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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