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12/04/2013

Name: 02 03 04 051 1 1 11 TotalNumber: 1 1 1

EENG 451HighVo ltageTechniques

EasternMediterraneanUniversity

DepartmentofElectricalandElectronicEngineering

MidtermExam1,2013

Q1.ln an electrode system potential equation is given as ~ = A + ~ ) , the bou ndary

conditionsareas follows;forx=0.4 cm,V=V1 =20 kV and x=1cm V=V2 =0 V.

a. DeterminetheconstantAand B. Drawthechangeinpotentialwithrespectto x.

b. Determine electric field equation. Determine maximum and minimum electric field strength

values.

c\ , RoVl'\dcvj t.c"cli+iol\s :

X 4eM V: A ... ! . ;. 20 it. V0.4

V ; A i ~ = O V1

A _ A _ 2-D It V A o . 4 ~ b . 4 X.z-o

D.4A   - 13.334- :; - Y.f

g = 1 ~ . 3 3 4 :: 4f

v x) ; : -1 - i ) ;   ( ~ _   \-.

X  3

(ltV)

fp

20

UV\0 4

1

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f ( ( O\)< ::: t 0 -  -4 ~ 3. 3 s ~ V c W \ '3p

EM i I :. E 1 ) ~ ·1   ·  S k v {Wl 3(>

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- - -

Q2 . ln a two layered, cylindrical system, radii of cylinders are: r1 = 2 cm, r2 = 2.4 cm, r3 = 2.6 cm, and

relative dielectric constants of layers are 1':r1 = 2.2 Er2 = 4. If the applied voltage is 70 kV and the

breakdown strength of the layers are Eb1=70 kV/cm and Eb2 =80 kV/cm respectively;

a. Calculate voltages across each layer.

b. Calculate maximum field strengths across each layer.

c. Calculate the capacitance of the system.d. Determine whether the system can withstand to this applied voltage.

1 -  L r.i·.e,. Cl + i .e. -fl Jc 2IT e,{ . T I f ~ e 2n-e E 1 t. l r2

-  ( i i . ) tn( J. )(" A

r

1 _ .._ 2.ne ~ i [_  ~  s.. :I. )d £r{ : ~c- zne - If 2. 2 Z U LDA-;_f 0--10 2.9 2 ~c U ~ L.'\.U :=. C'2.Uz f\)

U1C.l) --   U - u t ~ -- •

fO k-V e.f \ 2 lI\

[" - f'

- 22 1 l ~ r A ~ 1 0.10'29 . E 2 '2

LA rl - - fll, (1

1.fA r - A.

U~ . . ,A

r '

Zp=154. b k-V(CWl

-13 . b2.4-10 .02kV

-C,.VY'l

- 2 en 2. b2u

<-.

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c. . = 2ne _ 2  _ e 0_1D29

~ a

d .{ MO\X ::.

154--b kM

,>.f..b1 :': :fa k VI('/<.1

p

,st l o . ~ e ( .breoksdow/

t 7 V'J'M = "1-0 ..92 k.V/CM . £b 1 : : . ~ O kv LWl 2 f

.9 Ie e ( \.0I ltt s  t:"d s

IJ"I=U ~ +0 k.,V

I U~ N \ C \ > < : .

r'2 .-ev fir;;

~ . s ~ . s + e ; M COI{\ \oi s+-cl\d +t:, ~ \ s v o l ~ e11

f

::; 3b4·39 kVI  VVt

) £b,: D ~ v l c W l

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Q3 ln a co-axial cylindrical system, the insulation material th t is going to be used has a breakdown

strength of b = 80 kV/cm. In this configuration, 10 kV is applied to the inner conductor.

Determine the inner and outer conductor s radii considering the optimum geometry with regard

to breakdown. Determine maximum and minimum electric field strengths and system

capacitance so = 8.854xlO-  2F/m).

Df{:IMVM gec etr j

w \ ~ ' e 9 0 r c ~ 0

,6-\.( 01( clDWn

1 ~ V0.339 ,w)

--

c = 2n 0 . e( e

-t \ ,z.

r ,

C 5.5 ;,3 X r X e { 10 =

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Q4.An insulation environment consists of polyethylene (Crl = 2.2 and earth (Cr = 10). Electric fie ld

vector comes to the interface with an angle of 6 °through polyethylene.

3. Determine the refract ion angles of electric field lines and equipotential lines. e \ r t ~b. If electric field strength in polyethylene is El = 100 kV /cm determine electric field strength in. ,

E2 normal and tangential components of El and E2.

c. Determine whether the earth has a breakdown considering breakdown strength of earth is 5

kV/cm?

\\A i . e ~ {p O \ ~ e

£ll: ' bolle ( l - ~ \ O EI

~ o ~ ~&(lr bt -:. ~ I

t ~ rl

cZ , ~ 90 - £5 = 5 C> p

r A ? = ~f\ 0 .1\ ::: -E ~ f

til 1:1 '\., 4 CA/\cX 2-  ::.

tn +C1nd....,

~ 1 =90_0/. -1: G5 0

~ ~ o ol. := 25. 25 1- 

t

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1 x £Jt lb = 1£1 \ x SIr ) bU 1 u

UV

t t 1 tt. :. 100 k V )( s.ln 25 =42 2 b It v/cW ' ) fc.f/V\

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strengt of air is 30 kV/cm (k = 0.9).

J I.

t ,N \0\)c -::. '£. •

QS.

a. In a sphere-sphere (identical spheres) electrode system, using approximated field calculation

determine the breakdown voltage. Radii of the spheres are 3.5 cm, electrode separation is 0.3

r 2 :: r .sL:U/2 z

_U_,_ 2.  _ ~ . 2 i 4- J- 2 tJ X

- :..(V-"+cl -r

\-\d/Z ;2

IJb I ~ '3 ,S )4 0 .330ltV = 0 9

;rt l '2 3.5 0 . 3b. In the region given as Figure 2, V4 =20 kV ; Vs =10 kV ; V6=V7 =40 kV; Vg =Vg =50 kV; V10 =80

kV. Determine the potentials at the points I , 2 and 3 by Finite Element Method.

9

2

104

•f 1 m5

8

3

17

6

\It ~ L (V34- V[j VS V<, )4 --,v .1 .:> LI '

V, ': 1- (V3.f V4 + Vr:; ~ V10;:, )

4 - )

-- ;-'

V3--

1-4

(V1 + 2 - V ~ - - Vfr )

Figure 2

V,-\J3. = i-o

Lj-V2 -\13

-=-15 U

-V-1. _ V 2 ~ ~ V 3 = o