25-04-2012, 03:09 PM
AC Motor & AC Drive
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General
AC Motor is the main workhorse of the industry. These motors are the prime movers for various Mechanical, Pneumatic, Hydraulic & HVAC systems. 3Ø AC Induction motors contribute the 80% population of motors in industries. 1Ø AC Induction / Commutator motors hold the major share in hand held tools & automation accessories.
Motor Principle
Electric motor is a machine, which converts an electrical energy in to mechanical energy. Its action is based on the principle that, when the current carrying conductor is placed in the magnetic field, it experiences the force which may move the conductor provided a suitable engineering is applied for that. Fleming’s Left Hand Rule decides the direction of force.
In DC Motors electrical power is conducted directly to the armature whereas in Induction motor the rotor receives the electric power not by conduction but by induction.
3Ø AC Induction Motor Construction
Stator
The stator carries the 3Ø winding. It is wound for definite number of poles. When fed with 3Ø supply, produces a magnetic flux of constant magnitude, revolving at a speed known as Synchronous Speed.
Ns = (120 * F) / P, where as
Ns Synchronous Speed (Motor Input Speed)
F Frequency if Input Power Supply
P No. Of Poles
Rotor
Squirrel Cage Rotor:
The rotor consists of cylindrical laminated core with parallel slots for carrying rotor conductors. These are heavy bars of Copper or Aluminum. The rotor bars are short-circuited at both ends with the help of heavy & stout end rings.
Slip Ring Rotor (Phase wound rotor):
It comprises of three coils connected in Star fashion. These coils have variable resistances connected in a star fashion. While starting the motor this resistance is kept at highest value to get high starting torque and subsequently this is reduced to zero in the normal running condition.
Principle of Motor operation
When stator is fed with 3-phase power supply, a magnetic flux of constant magnitude and rotating with synchronous speed is set up. This flux passes through the air gap & cuts the rotor conductors. A relative speed between rotating stator flux & stationary conductors of Rotor induces an emf in rotor conductors. Frequency of induced emf is same as that of supply frequency. Its magnitude is proportional to the relative velocity between rotating flux & the conductors. Since the Rotor conductors are short-circuited, rotor current is produced whose direction as given by Lenz’s law in such a way so as to opposes the cause that produces it. So the rotor starts rotating in the direction of the rotating flux to reduce the relative speed between the stator flux & rotor.
Slip:
The difference between synchronous speed of stator field & rotor speed or actual speed is known as slip.
% Slip S = (Ns-Nr) / Ns x 100
• Ns Synchronous speed of rotating flux produced by ‘3 Phase’ supply given to Stator Coils
• Nr Rotor speed
• (Ns – Nr) Slip speed.
Exercise 1
A ‘3Phase’ 4-pole induction motor is supplied with 50 Hz supply, calculate the synchronous speed of the motor. If slip is 4% calculate the rotor speed.
Ans:
Ns= (120*f)/P = (120 * 50)/4 = 1500 RPM. It is the synchronous speed.
If slip is 4% then
0.04 = (1500 – N) / 1500
60 = 1500 – N
Hence,
N = 1500 – 60 = 1440 RPM is the rotor speed.
Torque:
Torque is turning or twisting moment of a force about an axis.
T = F x r Newton-meter (N-m)
Work done by this force in one revolution
W = Force x distance
= F x 2πr (joule).
Power developed
P = F * 2πr * m (Joules / second = Watts)
P = (F * r) * 2π * m (watt). (Where m is r.p.s.)
P = T * 2π * m (watt). (Where m is r.p.s.)
P = T * 2π * (N / 60) (watt). (Where N is r.p.m.)
T = (60 * P) / (2πN)
T = 9.55 * P / N (N-m)
Torque in AC induction motor is proportional to the product of flux & rotor current.
T α Ф I2 CosØ
Starting torque of the squirrel cage motor is poor i.e. 1.5 times its full load torque. Though the starting rotor current I2 is high, it lags behind induced rotor emf by a very large angle.
Torque is sensitive to any changes in the supply voltage.
T α V2
Torque-Speed Curve
1: Locked rotor torque
2: Pull up torque
3: Break down torque
4: Full load
Power stages in An Induction Motor
Stator Input Pi = Stator output Pr + Stator losses (i.e. Iron losses & Cu Losses)
Rotor Input Pr = Mechanical Power Pm + Rotor Cu Losses
Mechanical Power at Shaft Pout = Mechanical power Pm – Friction & Windage Losses
T = 9.55 x P / N
If the gross output of the rotor of the induction motor is Pm, & its speed is N rpm, then gross torque Tg developed is
Tg = 9.55 x Pm / N (N-m)
Tsh = 9.55 x Pout / N (N-m)
Exercise 2:
A 50Hz, 8-pole, 3-phase induction motor delivers an output of 3.1kW, at the slip of 0.02. Friction & Windage losses are of 100W. Find the value of gross torque in N-m.
Ans:
Ns = (120 * 50) / 8 =750 RPM.
The rotor speed will be 0.02 * 750 = 750 – N
N = 735 RPM
Pg = Pout + (Windage & friction loss)
= 3100 + 100 = 3200
Hence
Tg = 9.55 * 3200/735 = 41.58 N-m
Mechanical Power
1 H.P = 550 ft. lb / sec = 746 watts (FPS System)
1 K.W = 1.34 H.P.
1 H.P = 75 Kg. meter / sec = 735 watts (MKS System)
1 K.W = 1.36 H.P.
Electrical Power
Single phase: P = VL * IL * CosØ watts
Three Phase: P = √3 * VL * IL * CosØ watts
Efficiency
It is expressed in %(percentage)
It is given as Efficiency = (Output Power / Input power) * 100
Exercise 3:
A pump is driven by a 3-phase 400V AC Motor. It is throwing 1600 liters of water at a height of 70 meters per 25 minutes. If the efficiency of the pump is 70%, find out the HP of the motor.
Ans:
Pump is doing 1600 * 70 Kg M work in 25 * 60 sec
So in one sec., it will do (1600 * 70) / (25 * 60 = 74.67Kg.m work.
Now to do 75 Kg.m work per second 1 HP power is required
So to do 74.67 Kg.m work in one sec 74.67 / 75 = 0.9956 HP power is necessary.
Now, as pump efficiency is 70%
Input provided by motor to pump will be
Output / Input *100 = 70
Hence Input to pump = 0.9956 *100 / 70 = 1.42HP
So 1.42HP motor will be required to carry out this work.
Exercise 4:
A 3-phase 440V induction motor takes a line current of 30A at power factor of 0.9. If the efficiency of the motor is 0.8, find the motor output in H.P.
Exercise 5:
A 3 Phase 50Hz induction motor has eight poles. If full load slip is 2.55%, find the synchronous speed & Rotor speed.
Exercise 6:
A pump is to throw 700 liters of water per minute at the height of 60 meters. Find out the HP of the motor if the pump efficiency is 80% and motor efficiency is 85%. Also if the power factor is 0.8, & voltage is 400V, find the full load current.
Operating Features of Induction Motor
•Starting
•Speed Control
•Direction Change
•Stopping
•Braking
Starting of Induction Motor
When the voltage is aplied to the stationary motor, a very large initial current is taken by short circuited rotor conductors as there is no opposition at the start for a short time . It is normaly 5 to 7 times of full load current. This initial inrush current is not desirable as it produces a large line voltage drop, which in turn would affect the other electrical equipments connected to the same line.
So higher HP rating motors should not be switched on directly. In case of Slip Ring motors , resistance can be added to the rotor cuicuit that will limit the inrush current, but in case of squirrel cage motor, this initial current can be controlled by appling a reduced voltage to the stator at the start & full votage is applied as the motor gains the sufficient speed. There are certain starting methods that are used for this purpose.
Starting Methods
• DOL(Direct On Line) starting:This method is used for the motors up to 5HP rating.
• Star Delta Starting:This method is used for the motors above 5HP.
• Auto Transformer.
• Soft Starters.
• AC Drives.