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BEK 4113/BEX 44503 High Voltage Engineering ‘‘Breakdown Breakdown in Dielectrics Dielectrics’’ Quiz
Dr Rie
1
BEE 3243 Electric Power Systems – Module 1
Townsend’s Mechanism: Tutorial 1
Problem 1: Estimate the sta static tic bre breakdo akdown wn volt voltage age V in kV of an ‘air ‘air gap’ at 100 mm.Hg pressure between two parallel plates that ensure a uniform field. field. α/p as a function of E/p of E/p is shown in next figure. figure. Assume γ = 10-3 electron/incident positive ion. The gap distance is 1 cm. cm. Estimate the static breakdown voltages for N for N2, H2, A and Ne gases as well. Neglect recombination and attachment.
2
BEE 3243 Electric Power Systems – Module 1
Townsend’s Mechanism: Tutorial
3
BEE 3243 Electric Power Systems – Module 1
Townsend’s Mechanism: Tutorial 1 Solution: Here, pd = gap pressure multiply gap distance = 100 mm.Hg/cm. At this pd value the Townsend’s mechanism holds good. The Townsend’s criterion is
= 6.908ion. pairs / cm 1.0cm 6.908ion. pairs / cm = 0.06908ion. pairs / cm.mm. Hg 100
= 0.069ion. pairs / cm.mm. Hg
Refer from the graph, for Air: E b p
= 54V / cm.mm. Hg
= 54 × pressure = 54 ×100 = 5400V / cm = 5.4kV / cm V b = E b × d = 5.4kV / cm × 1.0cm = 5.4kV
E b
4
BEE 3243 Electric Power Systems – Module 1
Townsend’s Mechanism: Tutorial 1 For other gases, the breakdown voltages are calculated likewise and tabulated in table below: Gas
N2
H2
A
Ne
E/p (V/cm.mm.Hg)
60
29
19.5
12
Eb (V/cm)
6000
2900
1950
1200
Vb (V)
6000
2900
1950
1200
5
BEE 3243 Electric Power Systems – Module 1
Townsend’s Mechanism: Tutorial 1
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.
= 1.974ion. pairs / cm 3.5cm 1.974ion. pairs / cm α = = 0.0263ion. pairs / cm.mm. Hg 75 p
α
α
p
= 0.03ion. pairs / cm.mm. Hg
Refer from the graph, for Air:
E b p
= 44V / cm.mm. Hg
= 44 × pressure = 44 × 75 = 3300V / cm = 3.3kV / cm V b = Eb × d = 3.3kV / cm × 3.5cm = 11.55kV
E b
8
BEE 3243 Electric Power Systems – Module 1
Townsend’s Mechanism: Tutorial 1 Continue: For N2: E b p
= 50V / cm.mm. Hg
= 50 × pressure = 50 × 75 = 3750V / cm V b = Eb × d = 3750V / cm × 3.5cm = 13125V = 13.13kV
E b
For A: E b p
= 14V / cm.mm. Hg
= 14 × pressure = 44 × 75 = 1050V / cm V b = Eb × d = 1050V / cm × 3.5cm = 3675V = 3.68kV
E b
For Ne and H2 are out of range.
9
BEE 3243 Electric Power Systems – Module 1
Townsend’s Mechanism: Quiz 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. Neglect recombination and attachment.
= 1.535ion. pairs / cm 4.5cm α 1.535ion. pairs / cm = = 0.171ion. pairs / cm.mm. Hg 9 p
α
= 0.17ion. pairs / cm.mm. Hg
E b _ Ne p
= 18V / cm.mm. Hg
= 18 × pressure = 18 × 9 = 162V / cm = 0.162kV / cm V b _ Ne = E b _ Ne × d = 0.162kV / cm × 4.5cm = 0.729kV
E b _ Ne
Refer from the graph, for A:
E b _ A p
= 25V / cm.mm. Hg
= 25 × pressure = 25 × 9 = 225V / cm = 0.225kV / cm V b _ A = E b _ A × d = 0.225kV / cm × 4.5cm = 1.01kV
E b _ A
12
BEE 3243 Electric Power Systems – Module 1
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 E 0 of 3.9 kV/cm, it was found that the space charge created by an avalanche lay nearly in a sphere of radius r d = 0.08 cm. Estimate the value of d for favourable condition for the formation of streamers in the methane gap.
13
BEE 3243 Electric Power Systems – Module 1
Tutorial 2 Streamers Mechanism: Answer Solution: If the space charge in an avalanche is assumed to be in a sphere of radius r d, then the electric field of this charged sphere at its surface is
E r
=
Where
q
= ε
4π K 0 rd2
α d
e
4π K 0 rd 2
q = charge in this sphere =
ε x (no. of charged particles in sphere) = e
d
ε = 1.6 x 10-19 Coulomb (charge of an electron) Favourable condition for the formation of streamer is E r ≈ external applied field ≈ E0 Absolute permittivity of air or vacuum = K 0
= ε 0 =
1 36π ×10
= Calculator CONST 32
9
= 8.854 ×10−12 Farad / m 14
BEE 3243 Electric Power Systems – Module 1
Tutorial 2 Streamers Mechanism: Answer Therefore,
1.6 ×10 −19 eα d = r E 1 4π × 36π ×109
= 2250 ×10 −6 e
αd
×
1 −2
( 0.08 ×10 )
2
/ V m
−8
V / m = 2250 ×10 e V / cm α d
Favourable condition for the formation of streamer is E r ≈ external applied field ≈ E0, i.e. 2250 x 10-8 eαd = 3.9 x 10 3 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
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BEE 3243 Electric Power Systems – Module 1
Tutorial 2 Streamers Mechanism: Answer Actual solution:
E r = ε E r =
e
α d
4π × ε 0 r d
2
1.6 ×10 − e 19
1.1127 × 10
=
α d
−10
−9
1.6 × 10 −19 eα d
E r = 1.438 ×10 e
4π × 8.8542
×
α d
1 6.4 × 10 −7
× −12
1
(0.08 ×10− )
2 2
V / m
V / m
×1.5625 ×106 V / m
− E r = 2.2469 × 10 e V / m 3
E r =
α d
2.2469 × 10 −3 eα d 1× 10 2
V / cm
− E r = 2.2469 × 10 e V / cm 5
E 0 e
α d
= E r
α d
=
3.9 × 10
3
2.2469 × 10 −5
= 173.57 ×106 6 α d = ln(173.57 × 10 ) α d = 18.97ion _ pairs e
α d
α
=
18.97 3
= 6.32ion _ pairs / cm
16
BEE 3243 Electric Power Systems – Module 1
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 E 0 of 2.5 kV/cm, it was found that the space charge created by an avalanche lay nearly in a sphere of radius r d = 0.5cm. Estimate the value of d 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 ε is 1.6x10-19 17
BEE 3243 Electric Power Systems – Module 1
Tutorial 2 Streamers Mechanism: Answer Actual solution:
E r = ε E r =
e
α d
4π × ε 0 r d
2
1.6 × 10 − e 19
1.1127 × 10
=
α d
−10
−9
1.6 × 10 −19 eα d
E r = 1.438 × 10 e
4π × 8.8542
×
α d
1 2.5 × 10 −5
× −12
1
(0.5 ×10− )
2 2
V / m
V / m
× 40 ×103V / m
− E r = 5.752 ×10 e V / m 5
E r =
α d
5.752 × 10 −5 eα d 1× 10 2
V / cm
− E r = 5.752 ×10 e V / cm 7
E 0 e
α d
= E r
α d
=
2.5 × 10
3
5.752 × 10 −7
= 4.35 ×109 9 α d = ln(4.35 × 10 ) α d = 22.19ion _ pairs e
α d
α
=
22.19 4
= 5.55ion _ pairs / cm
18
BEE 3243 Electric Power Systems – Module 1
Tutorial 1 Paschen’ Law: 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. V b _ kV
= 24.22(3.8629 ) + 6.08 (3.8629 ) V b _ kV = 93.559 + 11.9498 V b _ kV = 105.51kV V b _ kV = 106kV V b _ kV
20
BEE 3243 Electric Power Systems – Module 1
Tutorial Paschen’ Law: Bring Home Quiz: The test was conducted inside a pressurised chamber at p 1 = 1bar and p 2 = 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 Vb 1 and Vb 2 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. V b (kV). V b _ kV
= 24.22
293 p 760T
d + 6.08
⎛ 293 p d ⎞ ⎜ ⎟ 760 T ⎝ ⎠
21
BEE 3243 Electric Power Systems – Module 1
Tutorial Paschen’ Law: Bring Home Quiz Answer Solution:
Info: 1 bar = 750.06 mm.Hg. p1 = 1bar x 750.06mm.Hg = 750.06mm.Hg p2 = 2.5bar x 750.06mm.Hg = 1875.15mm.Hg d=0.35m=35cm T= 80°C +273 = 353 °K The Paschen’s Law equation;
V b _ kV
Answer a)
= 24.22
293 p 760T
d + 6.08
293 p 760T
d
Calculation of breakdown voltage V b1 (kV) using p 1 value, V b1_ kV
= 24.22
293 × 750.06 × (35) 760(353)
+ 6.08
293 × 750.06 × (35) 760(353)
= 694.41 + 32.56 V b1_ kV = 726.97kV = 727kV V b1_ kV
Calculation of breakdown voltage V b2 (kV) using p 2 value, 293 × 1875.15 × (35) 293 × 1875.15 × (35) V b 2 _ kV = 24.22 + 6.08 760(353)
= 1736 + 51.47 kV = 1787.47 kV = 1787 kV
760(353)
V b 2 _ kV V b 2
22
Answer b) The breakdown voltage increasing from V b1(kV)=727kV to Vb2(kV)=1787kV. Increasing value in percentage:
