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Energy Efficient Motors and Transformers Workshop MOTORS LIEN 7 May 2003. Dr Hugh Falkner MIEE CEng Future Energy Solutions. Contents. The System Perspective Higher Efficiency Motors Motor Management Policy VSDs Identifying energy saving opportunities Maintenance and Energy Saving
Dr Hugh Falkner MIEE CEng
Future Energy Solutions
Don’t waste time making a useless system more efficient!
Some likely candidates:
Keep in mind the 80/20 rule of thumb:
Average savings of 3% don’t sound very impressive compared to VSDs.
But, if all your motors were HEMs, the savings would be huge
Characteristic Voltage Voltage 6% High6% Low
Depends on many factors...
Typical faults from a sub-standard repair
Original - 90.5%
Repair - 90.0%
HEM - 92.5%
Getting everybody involved is essential - everyone has different motivations
Before deciding to alter the speed, make sure that you understand the system. Otherwise you could make things worse!
If the machine can always turn at a lower speed, then you could alter the speed in many ways, you don’t have to use a VSD.
However, as the price of VSDs gets lower, they are being used in more and more applications
Input Power (%)
Fan Affinity Laws
(Applies to all centrifugal loads)
AVSD is only as good as the way it is commissioned.
There are two methods of control:
Some symptoms of interest for pumping systems:
This is based on the fact that the torque:slip characteristics in the normal operating region of an induction motor are very straight. So, at full rated load the motor will be at its maximum slip (and hence minimum speed.) By comparing the actual measured speed with the nameplate rated slip, the power can be estimated.
Mechanical Power =
(nno load - n meas) x kW (rated)
(nno load - nrated)
nno load = No load (synchronous) speed
n meas = Measured speed
n rated = Nameplate rated speed
From the nameplate, 1,470rpm at 55kW (rated) load
Using the stroboscope it is measured at 1,480 rpm.
What is the load?
Load = 1,500 - 1,480 = 20 = 67% of rated power
1,500 - 1,470 30
Mechanical (output) power is 67% x 55kW = 37kW
This is only approximate, but it does give a very quick indication of power consumption
saving energy and maintenance costs.
You’ll struggle to save energy and not reduce maintenance costs, and vice versa.
Don’t jump in and just try to fix the reported problem, find out about the whole related system.
Energy Used = 100kW x 24 hours = 2400kWh per day
Useful Work done (assuming 60% efficiency)
= 1140kWh. Where does the 960kWh go!?
It is this surplus energy, ie energy that is doing nothing useful, that causes maintenance problems.
Principally Direct heat, friction and their resultant effects.
Equipment that is 100% efficient has no surplus energy for causing maintenance problems.
The condition of chemical
And fluid properties
Presence of fluid and
surface destructive contaminants
Wear Debris Analysis
Presence of machine