EE-362 ELECTROMECHANICAL ENERGYCONVERSION-II

Equivalent Circuit of Induction Machines

Ozan Keysan

keysan.me

Office: C-113 • Tel: 210 7586

1 / 18

Assume the rotor is stationary (s=1):

Then the frequency of rotor currents:

Machine just behaves just like a transformer(secondary short-circuited)

Stator winding: Primary winding of the transformer

Rotor bars: Secondary winding of the transformer

=fr fs

2 / 18

Equivalent Circuit, Rotor Stationary (s=1)

Same as a transformer with secondary side short-circuited

3 / 18

Assume now the rotor starts rotating (with )

Rotor induced voltage reduces (as the frequencydi�erence between rotor and stator reduces)

Nr

4 / 18

Assume now the rotor starts rotating (with )

Rotor induced voltage reduces (as the frequencydi�erence between rotor and stator reduces)

Therefore current reduces,

but not that much, because the rotor sideimpedance reduces with reduced rotor currentfrequency ( jwL)

Nr

4 / 18

Stator Side Equivalent Circuit

5 / 18

Rotor Side Equivalent Circuit

Rotor side impedances can be modi�ed to transfer these to the statorside

For curious students: P.C.Sen, Principles of Electrical Machines and PowerElectronics, Section 5.7, Derivation of the equivalent circuit of inductionmachines

6 / 18

Equivalent Circuit with Referred Rotor

7 / 18

Determination of Equivalent Circuit Parameters

Equivalent tests for induction machines:

Open-circuit Test = No-Load Test

Short-circuit test = Locked Rotor Test

9 / 18

Locked-Rotor Test

Rotor kept stationary with a mechanical locking system

Apply a small voltage (10-15% of the rated) to obtain rated currentand measure:

Input Power

Input Voltage

Current

= 0 → s = 1nr

10 / 18

Locked-Rotor Test

Neglect the parallel branch and s = 1, the circuit becomes:

11 / 18

Locked-Rotor Test

Measure using an ohm-meter, assume

Assume

= = ( + )Pphase

Ptotal

3I21r1 r′

2

r1(dc)

= 1.1r1(ac) r1(dc)

= =V1

I1

Zeq ( + + ( +r1 r′

2)2 X1 X ′

2 )2

‾ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

=X1 X ′

2

12 / 18

No-Load Test

Run the motor at no-load, applying rated voltage

Again measure:

Input Power

Input Voltage

Current

≃ → s ≃nr ns0

13 / 18

No-Load Test

Rotor-side is open-circuited (s=0), the series branch (R1,X1) can alsobe neglected:

14 / 18

No-Load Test

, �nd

, �nd

But, how about mechanical friction and windagelosses?

Get a few measurements at di�erent voltages and speeds toestimate the friction.

= =Pphase

Ptotal

3

V2

1

Rc

Rc

= = //V1

I1

Zeq Rc Xm Xm

15 / 18

Example:

Estimate the parameters of a 30 kW, 50 Hz, Delta-connected, 415 V,3-phase machine, if the test results are as follows:

Locked-Rotor Test:

Input Power: 6.4 kW

Line Current: 77 A

Line Voltage: 130 V

Resistance between two lines: 0.293 Ω

16 / 18

Example:

Estimate the parameters of a 30 kW, 50 Hz, Delta-connected, 415 V,3-phase machine, if the test results are as follows:

No-Load Test:

Input Power: 1.65 kW

Line Current: 22.8 A

Line Voltage: 415 V

Friction and windage losses: 1.15 kW

Assume =X1 X ′

2

17 / 18