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ECET 211 Electric Machines & Controls
Lecture 5-4 Electric Motors(4 of 4)
Text Book: Chapter 5 Electric Motors, Electric Motors and Control Systems, by Frank D. Petruzella, published by McGraw Hill,
2015.
Paul I-Hai Lin, Professor of Electrical and ComputerP.E. States of Indiana & California
Dept. of Computer, Electrical and Information Technology
Purdue University Fort Wayne Campus
Prof. Paul Lin
Lecture 5-1 Electric Motors
Chapter 5. Electric Motors• Part 1. Motor Principles
• Part 2. Direct Current Motors
• Part 3. Three-Phase Alternating Current Motors
• Part 4. Single-Phase Alternating Current Motors
• Part 5. Alternating Current Motor Drives
• Part 6. Motor Selection
• Part 7. Motor Installation
• Part 8. Motor Maintenance and Troubleshooting
Prof. Paul Lin 2
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Part 6 Motor Selection
Motors Selection based on
• General-purpose applications vs Specific tasks
General purpose motor, Inverter-duty motors
• Technical Requirements:
Mechanical load, Motor output horsepower, Torque, Speed
Voltage, Frequency, Phase, Starting (current), Efficiency,
Power factor, Motor temperature, Service Factor, Duty cycle,
Frame size
• Government mandates
EPAct (Energy Policy Act) 1992
Premium Efficiency - EISA (Energy Independent and Security Act) 2007: all applicable motors manufactured or imported into the U.S. after Dec. 2010 must meet the Premium Efficiency guidelines
Prof. Paul Lin 3
Part 6 Motor SelectionReferences
Premium Efficiency Motor Selection and Application Guide – A Handbook for Industry (136 pages), Advanced Manufacturing Office, U.S. Dept. of Energy, http://energy.gov/sites/prod/files/2014/04/f15/amo_motors_handbook_web.pdf EISA 2007 (Energy Independent and Security Act)
Motor losses and loss reduction techniques
MotorMaster+ software
The Motor Guide (Low-power standard motors) – ABB Group, (135 pages), http://www04.abb.com/global/seitp/seitp202.nsf/0/12c3580f179a9d58c125761f0057ca5c/$file/motor+guide+gb+02_2005.pdf
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Part 6 Motor SelectionReferences
Guide for AC Motor Selection (Small size, standard AC motors),https://www.orientalmotor.de/media/files/17112005104817.pdf
Selection Procedure:
(1) Required specifications
(2) Calculate the operating speed
(3) Calculate the required torque
(4) Select a motor and gearhead
(5) Confirm the speed
Prof. Paul Lin 5
Part 6 Motor Selection
Mechanical Power Rating
Torque
Current
Code Letter
Efficiency
Energy-Efficient Motors
Frame Size
Full-Load Speed
Load Requirements
Motor Temperature Ratings
Duty Cycle
Motor Enclosure
Metric Motors
Prof. Paul Lin 6
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Part 6 Motor Selection Mechanical Power Rating
• is expressed in either horsepower (hp) or watts (W)
• 1 hp = 746 W
• P = ωr*T
• Horsepower = Torque x Speed/5252
Torque in lb-ft ; Speed in rpm
Torque
Motor torque is the twisting force exerted by the shaft of a motor.
Figure 5-70 motor’s torque-speed shows how a motor’s torque production varies throughout the different phases of its operation.
Prof. Paul Lin 7
Part 6 Motor Selection Torque
Figure 5-70 motor’s torque-speed curve
Locked-rotor torque (LTR) or Starting torque is produced by a motor when it is initially energized at full voltage.
Pull-up torque (PUT) is the minimum torque generated by a motor as it accelerates from standstill to operation speed.
Breakdown torque or pull-out torque is the maximum amount of torque a motor can attain without stalling.
Full-load torque is produced by a motor functioning at rated speed and horsepower.
Prof. Paul Lin 8
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Part 6 Motor Selection
Example 1. A motor delivers 300 hp at 1800 rpm. How much torque does it produce?
Answer:
P = ωr T
ωr = 2π nr / 60; nr in revolution/minute
T = P/ωr =60 P/(2πnr ) N·m
T = (5252 x hp)/ nr lb ft
T = 5252 x 300 hp/1800 = 875.3 lb ft
T = P/ωr =60 P/(2πnr ) N·m
1 HP = 0.746 kW
P = 300 hp = 223.8 kW
T = P/ωr =60 P/(2πnr ) = 60 * 223,800/(2π * 1800) = 1187N·m
Prof. Paul Lin 9
Part 6 Motor Selection
Example 2: An elevator is required to lift a load of 1,000 kg to an altitude of 30 m. (a) How much energy must the motor provide? (neglecting losses in the hoist assembly)
Answer (a):
Wout = m·g·h
= 1000 kg · 9.81N/kg ·30m
= 295,300 N·m ≈ 0.3 MJ
= 0.083 kwh
1 N·m = 1 Joule
1 kwh = 3.6 MJ
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(b) What size horsepower would be required to lift the elevator from ground up for 30 meters if the total time required is to be 45 seconds.Answer (b):Win = Wout/η = Wout = 0.083 kwhPavg
= Win /∆t = 0.083 kwh/(45 sec/3600sec) = 0.083/0.0125= 6.64 kw
kw => hp conversion= 6.64 kw/0.745 kw/hp= 8.9 hp=> chose 9-10 hp motor
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Part 6 Motor Selection
Example 3: Assume the hoist assembly is 66% efficient. (a) If the time required to lift from bottom to the top is to be 45 sec. What size horsepower would be required.
Answer (a):
Win = Wout/η = 0.083/0.66 = 0.126 kwh
Pavg = Win /∆t
= 0.126 kwh/(45 sec/3600sec)
= 0.126/0.0125 = 10.08 kw
kw => hp conversion= 10.08 kw/0.745 kw/hp
= 13.5 hp
Prof. Paul Lin 11
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(b) If the time required to lift from bottom to the top is to be 60 sec. What size horsepower would be required.
Answer (b):
Pavg = Win /∆t
= 0.126 kwh/(60 sec/3600sec)
= 0.126/0.0167 = 7.55 kw
kw => hp conversion
= 7.55 kw/0.745 kw/hp
= 10.1 hp
Part 6 Motor Selection
Exercise Question 1: This exercise question is about he potential energy and horsepower calculation for a hydraulic car-lifting system for use in a auto repair shop. If the maximum weight of the lifting system is 5,000 pound, and it is expected to lift up 6 ft in 10 second, find
(a) The potential energy need to lift the car.
(b) Power needed to lift in 10 sec.
(c) Input power to the motor for used in the hydraulic system. Neglecting all the losses.
Answer
(a) W= m*g*h (joule)(b) Plift = W/t (watts)
(c) Pin_lift = Plift/η
Convert Pin from kW to hp : 18.2 hp, chose 20 hp
Prof. Paul Lin 12
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Part 6 Motor Selection
Mechanical Power Rating
Torque
Current
Code Letter
Efficiency
Energy-Efficient Motors
Frame Size
Full-Load Speed
Load Requirements
Motor Temperature Ratings
Duty Cycle
Motor Enclosure
Metric Motors
Prof. Paul Lin 13
Part 6 Motor Selection
Current
• Full-load current (nameplate current)
The amount of current (amperes) the motor can be expected to draw under full load (torque) condition.
Used to determine the size of overload sensing element for motor protection
• Lock-rotor current: starting inrush current
• Service-factor current
The amount of current the motor will draw when it is subjected to a overload equal to the service factor on the nameplate of the motor.
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Part 6 Motor Selection
Code Letter (NEMA)
• Assigned to motors for calculating the lock-rotor current based on the kilovolt-amperes per nameplate horsepower.
LR current (single-phase motors)
• Code letter value x hp x 1000/Rated voltage
LR current (three-phase motors)
• Code letter value x hp x 577/Rated voltage
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Locked-Rotor Code, kVA/hpA 0.01-3.14 G 5.6-6.3B 3.15-3.55 H 6.3-7.1C 3.55-4.0 J 7.1-8.0D 4.0-4.5 K 8.0-9.0E 4.5-5.0 L 9.0-10.0F 5.0-5.6 M 10.0-11.2
Part 6 Motor Selection
Design Letter (NEMA)
• NEMA defines four standard motor designs for AC motor: A, B, C, and D.
• The design letter denotes the motor’s performance characteristics relating to torque, starting current, and slip.
Efficiency
• η = mechanical power output / electrical power input
• Power losses = core loss + stator and rotor resistance loss (copper losses) + mechanical losses + stray loss
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Part 6 Motor Selection
Energy-Efficient Motors
• Efficiencies ranges between 75 and 98 percent
• Energy-efficient motors are manufactured with higher-quality materials and techniques.
• Figure 5-66 Typical energy efficient motor
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Part 6 Motor Selection Load Requirements
• Must be considered in selecting the correct motor for a given application.
Figure 5-67 Constant-torque load
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Part 6 Motor Selection Load Requirements
Figure 5-68 Variable-torque load
Figure 5-69 Constant-horsepower load
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Part 6 Motor Selection
Motor Temperature Ratings: Ambient temperature, Temperature rise, Hot-spot allowance, and insulation class
Duty Cycle: Continuous duty, Intermittent duty
Motor Enclosure
• Figure 5-71 Motor enclosures
• Open drip- proof (ODP), Totally enclosed, fan-cooled (TEFC)
• Totally enclosed, non-ventilated (TENV), Hazardous location
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Figure 5.72 Explosion-proof motor
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Part 6 Motor Selection Metric Motors
• Replacement for a metric (IEC)
motor installed on imported equipment
1) Get an exact replacement
2) Other considerations
• IEC (kw) vs HP
• IEC frame size – metric dimension
• Frequency may be 50 Hz vs 60 Hz
Prof. Paul Lin 21
Speed, rpmFrequency 50 Hz Frequency 60 Hz
Poles Sync Ns Full-load Nr
Sync Ns Full-load Nr
2 3,000 2,850 3,600 3,4504 1,500 1,425 1,800 1,7256 1,000 950 1,200 1,1508 750 700 900 850
Figure 5.73 IEC motors
Part 7 Motor InstallationProcedure and Checklist References: Horizontal AC Small Industrial Motor –
Motor Installation and Maintenance Instruction, http://www.gepowerconversion.com/sites/gepc/files/product/GEI-56128(NEMA_140-500_HorizMotor).pdf
AC & DC Motor Installation & Maintenance Instruction, http://www.baldor.com/mvc/DownloadCenter/Files/LB5001
Installation
• Foundation
• Mounting
• Motor and Load Alignment
• Motor Bearings
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Procedure and Checklist
Power Supply, Conductor Sizing, Wiring and Connections
• Electrical Wiring and Connections
• Grounding
• Conductor Size
• Voltage Levels and Balance
• Built-in Thermal Protection
Operation
• Steps Prior to Starting
• Initial Start
• Jogging and Repeated Starts
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Part 7 Motor Installation Foundation
• Minimum vibration and proper alignment between motor and load
• Concrete – for large motors and driven loads
Mounting
• Figure 5-74 Common type of motor mounting
Motor and Load Alignment
• Figure 5-75 Laser alignment kit
• Direct-drive motors: 1:1 speed ratio
• Coupling - gears or pulley/belts
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Part 7 Motor Installation Motor and Load Alignment
(continue)
• Formula for calculating speed and pulley sizes for Belt-Driven System:
Motor rpm/ Equipment rpm
=
Equipment pulley diameter/Motor pulley diameter
Example 5-5: What size of pulley is needed for the load?
• Figure 5-76
Solution:
1725/1150 = Equipment-pulley/2
Equipment-pulley = 3-inch
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Part 7 Motor Installation Motor and Load Alignment
(continue)
• Figure 5-77 Servicing a V belt-drive system
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Part 7 Motor Installation Motor Bearings
• Figure 5-78
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Part 7 Motor Installation
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Procedure and Checklist
Power Supply, Conductor Sizing, Wiring and Connections
• Electrical Connections:
NEMA standards
NEC Article 430
State & Local Code
• Grounding
Equipment grounding conductor
Figure 5-79 Motor shaft grounding ring
Part 7 Motor Installation
Prof. Paul Lin 28
Procedure and Checklist
Power Supply, Conductor Sizing, Wiring and Connections
• Conductor Size (motor branch circuit conductor)
Article 430 of the NEC
Based on the motor full-load current, and increased where required to limit voltage drop.
Undersized wire between the motor and the power source will limit starting abilities and cause overheating of the motor.
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Part 7 Motor Installation
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Procedure and Checklist
Power Supply, Conductor Sizing, Wiring and Connections
• Conductor Size (motor branch circuit conductor)
Example 5-6. What size THW CU (Thermoplastic Insulated Wire, Copper wire) conductors are required for a single 15 hp, three-phase, 230 V squirrel-cage motor?
Step 1. Full-load current (FLC) rating of the motor => conductor size. NEC 2008 Table 430.250: 230V, 15 hp => FLC 42 amperes.
Step 2. NEC 430.22 required branch circuit conductor supplying a single motor to have an ampacity not less than 125 percent of the motor FLC.
Rated ampacity = 42 A x 125% = 52.5A
Step 3. According to table 310.15(B)(16) => conductor size
6 AWG THW CU (55A with 60°C insulation)
Part 7 Motor Installation
Prof. Paul Lin 30
Voltage Levels and Balance
• Voltage Levels
Voltage level with maximum deviation of 5 to 10 percent.
Large voltage variation can have negative effects on torque, slip, current, efficiency, power factor, temperature, and service life.
• Unbalanced motor voltage
Unbalanced current => overheating of the motor’s stator windings and rotor bars, shorter insulation life, and wasted energy.
Acceptable voltage unbalanced no more than 1 percent
Voltage unbalance exceeds 5 percent => not to operate the motor.
Percent voltage unbalance = (Max voltage deviation from the voltage average)/ Average voltage x 100
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Part 7 Motor Installation
Prof. Paul Lin 31
Example 5-7 What is the percent voltage unbalance for a three-phase supply voltage of 480V, 435V, and 445V (Figure 5-80)
Solution:
Average voltage = (480+435+445)/3 = 453V
Maxi deviation from the average voltage = 480 – 453 = 27V
Percentage voltage unbalance = Max voltage deviation/Average x 100 = 27/453 x 100 = 5.96%
Part 7 Motor Installation
Prof. Paul Lin 32
Built-in Thermal Protection
• Overload relay
• Thermal protectors inside the motor that sense motor windings temperature
• Figure 5-81 Built-in thermal motor protection
Automatic reset
Manual reset
Resistance temp. detectors
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Part 8 Motor Maintenance and TroubleshootingMotor Maintenance
Schedule Periodic Inspections
Brush and Commutator Care
Testing Windings Insulation
• 600V and below 1.5 MΩ
• 2,300 V 3.5 MΩ
• 4000V 5.0 MΩ
Keep Your Motor Clean
Keep Your Motor Dry
Check Lubrication
Check for Excessive Heat, Noise, and Vibration
Excessive Starting is a Prime Cause of Motor Failures
Prof. Paul Lin 33
Part 8 Motor Maintenance and TroubleshootingTroubleshooting Motors
Digital Multimeter (DMM)
Clamp-on ammeter
Mega-ohmmeter
Infrared (IR) thermometer
Tachometer
Oscilloscope
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Summary & Conclusion
Questions?
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