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Replacement Of Old Vs Motors With New Generation


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#1 AB2005

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Posted 11 September 2011 - 06:42 AM

Dear Friends;

I have started working on an "Energy Saving" project of our department. First I have selected two machines one is ISOWA brand Japanese Flexo Graphic Printing Machine about 31 years old and other is 20 years old. Both machines have YASKAWA brand variable speed eddy current motors (Name plate image attached). You may know the basic principle of this motor that inside this, motor rooter couples with output shaft via electro magnetic coupling. Due to this reason, there is wastage of energy occurs in this motor. So I want to install simple motors with inverters. Each machine have one motor of 37KW and two motors of 15KW. If some body has already done such job, please share his experience with that how much energy he prevented to waste after installation of simple motor and inverter or any other tip for this job.

Motor and Nameplate Image
"Don't assume any thing, always check/ask and clear yourself".

#2 marke

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Posted 12 September 2011 - 07:48 AM

Hello AB2005

The eddie current drive is a torque transfer system whereby the torque out is equal to the torque in.
The speed of the load is a function of the torque in and the torque speed curve of the load.

If you replace the eddie current coupling with a VFD, then you will definnitely save energy, but the amount of energy saved depends on the amount of energy wasted.

Slip loss is the slip speed mulitplied by the torque. If you have full rated torque and the load is operating at half load, then the power in the load is equal to half rated power and the slip loss is equal to half rated power.

Best regards,
Mark.

#3 Tractor

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Posted 10 February 2012 - 03:31 PM

This is worth the effort, I have done this with an American mag coupling on a printing press with 6 print modules. The motor & coupling was driving the modules through a layshaft. These things are left running for a lot of the time, possibly 24 hours / day,
and generate sufficient heat to keep the press shop warm.

I replaced the coupling and motor with a standard asynchronous motor, driven with an inverter drive with regeneration, this provided rapid stop without requiring a brake resistor this was useful for press set up; this drive also has a 'Safe Low Speed input' for machine set up, suitable for SIL2 Cat 3, here a digital input can set the safe low speed conforming to a safety category.

The payback was about 3 Months, because the motor now only runs when required, and never gets hot.

Before you do this it is important to gather energy consumption data first ,over say a month at least; because it is this detail which provides the evidence of the good work you are doing. If you start the modification before gathering this data you have no proof.

Tractor..

#4 AB2005

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Posted 13 January 2013 - 04:52 AM

Hello Marke;

 

That job had been postponed due to a project and now I again started working on it. Now I am working to get data before installation of VFD. I installed an energy analyzer on a 37KW VS motor and measured the energy which was as following;

 

Only motor rotor spinning.

380V, 23A, 2.3KW, cos0.15, 16.6kva.

 

Motor shaft turning (machine running)

 380V, 31A, 13.8KW, cos0.68, 20.5kva.

 

I astonished to see the very low power factor when only motor rotor was spinning. Is that power factor real? It happens many times that only motor rotor is sinning but machine is stopped. S I want to measure the energy during that period but as mentioned in given figures, it is just 2.3KW.


"Don't assume any thing, always check/ask and clear yourself".

#5 marke

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Posted 13 January 2013 - 05:14 AM

Hello AB2005

 

The power factor on an unloaded motor is very low, it can go as low as 0.1.

Under open shaft conditions, a large motor will draw around 20% of it's rated current and the only power consumed is the iron loss, the copper loss and the fan losses. These are very small. The current flowing is primarily magnetising current which is totally inductive.

 

Your major potential energy saving, is not in the motor, it is the slip losses in the coupling. The torque out of the motor shaft is equal to the torque into the driven load.

Power is torque time speed, so if the load is running at half speed, then the power in the load is equal to the slip losses. Elimminate the slip losses and the savings could be considerable depending on the speed of  the load.

 

Best regards,

Mark.



#6 AB2005

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Posted 16 January 2013 - 07:01 AM

Understand Marke.

It would be very hard to know/measure the wastage of energy inside of motor/coupling. Only way is to measure the energy before and after installation of simple motor and vfd. So I think we should measure the total energy consumption of motor during machine running.


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#7 marke

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Posted 16 January 2013 - 05:56 PM

At any one time, you can measure the total power consumed, measure the speed of the load and the speed of the motor, and at that time, the energy wasted by the coupling, is the total energy times the difference in speed divided by the motor speed.



#8 AB2005

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Posted 27 January 2013 - 04:28 AM

Hello Marke;

 

I did arrangement and got data as you suggested;

  • Power measured by energy meter = 10.3kw.
  • 380V, 27.6A, Cos0.55.
  • Load speed = 558rpm.
  • Motor speed  = It could not measure as there was no space to measure. Neverthless, it is a 4 pole 37KW motor (Name plate image attached) and we assume 1450rpm.
  • Coupling 12.8V, 1A.

 IMG00912-20110911-0807_zps130b9e9e.jpg


"Don't assume any thing, always check/ask and clear yourself".

#9 marke

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Posted 27 January 2013 - 09:26 AM

Hello AB2005

 

Based on the information you ahve given,

Total Power in = 10.3KW at 1450RPM

 

> Load power = 10.3 x 558/1450 = 3.96KW

 

> Power loss in coupling = 10.3 - 3.96 = 6.34KW

 

There is considerable energy wasted in this application at this speed.

 

NB as the load speed is increased, the wasted energy as a percentage, reduces.

 

Best regards,

Mark.



#10 AB2005

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Posted 02 February 2013 - 07:15 AM

Dear Marke;

 

I am sorry but I disagree with your "Load Power" calculation method that is only 3.96KW??. Machine is too heavy and I am not certain that only 3.98KW require to accelerate it at 558rpm of motor. So the power loss of 6.34KW is too big.


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#11 marke

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Posted 02 February 2013 - 08:13 AM

Hi AB2005

 

Power = torque x speed.

 

If the slip speed of  the coupling was equal to the speed of the load, then half the power would be absorbed by the load, and half lost in the coupling.

In this case, from you figures, the speed of the load, is approximately 1/3 of the motor speed, so the power dissipated in the load will be about one third of the total.

The torque applied to the load is equal to the torque from the motor shaft. A gearbox would change the torque and the equation would be much different.

 

Best regards,

Mark






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