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AC Drives Speed control for incline belt conveyor in Coal-Mining industry


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#1 venuballa

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Posted 07 July 2006 - 12:12 PM

Can anyone send me solution for below query:



#2 mariomaggi

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Posted 07 July 2006 - 12:34 PM

You can measure the output current, probably the same inverter has an analog output "output current".

Inverting this signal, and using the inverter signal as "frequency reference" you can have a low frequency output when the load is maximum.
When the current decreases, the frequency increases. A PID could help to have a stable frequency.

Regards
Mario

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#3 jraef

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Posted 07 July 2006 - 07:59 PM

QUOTE(mariomaggi @ Jul 7 2006, 05:34 AM) View Post

You can measure the output current, probably the same inverter has an analog output "output current".

Inverting this signal, and using the inverter signal as "frequency reference" you can have a low frequency output when the load is maximum.
When the current decreases, the frequency increases. A PID could help to have a stable frequency.

Regards
Mario


Mario,
That is a really interesting concept. Have you ever done it? It seems too simple to me, like maybe there is something I'm not thinking of that would interfere with it working. For instance, if you slow down with increased speed, you would run a greater risk of stall wouldn't you? After all, you are not increasing torque with the lower speed, you are lowering output power to the conveyor. Maybe it would mean sizing the motor for the load at lowest speed then. Hmmm...


venuballa,
There are devices called "Belt Scales" used in the aggregate industry that weight the material on the belt and provide an analog output that can be fed into a PID controller or the PID function of the VFD. Here is one that I have used successfully over the years.

http://www.beltwaysc...com/default.htm
"He's not dead, he's just pinin' for the fjords!"

#4 mariomaggi

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Posted 07 July 2006 - 08:45 PM

jraef,
I've tested it many years ago, the system is not perfect, but it runs.

QUOTE
After all, you are not increasing torque with the lower speed, you are lowering output power to the conveyor.

No, simply I consider a normal "constant torque" characteristic. At low speed there is the nominal torque, if the load ask less torque the output current signal increases the speed.

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Mario

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#5 jOmega

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Posted 08 July 2006 - 03:49 PM

QUOTE
..... if you slow down with increased speed, you would run a greater risk of stall wouldn't you? After all, you are not increasing torque with the lower speed, you are lowering output power to the conveyor. Maybe it would mean sizing the motor for the load at lowest speed then....



Hi jraef,

The VFD/AFD is not allowed to operate below 1.5 - 2x slip speed of the motor. The MINIMUM SPEED parameter would need to be set to whatever that value is determined to be. Also, this is a Constant Torque application; not constant power.

Let's look at a typical application with this same requirement; i.e., reduce speed as a function of increasing load.

Consider a conveyor that is hauling a product and delivering it to some other apparatus in the process.
That next apparatus may have a limitation on how much through-put it can handle..... Therefore, would it not make sense to reduce the speed of the "in-feed" conveyor so as to stay within the working limitations of the next stage apparatus?

As an example, consider a conveyor in a Rock Quarry application. The huge chunks of rock which are dislodged from the earth are placed on the conveyor belt that transports the large pieces to the Crusher.

It takes time for the Crusher to break the large pieces of rock into much smaller pieces.

Lets say that the Crusher can handle 15 tons of material an hour.... (15t/h), and the conveyor belt, fully loaded and operated at top speed can deliver 60 tons of material an hour ...(60t/h).

Wouldn't take very long before there is a major problem at the input of the Crusher. So, the solution would be to slow down the conveyor as a function of how much material is placed on the belt ....i.e.. load.

In this way, the through-put can be controlled so as to stay within the limitations of the Crusher, and the process of converting large pieces of rock and boulders to smaller sized rocks or gravel, can be optimized.

(As an aside, some years ago, I did an application of this type where I used the audible noise of the Crusher to control the speed of the conveyor. As the Crusher loaded up with product, the sound level increased. The inverse of the sound amplitude was used to control the speed of the feed conveyor.)


I liked your Belt Scales suggestion.
A signal processor that would invert the sense of the analog output would be required.........
i.e., changes the analog output voltage as the inverse of the load on the belt.
(most VFDs or AFDs ... do not accept a bipolar analog signal - so the analog signal must conform to 0 to +10v or 0 to +5V or +4 to +20 mA.........simply changing the polarity of the analog signal by swapping the leads so that the + side of the signal is connected to input common of the VFD/AFD......won't work.)

I would expect that this would be a somewhat expensive solution; one best considered during the design phase of the conveyor. At this stage of "venuballa's" application, the conveyor undoubtedly exists (is installed) so that the material cost of the Belt Scales equipment will be less than the cost of modification and installation of this solution on the conveyor.

As to Mario's Inverse current control proposal, you are quite correct in your assessment thereof; i.e.,
"It seems too simple to me, like maybe there is something I'm not thinking of that would interfere with it working."
I'll go into that in a subsequent post.

Kind regards,





#6 jraef

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Posted 08 July 2006 - 11:49 PM

JOmega,
No disrespect, but I understand why people put VFDs on conveyors, I do that for a living.

Your point about it being a constant torque application is exactly the point I was concerned about. Read the OP's post again; he wants to decrease speed based on the belt weight, not the product flow. If the conveyor belt is more loaded, friction and inertia increases and you will need more torque to keep it moving, not less, or even the same. If the VFD is allowing the motor to give out constant torque, the HP is decreasing at the motor shaft. If however the reason you were lowering the speed is because the belt is more loaded (therefore more motor load), then it seems to me that you are defeating your purpose. Even though torque is staying the same, you are decreasing shaft power at the motor right when you ostensibly need more power to the shaft. That's why I'm a little concerned with Mario's suggestion; using the VFD to detect that it is being called upon to do more work, and responding by doing less. In fact, even my suggestion of a belt weigh system carries that problem with it as well now that I think it through.

Here is the problem I had in my mind with this scenario, using your example.

60t/h = 1t/min or 2000lb/min, using a 125HP motor, assuming 1750RPM, so Tq = 125*5250/1750 = 375lb-ft.
If the belt is moving 100ft/min, then 100ft of belt must hold 2000lbs of material or 20llbs/ft.
If that material were allowed to build up to 40lbs/ft and that causes the VFD to slow the belt down to 50ft/min, then you have 40 * 50 = 2000lbs or 1t/min so the tonnage stays the same.
The motor however has been reduced to 875RPM with torque remaining constant.
So since HP = 375lb-ft * 875/5250 = 62.5HP to move the same amount of material, I see that as the motor being overloaded!

I agree with you that if the conveyor motor were designed for a lot more capacity that you are currently using it for, as in your example of a 60t/h conveyor only delivering 15t/r, then I can see it working. Even if the conveyor was doing only 50% of the work it could do, then the scenario should work. For example (from the above), if it was using the 62.5HP out of the 125HP to move 30t/h but it was designed for 60t/h, then if the belt load increases, the VFD slows, the tonnage goes up to 60t/h, but the motor is still only 62.5HP so it works. As long as you never exceed the maximum system design limits.

What this really means is that it depends totally on the system design.

In the past his type of application worked great with varidrives (mechanical variable speed drives) because they used pulley or gear ratios to decrease speed, but in that case torque was INCREASED as speed decreased and HP remained the same. Often times people who used varidrives in the past assume that VFDs can directly replace them, which they can't in every circumstance. That was what was behind my apprehension.
"He's not dead, he's just pinin' for the fjords!"

#7 venuballa

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Posted 12 July 2006 - 04:24 AM

QUOTE(mariomaggi @ Jul 7 2006, 06:04 PM) View Post

You can measure the output current, probably the same inverter has an analog output "output current".

Inverting this signal, and using the inverter signal as "frequency reference" you can have a low frequency output when the load is maximum.
When the current decreases, the frequency increases. A PID could help to have a stable frequency.

Regards
Mario


Mario,
Should we take a linear 4 to 20 mA o/p from a Current transducer which senses one of the three phases using CT of ratio 250:5 Amps ? or inverted o/p from Current transducer ?
If we take linear then a PID function can be used else direct inverted ref. can be fed as a speed ref. i/p to drive control terminals. Is my understanding correct ?
Can you elaborate about sensing of drive o/p current in all 3 phases ? is it ok if we sense only one of the phases.
Venugopal Balla,
India.

#8 mariomaggi

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Posted 12 July 2006 - 03:58 PM

Dear venuballa,
pay attention, this application is critical for many reasons.

1st - I understood that you cannot stop the belt to make tests, the solution must we winning at 1st time

2nd - you have to manage a non-sinus current signal coming from an inverter output, a standard CT could introduce a measuring error

3rd - you need an inverted output, probably it is better to use analogue voltage signals instead of current signal, it will be easier to reverse and compensate them.

4th - I agree with jraef, but here I assume that the VSD+motor+gearbox is designed with enough torque to drive the belt at low speed at maximum load. The same group must be in position to restart the belt in case of undesired stops. If this condition is not verified, you cannot make this modification.

Regards
Mario

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#9 venuballa

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Posted 13 July 2006 - 03:13 AM

QUOTE(mariomaggi @ Jul 12 2006, 09:28 PM) View Post

Dear venuballa,
pay attention, this application is critical for many reasons.

1st - I understood that you cannot stop the belt to make tests, the solution must we winning at 1st time

2nd - you have to manage a non-sinus current signal coming from an inverter output, a standard CT could introduce a measuring error

3rd - you need an inverted output, probably it is better to use analogue voltage signals instead of current signal, it will be easier to reverse and compensate them.

4th - I agree with jraef, but here I assume that the VSD+motor+gearbox is designed with enough torque to drive the belt at low speed at maximum load. The same group must be in position to restart the belt in case of undesired stops. If this condition is not verified, you cannot make this modification.

Regards
Mario



Mario,
Thanks for the precautionary vision.
There is a Thruster Brake Motor installed in the system and also there is a BYPASS arrangement for INVERTER BYPASS ( DOL STARTER, in case VFD is taken for maintenance). There is an analog output available in the VFD but it is used for displaying the Speed ( RPM) indicator on the Remote Control Station(box).
At the input of VFD there is an 3-phase, 3wire MCCB LM250B coupled with Earth Fault Module - GE make & customer also needs a modification to detect a current sensitivity for earth leakage relay ( to be installed) of 0.5 to 2 A with test, reset & aux. NO/NC contact.
If you have any more suggestions ( practical), please send it across.
Regards,
Venugopal

#10 mariomaggi

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Posted 13 July 2006 - 05:31 AM

Dear Venugopal,
if your inverter has an analogue output "output current", this signal will be probably more reliable than a signal coming from external CTs.

Has your inverter an RFI (Radio Frequency Interference) filter at the line input? In this case a 0.5 A earth leakage relay could trip due to additional leakages due to capacitors-to-earth in the filter section.
Please check that your new relay will be OK for non-sinusoidal currents.

Regards
Mario

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#11 kens

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Posted 14 July 2006 - 12:06 AM

QUOTE(mariomaggi @ Jul 13 2006, 05:31 PM) View Post

Dear Venugopal,
if your inverter has an analogue output "output current", this signal will be probably more reliable than a signal coming from external CTs.

Has your inverter an RFI (Radio Frequency Interference) filter at the line input? In this case a 0.5 A earth leakage relay could trip due to additional leakages due to capacitors-to-earth in the filter section.
Please check that your new relay will be OK for non-sinusoidal currents.

Regards
Mario


Mario makes a very important point regarding the RFI filters and the ELR. You will need to ensure that the earthing arrangements are all correct and that the relay used is suitable otherwise the conveyer will become very unreliable.

Ken
An expert is one who knows more and more about less and less until he knows absolutely everything about nothing

#12 venuballa

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Posted 18 July 2006 - 11:06 AM

Thanks for your repsonse




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