High Inertia Load during startup

[santu398]

Dear all,

We are having sand mills with a motor capacity of 30 HP. As the sand settles during shutdown, it requires a huge initial torque to start the sand mill. Presently, it is having star-delta starter.

The motor loading percentage in delta is around ~ 50 percent. The sand mill runs for a period of 20 hrs in a day in delta connection.
The motor loading percentage in star is around ~ 126 percent for a time period of 10-15 seconds. The sand mill runs for 4 min in star connection in a day.


Considering the losses at 50 % motor loading percentage, we are planning to change the 30 HP motor with lower capacity motor in such a way that motor loading percentage is 85 % in delta connection.

Can you please suggest what kind of starter is to be used as the motor loading percentage in star connection with a smaller capacity motor would go very high and damage the motor?

Request you to please recommend the different types of starters which can be used in this application. Can VFD's be used?


Regards,
Santosh Kumar

 

[jraef]

Do not change the motor size. It is ALREADY struggling to accelerate that load in Star, if you use a smaller motor it will likely stall or burn up.

Just because your motor is only 50% loaded does NOT mean there is a lot of wasted energy. Motor energy consumption is made up of the energy required by the LOAD on it, plus a small amount of losses inside of the motor to make it a motor (as opposed to a boat anchor). Your motor is likely in the area of 75% efficient at 50% load. Of that 25% losses, only maybe 3-5% of it would change by going with a smaller motor, but you now run the risk of it not working at all or failing sooner, which negates and possibly far exceeds any potential energy savings in unscheduled down time and loss of production. Some things just need to be looked at as the cost of operations.

If you do persist on changing the motor to be smaller, then ALSO change your procedure to ALWAYS make ABSOLUTELY SURE that all sand is cleared out EVERY TIME it is shut down. Most likely that is impossible, and the original designers knew that, hence the seemingly over sized motor. In other words there may have been a method to their madness.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

 

[JuanBC]

Agree with jraef.

As an example, a 30 HP / Standard efficiency / IEC WEG motor has a 90.7% efficiency at 50% load (you can check it on their website)

If you want a better efficiency, you can consider buying a Premium Efficiency (for example, 92.5% efficiency at 50% load)

JBC
.......
"The more I read, the more I acquire, the more certain I am that I know nothing"

 

[HCBFlash]

More agreement with jraef. The idea of changing to a smaller motor is unwise and ill-considered. If someone with such little technical knowledge insists on invading and changing a functioning mechanism, properly designed and implemented in its current state, additional changes will also be required. Some might include additional mechanisms to remove and re-introduce the sand load, across-the-line starting (and more frequent equipment failures, and changes being required in other wiring)

Keep unqualified parties away from functioning equipment!

It's NOT broken, DON'T FIX IT!

.

(Me,,,wrong? ...aw, just fine-tuning my sarcasm!)

 

[waross]

If it works don't fix it.
Don't change the motor.

The motor loading percentage in star is around ~ 126 percent for a time period of 10-15 seconds. The sand mill runs for 4 min in star connection in a day.

I don't see anything wrong with that. It is a lot less than the starting current that many motors see when started Direct On line.

The sand mill runs for 4 min in star connection in a day.

That seems like a needlessly long time on the star connection. Best practice is to change to the delta connection as soon as the motor is no longer accelerating. You say a high inertia load. It sounds more like high sticktion than high inertia.

126 percent for a time period of 10-15 seconds.

You possibly should be changing to the delta connection after 15-20 Seconds.


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

 

[JuanBC]

It's NOT broken, DON'T FIX IT!

+1

 

[waross]

" If someone with such little technical knowledge insists on invading and changing a functioning mechanism,......"
Tell them to "Pound Sand".

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

 

[itsmoked]

If you want to use a smaller motor then you have to have no sand present at startup. Figure out a way to clear the system. Compressed air perhaps?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[LionelHutz]

Can you post the 50%, 75% and 100% efficiency data for your motor? I'd like to see how bad this motor is. You must have this data because you you can't make an informed decision about motor efficiency improvements without it.

How are you measuring motor loading percentage? Power meter?

Typical motor datasheets I look at have a 50% efficiency which is within 1.5% of the peak motor efficiency. Both are typically over 90%. As an example of typical motor efficiency data that I see,
50% = 91.8%
75% = 93%
100% = 93%

So, is it worth the issues you will create to save maybe 1.5% of the energy going into the motor? Is it even worth the effort to spend the money on the change?

Assuming it's a fairly typical motor, I wouldn't even expect a reasonable payback on the new motor and installation costs. Lets do a quick number crunch. 50% load on that motor would be 15hp. Lets call it 10kW of power at the motor shaft. A gain of 1.5% on the motor efficiency would reduce the input power by a little over 150W. This means you save 1kWh of electricity every 6.5hrs of operation. For a 20 hour shift, you would save about 3kWh of electricity. This would be a savings of 60 cents per day of operation if electricity costs your plant 20 cents per kWh. The payback would be years.

 

[jraef]

And as I said earlier, the first time your machine fails to start, which becomes MUCH more likely, any saved energy value has been wiped out by the value of your lost production.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

 

[GroovyGuy]

Howdy Santu,
In your blurb above you stated;
The motor loading percentage in delta is around ~ 50 percent
The motor loading percentage in star is around ~ 126 percent
What are the motor loading units [ ie Amperes (A), real-power (kW), apparent-power (kVA) or torque (ft-lbs) ]?
Regards,
GG

"I have not failed. I've just found 10,000 ways that won't work." Thomas Alva Edison (1847-1931)

 

[LionelHutz]

The star of star-delta starting produces 33% of the motor rated locked rotor torque.
Reduction the motor size to 50% will reduce the LRT by 50%, assuming the motors have similar characteristics.

The available full-voltage starting torque is 67% more than the torque presently used while the reduction in motor size decreases the torque by 50%. So, there is the possibility for a net gain in starting torque even with the smaller motor. You just can't continue to use the star-delta starter.

But there is no point in arguing about the application before looking at the power numbers and proving money could be saved. There is no point considering anything else until there is an acceptable accounting payback on the change.

 

[waross]

The sweet spot for best efficiency is often between 66& and 75% of rated load.

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

 

[santu398]

Hello,

What are the motor loading units [ ie Amperes (A), real-power (kW), apparent-power (kVA) or torque (ft-lbs) ]?

KW

Regards,
Santosh

 

[jraef]

All of the above, from different perspectives. kW is representative of work being performed and is generally the accepted term for "load". But kVA is the "load" to a transformer, Amps are the "load" to an overload ad relay, torque is the "load" to the mechanical components.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

 

[waross]

Are you using a power meter?
If you are measuring current, you are getting KV, not kW.
It is not unusual for an oversized motor to use less energy than a fully loaded motor.
For example, if the load demands 7.5 HP, a 10 HP motor may actually be a little cheaper to run than a 7.5 HP motor.
Virtually all induction motors reach best efficiency at less than 100% load.


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