Gear changes' effect on motor current

[OkiePC]

I have an auger that has had problems with motor failures.

Fully loaded it draws just above the FLA of the 7.5 HP motor, let's call it an even 10amps.

It runs at 95Hz.

To solve the problem we intend to upgrade to a 10HP motor, but would like to address the gearing as well, since the reducer will need to be replaced anyway, we'd like to get the motor command down closer to 60Hz without changing the speed of the auger itself.

So, if the auger is turning 50 rpm and requires 10 amps, if I change the gearing and the motor speed so that the auger is still turning at 50 rpm when all is said and done, would the current requirement stay about the same?

 

[dahnuguy]

current in work out.

A certain amount of power is required to do the work you want to do.

Increasing the motor size will make the work less stressful for the motor.
Increasing the gear box ratio could reduce the current draw on the motor, but would slow down the auger.

If you want this same amount of work done at the same rate, it will always require about the same amount of power.

A motor will only draw the amount of current needed to do the work.

Even a 50HP motor would do this work by converting about 10 amps into the needed torque.

If you want to use less current, slow down. If you can't slow down, pick the components that best work at that speed and power draw.

For simplification, I have ignored the possibility of using less power due to changes in efficiency.

 

[jrwb4gbm]

Simple answer is yes, the current should stay about the same.
Longer answer is as follows. I was told a long time ago by an old electric motor rebuilder that horsepower demand for a load is the same whether the motor itself was oversized or undersized, overloaded or underloaded. What he was saying for a given load was that if the input voltage was not changed, the amps would be the same for motors either too large or too small as long as the load remained the same.

 

[DickDV]

As you already know, OkiePC, I am an advocate for the use of overspeed in VFD/Induction motor systems. So I would hate to see you drop the top motor speed down to 60hz.

It's better, in my opinion, to size motors, gearboxes, and VFD's on torque and speed (amps and hz) rather than on hp. Hp is just a calulated number while the others are real.

If the existing motor is 4 pole then it develops about 7.5x3=22.5ft-lbs of torque at base speed. At 95hz, the torque would be down to 22.5x60/95=14.2ft-lbs. Let's just say 14.8ft-lbs to be safe.

If you were to just change to a 10hp 4 pole motor, the available torque would be 10x3=30ft-lbs at base speed and 30x60/95=18.9ft-lbs. That would be adequate to do the work and you would have a system that cools the motor very well due to the higher motor speed and a system with lots of starting and low-speed torque if that is needed. On the other hand, 95hz is reaching a bit and dropping to 80-85hz would be better. I wouldn't drop to 60hz even if there is enough torque to do the job.

If the power train is a gearbox, ratio changes don't come cheaply. If its a belt or chain drive, the change would be much easier and cheaper.

How about letting us know what you actually decided to do and how it worked out.

 

[OkiePC]

Thanks for confirming what I thought "seemed right". I think that dropping the motor required speed to 60Hz would probably result in increasing it back up to around 80Hz at some point as our overall throughput of this system increases. I am uncomfortable going above 90Hz due to past experiences with surge loads causing sudden stalls and product pile-ups at those speeds (less available torque).

If I find out that this reducer can handle the increase in HP, I will probably leave the ratio and command speed alone and just spec a new motor. I will do more research this weekend with the equipment down so I can get a good look at all the data plates and verify the reducer ratio by hand if necessary.

The Sew Eurodrive reducer does have a shaft output, chain and sprockets to the load, but I can't look at any of that stuff today to see if we have any wiggle room on sprocket sizes.

 

[The Plc Kid]

If the existing motor is 4 pole then it develops about 7.5x3=22.5ft-lbs of torque at base speed. At 95hz, the torque would be down to 22.5x60/95=14.2ft-lbs. Let's just say 14.8ft-lbs to be safe.

Dick can you explain this calculation? I understand the first part but not calculating drop off torque above 60hz.

I wouldn't drop to 60hz even if there is enough torque to do the job.

Why not? Because of better Cooling at the higher speed?

 

[L D[AR2P#0.0]]

Dick can you explain this calculation? I understand the first part but not calculating drop off torque above 60hz.

Above base speed you get constant power, so if you double the speed for example, you get half the torque.

 

[milldrone]

I would hate to see you drop the top motor speed down to 60hz.

Why not? Because of better Cooling at the higher speed?

Actually it's the low speed cooling. This next quote is from a previous DickDV post (February 11th, 2003).

Second, at all operating points, the motor is turning faster than normal for the same load speed due to the increased mechanical reduction so the motor fan cooling is better. This is especially handy if you have a TEFC or ODP motor and need to operate at the low speed end of the motor's thermal capacity.

 

[DickDV]

PLC Kid, as already mentioned, as you take a motor over its base or nameplate speed, the torque falls off as a function of the inverse of the overspeed.

So, if you make 30ft-lbs of torque at base speed and you are going overspeed to 95hz, then the overspeed ratio inverted is 60/95 which is .6315. That ratio times the base speed torque of 30 gives you the torque available at the overspeed point.

Also, as mentioned above, most NEMA motors can hold to this rule from 60 to 90hz. Above 90hz, the torque starts to fall off faster than the speed increases which results in a decrease in hp. The exception would be inverter-duty motors that are nameplated for 2/1 overspeed operation. These can be expected to follow the inverse speed rule and hold their hp constant up to 120hz.

 

[Kev77]

You don't get something for nothing.

Everything comes back to "work". If you have a conveyor that is running overloaded you can do two things, more horsepower or higher ratio (slow it down). You stated that you went up in horsepower and this corrected the issue. Now if you change the ratio to slow down the motor, the change in ratio will cause the amps to rise. I have attached a spreadsheet that might help you. The first tab you can enter amps and see the change in in/lbs "work". It is a little crude but it works.

A good rule of thumb is a 4 pole motor devolops 3 ft/lbs per horsepower and a 2 pole motor devolops 1.5 ft/lbs per horsepower.

 

[DickDV]

Actually, kev77, I don't think your spreadsheet works with a motor in the overspeed range. I base this on your statement that reducing the power train ratio will result in an increase in amps. That is not true.

When a motor operates in the overspeed range, it is capable of constant hp over the speed range. As a result, the motor also needs about the same input kw over the range. Since voltage is constant, the amps would also be about constant.

In overspeed, the motor simply trades off torque for speed. So, as the ratio goes down, and the load speed stays the same, the motor trades off the speed reduction for a torque increase. The resulting input amps to the motor is about constant at either speed point.

Below motor base speed, the behavior is entirely different due to the output hp varying with speed. I think your spreadsheet would be ok in that range.

 

[Kev77]

Maybe I was not clear.

Here is a formula that was provided to my by an engineer at Baldor. I was trying to find out what work was being done on some press rolls that I work on.

[FONT=Calibri,sans-serif]HP = Tq(Lb-Ft) X RPM / 5252[/FONT]

[FONT=Calibri,sans-serif]5252 X HP / RPM = Tq (Lb-Ft); HP = 15, RPM = (102 Hz – slip Hz)/60 Hz X Sync speed = (102 Hz – 1.2 Hz)/60 Hz X 1800 = 3024[/FONT]

[FONT=Calibri,sans-serif]5252 X 15 / 3024 = Tq (Lb-Ft)[/FONT]

By raising the horsepower that will take care of the high amps issue. If you just slow down the motor the amps will drop. When you change the ratio it will bring them back up. You are correct that the work is the same but if the changes you make with the horsepower fix the high amp issue and now you are at full load amps, that is the fastest you can go without causing another over amp issue.

As far as the spreadsheet working on overspeed conditions if you enter say 15 hp motor at 1750 rpm, that is 45 ft/lbs of work and if you change that to 3000 rpm, the torque will drop to 26 ft/lbs. That is the trade off.

The point I was trying to make is that the work being done is split between the motor and ratio. Once you know the work that needs to be done the spreadsheet will help you make sure that you don't have an over amp condtion.

 

[DickDV]

Let's take a conveyor with the head pulley running at 175rpm and needing 300ft-lbs of torque to function. Using the formula you provide, hp will be 300 x 175rpm/5252 = 10hp.

Assuming no losses in the gearbox, we could take a 10hp four pole motor and connect it to the head pulley thru a 10/1 ratio gearbox. The motor will turn 1750rpm which is full load base speed and will need to produce 30ft-lbs of torque. That's 10hp or full load which means that with a nameplate FLA of 17amps at 460V we would be at 17amps.

Or, we could take the same motor and use a 15/1 ratio gearbox and run the motor to 90hz. At that frequency, the motor shaft speed will be 2700rpm and the full load torque on the motor will be down to 20ft-lbs. That also calculates to 10hp and the resulting amps into the motor will be 17amps.

So, what if anything have we accomplished by using the higher ratio?
1. The motor speed is 50% higher at all conveyor speeds so the motor cooling can be expected to be better.
2. At all speeds the motor has to produce 20ft-lbs of torque but, except at 90hz, it can produce more than that (actually 30ft-lbs from 60hz down to 0hz). This further reduces motor heating due to reduced torque demand.
3. Because the motor torque demand is less than full torque at all frequencies under 90hz, we can expect that the speed error (actually the slip speed which is proportional to torque) will be less than full slip making the system more accurate than the 60hz solution.
Those are all pluses from a performance standpoint. I know of only one negative consideration and that is a slight drop in efficiency due to the higher fan speed. Of course, that's what powers the improved cooling.

I have no problem stating that the above is 100% trustworthy. I've done it more times than I can count on a huge range of hp's and motor speeds. If this were not true, why would many inverter duty motor manufacturers go to the trouble of designing their motors with a 60hz base speed and a guaranteed constant hp range up to 120hz?

 

[OkiePC]

Don't let my update to the original problem interrupt an excellent discussion.

Today, after looking at the painfully obvious root cause of the motor mismatch, we are going to try a supposedly simpler mechanical avoidance of the correct fix.

This machine is an inclined twin screw auger (2 augers, one mostly shared trough) delivering ground (chopped) meat toward a large (10,000 lb ) blender at a pretty quick rate (~30-40 pounds per second until approaching overload via devicenet PF700 in sensorless vector mode) .

One of the screws is backwards. LEt me emphasize that...It has two left feet...

They both drive the same direction unlike any other twin screw auger I have seen. It is long, has a large diameter core and pushing meat uphill at about 30 degree incline. All the meat ends up on the south auger. The two screws are mounted in a more or less common trough with independent drve trains.

All of this is from memory, and I didn't measure anything or open the mechanical drawings yet, but the directors' solution is to (rather than buy a new motor and adapter or a counter closkwise screw) weld a divider between the augers to make them independent, hope we can meter the meat in more evenly and keep t that way.

It will probably work. I would just gritch to the manufacturer of the conveyor and insist that they fix their mechanical error or make it cheap...buy the right screw and be done with it, but I have not attained complete control of these simple, obvious decisions.

I got the data from the gearbox and researched it. It's a Sew Euridrive gearmotor, but I havent ID'd the ratio yet. I could fix the problem with a new gearmotor, possibly just a new motor, better yet, an adapter plate to a NEMA motor and a smooth stainless Sterling motor, VFD rated 10HP If the south screw fills up the meat will gradully fill up the north side toward the top like it does now only we could take out the "fault avoidance production restriction logic"...

But, that would make us right and them wrong...I shake my head...
What could be simpler to pull off in a few hours...they think they can weld in a piece of metal on a weekend and put the screws back in, cheaper than bolting on a new drivetrain? They never seem to guesstimate labor even close on these jobs, so I can expect we'll do that in-house too.

Tonight I am watching the free UFC and listening to the Thunder as they swept the Mavs, but in the a.m. after I upgrade a 75HP soft starter that blew up Friday night, I will try to post the details...I am, getting quotes for the adapter, ten horse Sterling option Monday too. Surely they will listen to common $ense this time...

If you want to use less current, slow down. If you can't slow down, pick the components that best work at that speed and power draw.

For simplification, I have ignored the possibility of using less power due to changes in efficiency.

Can't slow down...being the "middle man" between two other conveyors, slowing down will increase the product load at any given time. I think they sped up to avoid faults in the first place (I think this screw was installed in '05), as well as to decrease transfer times. I am going to discuss with my counterpart, slowing down to 90, then 85 , then to 80Hz while trending current and power for a few hours. I am sure he has already tried that though...

 

[Kev77]

I don't dispute any of the math and I think we are saying the same thing. I think I have a bad way of saying what I am thinking.

If the motor runs at 30 ft/lbs @ 1750 or 20 ft/ lbs @ 2700, that is still 100% of the motor capicity. The only difference is that the gearbox is making up for the lost torque because of the ratio.

Lets say that you go with the same 10HP motor, 1750 rpm motor and 10:1 gearbox and you decide to take it to 2000 rpm, you will over work the drive because you go from a system that required 10 hp to one that requires 11.5 hp.

I was simply just stating that you don't get something for nothing and the calculations are the same no matter if it is at 60hz or above.

@Okiepc, I can't wait for the OKC and the Lakers, sending Kobe home early will be sweet! But then the Heat will take it all...