EE321 Spring 2015
Homework 11
Problem 1 Reluctance Machine (ECE595 Only)
In qd0 variables in the rotor reference frame, the flux linkage equations of a
certain reluctance machine may be expressed:
r
r
qs = 4iqs
r
r
ds = 10ids
0 s = i0 s
Express the mutual
Lecture Set 5
Distributed Windings and
Rotating MMF
S.D. Sudhoff
Spring 2015
Distributed Windings and
Rotating MMF
Objective
In this chapter, we will set the stage to study ac
machinery including permanent magnet
synchronous machines as well as induction
EE321 Spring 2015
Homework #12
Problem 1
500 e
Xphasor :=
5+
1
20
( 200j) +
Xphasor = 3.884
(
j
1
300
( 200j)
2
Xphasor 2 = 5.493
)
arg Xphasor = 2.064
x=3.884*sqrt(2)*cos(200t-2.064)
If the cosine term were a sine, I simply would have adjusted the phase
EE321/ECE595 Spring 2015
Homework 10
Problem 1 Unbalanced MMF
The conductor turns density of a two phase machine is given by
n as = 100 cos 2 sm
n bs = 100 sin 2 sm
The a- and b-phase currents are given by
i as = 5 cos(400t )
ibs = 4 sin( 400t )
Express t
EE321/595 HW#9
Problem 1
Note: I have adjusted array indices to start as 1 in this sheet
T
Nas := ( 0 4 4 0 4 4 0 4 4 0 4 4 )
T
Nbs := ( 4 4 0 4 4 0 4 4 0 4 4 0 )
T
Ncs := ( 4 0 4 4 0 4 4 0 4 4 0 4 )
By inspection
Nsslt := 12
P := 4
Now we have
Nsslt
1
Wa
ECE321/ECE595
Homework 9
Problem 1 Discrete Winding Function
The number of conductors in each slot of the a-phase of the stator of the machine
are as follows:
N as = [0 4 4 0 4 4 0 4 4 0 4 4]T
Compute and graph the winding function associated with this wi
EE321 Spring 2015 HW8
Problem 1
This example illustrates the torque speed curve of a permanent magnet dc machine fed from
a buck converter.
Parameters:
ra := 1
kv := 0.05
3
Laa := 3 10
vdc := 20
f := 1.0 10
vfsw := 1
3
Armature current
i acont( r , d ) :=
ECE321/ECE595 Spring 2015
Homework 6
Problem 1 Shunt Connected DC Machine
The parameters of a shunt connected dc machine are ra = 10 , Rf = 40 , and LAF = 0.5 H.
Neglect Bm. The applied voltage is 30 V. Calculate (a) the steady-state stall torque, (b) the
llm ;
ECE321/ECE595 Exam 1
Spring 2015
Notes: You must Show work for credit (except for problem 4).
This exam has 4 problems and 10 pages.
Note that problems 1 and 2 are different depending on if you are in ECE321 or
ECE595
Good luck! 1) 25 pts. Consider
Sowélw 7
Name:
Class: 321 or 595
ECE321/ECE595 Exam 3
Spring 2015
This exam is closed book, and closed lecture notes.
You are allowed to have both sides of one 8.5” by 11” crib sheet.
You may only use a Texas Instruments TI—30X IIS scientiﬁc calculator.
Y
EE321 Spring 2015
Homework 12
Problem 1 Phasor Analysis
Consider the differential equation
5x +
1 dx
1 d 2x
+
= 500 2 cos(200t + 1)
20 dt 300 dt 2
where all arguments of the cosine term are in radians. Use phasor analysis to find the steady-state solution
EE321 Spring 2015
Homework 13
Problem 1 Rotating MMF
Using the configuration we studied in class (Figure 6.2-1 in text), the stator currents of a 4 pole
machine are given by
ias = 50 sin( 200t )
ibs = 50 cos(200t )
The speed of the machine is 500 rpm in t
ECE321/595 Spring 2015
Homework 14
Problem 1 Steady-State Operation
Consider a delta-connected 3-phase machine with the following parameters: rs = 72.5 m ,
Lls = L'lr = 1.32 mH, Lms = 20.1 mH, rr' = 41.3 m , and P = 4 . The load torque varies with the spe
Lecture Set 1
Introduction to
Magnetic Circuits
S.D. Sudhoff
Spring 2015
1
Goals
Review physical laws pertaining to analysis
of magnetic systems
Introduce concept of magnetic circuit as
means to obtain an electric circuit
Learn how to determine the flu
Lecture Set 6
Brushless DC Machines
S.D. Sudhoff
Spring 2015
Reading
Chapter 8, Electromechanical Motion
Devices, 2nd Edition
2
A Brushless DC Machine
3
Sample Applications
Low Power:
Disk drive motors
Medium Power:
Robot manipulators
Servo systems
Hyb
EE321 Spring 2015
HW#11
Problem 1
Problem 2
Machine Parameters
rs := 2
m := 0.15
3
Lss := 10 10
N := 3
P := 4
Load
2
r P
TL( r) := 2
250
3.2
Source
vs := 100
v := 0
vq :=
( )
2 vs cos v
( )
vd := 2 vs sin v
Ok - lets solve the problem
(
)
rs vq r m
EE321 Spring 2015
Homework #13
Problem 1
We have:
i as = 50 sin( 200 t )
i bs = 50 cos( 200 t )
Now, we will also have
(
)
(
)
was = W cos 2sm
wbs = W sin 2 sm
Thus, the stator MMF is
(
)
Fs = 50 W sin 200 t 2 sm
Whereupon the stator MMF is moving at 100
Lecture Set 7
Transformers
S.D. Sudhoff
Spring 2015
Reading
Chapter 1, Electromechanical Motion
Devices, 2nd Edition Section 1.5
2
Review of Phasor Analysis
3
Review of Phasor Analysis
Basis of phasor analysis:
d au
e = ae au
du
e j = cos + j sin
4
Rev
Lecture Set 4
DC Machines and Drives
S.D. Sudhoff
Spring 2015
Reading
Electromechanical Motion Devices, 2nd
Edition
Sections 3.1-3.9
2
General Comments on DC Machines
Attractive Features
Drawbacks
3
DC Machine Cutaway View
4
DC Machine Cutaway View
5
A
Lecture Set 2
Energy Conversion
S.D. Sudhoff
Spring 2015
Reading
Sections 2.1-2.6
2
Overall Goal
We know how to
(1) flux linkage in terms of current and position
(2) current in terms of flux linkage and position
Using either (1) or (2), we can find an
ECE321/ECE595 Spring 2015 HW#10
Problem 1
The winding functions are
(
)
(
)
was = 50 sin 2 s
wbs = 50 cos 2 s
but now the currents are
ias = 5 cos( 400t)
ibs = 4 sin( 400t)
So the total MMF is given by
(
)
(
)
F = 250 cos( 400 t) sin 2 s 200 sin( 400 t) c
Lecture Set 0
ECE321/ECE595
Electromechanical Motion Devices/
Electromechanics
S.D. Sudhoff
Spring 2015
Courses Meeting Together
Courses
ECE321 Live cfw_321L
ECE321 Video cfw_321V
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ECE595 Off Campus Pro Ed cfw_59