Motor

RPM=

(120 x frequency) / # of field poles

Describe some characteristics of Induction Motor.

* Induction motors rotates at asynchronous speed. Asynchronous speed depends on a fraction of the synchronous speed. The fraction is called the "slip". * Watch out for excessive voltage drop during start up * Size the supply transformers for handling the inrush and full load current. * Stator of an induction motor is built in the same way as the stator of a synchronous motor. However, the rotor is constructed differently.

Induction Motors Design Classes

* NEMA has assigned letters to each motor design class. The design classes are: A, B, C, D, and E and F. * Within each design classification there are other codes and classes, such as locked rotor codes and insulation classes.

Describe some characteristics of Synchronous Motor.

* Rotates in synchronous speed. * Stator is excited by AC and rotor is by DC current through brush contacts. * Rotor is pushed and pulled by stator field. * Rotor must be brought up to synchronous speed by external means. * The rotor turns because the rotor and stator magnetic fields tend to align themselves while rotating at synchronous speed.

Types E and F

- Called soft-start induction motors have very low starting currents. They are used for low starting-torque loads. These designs are now obsolete.

Type D

- Main features are: 1. Very high starting torque (more than 275% of rated torque) 2. Low starting current 3. High slip at full load 4. Used in special applications requiring the acceleration of high-inertia loads (flywheels in punch presses, elevators and hoists)

Example 2 (contd)

- Using the same parameters from the previous examples, the inrush current is calculated to be 1,513 amps [= (1258x103)/(√3·480)] - Assume the feeder load current prior to the start of the motor is 500 amps - Assume that the initial voltage is 480V (3Φ)

Type C

-More expensive than Class A or B. Main features are: 1. High starting torque but slightly lower than for Class A 2. Low starting current (250% of the full load torque) 3. Low slip (less than 5%) at full load 4. Typical applications: high starting torque loads, loaded pumps, compressors, conveyors

1 Horse power (HP) is how many foot-pound/sec or Watts?

1 HP= 550 ft-lb / s = 746 Watts

Approximate voltage-drop calculation due to motor inrush.

One approximate way to calculate the voltage drop is to use the impedance of the service transformer, service conductor, and inrush kVA of the motor − R is the total resistance per phase of the transformer in series with the service conductor in ohms − X is the total reactance per phase of the transformer in series with the service conductor in ohms − The pf is assumed to be 0.6 (rf = 0.8) at start up − V is the voltage at the motor terminals − VAinrush is the locked rotor (kVA/HP) value X motor HP X 1000

With direct mechanical drive, motor speed is determined by mechanical speed and physical dimensions

Speed (RPM) = v (ft/min) / (r (ft) x 2 x pi)

Slip=

Synchronous RPM - Actual RPM − At no load the slip is almost one − As the load increases, so does the slip − Slip can be given in rpm or percentage

What are the types of Induction Motors?

Two types: depends on how field in the rotor is constructed: * Squirrel-cage: Squirrel-Cage rotor consists of a series of conducting bars laid into slots carved in the face of the rotor and shorted out at each end by shorting rings. Usually, the bars are skewed to minimize harmonics. * Wound rotor: A wound rotor has a complete set of 3-PH windings that mirrors the stator windings. The rotor windings are usually Y-connected. The ends of the rotor windings are attached to brushes sliding on slip rings. -- Wound rotors have their rotor circuits available to the exterior. Extra resistances can then be inserted in the rotor circuits and the torque-speed characteristics of the motor can be modified.

1 Horse Power (HP) =

{Torque (lb-ft) X Speed (rpm)} / 5250

Example 2:

− 200 hp motor, Type B, Locked rotor code G − Transformer is 2.5 MVA, secondary volts at 480/277GRDY - Assume that R = 0.00395 and X = 0.03715 ohms for the transformer and the cable - Pf is assumed to be 0.6 at time of motor start - VAinrush = 1,258,000 based upon locked rotor code (6.29)

Voltage Drop in a Radial Feeder Due to Motor Start - Taking into Account Previous Feeder Load Current

− Determine the load current prior to the start of the motor − Determine the total resistance and reactance − Determine the voltage prior to the motor start − Calculate the starting current from the locked rotor equation − Substitute these values in the following formulas

1-phase LRA (Starting)=

− Ilr = Locked rotor Current − kVA/hp = from letter code table − hp = motor rating − V = line-to-ground volts

Type B

− Newer version of Type A, general purpose motor. Main features are: 1. Normal starting torque 2. Lower starting current (25% less current than Class A) 3. Pullout torque 200% or more than the rated load torque (less than Class A) 4. Rotor slip at full load is less than 5 % 5. Same applications as Class A

Type A

− Standard motor design for general purpose motor with normal starting torque, normal starting current and low slip. Main features are: 1. Full load slip must be less than 5%. 2. Pullout torque is 200 to 300% of the full load torque and occurs at low slip (less than 20%). 3. Inrush currents are from 500% to 800% of the rated current (this is the main problem) 4. Typical applications: Fans, blowers, pumps, lathes, machine tools.

3-phase LRA (Starting)=

− kVA/hp = from letter code table