LOAD RPM

Status
Not open for further replies.
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
I have augers that I monitor the RPM. Priority is no rotation, then drive belt slippage. The overloads are very good at detection of LRA so that is not an issue.
What reduction in RPM should I expect from an unloaded motor to fully loaded? My suspicion is none, but I can set a percentage.
One auger is sheave reduction only the others are sheave to gear reduction.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
210311-0809 EST

ptonsparky:

Your suspicion is wrong unless the motor is a synchronous AC motor inherently.

A DC permanent magnet or fixed field motor is approximately a straight line with respect to torque load.
A series wound DC motor, armature is in series with field, has a very nonlinear curve. High torque at zero speed.

An AC induction motor is a more nonlinear speed torque relationship. Usually has a foldback characteristic.

.
 

Besoeker3

Senior Member
Location
UK
Occupation
Retired Electrical Engineer
I have augers that I monitor the RPM. Priority is no rotation, then drive belt slippage. The overloads are very good at detection of LRA so that is not an issue.
What reduction in RPM should I expect from an unloaded motor to fully loaded? My suspicion is none, but I can set a percentage.
One auger is sheave reduction only the others are sheave to gear reduction.
What sort of motor?
Three phase AC cage probably?
 

Besoeker3

Senior Member
Location
UK
Occupation
Retired Electrical Engineer
Grain legs with about 60 rpm, we use 10 and 20% levels. I'm dealing with about 500 and. 300 RPM.
So the motor would be about 1750 rpm full load to about 1795 rpm no load. The auger is not really an electrical issue.
 

kwired

Electron manager
Location
NE Nebraska
Isn't it always.............................:(
Maybe you have better mechanics over there? Around here if a motor overload is tripping it is an electrical problem - call the electrician. Electrician evaluates things and says this or that is worn, gummed up, has bad bearing, etc. and once you resolve that mechanical issue the motor won't be overloaded anymore. Happens a lot.

Or the not so knowledgeable mechanic decides to turn up setting of motor overload then wonders why the motor burns out eventually.
 

kwired

Electron manager
Location
NE Nebraska
So the motor would be about 1750 rpm full load to about 1795 rpm no load. The auger is not really an electrical issue.
I've seen speed sensors like he probably has. They are usually attached to something on the driven load and not the motor shaft.

Grain elevator, the motor and main drive unit is at the top, speed sensor is on bottom idler pulley of the elevator belt. If that idler slows down they want the sensor to trigger an output whether it be an alarm or a shutdown, usually an alarm or if a shutdown you shut down the grain supply first you don't want to shut down the elevator itself if loaded, that can become a big problem getting it started again or can be damaging if it does start again under load.
 

Besoeker3

Senior Member
Location
UK
Occupation
Retired Electrical Engineer
Maybe you have better mechanics over there? Around here if a motor overload is tripping it is an electrical problem - call the electrician. Electrician evaluates things and says this or that is worn, gummed up, has bad bearing, etc. and once you resolve that mechanical issue the motor won't be overloaded anymore. Happens a lot.

Or the not so knowledgeable mechanic decides to turn up setting of motor overload then wonders why the motor burns out eventually.
Oh, I think some are very good and some are, at best, indifferent putting it nicely.
Can be trying at times.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
Get a more sophisticated OL relay that looks at kW in addition to current. kW is always going to reflect true motor shaft loading, it doesn't vary with line voltage fluctuations like current does. Uncouple the motor, read the kW, that's your "Broken Shaft" threshold point. Couple it to the transmission but remove the belt, read the kW and that's your "Broken Belt" threshold. Couple it to the auger, run it unloaded, read the kW, that's your "Unloaded Auger" threshold point.
 
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
Get a more sophisticated OL relay that looks at kW in addition to current. kW is always going to reflect true motor shaft loading, it doesn't vary with line voltage fluctuations like current does. Uncouple the motor, read the kW, that's your "Broken Shaft" threshold point. Couple it to the transmission but remove the belt, read the kW and that's your "Broken Belt" threshold. Couple it to the auger, run it unloaded, read the kW, that's your "Unloaded Auger" threshold point.
I agree. Hindsight I should have gone with the higher end HMI that could log the data as well. Only $600 difference. This was supposed to be simple 5 motor project. Knowing no one would be monitoring the process we fancied up the control a bit. The millwrights got almost nothing right as far as I can see.
 

kwired

Electron manager
Location
NE Nebraska
In our plants, the millwrights would set machine tools. After they anchored them, they’d snap a chalk line on all four sides. To the casual observer, they were perfectly placed.
Extending a line as a reference for something else or just chalking the perimeter only of the machine they just set?
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
The motor nameplate should have the full load RPM listed. The no load speed will be a simple fraction of 3600 above that. The actual speed vs load is approximately linear between those two points.

The 1750 RPM number in post 7 is a good value for a small 4 pole induction motor at full load, so a 2.7% drop in speed from no load to full load is a good start. But the actual nameplate value should be your guide since different motor designs will have different full load 'slip'.

Jon
 

kwired

Electron manager
Location
NE Nebraska
The motor nameplate should have the full load RPM listed. The no load speed will be a simple fraction of 3600 above that. The actual speed vs load is approximately linear between those two points.

The 1750 RPM number in post 7 is a good value for a small 4 pole induction motor at full load, so a 2.7% drop in speed from no load to full load is a good start. But the actual nameplate value should be your guide since different motor designs will have different full load 'slip'.

Jon
The rpm monitoring devices I have encountered have usually been on the driven load somewhere, so you would need to factor in speed increase or reduction factors from gearboxes, pulleys, etc.
 

retirede

Senior Member
Location
Illinois
Extending a line as a reference for something else or just chalking the perimeter only of the machine they just set?

Chalking the perimeter of the machine they just set. People would assume that the lines were there first and that the machine was perfectly positioned by said millwrights.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
The rpm monitoring devices I have encountered have usually been on the driven load somewhere, so you would need to factor in speed increase or reduction factors from gearboxes, pulleys, etc.

Agreed. However the % change should be similar, at least if you are considering the friction in the gearing to be part of the load on the motor.

Say the motor has a synchronous speed of 1800 RPM and a full load speed of 1746 RPM. The motor fully loaded shows a speed reduction of 3%. Now you add a 6:1 gear reduction. The synchronous speed of the system should be 300 RPM, and when the motor is fully loaded the speed would be 291 RPM with the same 3% drop. Say with nothing in the auger the speed is 297 RPM. That 1% drop is caused by the system friction and is loading the motor at about 33% of capacity.

-Jon
 
Status
Not open for further replies.
Top