Lecture 1 Magnetic Coupling of Multiple Coils In this lecture, we shall develop the space dependent NxN inductance matrix [L(m)] which embeds the magnetic coupling of N coils in an electric machine. The magnetic energy Wm(m) is computed and since the syst

Lecture 3
dependence of Inductance due to Geometry of Iron
Case 1. Cylindrical stator and rotor; Stator winding has inductance LS ( m )
Point A on the round rotor makes angle m with stator winding axis.
Ls ( m )
m
A
m
The stator inductance LS ( m ) is in

Set A
Notation: Lower case, such as v, represents instantaneous quantity and bold upper case, such as V, represents phasor quantity. 1. A 3-phase balanced load is connected to a 3-phase balanced power supply through identical transformers of 1:1 turns rat

Set A
Notation: Lower case, such as v, represents instantaneous quantity and bold upper case, such as V, represents phasor quantity. 1. A 3-phase balanced load is connected to a 3-phase balanced power supply through identical transformers of 1:1 turns rat

PROJECT 2 The motor parameters of an induction machine are: RS=0.3 ohm, RR=0.15 ohm, M=0.035 H, S =0.0015H, R =0.0007H. Part I In the - frame, the electrical equations can be put in the form: e = [ R ]i + [ L] pi + nm [G ]i + [ F ]i where
RS 0 [R] = 0 0

PROJECT 1 From the d-q model of the 2-pole induction motor, predict the currents, speed and torque of the induction motor. The motor parameters are: RS=0.3 ohm, RR=0.15 ohm, M=0.035 H, s =0.0015H, R =0.0007H. At t=0, the speed is zero and the currents are

1
ECSE 462
PRACTICE SESSION IV
Trigonometric Identities cos(A+B)=cosAcosB-sinAsinB cos(A-B)=cosAcosB+sinAsinB sin(A+B)=sinAcosB+cosAsinB sin(A-B)=sinAcosB-cosAsinB In Lecture 6, a current ia of the a-phase winding having an airgap B-field Ba = ia bm cos a

ECSE 462
PRACTICE SESSION III October 2009
Refer to Lecture 12. Question 1: From Fig. 1 below show that airgap power is synchronous speed multiplied by the torque.
RR m
RS
j
S
j
R
+
~ ES
~ IS
j M
~ IR
+
~ ER m
Fig. 1 is developed from (10)
~ ~ ~ E S = ( R

ECSE 462 PRACTICE SESSION II
Question 1
r axis
s axis
r axis
m
s axis
Fig. 1 electric machine with - frame windings. The current vector of the electric machine in the - frame is i
iS i S = iR iR
From equation 14 of Lecture 9, the mutual inductance matri

ECSE 462
PRACTICE SESSION I (2009)
The exercises are for practice only. Do NOT submit.
Trigonometric Identities cos(A+B)=cosAcosB-sinAsinB cos(A-B)=cosAcosB+sinAsinB sin(A+B)=sinAcosB+cosAsinB sin(A-B)=sinAcosB-cosAsinB e jx = cos x + j sin x
1. A voltage

1 DEPARTMENT OF ELECTRICAL AND COMPUTING ENGINEERING FINAL EXAMINATION
ELECTROMECHANICAL ENERGY CONVERSION
ECSE-462B
Examiner: Professor B.T.Ooi Associate Examiner: Professor F.D.Galiana Time: AM (3 hours) INSTRUCTIONS: OPEN BOOK EXAMINATION Answer all si

Solution of Final Exam of ECSE 462 / fall 2004
Prepared by Changling Luo
Question 1 (a) K S = u z i S B S sin(2 )
H d l = J dA =
0
K S u z Rd = u z i S B S sin( 2 ) u z Rd = Ri S B S sin( 2 )d
0
0
1 1 = Ri S B S cos(2 ) = Ri S B S [1 cos(2 )] = g [H (

DEPARTMENT OF ELECTRICAL AND COMPUTING ENGINEERING
FINAL EXAMINATION
ELECTROMECHANICAL ENERGY CONVERSION
304-462B
Examiner: Professor B.T.Ooi Time: PM (3 hours) INSTRUCTIONS: OPEN BOOK EXAMINATION Answer all six questions. Trigonometric Identities cos(A+B

ASSIGNMENT 4 (2009) Question 1. The inductance matrix of a 3-phase stator and a single-phase rotor is:
Sa L11 L Sb = 21 Sc L31 L R 41
L12 L22 L32 L42
L13 L23 L33 L43
L14 iSa L24 iSb L34 iSc L44 iR
The inductance is:
1 1 3 2 [L] = 0 LR (BSk )2 2 1 gk 2

1 ASSIGNMENT 3 Trigonometric Identities cos(A+B)=cosAcosB-sinAsinB cos(A-B)=cosAcosB+sinAsinB sin(A+B)=sinAcosB+cosAsinB sin(A-B)=sinAcosB-cosAsinB
Question 1 The inductance of a single-phase reluctance machine is:
La = [ L A + LB cos 2 d ]
The single-pha

1 ASSIGNMENT 2
Question 1
A where A is the area of the plates enclosing the x dielectric between the separation x . Derive the formula of the force f e acting on the plates based on electrostatic energy We. Consider a conservative surface enclosing the ca

ECSE -462B Assignment 1 1. A balanced 500V, 60 Hz 3-phase voltage supply feed a balanced 3-phase resistive load of 2 ohm each. The voltage across the 2 ohm resistor has to be 110V. Therefore, 3-phase voltage step-down transformers are needed. Three single

Lecture 4 RELUCTANCE MACHINES The inductances in the 2x2 inductance matrix are:
Lxx = [ L A + LB cos 2( d x )]
L yy = [ L A + LB cos 2( d y )]
L xy = L yx = [ L A (cos( x y ) + LB cos(2 d x y )]
Single-Phase Reluctance Machine Consider the x-winding as th

1 LECTURE 5 This lecture moves from reluctance machines to induction machines. The formulae of the coupling between winding-x and winding-y which have been developed in chapter 3 will be applied. The resultant equations are simpler because induction machi

Lecture 6 Induction Machine Equations in a-b-c frame Having developed the 6x6 inductance matrix of the 3 stator windings magnetically coupled to the 3 rotor windings, they are assembled here with the 3x3 matrices of the stator and rotor resistances. The s

Lecture 2. Variable Reluctance Machine
Example 1
i
N turns
x
fx
armature
A C-core electromagnet has a winding of N-turns. What is the force acting in the xdirection on an iron armature? Assume the iron of the C-core and the armature to have very high rela

1
Lecture 21
From Lecture 20
i S i S x = iR (8) i R m
i S i S and i = iR iR
(9)
equations (6) and (7) can be rewritten as
i A11 p = m A21
A12 i A22 m
(10)
where the 4x4 [A11] matrix is:
[ A11 ] = [L]1cfw_[R] + nm [G] + [F ] where the 4x1 matrix [A12

1
ECSE 462 Lecture 20 Small Perturbation Linearization of Induction Machine Equations From Lecture 11, the - frame electrical equations are:
e = [R]i + [L] pi + nm [G]i + [F ]i where
e S e e = S e R eR RS 0 [ R] = 0 0 0 RS 0 0 0 0 RR 0 0 0 0 RR iS i i =

ECSE 462 Lecture 19 Revisions of Linear Algebra Consider an N-tuple vector
x1 x 2 x= . x N 1 xN Let the linear equation to be solved be
dx = [A]x + [ B ]u dt (2)
(1)
where [A] is a NxN real constant matrix and [B] is a NxM real constant matrix and u is a

Lecture 18 Induction Machine Torque Transients at Zero Speed
From (28) of Lecture 29, the solutions of the stator and rotor currents of the d-phase are of the form:
R 0 C1 Q1SS cos(t + 1 ) iSd (t ) 1 1 1 exp( 2 M + l t ) + i (t ) = 2 1 1cfw_ R C2 Q2 SS c

Lecture 17 Engineering practice uses many numerical methods and software tools for the analysis. Therefore the emphasis is to make students understand the theory in order to apply the available software tools. This lecture introduces linear analysis of th

Lecture 16 Although numerical integration can solve nonlinear equations, the solutions are not easy to understand, even after post-processing. The approach is to find ways to approximate the nonlinear equations with linear equations. This course presents:

1 Lecture 15 SIMULATION RESULTS
The results in this lecture are based on numerically integrating dx = f ( x , u ) , modeling the
dt
induction motor with short-circuited rotor windings. The initial states are x (0) = 0 . Current Pattern Fig. 1 shows the fi

LECTURE 14 Numerical Integration Numerical Integration proceeds from time t 0 to the next time instant t = t0 + h where h is the step-size. For high accuracy, the step-size must be kept small. In the Runge-Kutta method of integrating dx = f ( x , u ) for