ECE 330 homework assignment #3, due on Friday, March 21, 2014
Problem 1
A 15 kVA, 2200/220 V transformer has Rc = 195 and Xm = 170 referred to the LV side. Req
and Xeq referred to the HV side are 6 and 10 , respectively. The transformer is supplying a loa
ECE330 Spring 2014
Problem 1 (60 pts)
i
a
a
a
N
Depth d into page
Riron
Ni
g
x
Rgap
a
Riron =
lc
R gap =
r 0 (ad )
2x
(x < a, d)
0 (a + x )(d + x )
R ( x ) = Riron + R gap
a)
N 2i
=
=
R(x )
Ni
=
R (x )
N 2i
lc
r 0 (ad )
2x
0 (a + x )(d + x )
R( x ) 2
ECE330 Spring 2014
Problem 1
+ I1
V1
Req
Rc
I2/a
jXeq
V2
jXm
Load
_
10:1
Turns ratio a = 2200/220 = 10.
Rc' = a 2 Rc = 19500
'
X m = a 2 X m = 17000
Assume V2 = 2150 0 V,
Load current:
10000
I2 =
cos 1 (0.96 ) = 46.51 16.26 0 A
215
Primary voltage:
I
V
ECE330
Power Circuits and Electromechanics
Dr. Nam Nguyen-Quang
Spring 2014
http:/www4.hcmut.edu.vn/~nqnam/lecture.php
[email protected]
Lecture 1
1
Review of power
Assuming sinusoidal voltage and current, i.e.
v(t ) = Vm cos(t + v )
i (t ) = I m cos(t +
ECE330
Power Circuits and Electromechanics
Dr. Nam Nguyen-Quang
Spring 2014
http:/www4.hcmut.edu.vn/~nqnam/lecture.php
Lecture 3
1
Transformers Introduction
Transferring electrical energy from one circuit to another through
time-varying magnetic field.
Ap
ECE330
Power Circuits and Electromechanics
Dr. Nam Nguyen-Quang
Spring 2014
http:/www4.hcmut.edu.vn/~nqnam/lecture.php
Lecture 4
1
Electromechanical system Introduction
Magnetic circuits with one moving member will be studied.
Systematic derivation of mat
ECE330
Power Circuits and Electromechanics
Dr. Nam Nguyen-Quang
Spring 2014
http:/www4.hcmut.edu.vn/~nqnam/lecture.php
Lecture 9
1
DC machines Introduction
DC machine is a versatile device having superior torque-speed characteristics.
The speed control is
ECE330
Power Circuits and Electromechanics
Dr. Nam Nguyen-Quang
Spring 2014
http:/www4.hcmut.edu.vn/~nqnam/lecture.php
Lecture 7
1
Synchronous machines Introduction
Synchronous machines are mainly used as three-phase generators in power
systems. Power can
ECE330
Power Circuits and Electromechanics
Dr. Nam Nguyen-Quang
Spring 2014
http:/www4.hcmut.edu.vn/~nqnam/lecture.php
Lecture 8
1
Induction machines - Introduction
The most widely used machine in the industry, mainly as a motor. Both stator
and rotor car
ECE330
Power Circuits and Electromechanics
Dr. Nam Nguyen-Quang
Spring 2014
http:/www4.hcmut.edu.vn/~nqnam/lecture.php
Lecture 2
1
Introduction
Electromagnetic theory: basis for explaining the operation of all
electrical and electromechanical systems.
The
ECE 330 homework assignment #2, due on Friday, March 7, 2014
Problem 1
A coil of 180 turns is wound uniformly over a wooden ring having a mean circumference of 700
mm, and a uniform cross sectional area of 400 mm2. If the current through the coil is 6 A,
ECE330 Spring 2014
Problem 1
A wooden ring would have the permeability of the air.
a)
Magnetic field strength:
NI 180 6
H=
=
= 1543 (A/m)
l
0 .7
b)
Flux density:
B = 0 H = 4 10 7 1543 = 1.939 (mT or mWb/m2)
c)
Total flux:
= B A = 1.939 10 3 400 10 6 = 0.
ECE330 Spring 2014
2.1
(
)
v(t ) = 100 cos 377t + 10 0 V => V =
(
)
i (t ) = cos 377t + 55 0 A => I =
1
2
100
2
10 0 V
55 0 A
Average power: P = VI cos( v i ) =
100 1
(
)
(
2 2
Power factor: PF = cos( v i ) = cos( 45 0 ) = 0.7071 leading
2.2
(
)
a) v(t )
ECE 330 homework assignment #4, due in Mid-term examination
Problem 1
For the magnetic circuit in Fig. 1 below, the upper part with winding is fixed, and the lower part is
movable (up and down only). Take into account the effect of fringing in the air gap
ECE 330 homework assignment #6, due on Friday, May 30, 2014
Text problem 7.5
Text problem 7.16 (Use the exact equivalent circuit)
Text problem 7.17
Text problem 7.22 (Use the exact equivalent circuit)
Text problem 7.25
ECE330 Spring 2014
6.2
Total per-phase complex power is
S Tp = (1500 + j 300) 3 = 500 + j100 kVA
Assuming VaLp = VaMp = VaLp 0 0 , the per-phase load complex power is
*
S Lp = VaLp I aLp =
2300
0 0 20045 0 = 265.5845 0 = 187.79 + j187.79 kVA
3
The per-pha
ECE330 Spring 2014
7.5
Rr' 0.048
=
= 1.6
s
0.03
a)
440
0 = 2540 , the rotor current referred to stator side is
3
2540
=
= 146.02 17.39 A
(0.06 + 1.6) + j (0.26 + 0.26)
Assuming phase voltage is Vap =
I r' =
(R
a
Vap
) (
'
+ R s + j xls + xlr
'
r
)
I = 13
ECE 330 homework assignment #1, due on Friday, February 21, 2014
Text problem 2.1 (Partial answer: 0.707 leading)
Text problem 2.2 (Partial answers: -250, -500, 500, 0, 150, 1900, -150)
Text problem 2.8
Text problem 2.9
Text problem 2.10 (Partial answer: