- Electrical Motors Basic Components
- Classification of Electric Motors – Part One
- Classification of Electric Motors - Part Two
- Classification of Electric Motors - Part Three
- Classification of Electric Motors - Part Four
- Classification of Electric Motors - Part Five

Today, I will explain

**How to select a Motor for a given Applications**as follows.

__First: AC Motors Selection Procedures__

The type of motor chosen for an application depends on the characteristics needed in that application which include:

- The power supply,
- System requirements,
- Motor class,
- Motor insulation type,
- Motor Duty Cycle,
- Bearing type,
- Method of mounting the motor,
- The cost and size of the motor,
- Method of speed control,
- Environmental conditions.

__1- The Power Supply__

The power supply is distinguished by its number of phases, rated voltage and frequency as follows:

__1.1 Number of phases__

- A power system can be either single-phase or poly-phase.
- Single-phase power is most commonly found in homes, rural areas and in small commercial establishments.
- A poly-phase power system consists of 2 or more alternating currents of equal frequency and amplitude but offset from each other by a phase angle.
- For motors, an advantage of three-phase power is simpler construction which requires less maintenance. Also, a more powerful machine can be built into a smaller frame and will generally operate at a higher efficiency than single-phase motors of the same rating.

__1.2 Voltage:__

**1.2. A- Motor Nameplate Voltage**

The motor nameplate voltage is determined by the available power supply which must be known in order to properly select a motor for a given application. The nameplate voltage will normally be less than the nominal distribution system voltage to allow for a voltage drop in the system between the power source and the motor leads.

The bellow image lists motor nameplate voltages and provides the best match to distribution system voltages and meets current motor design practices.

**1.2. B- Dual Voltage Motors**

Poly-phase and single-phase motors may be furnished as dual voltage ratings under the following conditions:

- Both voltages are standard for the particular rating as listed in the above image.
- The two voltages are in a ratio of either 1:2 or 1:Ã3 (e.g. 230/460, 60 Hz; 2300/4000, 60 Hz; or 220/380, 50 Hz).
- Single-phase voltage ratios are 1:2 only.

**1.2. C- Voltage Unbalance**

Unbalanced line voltages applied to a poly-phase motor result in unbalanced currents in the stator windings. Even a small percentage of voltage unbalance will result in a larger percentage of current unbalance, thus increasing temperature rise and possibly result in nuisance tripping.

**Percent voltage unbalance is calculated as follows:**

Percent Unbalance = (100 x Maximum Voltage Deviation from Average Voltage) / Average Voltage

**Note:**Motor operation above 5% voltage unbalance is not recommended.

**Unbalanced voltages will produce the following effects on performance characteristics:**

**Torques**: Unbalanced voltage results in reduced locked-rotor and breakdown torques for the application.**Full-Load Speed:**Unbalanced voltage results in a slight reduction of full-load speed.**Current:**Locked-rotor current will be unbalanced to the same degree that voltages are unbalanced but locked-rotor KVA will increase only slightly. Full-load current at unbalanced voltage will be unbalanced in the order of six to ten times the voltage unbalance.**Temperature Rise:**A 3.5% voltage unbalance will cause an approximate 25% increase in temperature rise.

__1.3 Frequency__

**1.3. A- Standard Frequency**

The predominant frequency in the United States is 60 hertz. However, 50 hertz systems are common in other countries. Other systems, such as 40 and 25 hertz are isolated and relatively few in number.

**1.3. B- 50 Hz Operation of 60 Hz Motors**

General Electric standard motors rated at 60 Hz may be successfully operated at 50 Hz at reduced voltage and horsepower as shown in the following table:

- Rated Hp at 50 Hz = Nameplate Hp x Derate Factor.
- Allowable voltage variation at derated Hp = ±5%.
- Select motor overload protection for 60 Hz Amps and 1.0 Service Factor.
- Motor speed = 5/6 nameplate rated speed.
- Service Factor = 1.0
- Sixty hertz motors intended for use as shown above should be ordered as 60 Hz motors with no reference to 50 Hz operation.

**1.3. C- Dual Frequency**

Motors that require 50 and 60 Hz operation of the same motor are non-NEMA defined motors and will be nameplated as such. When this is a motor requirement, it must be specified with the order.

__1.4 Voltage and Frequency Variation__

All motors are designed to operate successfully with limited voltage and frequency variations. However, voltage variation with rated frequency must be limited to ±10% and frequency variations with rated voltage must be limited to ±5%. The combined variation of voltage and frequency must be limited to the arithmetic sum of 10%.

Variations are expressed as deviation from motor nameplate values, not necessarily system nominal values. The allowable ±10% voltage variation is based upon the assumption that horsepower will not exceed nameplate rating and that motor temperature may increase.

**The following conditions are likely to occur with variations in voltage:**

- An increase or decrease in voltage may result in increased heating at rated horsepower load. Under extended operation this may accelerate insulation deterioration and shorten motor insulation life.
- An increase in voltage will usually result in a noticeable decrease in power factor. Conversely, a decrease in voltage will result in an increase in power factor.
- Locked-rotor and breakdown torque will be proportional to the square of the voltage. Therefore, a decrease in voltage will result in a decrease in available torque.
- An increase of 10% in voltage will result in a reduction of slip of approximately 17%. A voltage reduction of 10% would increase slip by about 21%.

**The following conditions are likely to occur with variations in frequency:**

- Frequency greater than rated frequency normally improves power factor but decreases locked rotor and maximum torque. This condition also increases speed, and therefore, friction and winding losses.
- Conversely, a decrease in frequency will usually lower power factor and speed while increasing locked-rotor maximum torque and locked-rotor current.

__1.5 Variable Frequency Operation__

Motors are available for use on variable frequency inverters. Generally speaking, there are three different types of inverters:

- VVI is a square wave inverter in which voltage and frequency vary in proportion (constant volts per hertz).
- PWI is a pulse width modulated inverter and the same as the VVI type except pulses are varied in time to simulate a sine wave.
- CCI is a constant current inverter, which utilizes a square wave current supply as opposed to voltage.

__2- System requirements:__

This will include:

- — Rated Speed (Speed measured in shaft revolutions per minute (RPM)).
- — Torque.
- — Horsepower.
- — Torque-Speed performance of a motor.
- — Torque, Speed and Current Relation of a motor.

__2.1 Rated speed__

The speed at which an induction motor operates is dependent upon the input power frequency and the number of electrical magnetic poles for which the motor is wound.

The higher the frequency, the faster the motor runs. The more poles the motor has, the slower it runs.

The speed of the rotating magnetic field in the stator is called synchronous speed. To determine the synchronous speed of an induction motor, the following equation is used:

**Synchronous Speed (rpm) = (60 x 2 x Frequency) / Number of poles**

Actual full-load speed (the speed at which an induction motor will operate at nameplate rated load) will be less than synchronous speed.

The difference between synchronous speed and full-load speed is called slip. Percent slip is defined as follows:

**Percent Slip = (Synchronous Speed - Full Load Speed) X 100 / Synchronous Speed**

Induction motors are built having rated slip ranging from less than 5% to as much as 20%. A motor with a slip of less than 5% is called a normal slip motor. Motors with a slip of 5% or more are used for applications requiring high starting torque (conveyor) and/or higher than normal slip (punch press) where, as the motor slows down, increased torque allows for flywheel energy release.

__2.2 Torque__

Torque is one key motor characteristic (in addition to horsepower) that determine the size of motor for an application. Torque is merely a turning effort or force acting through a radius.

__2.3 Horsepower__

Horsepower take into account how fast the motor shaft is turned. Turning the shaft rapidly requires more horsepower than turning it slowly. Thus, horsepower is a measure of the rate at which work is done. By definition, the relationship between torque and horsepower is as follows:

**Full-load torque in lb-ft = (Hp x 5252) / Full-Load rpm**

__2.4 Torque-Speed performance of a motor__

The following graph illustrates a typical speed torque curve for a NEMA design B induction motor. An understanding of several points on this curve will aid in properly applying motors.

**2.4. A- Locked-Rotor Torque**

Locked-rotor torque is the torque which the motor will develop at rest (for all angular positions of the rotor) with rated voltage at rated frequency applied. It is also sometimes known as “starting torque” and is usually expressed as a percentage of full-load torque.

**2.4. B- Pull-Up Torque**

Pull-up torque is the minimum torque developed during the period of acceleration from locked rotor to the speed at which breakdown torque occurs. For motors which do not have a definite breakdown torque (such as NEMA design D) pull-up torque is the minimum torque developed up to rated full-load speed. It is usually expressed as a percentage of full-load torque.

**2.4. C- Breakdown Torque**

Breakdown torque is the maximum torque the motor will develop with rated voltage applied at rated frequency without an abrupt drop in speed. Breakdown torque is usually expressed as a percentage of full-load torque.

**2.4. D- Full-Load Torque**

Full-load torque is the torque necessary to produce rated horsepower at full-load speed. In pound-feet, it is equal to the rated horsepower times 5252 divided by the full-load speed in rpm.

**Full-load torque in lb-ft = (Hp x 5252) / Full-Load rpm**

__2.5 Torque, Speed and Current Relation of a motor__

In addition to the relationship between speed and torque, the relationship of motor current to these two values is an important application consideration. The speed/torque curve is repeated below with the current curve added to demonstrate a typical relationship.

Two important points on this current curve need to be examined:

**2.5. A- Full-Load Current**

The full-load current of an induction motor is the steady-state current taken from the power line when the motor is operating at full-load torque with rated voltage and rated frequency applied.

**2.5. B- Locked-Rotor Current**

Locked-rotor current is the steady-state current of a motor with the rotor locked and with rated voltage applied at rated frequency. NEMA has designated a set of code letters to define locked-rotor KVA/HP. This code letter appears on the nameplate of all AC squirrel-cage induction motors. KVA per horsepower is calculated as follows:

**For three-phase motors:**

KVA/HP = √3 x current (in amperes) x volts / (1000 x Hp)

**For single-phase motors:**

KVA/Hp = current (in amperes) x volts / (1000 x Hp)

The locked-rotor kilovolt-ampere-per-horsepower range includes the lower figure up to, but not including, the higher figure. For example, 3.14 is letter “A” and 3.15 is letter “B”.

By manipulating the preceding equation for KVA/Hp for three-phase motors the following equation can be derived for calculating locked-rotor current:

**LRA = (1000 x Hp x Locked-Rotor KVA/Hp) / (√3 x Volts)**

This equation can then be used to determine approximate starting current of any particular motor. For instance, the approximate starting current for a 7 1/2 Hp, 230 volt motor with a locked-rotor KVA code letter G would be:

**LRA = (1000 x 7.5 x 6.0) / (√3 x230) = 113 Amps**

In the next Topic, I will continue explaining the

**AC Motors Selection Procedures**. So, please keep following.

**Note:**these topics about Motors in this course EE-1: Beginner's electrical design course is an introduction only for beginners to know general basic information about Motors and Pumps as a type of Power loads. But in other levels of our electrical design courses, we will show and explain in detail the Motor and Pumps Loads calculations.

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