EEL4205 Fall2012 Exam 1
Name: _
Problem 1. (50 points)
Consider the magnetic circuit shown below. Here, we neglect any flux outside the core area and
suppose that the air gap has 5% fringing.
(1) Find the reluctance of the core;
(2) Find the reluctance of
EEL4205 Exam 2
Name:_
Problem 1. (50 points)
A 60 Hz single phase transformers equivalent circuit is shown in the Figure. It has a nameplate
voltage rating of 5-kV:50-kV (turns ratio is 1:10). An open-circuit test is conducted from the
low-voltage primary
Magnetic Circuits and Materials
Ampere's Circuital Law
dl
I L
H
H dl I
L
enclosed
length d
H dl
NI
Hd NI NI H d NI B d
Constitutive Relationship
In isotropic media
r
B H 0 r H
permeability relative permeability
r 1, 1 : diamagneti c material r 1 : v
Magnetic Circuits (II)
Example 3
r=2000
5% cross-section increase for
fringing in airgap
Find: (a) total reluctance of the flux path;
(b) current required to produce B = 0.5 T in the air gap;
(c) inductance of the coil.
Example 4
N = 400, A = 150 cm2
lc =
Three Phase Circuits (II)
Analysis of Three Phase Balanced
Circuits: Y-Y Connection (1)
Analysis of Three Phase Balanced
Circuits: Y-Y Connection (2)
A
a
B
VAN
b
VBN
N
n
VCN
C
c
Analysis of Three Phase Balanced
Circuits: Y-Y Connection (3)
Per phase circu
Faradays Law
Faradays Law (I)
d
d
E dl = dt B dS = dt
C
S
B
dl
C
n
E
S
Electromotive Force (emf)
- V
emf = E dl
C
Total Magnetic Flux
=
- Wb
B dS
S
Faradays Law (II)
d
d
E dl = dt B dS = dt
C
S
B
dl
C
n
E
S
The emf generated around a closed contour C
Introduction to Transformer
UCF
Introduction to Transformer (1)
Transformer: A device that changes AC electric power at one
voltage level to ac electric power at another voltage level
through the action of a magnetic field.
Shell-form transformer
UCF
Core
Non-ideal Transformer
UCF
Non-Ideal Transformer (1)
(1) r
Rcore = N pi p N sis
=
N pi p N sis
d
d N pi p N sis
= Np
= Np
vp
Rcore
dt
dt
2
Np d
Ns
is
ip
Rcore dt
Np
Rcore
Non-Ideal Transformer (2)
UCF
2
Np d
Ns
ip
is
vp =
Np
Rcore dt
Defi
Transformer Voltage Regulation
UCF
Transformer Voltage Regulation (1)
Ideal:
VP
n
VS
n
Non-ideal:
NP
NS
UCF
Transformer Voltage Regulation (2)
No load (open circuit):
Full load : VP
n
VP
VSnl
n
VSfl ( Reqs jX eqs )I S
Voltage regulation: VR
VSnl VSfl
V
Autotransformer
UCF
Autotransformer (1)
Autotransformer (2)
UCF
VC
= n,
VSE
VL = VC
VH = VC + VSE
I L = I C + I SE
I H = I SE
I SE
=n
IC
VC
n
VL
=
=
= n'
VH
VC + VSE n + 1
I C + I SE n + 1 1
IL
=
=
=
IH
I SE
n
n'
n' =
n
n +1
Autotransformer (3)
UCF
VL = V
Cheat Sheetfor Exam 1
clear all; close all; clc;
cm = 0.01; % 1 cm = 0.1 m
mu0 = 4*pi*10^(-7);
-Current that will produce a flux of _ Wb.
Find the Reluctance R for each side:
R = l/(mu*A)
*mu = mur*mu0
*A = depth*thickness
Find total Reluctance R
% Add at beginning of code:
% d2r = @(x) (x*pi/180);
% r2d = @(x) (x*180/pi);
To convert polar to rectangular
Given: 7.07 angle -36.87 = 7.07cos(377t-36.87)
-Matlab: 7.07*exp(1j*d2r(-36.87) and then run the code.
-Matlab: [real(Is) imag(Is)]
-Calc.: (
% Fall 2013 Exam 2 Problem 1
clear all; close all; clc;
% Input open circuit and short circuit testing results:
kV = 1000;
kW = 1000;
absVOC = 7.2*kV/sqrt(3); % Open circuit phase voltage
absIOC = 16; % For Y connection, the line current is equal to phase
Chapter 3: Transformers
3-1.
The secondary winding of a transformer has a terminal voltage of v s (t ) = 282.8 sin 377t V . The turns
ratio of the transformer is 50:200 (a = 0.25). If the secondary current of the transformer is
i s (t ) = 7.07 sin ( 377t
Three Phase Circuits (I)
Time-harmonic Signal
v (t ) 2V cos(t v ) V and I are rms values.
i (t )
i (t ) 2 I cos(t i )
j v
v (t ) Re[ 2Ve e
jt
+
]
v(t )
_
Phasor Domain
V Ve
ji
i (t ) Re[ 2 Ie e
I Ie
j v
jt
ji
or
V V v
rms
I Ii
rms
]
or
Z (t )
Instantaneou
Magnetic Circuits (I)
Amperes Circuital Law
I
L
dl
H dl I
H
enclosed
L
length d
H dl
NI
Hd NI
NI
H
d
NI
B
d
Constitutive Relationship
In isotropic media
B H 0 r H
r
permeability
relative permeability
r 1, 1 : diamagneti c material
r 1 : vacuum
EEL4205 Fall2012 Exam 1 Problem 1. (50 points)
Name: _
Consider the magnetic circuit shown below. Here, we neglect any flux outside the core area and suppose that the air gap has 5% fringing. (1) Find the reluctance of the core; (2) Find the reluctance of
EEL4205 Exam 2
Name:_
Problem 1. (50 points) A 60 Hz single phase transformer's equivalent circuit is shown in the Figure. It has a nameplate voltage rating of 5-kV:50-kV (turns ratio is 1:10). An open-circuit test is conducted from the low-voltage primar
EEL4205 Exam 3 Problem 1. (50 points) A 480 V (terminal voltage), 500 kVA, 60 Hz, four pole, Y connected synchronous generator has a synchronous reactance of 0.08 and an armature resistance of 0.02 . At 60Hz, its friction and windage losses are 8 kW, and
EEL4205 Final Exam
Fall2012
1. (20 points) A 480 V (terminal voltage), Y-connected, 60 Hz, synchronous motor has a
synchronous reactance of 2.5 and a negligible armature resistance. Its friction and windage losses are 2.2kW and its core losses are 0.8kW.
Department of Electrical Engineering and Computer Science
EEL4205 Electric Machinery
Homework Set No. 5 Names: (1)_; (2)_ (3) _
Problem 1. A 4 pole motor is supplying 60 N.m torque to its load. The shaft speed of the motor is 4000 rpm. The rotor outer dia
Department of Electrical Engineering and Computer Science
EEL4205 Electric Machinery
Homework Set No. 6 Names: (1)_; (2)_ (3) _
Problem 1. A 2300 V (terminal voltage), 1000 kVA, 0.8 PF lagging, 60-Hz, two pole, Y connected synchronous generator has a sync
Department of Electrical Engineering and Computer Science
EEL4205 Electric Machinery
Homework Set No. 7 Names: (1)_; (2)_ (3) _
Problem 1. A 480 V, 60 Hz, 400 hp, six pole, connected synchronous motor has a synchronous reactance of 1.1 and negligible arma
Department of Electrical Engineering and Computer Science
EEL4205 Electric Machinery
Homework Set No. 8 Names: (1)_; (2)_ (3) _
Problem 1. A 220 V, three phase, two pole, 50 Hz induction motor is running at a slip of 5 percent. Find: (1) the speed of the
EEL4205 Exam 3
Problem 1. (50 points)
A 480 V (terminal voltage), 500 kVA, 60 Hz, four pole, Y connected synchronous generator has a
synchronous reactance of 0.08 and an armature resistance of 0.02 . At 60Hz, its friction and
windage losses are 8 kW, and
EEL4205 Final Exam
Fall2012
1. (20 points) A 480 V (terminal voltage), Y-connected, 60 Hz, synchronous motor has a
synchronous reactance of 2.5 and a negligible armature resistance. Its friction and windage
losses are 2.2kW and its core losses are 0.8kW.
Introduction
UCF
EEL4205 Electric Machinery
Course website:
http:/www.eecs.ucf.edu/~tomwu/course/eel4205/eel4205.htm
Or Google: Thomas Wu
Required Software:
MatLab/SimuLink Student Version (2010a and later)
MatLab tutorial website:
http:/www.mathworks.com