be (Rated Current)*0.66 = 0.95 A * 0.66 = 0.62 A

Am I doing things right ? If yes, I suppose that if I configure my VFD to
1) Motor rated Voltage = 0.66*440 2) Motor rated speed = 60 Hz 3) Required output speed = 41 Hz
Then I will get 0.62 Amps of current !

 

[LiteYear]

Point 1->2 looks good.
Point 2->3 not so.

Reducing supply current is not a reliable way of reducing torque unfortunately. Not only is the relationship between supply current and torque difficult to determine, the relationship changes over the speed range so starting the motor can be an issue. Instead, you're better using a VFD that has torque limit control and setting that correctly (factoring in any torque multiplication/loss due to the gearbox). Even better, if I'm to interpret the reason for the torque limit as safety (I don't understand how "security" is relevant), then mechanical means could be considered. For example, a torque limiting coupling or a sacrificial coupling.

Back to the supply calculations: you can't apply two voltage reductions, otherwise the speed reduction will be duplicated. So instead of:

V=0.68*0.66*440V = 197 V

Should be:

V=0.68*440V = 300 V

But you should let your VFD handle this. So instead, make these configurations:
1) Motor rated voltage = 440V, 2) Motor rated speed = 60Hz, 3) Required output speed = 41Hz. 4) Torque limit = rated (motor, not output) torque * 25/59.

 

[jraef]

Generally, inexpensive "Sensorless Vector Control" drives can give you torque limit, but at the cost of speed control. In other works HOW they limit torque is to artificially reduce the commanded speed. So you can have controlled speed, OR limited torque, pick one. So for example when you set a speed command at 40RPM and then set a torque limit and reach it, the VFD may respond by driving the speed command down to 29RPM in spite of what you want.

More advanced VFDs have the capability to do what is called Flux Vector Control (or Direct Torque Control or Field Oriented Control) and can separate the the torque producing flux from the magnetizing flux, which means that you can limit torque WHILE controlling speed at any point. Some VFDs can only accomplish this with the addition of a shaft encoder feedback on the motor, then the more advanced ones (translate more expensive) are able to accomplish this without external sensor feedback (the sensors are internal to the VFD). If you go that route, it really isn't necessary for you to do the calculations, the VFD will do that for you. You just need to program it for what you want to accomplish.

"Will work for (the memory of) salami"

 

[waross]

But Jeff, if you use the more expensive drive and it limits the torque at the commanded speed, won't the motor then slow down on its own due to torque limit? If driving the load at 40 RPM requires more torque than the set limit, the motor should never reach 40 RPM.
Am I missing something? Does this have to do with the accuracy of the torque limit?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jraef]

Well, he didn't mention load, so yes, this would assume that the load does not REQUIRE the full torque output of the motor.

"Will work for (the memory of) salami"

 

[Skyba]

Thank you for your answers.

Instead, you're better using a VFD that has torque limit control and setting that correctly

Yes I agree, sorry for not saying that in the original post, but that is not an option : I want to control multiple motors through one VFD. Reason is cost and simplicity. I have gone through this discussion already (on this forum ). In short: The motors would share a load and I would like to reduce maximum torque in order to enhance the slip effect. This allows a better (horizontal) load distribution since lightly motors speed up while heavily loaded motors slow down. Several people have told me out of experience that this is achievable through reducing total voltage output.
By security I meant safety of course, yes ^^. The thing is that the chain hoists and the jacks that work with the motors only support 5 tons which is equivalent to a force of 25 N.m.

Anyway, the initial request is : How do I calculate the current that will flow through the motors in steady state ?

Not only is the relationship between supply current and torque difficult to determine

Is the rhe relation torque = voltage squared * contstant not a good approximation ? I have found this on different sources, like

http://www.toshont.com/ag/mtrldesign/AG09 (Speed Torque Curves).pdf

(look for "squared")

http://electromotores.org/PDF/InfoT...ifferences Estrella Delta vs Part Winding.pdf

(End of 1st paragraph)
And since current/voltage is linear for a given speed.. If this approximation is wrong, I could also try and ask for a test curve from manufactor, with torque on X-axis, and Amps and other parameters on Y-axis. So I could pinpoint my point 3 directly on the chart and find my point 2 and 1 from there, to configure the VFD.

relationship changes over the speed range so starting the motor can be an issue

This is indeed a problem. The speed will be limited to a narrow range, like 40, 50 and 60RPM. I could do some tests with these values if it is impossible to determine them theoretically. But how do people configure their overload protection anyway ? They probably enter motor rated current, but what if you do not work at rated speed ? All the overloads have a very narrow working range like 0.7->1 A, so I figured i HAVE to determine the current. Worst case I buy them after I have done some initial tests but this would only be my last option.

 

[waross]

Try programming the drive for a motor voltage of 68% rated voltage at rated frequency and let us know how it works.
For your first lift with the 25 hoists, lift something cheap and clear the area just in case.
By the way, FORGET OHM'S LAW when working with motors. Too many variables; Ohm's law does not take into account the load on the motor or that the magnetizing current is close to 90 degrees out of phase with the load current. Load current and magnetizing current can not be added arithmetically. The current to magnetize the air gap is another complication. The motor either produces the torque that the load demands, runs slow or stalls. Even though the motor may continue to produce more torque than rated and run a little slow, it may be overheating.
Most motors will produce enough torque and HP to damage or destroy themselves. The ratings are not the absolute capability of the motor, they are the safe limits, within which the motor may operate without self damage.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LiteYear]

Anyway, the initial request is : How do I calculate the current that will flow through the motors in steady state ?

One way is to balance power in to power output, and adjust for losses:

Power in electrical (Pi) = Vll * Irms * sqrt(3) = 300V * Irms * sqrt(3)
Power out mechanical (Po) = w * T = (2*pi* 40 / 60) * (25) = 105W

Estimate efficiency of motor and gearbox... 90%?
Then: 300V * Irms * sqrt(3) * 0.9 = 105 => Irms = 0.22A per phase

Is the rhe relation torque = voltage squared * contstant not a good approximation ? I have found this on different sources, like

http://www.toshont.com/ag/mtrldesign/AG09 (Speed...

(look for "squared")

http://electromotores.org/PDF/InfoTécnica/EAS...

(End of 1st paragraph)

I'm afraid you've been misled, and given you references that's quite understandable! In the first reference notice dot points 3 and 5 on the 2nd page - torque is actually "close to" to the current to the motor. Even then, it must be adjusted for magnetising current. The keyword in the second reference is "Available" - the torque that the motor could produce at any fixed speed is proportional to the voltage. But think of what happens in an unloaded motor - if I apply x volts the motor spins but torque is very low (there's nothing to drive). If I apply 2x volts, the motor spins faster but torque hasn't changed much. Conversely, if I apply x volts and then load the motor, torque goes up while volts stays the same. Torque will continue to rise while volts is constant, until I reach the stall point of the motor - ie. the "available" torque of the motor. If I were to do the same experiment starting with higher volts, then I could apply more torque before stalling the motor - the "available" torque has gone up.

And since current/voltage is linear for a given speed

Unfortunately, as waross says, Ohms law is not particularly useful in an induction motor. It is for a permanent magnet motor, but an induction motor has the extra complication of magnetising current.

If this approximation is wrong, I could also try and ask for a test curve from manufactor, with torque on X-axis, and Amps and other parameters on Y-axis. So I could pinpoint my point 3 directly on the chart and find my point 2 and 1 from there, to configure the VFD.

Yep, it's perfectly reasonable to expect a series of "power curves" from a manufacturer. Not however, that if the manufacturer only designed for fixed speed operation, they might only be able to provide you with one power curve! You might need to experimentally determine others.

But how do people configure their overload protection anyway ? They probably enter motor rated current, but what if you do not work at rated speed ?

The motor will handle rated current even at the lower speed. If you mean you want to set an overload on torque, then you might need to try one of the existing suggestions - eg. use a VFD with torque control.

 

[LionelHutz]

Your whole torque limiting theory appears to hinge on the idea that a motor rated for 59Nm will only produce a maximum of 59Nm. This simply isn't true. Typical induction motors are capable of 1.5 to 2.5 times their rated torque before they stall. This is called breakdown torque. You really need to find the speed-torque curve of the motor.

Note that you also don't want the motors to ever slip below the breakdown torque when you're using them with a VFD. You always want to operate the motor between the breakdown speed and synchronous speed.

Torque limiting was one of the reasons I recommended you use 1 VFD per motor in your other thread....

 

[jraef]

In addition to Lionel's post, another reason for one drive per motor, you cannot do vector control on multiple motors behind one drive. The drive accomplishes this by creating a mathematical model of the motor circuit to determine rotor position at any given point. With multiple motors, there is no way to do that.

Honestly, i think you have now expended more of your time thinking about how to avoid buying 5 (?) cheaper drives vs one large one than what the difference in cost is worth.

"Will work for (the memory of) salami"

 

[LionelHutz]

Yes, the time wasted trying to figure out how to save a few bucks on the VFDs is getting rather humorous.

Depending on where you are in the world, get a 1/2hp VFD that either takes 120VAC or 220VAC 1-phase input, matching your standard wall power. Mount the VFD on the jack. Make sure it's shielded against damage. Possibly mount it on rubber isolators with some metal protection around it. Get male panel mount receptacles, 120V or 220V as used in your area, for the VFD end. Make a power distribution panel with enough 120V or 220V receptacles for the maximum number of jacks. Use standard extension cords to connect power to each jack.

As for the controls, run another cable with the appropriate start/stop and speed reference wires. Should be about 4 conductors. Use some kind of rugged connectors for this cable. Build another panel with the controls, again with enough receptacles for the maximum number of jacks. Just as an idea, you could find a rugged panel mount Ethernet jack and use that on both ends. Then, use standard cat5 patch cables. As an alternative to hard-wired controls, use a VFD with Ethernet and then just use Ethernet communications to control the VFDs.

 

[iop995]

Motor torque is something that motor can react when is loaded or in other words, motor torque don't exist without load. In your application don't need (and I think is not good) to use VFD on each motor. You need a single VFD (simple V/f, not need for vectorial or other enhanced contolling technique) and motor with higest rpm possible to reduce final jacket speed difference between them. If load torque will be at 1/3 - 1/2 of nominal motor torque, hoists speed will be very close. Torque may be limited from VFD but why need to "control" it?

 

[Skyba]

Honestly, i think you have now expended more of your time thinking about how to avoid buying 5 (?) cheaper drives vs one large one than what the difference in cost is worth.

Haha, i totally agree. This is getting ridiculous... I think i´ll go for the sensorless Vector control. I´ll add the encoders later if the load crashes xD.

One quetion though : Is it necessary to connect the VFDs between each other for torque feedback (master/slave) ?