Selecting efficiency and estimating savings

1. 1
Selecting Efficiency and Estimating Savings
High efficiency motors can help minimize your motor operating costs. Because many
motors operate 40-80 hours per week or more, even small increases in efficiency can
substantial energy and money savings.
Efficiency
: the ratio (in percent) of mechanical power output to the electrical power input.
Fig1 : Lifetime Motor Costs
Equation 1.
Motor Efficiency Energy Cost Savings Equation
kWhmotormotoryear
hours $%100%100
hp
Kw
0.746LFHPSavingEnergyAnnual
21







Example Annual Energy Cost Savings Calculation with Upgrade to NEMA Premium
Estimated annual dollar savings associated with replacement of a totally enclosed fan
cooled (TEFC) 150hp,1800 rpm motor below EPAct efficiency standards with a NEMA
Premium motor of the same size and type.This calculation assumes both motors have the
same load factor, 75%

2. 2
Motor data
Motor Power (hp) 100 hp
Load Factor (LF)% 75%
Annual Operation hours 5,200 hours
Motor 1 efficiency 93.0%
Motor 2 Efficiency 95.8% premium
Electricity cost (Rm/ Kwh) 0.07/kwh
National Electrical Manufacturers Association (NEMA)
$640/year
kWh
$0.07
95.8%
100%
93.0%
100%
year
hours
5,200
hp
Kw
0.7460.75100hpSavingEnergyAnnual








Equation 2.
Motor Energy Demand Savings Equation







motor2motor1 E
hp
E
hp
hp
kw
0.746ED)(SavingDemandElectric
Example Energy Demand Savings Calculation with NEMA Premium Motor
Kw4
0.958
150hp
0.93
150hp
hp
kw
0.746ED)( 




The most common type of general-purpose motors found in industrial motor systems are
squirrel cage induction motors.These motors are generally referred to
as“general-purpose motors.”
Motor Speed
Equation
Synchronous Speed Calculations power consumption is related to speed,the higher
speed can lead the motor to draw more power.
 
   
   
rpm009
poles8
60Hz120
poles8forrpm8001
poles6
60Hz120
motorpole6for
1800rpm
poles2
60Hz120
poles4for3600rpm
poles2
60Hz120
motorpole2for
motortheofpolesofnumber
60Hz120
speedssynchronou













3. 3
Estimating Energy and Cost Savings with ASDs
Matching motor speed to application requirements through the use of ASDs,
also referred to as VFDs or inverters, can achieve significant electricity savings when
connected to motors in appropriate applications such as centrifugal pumps and fans
ASDs can provide additional benefits related to energy efficiency
 Improved process control, such as speeding up or slowing down a machine or process:
 Inherent power factor correction
 Bypass capability in the event of an emergency
 Protection from overload currents
.Equation
Annual Energy Cost Equation for Motor-Driven System with ASD
 
ASDE
1
kwh
$
hrs
2
decimalaasexpressedspeedrated%full
hp
Kw
0.746LF
motor
E
P[hp]
CostEnergyAnnual 
Potential Savings with ASD on Centrifugal Loads
Below,a 50hp centrifugal pump operating 4,067hours annually,with a 75% load factor, a
throttling valve to regulate flow to 70% on average, and primarily frictional losses and
negligible static head.
Pump System data
Motor Power (hp) 50 hp
Motor Efficiency (Emotor) 0.93 (1800 rpm)
Load Factor (LF)% 75%

4. 4
Percent full rated speed 100%
Annual operating hours 4,067 hours
Electricity cost 0.07/kwh
ASD Efficiency (EASD) 97%
Fig :Pump System Diagram with Throttling Valve
Annual Energy Cost Calculation with Throttling Valve in Pump System
  year
hp
/$564,8
kwh
0.07$
hrs067,4
2
0.70
hp
Kw
0.74675.0
93.0
50
CostEnergyAnnual 
The same system appears below, except an ASD replaces the throttling valve to achieve
the same flow regulation by varying the motor’s rotational speed
Fig :Pump System Diagram with ASD
Annual Energy Cost Calculation with ASD in Pump System

5. 5
  year
hp
/$326,4
97.0
1
kwh
0.07$
hrs067,4
2
0.70
hp
Kw
0.74675.0
93.0
50
CostEnergyAnnual 
Annual Energy Cost Savings Calculation Associated with ASD in Pump System
4,238$4,326$-8,564$
Cost(ASD)Energyannual-Valve)g(ThrottlinCostEnergyAnnual

Electric Demand Savings Equation with ASD in Pump System
   







ASD
2
2
motor E
e)speed(drivrated%full
r)speed(motorated%full
hp
0.746kw
LF
E
hp
ED)(
Electric Demand Savings Calculation with ASD in Pump System
    KW
hp
15
97.0
2
7.02
0.1
hp
0.746Kw
75.0
93.0
50
ED 








Using the information from each scenario, potential savings are calculated: replacing the
throttling valve with the ASD can achieve approximately $4,238 in annual energy cost
savings and saves approximately 15kW of electric demand

6. 6
Motor Energy Saving Calculation Form
Employee Name: Location:
Company: Process:
Date:
Motor Nameplate&Operating Information
Manufacture: Full Load speed:
Motor ID NO: Full load amperage:
Size(hp): Full-Load Power Factor (%)
Enclosure Type: Full-Load Efficiency (%)
Synchronous speed: Annual Operating Time ______ hours/year
Utility Rates
Energy Rate ($/kwh):………………………….

7. 7
Monthly Demand Charge ($/kW/mo.)
Annual Operating Hours (hrs/yr.)
Annual Energy Use and Cost
Input Power (kW)
Annual Energy Use < Input Power x Annual Operating Hours >
Annual Energy Cost: < Annual Energy Use x Energy Rate >
Total Annual Cost: < Annual Energy Cost + Annual Demand Cost >
Motor Load and Efficiency Determination
Load : < Input Power(kW) / [ Motor Size(hp) x 0.746 / Efficiency at Full Load ] >
Motor Efficiency at Operating Load:
Energy Savings and Value
kW saved=< Input Power - [ Load x hp x 0.746 / Efficiency of Replacement Motor at Load Point ]>
kWh saved < kW saved x Annual Operating Hours>
Total Annual Savings
Total Annual Savings < (kW saved x 12 x Monthly Demand Charge) + (kWh saved x Energy Rate)>
Economic Justification
Payback (years)

8. 8
Report on
Study on International Efficiency (IE) Efficiency Classes for AC Motors
Purpose
 To promote higher energy efficiency to reduce the energy consumption and the
energy cost of high voltage AC motors.
 To facilitate the procurement of new motors with higher energy efficiency, a study
on the IE and its application was conducted.
A case study on calculating the energy saving of replacement of energy efficient motor
is covered
a survey on the motor and its price of different classes from major suppliers in Malaysia
EFF Classification
The EFF has 3 classes, i.e. EFF1, EFF2 and EFF3 respectively. EFF1 is the most energy
efficient, while EFF3 is the least energy efficient.
Alignment of the energy efficiency standard of CEMEP and IEC
Efficiency Class IEC CEMEP
Super Premium Efficiency IE4
Premium Efficiency IE3
High Efficiency IE2 EFF1
Standard Efficiency IE1 EFF2
Below Standard Efficiency EFF3
The scope of classification is defined for single speed, single and three phase, and
continuous duty electric motor with 2, 4, 6 or 8 poles. The rated output ranges from
0.75kW to 375 kW; and the frequency is between 50Hz and 60Hz. The scope also
includes ambient temperature within the range of -20o C to +60o C and an altitude of up
to 4,000m above sea level.
In accordance with the Malaysia electricity supply context, 50Hz motor is the focus. It is
observed that the motors commonly used in Malaysia are 4-poles. A graphical
presentation of this type of motor is shown in Figure 1 while the minimum energy

9. 9
efficiency requirement of IE Efficiency Class of IE1 to IE4 of different poles is shown
in Table . .
Motors market and price
In Table . The motors are divided into 2 groups in accordance with their rated power of
37kW due to the significant difference of their shaft bearing design.
The listed prices are obtained from various manufacturers available in market The listed
price of IE2 4-poles 7.5 kW motor is around $6,000 market
The listed price of IE2 4-poles 45kW motor is around $28,000 in market.
Rated Power kw IE2 IE3 IE4
≤37 1.00 1.17 ratio 1.32 ratio
>37 1.00 1.13 1.21
Energy Saving by Energy Efficient Motor
The annual energy saving by upgrading to more efficient motor is calculated as the
formula below:
%SavingEnergyofPercentageyr)motor(kwh/ofnConsumptioEnergyAnnual
(kwh/yr)SavingEnergyAnnual

The percentage of energy saving can be calculated with the following formula:

10. 10
100%
Motor(%)NewofEfficiency
Motor(%)OldofEfficiency
-1%SavingEnergyofPercentage 






If the annual energy consumption of motor is not available, it can be estimated with the
following formula:
day/yearOpertingHours/dayOperting
EfficiencyEnergyMotorofPowerRated(kwh/yr)motorofnConsumptioEnergyAnnual


Case Study
To demonstrate the calculation of energy cost saving and payback period, a case study is
provided below. For Company A, the information regarding the old motor and the
operation pattern is shown in Table
Table Specification of old motor and operation pattern .
Rated Power 37 kW
No. of Poles 4
Efficiency 91.2 (IE1)
Operating Hours per Day 10
Operating Days per Year 360
The Annual Energy Consumption is calculated below:
146,053kwhdays/year360hours/day1091.2%37kw
(kwh/yr)motorofnConsumptioEnergyAnnual

Therefore, the annual energy consumption of old motor is 146,052kWh. 10 of 13
Company A is going to upgrade the motor to an IE3 motor as per the specification
shown in Table
Table Specification of new motor
Rated Power 37 kW
No. of Poles 4
Efficiency 93.9 (IE3)

11. 11
Cost of Motor (including installation) 40,000 $
Designed lifespan 400,000 hours
The percentage of energy saving and the anticipated annual energy saving is calculated
below:
ear4,206kwh/y2.88%yr146,056kw/(Kwh/year)SavingEnergyAnnual
%88.2%100
93.9%
91.2%
-1Saving(%)EnergyofPercentage








Therefore, the percentage of energy saving and annual energy saving is 2.88% and
4,206kWh per year, respectively.
To evaluate the cost effectiveness, the payback period is calculated as follows:
yryear /$206,44/$206,4)Motorof$(Cost40,000(year)PeriodPayback
$/yr4,2061$/kwhr4,206kwh/y/year)(1$SavingCostEnergyAnnual


Therefore, the payback period is 9.5 years, which equals to 34,675 operating hours. In
comparison with the designed lifespan of the motor (i.e. 400,000 hours), upgrading the
motor from IE1 to IE3 is considered to be cost effective.
Table Minimum Energy Efficiency Requirement of IE Efficiency Class IE1 to IE4 for
50 Hz motors

12. 12
‘How much electrical power may realistically be saved by using high or premium
efficiency motors’?
Let us compare energy and capital costs on two motor sizes and efficiencies. We will
consider 7.5kW IE1, IE2 and IE3 motors as well as a 75kW IE1, IE2 and IE3 motors.
Refernce
http://www.beeo.emsd.gov.hk/en/pee/BEC_2015.pdf
(code of practice for Energy Efficiency of Building Services Installation)
file:///C:/Users/user/Downloads/4E_Roadmap_for_Motors_and_VFD_231115_v3
.pdf
http://www.zvei.org/Publikationen/ZVEI%20Electric%20Motors%20and%20Vari
able%20Speed%20Drives%202nd%20Edition.pdf

13. 13