ecet 211 electric machines & controls lecture 5-4...

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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

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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

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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

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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

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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.

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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.

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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

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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

www.otisworldwide.com

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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

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www.otisworldwide.com

(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


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

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Mechanical Power Rating

Torque

Current

Code Letter

Efficiency


Frame Size

Full-Load Speed

Load Requirements

Motor Temperature Ratings

Duty Cycle

Motor Enclosure

Metric Motors

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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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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


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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• 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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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

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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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• Electrical Connections:

NEMA standards

NEC Article 430

State & Local Code

• Grounding

Equipment grounding conductor

Figure 5-79 Motor shaft grounding ring


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• 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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• 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)


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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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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%


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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

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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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