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Dr Rie Dr Rie BEK 4113/BEX 44503 High Voltage Engineering ‘Breakdown in Dielectrics’ Quiz BEK 4113/BEX 44503 BEK 4113/BEX 44503 High Voltage Engineering High Voltage Engineering Breakdown in Dielectrics Breakdown in Dielectrics Quiz Quiz

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### Transcript of Breakdown in Insulations 2011 Update - Quiz 1Dr RieDr Rie

BEK 4113/BEX 44503 High Voltage Engineering

‘Breakdown in Dielectrics’Quiz

BEK 4113/BEX 44503 BEK 4113/BEX 44503 High Voltage EngineeringHigh Voltage Engineering

‘‘Breakdown in DielectricsBreakdown in Dielectrics’’QuizQuiz 2

BEE 3243 Electric Power Systems – Module 1

TownsendTownsend’’s Mechanism: Tutorial 1s Mechanism: Tutorial 1

Problem 1:Problem 1:

Estimate the static breakdown voltage V in kV of an ‘air gap’ at 100 mm.Hg pressure between two parallel plates that ensure a uniform field. α/p as a function of E/p is shown in next figure. Assume γ = 10-3 electron/incident positive ion. The gap distance is 1 cm. Estimate the static breakdown voltages for N2, H2, A and Ne gases as well. Neglect recombination and attachment. 3

BEE 3243 Electric Power Systems – Module 1

TownsendTownsend’’s Mechanism: Tutorials Mechanism: Tutorial 4

BEE 3243 Electric Power Systems – Module 1

TownsendTownsend’’s Mechanism: Tutorial 1s Mechanism: Tutorial 1Solution:Solution:Here, pd = gap pressure multiply gap distance = 100 mm.Hg/cm. At this pdvalue the Townsend’s mechanism holds good. The Townsend’s criterion is

kVcmcmkVdEVcmkVcmVpressureE

HgmmcmVp

E

Hgmmcmpairsionp

Hgmmcmpairsioncmpairsionp

cmpairsioncm

pairsion

pairsiond

d

bb

b

b

4.50.1/4.5/4.5/54001005454

../54

../.069.0

../.06908.0100

/.908.6

/.908.60.1

.908.6

.908.610

11ln

11ln

3

=×=×===×=×=

=

=

==

==

=⎟⎠⎞

⎜⎝⎛ +=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

α

α

α

α

γα

Refer from the graph, for Air: 5

BEE 3243 Electric Power Systems – Module 1

TownsendTownsend’’s Mechanism: Tutorial 1s Mechanism: Tutorial 1

1200195029006000Vb (V)

1200195029006000Eb (V/cm)

1219.52960E/p (V/cm.mm.Hg)

NeAH2N2Gas

For other gases, the breakdown voltages are calculated likewise and tabulated in table below: 6

BEE 3243 Electric Power Systems – Module 1

TownsendTownsend’’s Mechanism: Tutorial 1s Mechanism: Tutorial 1

Problem 2:Problem 2:

Estimate the static breakdown voltage V in kV of an ‘air gap’ at 75 mm.Hg pressure between two parallel plates that ensure a uniform field. α/p as a function of E/p is shown in next figure. Assume γ = 10-3 electron/incident positive ion. The gap distance is 3.5 cm. Estimate the static breakdown voltages for N2, H2, A and Ne gases as well. Neglect recombination and attachment. 7

BEE 3243 Electric Power Systems – Module 1

TownsendTownsend’’s Mechanism: Tutorials Mechanism: Tutorial 8

BEE 3243 Electric Power Systems – Module 1

TownsendTownsend’’s Mechanism: Tutorial 1s Mechanism: Tutorial 1Solution:Solution:

kVcmcmkVdEbVcmkVcmVpressureE

HgmmcmVp

E

Hgmmcmpairsionp

Hgmmcmpairsioncmpairsionp

cmpairsioncm

pairsion

pairsiond

d

b

b

b

55.115.3/3.3/3.3/3300754444

../44

../.03.0

../.0263.075

/.974.1

/.974.15.3

.908.6

.908.610

11ln

11ln

3

=×=×===×=×=

=

=

==

==

=⎟⎠⎞

⎜⎝⎛ +=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

α

α

α

α

γα

Refer from the graph, for Air: 9

BEE 3243 Electric Power Systems – Module 1

TownsendTownsend’’s Mechanism: Tutorial 1s Mechanism: Tutorial 1Continue:Continue:

kVVcmcmVdEbVcmVpressureE

HgmmcmVp

E

kVVcmcmVdEbVcmVpressureE

HgmmcmVp

E

b

b

b

b

b

b

68.336755.3/1050/1050754414

../14

13.13131255.3/3750/3750755050

../50

==×=×==×=×=

=

==×=×==×=×=

=

For Ne and H2 are out of range.

For N2:

For A: 10

BEE 3243 Electric Power Systems – Module 1

TownsendTownsend’’s Mechanism: Quizs Mechanism: Quiz

Quiz 1:Quiz 1:

Estimate the static breakdown voltage V in kV of an ‘Ne’ and ‘A’ gases at 9 mm.Hg pressure between two parallel plates that ensure a uniform field. α/p as a function of E/p is shown in next figure. Assume γ = 10-3 electron/incident positive ion. The gap distance is 4.5 cm. Neglectrecombination and attachment. 11

BEE 3243 Electric Power Systems – Module 1

TownsendTownsend’’s Mechanism: Quiz 1s Mechanism: Quiz 1 12

BEE 3243 Electric Power Systems – Module 1

Solution:Solution:

kVcmcmkVdEV

cmkVcmVpressureE

HgmmcmVp

E

kVcmcmkVdEV

cmkVcmVpressureE

HgmmcmVp

E

Hgmmcmpairsionp

Hgmmcmpairsioncmpairsionp

cmpairsioncm

pairsion

pairsiond

d

AbAb

Ab

Ab

NebNeb

Neb

Neb

01.15.4/225.0

/225.0/22592525

../25

729.05.4/162.0

/162.0/16291818

../18

../.17.0

../.171.09

/.535.1

/.535.15.4

.908.6

.908.610

11ln

11ln

__

_

_

__

_

_

3

=×=×=

==×=×=

=

=×=×=

==×=×=

=

=

==

==

=⎟⎠⎞

⎜⎝⎛ +=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

α

α

α

α

γα

Refer from the graph, for Ne:

Refer from the graph, for A: 13

BEE 3243 Electric Power Systems – Module 1

Tutorial 2 Streamers Mechanism:Tutorial 2 Streamers Mechanism:

Problem 3:

A uniform static field was created in Methane at 102 mm.Hg pressure by a parallel plate electrode system with a gap distance of 3 cm. With an externally applied electric field E0 of 3.9 kV/cm, it was found that the space charge created by an avalanche lay nearly in a sphere of radius rd = 0.08 cm. Estimate the value of αd for favourable condition for the formation of streamers in the methane gap. 14

BEE 3243 Electric Power Systems – Module 1

Tutorial 2 Streamers Mechanism: AnswerTutorial 2 Streamers Mechanism: AnswerSolution:Solution:If the space charge in an avalanche is assumed to be in a sphere of radius rd, then the electric field of this charged sphere at its surface is

Whereq = charge in this sphere

= ε x (no. of charged particles in sphere) = εeαd

ε = 1.6 x 10-19 Coulomb (charge of an electron)

Favourable condition for the formation of streamer is Er ≈ external applied field ≈ E0

2 20 04 4

d

rd d

q eEK r K r

α

επ π

= =

1 12900

−×=×

==π

ε

Absolute permittivity of air or vacuum =

= Calculator CONST 32CONST 32 15

BEE 3243 Electric Power Systems – Module 1

Therefore,

Favourable condition for the formation of streamer is Er ≈ external applied field ≈ E0, i.e.

2250 x 10-8 eαd = 3.9 x 103

Giving eαd = 1.733 x 108

So αd = ln(1.733 x108) = 18.97 ion.pairs/cm

Therefore α =18.97/3cm = 6. 32 ion.pairs/cm

( )19

22

9

1.6 10 1 /1 0.08 10436 10

d

reE V m

α

ππ

×= ×

×××

6 82250 10 / 2250 10 /d de V m e V cmα α− −= × = × 16

BEE 3243 Electric Power Systems – Module 1

( )

cmpairsion

pairsiondd

e

e

EE

cmVeE

cmVeE

mVeE

mVeE

mVeE

mVer

eE

d

d

r

dr

d

r

dr

dr

d

r

d

d

d

r

/_32.6397.18

_97.18)1057.173ln(

1057.173102469.2

109.3

/102469.2

/101102469.2

/102469.2

/105625.110438.1

/104.6

1101127.1

106.1

/1008.0

18542.84

106.14

6

6

5

30

5

2

3

3

69

710

19

2212

19

20

==

=×=

×=×

×=

=

×=××

=

×=

×××=×

××

×=

××

××

=

−−

−−

α

αα

πεπε

α

α

α

α

α

α

α

ααActual solution:Actual solution: 17

BEE 3243 Electric Power Systems – Module 1

Tutorial 3 Streamers Mechanism:Tutorial 3 Streamers Mechanism:

Problem 4:

A uniform static field was created in Methane at 115 mm.Hg pressure by a parallel plate electrode system with a gap distance of 4 cm. With an externally applied electric field E0 of 2.5 kV/cm, it was found that the space charge created by an avalanche lay nearly in a sphere of radius rd = 0.5cm. Estimate the value of αd for favourable condition for the formation of streamers in the methane gap. The constant absolute permittivity K0 of the test area is 8.854x10-12 and the charge of electron εis 1.6x10-19 18

BEE 3243 Electric Power Systems – Module 1

( )

cmpairsion

pairsiondd

e

e

EE

cmVeE

cmVeE

mVeE

mVeE

mVeE

mVer

eE

d

d

r

dr

d

r

dr

dr

d

r

d

d

d

r

/_55.5419.22

_19.22)1035.4ln(

1035.410752.5

105.2

/10752.5

/10110752.5

/10752.5

/104010438.1

/105.2

1101127.1

106.1

/105.01

8542.84106.1

4

9

9

7

30

7

2

5

5

39

510

19

2212

19

20

==

=×=

×=×

×=

=

×=××

=

×=

×××=×

××

×=

××

××

=

−−

−−

α

αα

πεπε

α

α

α

α

α

α

α

ααActual solution:Actual solution: 19

BEE 3243 Electric Power Systems – Module 1

Tutorial 1 Tutorial 1 PaschenPaschen’’ Law:Law:

Problem 5:Problem 5:

Work out the estimate breakdown voltage Vb in kV during the breakdown process using the Paschen’s Law equation shown below. The test was conducted inside a pressurised chamber at p = 1.5 bar filled with normal air. The tests temperature area is 120°C. The electrodes gap is 3.5cm. Use 1 bar = 750.06 mm.Hg.

⎟⎠⎞

⎜⎝⎛+= d

Tpd

TpV kVb 760

29308.676029322.24_ 20

BEE 3243 Electric Power Systems – Module 1

Solution:Solution:

( ) ( )

( ) ( )

( ) ( )

kVV

kVV

V

V

V

V

dTpd

TpV

kVb

kVb

kVb

kVb

kVb

kVb

kVb

106

51.105

9498.11559.93

8629.308.68629.322.24

5.3)393(76009.112529308.65.3

)393(76009.112529322.24

5.3)273120(7605.106.75029308.65.3

)273120(7605.106.75029322.24

76029308.6

76029322.24

_

_

_

_

_

_

_

=

=

+=

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

×+

=

⎟⎠⎞

⎜⎝⎛+= 21

BEE 3243 Electric Power Systems – Module 1

Tutorial Tutorial PaschenPaschen’’ Law: Law: Bring Home Quiz:Bring Home Quiz:The test was conducted inside a pressurised chamberat p1 = 1bar and p2 = 2.5bars filled with normal air. Both tests temperature area are 80°C. The electrodes with a gap of 0.35m.

i. Work out the estimate breakdown voltage Vb1 andVb2 in (in kV) during the breakdown process using the Paschen’s Law equation shown below. Use 1 bar = 750.06 mm.Hg.

ii. Comment by comparing both results in terms of percentage increment / decrement.

iii. Also sketch the output graph of p(mm.Hg) vs. Vb(kV).

⎟⎠⎞

⎜⎝⎛+= d

Tpd

TpV kVb 760

29308.676029322.24_ 22

BEE 3243 Electric Power Systems – Module 1

Tutorial Tutorial PaschenPaschen’’ Law: Bring Law: Bring Home Quiz AnswerHome Quiz Answer

Solution:Solution:

dTpd

TpV kVb 760

29308.676029322.24_ +=

( )( )

( )( )

kVkVV

V

V

kVb

kVb

kVb

72797.726

56.3241.694353760

3506.75029308.6353760

3506.75029322.24

_1

_1

_1

==

+=

××+

××=

( )( )

( )( )

kVkVV

V

V

kVb

kVb

kVb

178747.1787

47.511736353760

3515.187529308.6353760

3515.187529322.24

_2

_2

_2

==

+=

××+

××=

Info:1 bar = 750.06 mm.Hg. p1 = 1bar x 750.06mm.Hg = 750.06mm.Hgp2 = 2.5bar x 750.06mm.Hg = 1875.15mm.Hgd=0.35m=35cmT= 80°C +273 = 353 °KThe Paschen’s Law equation;

Calculation of breakdown voltage Vb1 (kV) using p1 value,

Calculation of breakdown voltage Vb2 (kV) using p2 value, uporkVkV .46.2__%246%100

7271787

Answer b)The breakdown voltage increasing from Vb1(kV)=727kV to Vb2(kV)=1787kV.Increasing value in percentage:

727

1787

0

200

400

600

800

1000

1200

1400

1600

1800

2000

0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

p(bar)

Vb(k

V)