Electric motors convert electrical power into mechanical power within a motor driven system.
In industrial applications, electric motor driven systems are used for various applications such as pumping, compressed air, fans, conveyors etc. The system approach for optimizing energy efficiency of motor-driven system is recommended, which include the following:
Use of energy efficient motors;
- Selecting the driven equipment―like pumps, fans, compressors, transmissions,
- variable speed drives―right type and size, and high efficiency;
- Efficient operation of the complete system.
From the motor perspective, when buying a new motor, operating cost and not just the purchase cost should be the main consideration. In a single year the cost of energy can be up to 10 times the purchase cost. Over the life of the motor it is by far the most significant cost. Old motors, typically more than 15 years and operating for over 5000 hours in a year can be considered for replacement with energy-efficient motors to reduce energy costs.
IE Classification
International Efficiency (IE) is a new trend around the world in describing the energy efficiency of motors. The IE classes IE1 to IE4 are well developed, while the IE5 is under preparation. The classification method allows for further improvement in the energy efficiency of motors. The IE Classification as per IEC 60034-30-1 is shown as below:
IE Classes/Class Type Class Number
Standard efficiency / IE1
High efficiency / IE2
Premium efficiency / IE3
Super premium efficiency / IE4
Super premium efficiency / IE5
IE5 represents the highest energy efficiency while IE1 represents the least energy efficiency.
In other words, the higher the class number, the higher will be the motor efficiency. IE5 is to be incorporated in the next edition of IEC 60034-30-1, with a goal to obtain an energy loss reduction of 20% relative to IE4.
The annual energy saving by upgrading to more efficient motor is calculated as per the
following formula:
Annual Energy Saving (KWh/Year) = Annual Energy consumption of motor (KWh/year) x
Percentage energy saving (%)
Percentage of Energy Saving (%) = (1- Efficiency of old motor /Efficiency of new motor) x
100
If the annual energy consumption of motor is not available, it can be estimated with the
following formula:
Annual energy consumption of motor (Kwh/year) = Rated power of motor /rated efficiency x
Operating hours /day x operating days per year
Example : Replacement of IE1 motor with IE3 motor
The information regarding the old motor and the operation pattern is as follows:
Rated Power = 37 KW
No of Pole = 4
Efficiency = 91.2 (IE1)
Operating Hours per Day =10
Operating Days per year = 360
Annual Energy Consumption (kWh per Year) = 37 kW/0.912 x 10 hours per day x 360 days/
year = 1,46,052 kWh
It is proposed to replace IE1 is replaced with IE3 motor as per following specifications:
Rated Power = 37 KW
No of Pole = 4
Efficiency = 93.9 (IE3)
Operating Hours per Day =10
Operating Days per year = 360
Designed Life span = 400,000 hours
The percentage of energy-saving and the anticipated annual energy saving is calculated
below:
% Energy Saving = 1- (91.2/93.9) x 100% = 2.88%
Annual Energy Saving (KWh/year) = 1,46,052 (KWh/year) x 2.88% /100 = 4206 KWh/year
If power tariff = 8.00 Rs/KWh
Cost Saving per year = 33,648 Rs / year
If it is proposed to replace IE1 is replaced with IE4 motor as per following specifications:
Rated Power = 37 KW
No of Pole = 4
Efficiency = 95.2 (IE4)
Operating Hours per Day =10
Operating Days per year = 360
Designed Life span = 400,000 hours
% Energy Saving = 1- (91.2/95.2) x 100% = 4.2%
Annual Energy Saving (KWh/year) = 1,46,052 (KWh/year) x 4.2% /100 = 6134 KWh/year
If power tariff = 8.00 Rs/KWh
Cost Saving per year = 49072
Rs / year
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