2008 Final Exam Paper

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NAME:

SIGNATURE:

STUDENT NO:

The University of New South WalesSchool of Electrical Engineering & Telecommunications

FINAL EXAMINATION

June 2008

ELEC3115 - ELECTROMAGNETIC ENGINEERING

Time allowed: 3 hours

Answer FOUR questions only; Any TWO questions from Part A and any

TWO questions from Part B.

Answer each question in a separate answerbook. Write the question

number at the top of the front page of each answerbook.

Questions carry marks as indicated within the questions.

The Examinations Unit will provide the following:

Transmission Line (Smith) Chart, mm graph paper and approved electroniccalculator

Candidates may bring only drawing instruments into the examination room.

Answers must be written in ink. Except where they are expressly required, pencilsmay only be used for drawing, sketching or graphical work.

This question paper may NOT be retained by the candidate.

See back page for some constants and formulae.


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PART A

Answer any TWO questions from this part.

QUESTION 1 (15 marks)

A two-wire overhead power transmission line has two identical copper conductors

separated by air as the dielectric material, as indicated in Figure 1. Each conductor has a

circular cross-section of radius a = 2.5 mm and the separation between the conductors isD

= 7 mm.

aa

di di

d

D

Figure 1

Given: Relative permittivity of air, r 1

(i) Using Gausss law, derive an expression for the capacitance per meter of the twoconductors in terms of radius a of each conductor and their separation, D. Find

the capacitance Cof the cable-pair per meter using the dimensions given above.

[5 marks]

(ii) What are the limitations, or possible errors, of applying Gausss law in this casewhere the separation D is not large compared to the radius of the conductors?

Explain why the method of images could be used to find the capacitance

between the conductors with higher accuracy. To aid your explanation, sketch

the electric field distribution in and around the space between the conductors and

the charge distribution.

[3 marks]

Question 1 continues in the next page


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(iii) Find the energy stored in the air gap and in the steel core material for the coil

current of (i) using the formula:2

mv'

1 BW dv'

2 Joules. What fractions of the

total energy stored in the magnetic field is in the air gap and in the iron core?

[4 marks]

(iv) Find the total stored energy using 2m1

W LI2

whereL is the inductance of the

coil in Henry. Compare and comment on the values of Wm found using this

equation with the Wm found in (iii).

[4 marks]

Figure 2(b)

Please see over for Question 3


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A. What is meant by hysteresis and eddy-current losses in a magnetic core? How do these

relate to the frequency and maximum flux density in the core of a magnetic material?

[2 marks]

B. In order to find the hysteresis and eddy-current loss characteristics (coefficients) of a

laminated iron material, a ring specimen of the laminated iron, as indicated in figure 3,

is made with a cross section of 15 cm2. It is wound with a coil of 500 turns.

Figure 3

(i) Determine the RMS value of 50Hz sinusoidal input voltage that must be appliedto the coil so that the maximum flux density in the core of the ring specimen

becomes 1T.

[3 marks]

(ii) With this input voltage applied, the input power to the coil is measured to be12W. Where is this power absorbed and in what form? You may neglect the I2R

(copper) loss in the conductor of the coil.

[3 marks]

(iii) The applied voltage to the coil is then changed to half of the value found in (i)and its input frequency is also halved (i.e., to 25Hz). The input power is now

measured to be 5.5W. Find the hysteresis and eddy-current loss coefficients k1

and k2.

[4 marks]

(iv) Obtain the hysteresis and eddy-current losses of the iron specimen at the voltageand frequency of (i).

[3 marks]

Please see over for question 4.


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PART B

Answer any TWO questions from this part.


A 30 m long lossless transmission line withoZ 50 operating at 2 MHz is terminated

with a loadLZ ( 60 j40 ) . If the phase velocity is

0 6p

v . c (c is the speed of light

=8

3 10 m/sec) on the line find:

(a) The reflection coefficient L at the load.

(b) The standing wave ratio S.

(c) The input impedance,Zin.

(d) The reflection coefficientin at the input.


An air filled 5cm2cm waveguide has zjz ey50sinx40sin20E V/m at

15GHz.

(a) What mode is being propagated?

(b) Find phase constant .

(c) DetermineEy/Ex.


The radiation intensity of a certain antenna is

elsewhere

forU

0

0;0sinsin2),(

3

Determine

(a) The direction of maximum radiation.

(b) The directivity.


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Formulae Sheet

Maxwell`s Equations in Integral Form (Time Domain):

c

Edl= - sdt

dBds ;

s

Dds = v

dv

c

Hdl= s

Jds + s

dt

dDds ;

s

Bds = 0

Note that, boldface characters represent vector quantities and indicates scalar product.

Also note that for simple media: D = E and B = H

Stokes` Theorem: Divergence Theorem:

c Fdl= s (xF)ds s Fds = v (F) dvInput Impedance of a Lossless TL of Characteristic Impedance Z0, length , and

terminated in a load impedance of ZL:

tan

tan

0

00

L

Lin

jZZ

jZZZZ

where /2

Flux of a Vector Function F Intrinsic Impedance of a Lossless Medium:

Through a Surface S:

s

Fds

Boundary Conditions at Interface Between Perfect (Lossless) Dielectrics:

tt EE 12 ; tt HH 12 ; nn DD 12 ; nn BB 12

Some Constants That may be Useful: )/(1036

1 90 mF

; )/(104 70 mH

Snell`s Laws of Reflection & Refraction: (For non-magnetic media; 1 = 2 = 0)

ir sinsin ; it

sinsin

2

1

Wavenumber for a Lossless Medium: k

Reluctance of a magnetic circuit:l

S Inductance of a coil:

NL

i i

Transformer equation: max cV 4.44 fB NS


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Hysteresis Loss equation: nh 2 mP k B f

Eddy-current Loss equation: 2 2e 1 mP k B f

Cut off wave number:22

kkc ,22

bnamkc

Phase velocity: pv

ParametermnTE Mode mnTM Mode

Ez(x,y,z) 0 j zmn

m x n yB sin sin e

a b

Hz(x,y,z)

j zmn m x n yA cos cos ea b

0

Ex

2j z

mnc

j n m x n yA cos sin e

a bk b

2j z

mnc

j m m x n yB cos sin e

a bak

Ey

2j

mnc

j m m x n yA sin cos e

a bk a

2z

mnc

j n m x n yB sin cos e

a bbk

Hx

2 jmnc

j m m x n yA sin cos ea bk a

2 j zmnc

j n m x n yB sin cos ea bbk

Hy

2j z

mnc

j n m x n yA cos sin e

a bk b

2

j zmn

c

j .m m x n yB cos sin e

a bak

Z TEk

Z

TMZ

k

''

jkR

V

0 dVR

Je4

RA

'

'

max),(E

)(E),(F

3

4

0

320

22

rad dsin16

IdlkP

22

20

radrad

dl80

I

P2R


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d),(UPrad dRddsin.RdA

22

avav S

,S

U

,U,D

d,U4

1

,U,D

d),(F 2A

av

max

av

max

S

S

U

UD

d,U41,U av

d.

U

U

4D

max

A

4

t

rad

P

P

t

max

P

,U4),(G

DG

2

2

20

sin

2

klcoscos

2

klcos

R

I15)(S

i

inte

S

PA 83A

2

e 4

DA

2

e

2

rt22rtrt

t

rec

R4GG

R

AA

P

P

; 1rt

End of Paper

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2008 Final Exam Paper