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UNSW, Australia

School of Electrical Engineering & Telecommunications

FINAL EXAMINATION

Session 1, 2015

ELEC3115
Electromagnetic Engineering

TIME ALLOWED: 3 hours
TOTAL MARKS: 100
TOTAL NUMBER OF QUESTIONS: 5

THIS EXAM CONTRIBUTES 60% OF THE TOTAL COURSE ASSESSMENT.

This paper contains 9 pages, including a formulae sheet which can be found on page 8 and 9.
Candidates must answer ALL 5 questions: THREE from Part A and TWO from Part B.
Marks for each question are indicated beside the question.
This paper MAY NOT be retained by the candidate.
Authorised examination materials:
Drawing instruments may be brought into the examination room.
Candidates should use their own UNSW-approved electronic calculators.
This is a closed book examination.
All answers must be written in ink. Except where they are expressly required, pencils may only
be used for drawing, sketching or graphical work.
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PART A

QUESTION 1 [20 marks]

An insulated power cable with inner conductor diameter of 16 mm has two insulating layers:
XLPE (cross-linked polyethylene) with relative dielectric constant, r1 2.8  , and EPR
(Ethylene Polypropylene Rubber) with relative dielectric constant, r2 2  . The dielectric
breakdown strengths (Emax) of these two materials are:

XLPE : Emax = 20MV/m
EPR Emax = 15 MV/m

(i) Show that E field variation with radial distance around a long, isolated, cylindrical charged
conductor such as the power cable is LrE a 2 r



where r is the radial distance from the axis of the conductor
L is the line charge density of the conductor
 = or, is the total permittivity of the dielectric material; (o = 8.8541012 F/m),
where r is the relative permittivity of the material.

[3 marks]

(ii) Which dielectric material should form the inner layer? Discuss the reasons behind your
selection using boundary conditions for the normal components of the E fields between
material boundaries.
[2 marks]

(iii) Assume that the outside of the cable insulation is conducting ground and hence is at zero
potential. Find the minimum thicknesses of each insulation layer for operation of the cable
at 132kV, if the maximum E fields within both dielectric materials are not to exceed 50%
of their respective breakdown strengths (Emax).
[10 marks]

Question 1 continues in the next page
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(iv) Sketch neatly the maximum allowable E field variations with radial distance in the inner
and outer layers of insulation, indicating for each dielectric material the maximum and
minimum E field that will occur. Do not ignore the boundary condition between the outer
layer insulator and surrounding air.

[5 marks]

QUESTION 2 [20 marks]

A power transmission line consists of two bare copper conductors of circular cross-section, each
having a radius a = 12 mm, and they are spaced D = 1.2 m apart. The space between the
conductors is air with a dielectric strength of 3 MV/m.

(i) Assuming uniformly distributed charge on the conductor, show that the capacitance per
unit length of the line is
ln
C D a
a
  .
[5 marks]

(ii) Using the express of (i), calculate the capacitance/m of the line.

[2 marks]

(iii) Calculate the leakage resistance of the line (i.e. the resistance between the conductors) per
unit length, assuming that the conductivity of air is 510-10 S/m and that of copper is
5.8107 S/m.
[3 marks]

(iv) Taking help of expression of (i) and method of images, develop an expression for
capacitance between a bare copper conductor of the transmission line and ground which is
assumed to be a perfect conducting body. If the height h of the conductor above the ground
is 10m, calculate the value of this leakage capacitance/m.
[5 marks]

(v) A line trap is an air core coil used on high voltage power transmission lines to assist in
directing high frequency signals coupled to the line for communication purposes. A typical
coil has the following characteristics:
Question 2 continues in the next page
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Rated current, I = 400 A (this is the normal operating current)
Number of turns, n = 33
Rated voltage, V = 132,000 Volts (the device must be insulated from earth)
Maximum fault current = 10,000 A for 3 seconds
Radius of the coil, a = 17.5 cm
Length of the coil, l = 40 cm

The inductance of the coil is given by

2 2
on aL K
l
  (K = constant  0.7)
Determine the radial forces acting on the coil when the maximum allowable fault current of
10,000 Amps flows through the winding.
[5 marks]

QUESTION 3 [20 marks]

A toroidal core with a mean diameter of 250 mm and a cross-sectional area of 1000 mm2 has an
air gap of length 1.25 mm cut in it. Fringing and leakage of flux around the air gap may be
neglected.

(i) Determine the current required in a coil of 1500 turns to produce a flux density of 1 T. The
relative permeability r of the core material (iron) may be assumed to be 3000 for this
level of flux density. [Permeability H/m70 r 0, 4 10        ].
[5 marks]

(ii) The above coil is the primary of a single-phase ideal transformer. The AC supply voltage
to the primary winding is sinusoidal at 50 Hz. What is the highest RMS AC supply voltage
that can be applied to the coil, so the maximum flux density in the core does not exceed 1
T? [3 marks]
Question 3 continues in the next page
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(iii) If the RMS value of the available 50 Hz AC supply voltage is twice the value found in (ii),
what change(s) must be made to the coil so that the maximum flux density does not exceed
1 T?
[3 marks]
(iv) Can this transformer be safely used with a 400V (RMS), 60Hz AC supply? Support your
[4 marks]
(v) Suppose you do not wish to use an iron core for the transformer because of eddy current
and hysteresis loss. Can you use an air-core torus instead? If so, what is the number of
turns required if the torus is made of plastic with the same dimensions given above to
produce a flux density of 1T with the current calculated in (i)?
[5 marks]

PART B

QUESTION 4 [20 marks]

Consider a coaxial cable, filled with a non-magnetic dielectric with r = 4. With this dielectric,
the cable has a characteristic impedance Z0 = 50 . The cable is terminated by an impedance
ZL = 100 – j20 .

(i) Calculate the reflection coefficient  at the load.
[2 marks]

(ii) Calculate the Voltage Standing Wave Ratio (VSWR). Assuming a signal at frequency f =
400 MHz is applied to the line, calculate the distance between maxima of the standing
wave along the cable.
[3 marks]

Question 4 continues in the next page

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(iii) Using the supplied Smith chart, design a single short-circuited stub that matches the load
to the characteristic impedance of the cable (you may assume that you can cut the coaxial
cable at any point and insert a perfect T-piece to connect the stub). Explain in words each
step you take to arrive at the result. Once you have obtained the result, describe the stub in
terms of its length l, and its distance d from the load. Quote these lengths relative to the
signal wavelength. (Note: the stub is made of the same type of cable as the one that
[10 marks]

(iv) Now assume you replace the dielectric in the cables with a different one, that has r = 2.
Re-calculate  and VSWR at the load. (Note: the load ZL remains unchanged).
[3 marks]

(v) With the new dielectric in the cables, is the stub geometry you designed at point (iii) still
replaced everywhere, also in the stub).
[2 marks]

QUESTION 5 [20 marks]

Consider a rectangular waveguide, with transverse dimensions a = 10.668 mm and b = 4.318
mm. The waveguide is uniformly filled with a non-magnetic dielectric with r = 4.
We wish to use this waveguide to propagate a wave at frequency f = 9 GHz.

(i) Which propagation modes can be sustained by this waveguide? Explain in detail how you
[5 marks]

(ii) Calculate the propagation constant  for the allowed modes at f = 9 GHz.
[2 marks]

(iii) Assume that the 9 GHz signal along this waveguide is used to transmit data. At what speed
[3 marks]

Question 5 continues in the next page

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(iv) Now assume that the dielectric has broken perpendicularly to the length of the waveguide,
and that the two resulting pieces of dielectric have been pulled out slightly, leaving a 1 mm
gap (see figure below). Calculate the insertion loss (in dB) resulting from this fault. The
insertion loss is defined as:  10 Log(P2 / P1) where P1 is the incident power and P2 is the
transmitted power.
Explain the physical origin of this loss.
[10 marks]

******* End of Paper *******

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Formulae Sheet

Maxwell`s Equations in Integral Form :

c s v
c s s s
E.dl B.ds , D.ds dv
dH .dl J . ds D. ds; B.ds 0
dt
  
  
  
   
   
     
 
 

Stokes’ Theorem: Divergence Theorem:
 
c s
F .dl F . ds     
s v
F . ds ( . F )dv    
The operators in the three basic orthogonal systems
Coordinate system Relations
(u1, u2, u3)
Cartesian Coordinate
(x, y, z)
Cylindrical Coordinate
(r,, z)
Spherical Coordinate (R,,)
Base vectors (au1, au2, au3) (ax, ay, az) (ar, a, az) (aR, a, a)
Matric coefficients (h1,h2,h3) (1, 1, 1) (1, r, 1) (1, R, Rsin)
Differential length element u1 1 1 u2 2 2 u3 3 3d ( h du ) ( h du ) ( h du )  l a a a
Differential surface elements n
1 2 3 2 3 2 1 3 1 3 3 1 2 1 2
d ds
ds h h du du , ds h h du du , ds h h du du

  
s a

Differential volume element  1 2 3 1 2 3dv h h h du du du
Del operator
u1 u2 u3
1 1 2 3 3 3h u h u h u
          
a a a

Div
2 3 1 1 3 2 1 2 3
1 2 3 1 2 3
1.A ( h h A ) ( h h A ) ( h h A )
h h h u u u
          


Curl
u1 1 u2 2 u3 3
1 2 3 1 1 1
1 1 2 2 3 3
h h h
1A
h h h u u u
h A h A h A
      
a a a


Laplacian
2 2 3 1 3 1 2
1 2 3 1 1 1 2 2 2 3 3 3
h h h h h h1 V V VV
h h h u h u u h u u h u
                             

Boundary Conditions at Interface
s
2t 1t 0
E E   ; s2t 1t J 0H H  ; nn DD 12  ; nn BB 12  ;

Some Constants that may be useful:
)/(10
36
1 9
0 mF
  ; )/(104
7
0 mH
 
1 2
1 2
1 2
; t tn n
J JJ J   
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Useful Vector identities and properties:

 
 
 
 
2
2
. VA V .A A. V
VA V A V A
A .A A
. V V
    
     
    
  
    
   
      
  

Some useful formulas of Electromagnetics:

 
,
,
s
v
t
0
J u I J ds
lJ E, R , P E J dv
S
J e
t
J 0
RC

 
  

   
 
  
   
 

  
   
 
 

 V 0,     A 0   
 
s

enclosed
enclosed
s c
b
ba
a
0 0 e 0 r
N
e k k Q e V e
k 1 v'
2
QE ds , D.ds Q
E dl V , E V
D E P, D (1 )E E E, D ds Q
1 1W Q V D E dv , F W , F W
2 2
V

     

 
   
      
     
  
 

 
  
  
        
     

 

 
 
0 c enclosed
c c
b
0 R
2
a
o o 0 m 0 r
2
0
s
2 12
12
1
N N
m jk j k
j 1 k 1 v
B dl I , H .dl I
Idl aB( x,y,z )
4 R
B A, B H M , B 1 H H H
A J
lB ds , ,
S
N NL , L
i I
1 1 1W L I I = A.J dv
2 2

      

 
 
  
 

       
  
  
 
 
 

  
  
 
        
 
 
 
 
v
m m I m
c
H .Bdv
2
F I dl B, F W , F W 

    

 
     
 
 
C
rms m c
ac
n 2 2
h 2 m e 1 m
dEMF N , EMF u B .dl
dt
BE
t
V 4.44 fNB S
2 1 l, R
D
P k B f , P k B f

    
   
   

     
 
 
  