Increasing speed of 50hp motor

[rockman7892]

I have a 50hp (480V) motor used on a radial stacker conveyor to convey rock to the top of the conveyor. The motor shaft is coupled with belts (sheaves) to the head pully of the conveyor. People where want to increase the speed of the conveyor by mechanically changing the sheaves. Right now I am measuing 51A on the motor, which has a FLA of 62A when the belt is fully loaded.

My question is if we mechanically change the speed of the conveyor what effect this will have on the motor? I obviously expect the current to increase but am wondering what the factor causing it to increase will be and if there is a way to determine how much. Will the current increase because by mechanically changing the speed of the conveyor more torque will be required by the motor? Is there a way of determining how much of a current change can be expected from increasing the speed and therefore the amount of material on the belt?

 

[oldfieldguy]

You're asking for more out of it. You're going to have to put more into it.

I'm not sure of the calculations involved with belt conveyors though.

old field guy

 

[itsmoked]

Two things.
The more work you do the more HP that is demanded. Hence you will see the amps climb.

The second is that conveyors can be nasty to start as they can be loaded and present large inertial loads. So if the motor is not fully loaded while running it could be at its limits during starting. Hence increasing it's driving speed could prevent it from starting successfully. This may not be a problem with your system, can't say.

Now some clarification. Your statement could mean several things.

measuring 51A on the motor, which has a FLA of 62A when the belt is fully loaded

Are you saying the motor FLA nameplate is 62A or that you have measured the motor current at 62A when the belt was more loaded? This is a big difference. What does the name plate state?



Keith Cress
kcress -

http://www.flaminsystems.com

 

[rockman7892]

itsmoked

To answer your question, after looking at the actual motor nameplate data I found that the FLA for the motor is 57.7A with a service factor of 1.25%. I had our electricians go and measure the current on this motor while the conveyor was fully loaded and they measured a current of 51A. This tells me that there is a little room to get more horsepower out of this motor but not much. I understand what you are saying as well, that speeding the motor up mechanically may cause greater starting current.

Since we are mechanically changing the speed of the conveyor I'm assuing that we are not changing the rpm of the motor. I am looking at the following motor equation and trying to understand how mechanically speeding up the conveyor will increase the current:

hp= (Torque*rpm)/5250.

I'm assuming that the rpm of the motor is staying the same, and the 5250 constant is staying the same so the only thing changing by mechanically speeding up the motor is the torque thus equating to an increase in hp and therefore an increase in current? Am I understanding this relationship correctly? Does the same principle apply with the starting current due to increased starting torque. Is there anyway of calculating this torque?

 

[petronila]

Hello Rockman,

The Motor´s RPM Will remains constant and you only will change the pulley ratio, then the conveyor will run faster.So the change is on the conveyor not in the motor, if in this moments you have the motor loaded up to 90 % then you have a 10% of more current , Have the motor services Factor? this can help you too, due to the application is not a Fun or Pump, then you don´t have to apply the affinity laws. So I think you don´t gonne have problems to do that conveyors speed increase,and the motor will runs properly, the only I will suggest is make a good pulley balancing and excellent alignment and belt tension.

Regards

Carlos

Regards

 

[itsmoked]

Got a motor model number?

It is too hard for me to guess this one. Others more experienced might be willing to say you can increase your drive ratio by 10% or ? But the currents you measure are not totally attributed to the work your motor is actually doing. So I can't just calculate the power difference that we can then use as a horsepower difference.

As that non-fully loaded motor is more fully loaded a larger amount of the increasing current is actually providing more horsepower.

Here is my guesstimate, maybe someone will wade in and yeah or nae it.

Ignoring power factor:
42.4kW / 47.9kW = 88.5% of full load.
My numbers don't actually reflect the full truth because I am ignoring the powerfactor change and even it's existence.

However the way it actually changes means my number above is lower than reality in the conservative direction.

As petronila; correctly states use your equation above on the load side recognizing that the horsepower can increase at least 11.5%. I don't think this is even dipping into your service factor yet as that is on top of your stated FLA unless you included it(not clear to me).

That service factor also means that the motor can probably start the conveyor even if you raise your ratio that 11.5%.






Keith Cress
kcress -

http://www.flaminsystems.com

 

[rockman7892]

Ok so it sounds like my equation holds and the only thing that is changing is the mechanical torque leading to a larger horsepower. It sounds like I will be able to dip into the 125% service factor and still have no problem starting the motor.

I'm assuming the reason you disregarded the power factor in your calculation was due to the fact that as the load increases the power factor increases thus more work is done per unit increase of current. Therefore an 11% current increase may not corrolate directly to an 11% hp increase be may result in a larger hp increase due to the fact the the power factor will increase with loading and there will be more real power and less reactive power.

 

[itsmoked]

Yes.

and exactly!



Keith Cress
kcress -

http://www.flaminsystems.com

 

[armandh]

yes but.... besides the no free lunch calculation, is there increased friction?

 

[Ross0684]

Rockman
A "rough" calculation I use for HP on conveyors(horizontal) is:
(Total load(Lbs) x speed(FPM) x Friction)/33000=HP.

So therefore HP is directly proportional to speed.

The question of increase in friction as speed increases is valid, but not knowing your present speed or type of conveyor,it is hard to make a judgment, but assuming that the conveyor type is belt running over rollers, the friction factor is normally low (0.1), and any allowance for inclination will already have be taken into account in the original conveyor design.

Also,my experience on smaller type conveyors(up to say 10Hp)
indicates that the current draw in the unloaded state is anything up to 85% of the current draw when fully loaded.
(the application of the load only adds a small precentage to the total current draw)

Therefore, I believe we can safely say that HP will increase directly proportional to speed increase.

As previous respondents have already said, just be aware of the inrush start up current when starting fully loaded,as acceleration times etc will increase.
A good check would be a measure of existing inrush current on full load start-pass this past the electric motor gurus for their comments!
Good luck
Ross

 

[jraef]

There is a little more to this than meets the eye. A radial stacker is an inclined conveyor that is taking material from a source, usually another conveyor. If you simply increase the speed and do not increase the infeed, all you will do is fling the material off the end of the stacker a little farther. In other words, no increase in material on the belt will mean no increase in load. If they want to do this because they ARE increasing the infeed onto the belt and they don't want to let it pile up too high on the belt (legitimate reason by the way), then you will need more HP.

Use this simple calculator to determine your HP requirement.

http://www.sparksbelting.com/cd/calculator_rollerbed.html

 

[desertfox]

Hi rockman

Is it possible you can measure the motor torque during running and start up for the existing set up, then maybe you could look at the torque curves you have for the motor
and go from there.

regards

desertfox

 

[rockman7892]

Thanks for all of the great responses guys! Jraef that link to the conveyor calculator was great!

Here is some more information about the application for those that have asked:

- Production wants to speed up the conveyor because the amount of material that the upstream conveyor is dumping onto it is too much and therefore spilling over the sides. They hope that by increasing the speed of this conveyor the material will be moved faster and not spill over the sides.

- The conveyor I'm referring to is an incline radial stacker conveyor which has its head section (discharge section) 30ft high with reference to the ground. This conveyor is angled 16 degrees with reference to ground. I believe the conveyor is operating at 1200tph at 300ft/min but I have to confirm this.

- The motor is a 50 hp Weg motor with the following specs.

- Type: TEFC IP55
- FLA: 57.7A
- No Load Current: 18.5A
- Locked Rotor Current: 352A
- Speed 1770 rpm (4 pole)
- Locked Rotor and Breakdown Torque: 230%
- Full Load Torque: 146 ft/lbs
- service factor 1.25%

Unfortunately I do not have any way of measuring the torque for this motor, but I will measure the starting current when starting this belt.

Thanks for all the help!

- The motor being used on this conveyor

 

[oldfieldguy]

I know this may sound elementary, but if possible, sequence your system shutdowns to leave the conveyors empty for the next start. I've had to deal with systems that did this and systems that didn't. Those that didn't required larger motors to handle starting with loaded belts.

old field guy

 

[jraef]

Power failures are what usually necessitate sizing for loaded restart, otherwise you have to on put on a crew of flunkies with shovels to dig it out. The lost production time value the first time it happens usually exceeds the cost of sizing the motor for loaded restart.

rockman7892,
Use that calculator to prove my point. Plug in numbers that you have, guess on the others. I plugged in 50ft for the conveyor length, 36 in. for the width. It calculates that you need 58HP, close enough for this purpose. Now just change the speed, but not the TPH. Although it appears at first glance that the TPH would go up with speed (because the belt moves faster, so more material over time), it really isn't because the infeed, which is ALSO controlling the TPH, will remain the same. So change the speed all you want, you will see that the HP remains the same.

 

[waross]

I agree with jraef. A higher speed will mean more difficult starting, but the load on the motor will depend more on the amount of material delivered than on the speed.

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

 

[itsmoked]

Yes but if the point is to move the material faster because it falls off otherwise does this not infer carrying more or the same, just faster, which infers more work?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[rockman7892]

I had our electricians measure the starting current of this motor yesterday with the belt unloaded and they told me that the measured a starting current of 360A. This current is above the LRC value of 352A which I find confusing but have seen before? Can a starting current be higher than a LRC?

Jaref

I agree with what you are talking about referring to the online calculator and hp not changing as the speed increases however this doesn't agree with the above equation of

(Total load(Lbs) x speed(FPM) x Friction)/33000=HP

That someone else stated. According to this equation hp will increase with an increase in speed. Which calculation is correct?

 

[jraef]

First off, 360A starting current vs 352A LRC on the nameplate is not enough of a difference to write home about. Nameplate values are based on lab test conditions, you are in the real world. 2% is well within a reasonable margin of error. Don't sweat that.

I can't speak for where their formula came from, I am not an ME (but sometimes I play one on TV). I have used that calculator repeatedly and it has never let me down. In fact it is a little on the high side from my experience, which is not necessarily a bad thing. Here is another one.

http://www.superior-ind.com/app_ccalc/calc_hp_total.asp

If you notice, this one doesn't even have an input for belt speed at all.

As I see it, the speed would be a factor if you were pulling material from an infinite source; in other words, you will move more tonnage by moving faster. An example might be a trap conveyor at the bottom of a stockpile. In that case, faster speed means more tonnage per hour. But in this case, you are getting material from a finite source; the other feeder conveyor. So even if you moved the belt 3x the speed, the total tonnage per hour cannot change because you had no more material to move if the feeder didn't give it to you. It's just not going to pile up as high on the belt.

Maybe an easier way to think of it is by weight. If the pile on the belt is lower, the total weight on the belt at any given moment is lower. That means the required torque to lift it the weight is going to be lower. HP = Tq x RPM / 5250, so faster speed, but less Tq, nets out the same HP.

Keith, that hopefully answers your question too.

 

[LionelHutz]

I'd say it depends on how much is spilling. But, it's likely just a small amount which will cause little change in the total load.

The formula has Lbs x fpm. With a fixed feed rate the load on the conveyor (lbs) goes down as fpm goes up.