machines 3 - lec_004-power relations [compatibility mode]

Lecture Outlines

Quiz (2)

Power Losses in Induction Machines

PowerFlow Diagram

Tanta University Faculty of EngineeringElectrical Power and Machines Engineering Department EPM3215 Electrical Machines (3) Dr. Said M. Allam

PowerFlow Diagram

Power Relations

Examples

Quiz (2)1. A 3-phase, 4-pole, 50 Hz induction motor runs at a

speed of 1440 rpm. The rotating field produced by therotor rotates at a speed of rpm with respectto the rotor

a) 1500 b) 1440 c) 60 d) 0


a) 1500 b) 1440 c) 60 d) 0

2. In a 3-phase induction motor, the rotor field rotates atsynchronous speed with respect to

a) stator b) rotor c) stator flux d) none of them

Quiz (2)3. Which of the following rotor quantity in induction motor

does not depend on its slip?

a) reactance b) speed c) induced emfd) frequency e) none of them


4. A 3-phase, 6-pole, 50 Hz induction motor has a full loadspeed of 950 rpm. At half load, its speed would be rpm

a) 475 b) 500 c) 975 d) 1000

Quiz (2)5. A 3-phase, 8-pole, 60 Hz induction motor is driven at

1800 rpmby a prime mover in theopposite directionofrevolving magnetic field. The frequency of the rotorcurrent is

a) 60Hz b) 120 Hz c) 180Hz d) noneof them


a) 60Hz b) 120 Hz c) 180Hz d) noneof them

Power Losses in Induction Machines

Copper losses Copper loss in the stator (Pcu1) = I12R1 Copper loss in the rotor (Pcu2) = I22R2

Core loss (P )


Core loss (Pcore)

Mechanical power loss due to friction andwindage

Power Relations

Since the load resistance varies with the slip and the

slip adjusts itself to the mechanical load on the motor,

the power delivered to the load resistance is equivalent

to thepowerdevelopedby themotor.


to thepowerdevelopedby themotor.

For a balanced 3-phase induction motor, the powerinput is

Where is the input power factor

( )= cosI3VP 11in

Power Relations

The total stator copper loss is

The total core loss can be given by:

121cu1 R I 3P =


The net power that is crossing the air gap is transported

to the rotor by electromagnetic induction is calledthe

air gap power

c2cc R I 3P =

Power Relations

The air-gap power is

The air-gap power must also equal to the powerdeliveredto thehypotheticalresistance. Thatis

ccu1inag PPPP =


deliveredto thehypotheticalresistance. Thatis

The electrical power developed by the rotor is

sR I 3P 222ag =

ag222cu2 P s RI 3P ==

Power Relations

The power developed by the motor is

( ) ( ) ag222cu2agd

P s1ss1R3I

PPP

==

=


The electromagnetic torque developed by the motor is

ag22

s222sagmdd sR I 3PPT ===

Power Relations

From the previous power relations, the followingrelation can be obtained

P:P:P


s1:s:1

P:P:P dcu2ag

Power Relations

By subtracting the rotational power fromthe developed

power, the output power can be obtained

rotdout P -PP =


The per unit efficiency of the induction motor is then

given by:

rotdout P -PP =

in

out

P

P=

Example (1)A 480-V, 60 Hz, 50-hp, three phase induction motor is

drawing 60A at 0.85 PF lagging. The stator copper lossesare 2 kW, and the rotor copper losses are 700 W. Thefriction and windage losses are 600 W, the core losses are1800 W, and the stray losses are negligible.Find thefollowing quantities:


following quantities:

1. The air-gap power

2. The developed power

3. The output power

4. The efficiency of the motor

Example (2) A 460-V, 25-hp, 60 Hz, four-pole, Y-connected inductionmotor has the following impedances inohmsper phasereferred tothe statorcircuit:

R1= 0.641 R2= 0.332 X1= 1.106 X2= 0.464 Xm= 26.3 The total rotational lossesare 1100 W and are assumedto be


The total rotational lossesare 1100 W and are assumedto beconstant. The core loss is lumped in with the rotational losses. Fora rotor slip of 2.2 percent at the rated voltage and rated frequency,find the motors

1. Speed 4. Developed and output power

2. Stator current 5. Developed and load torque

3. Input Power factor 6. Efficiency

machines 3 - lec_004-power relations [compatibility mode]

Documents