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HIGH VOLTAGE breakdown behavior in concentric spheres

Assistant Professor Suna BOLAT

Eastern Mediterranean University

Electric and electronic department

Parameters (Definitions) for spherical electrode system

1. Actual electrode separation, d

2. Effective electrode separation,

3. Geometric characteristics, p&q

4. Utilization factor (Electrode efficiency),

Actual electrode separation, d

d = r2 − r1

d

d

d

It can be insulator thickness, electrode separation 𝐸𝑎𝑣𝑒 =

𝑈

𝑑→ 𝑑 =

𝑈

𝐸𝑎𝑣𝑒

Effective electrode separation,

𝛼 =𝑈

𝐸𝑚𝑎𝑥=𝑟1𝑟2(𝑟2 − 𝑟1)

r1

U

Non-uniform

electric field

d

uniform

electric field

𝐸𝑚𝑎𝑥 =𝑈

𝛼→ 𝑈 = 𝐸𝑚𝑎𝑥. 𝛼

Geometric characteristics, p&q

• In concentric spheres:

𝑝 =𝑟1 + 𝑑

𝑟1 𝑞 =

𝑟2𝑟1

𝑑 = 𝑟2 − 𝑟1

𝑝 =𝑟1 + 𝑟2 − 𝑟1

𝑟1=𝑟2𝑟1= 𝑞

Utilization factor (electrode efficiency)

𝜂 =𝐸𝑎𝑣𝑒𝐸𝑚𝑎𝑥

=𝑈𝑑

𝑈𝛼 =𝛼

𝑑≤ 1

𝑑 = 𝑟2 − 𝑟1 > 𝛼 =𝑟1𝑟2

𝑟2 − 𝑟1 𝑟2 > 𝑟1

Electrode efficiency

• Uniform field

𝐸 =𝑈

𝑑= 𝐸𝑎𝑣𝑒 = 𝐸𝑚𝑎𝑥 = 𝐸𝑚𝑖𝑛

𝜂 = 1

Electrode efficiency

• Non-Uniform field

𝑉 = 𝑉(𝑟)

𝐸 = 𝐸(𝑟) r2

r1

r

𝑉 = 𝑉(𝑟)

U

r2

r1

Emax

r

Emin

𝐸(𝑟)

E

𝜂 < 1

Breakdown behavior in concentric spheres

Goal: Analyzing the relationship between electrode radii and electric field and potential.

• Change in Emax with respect to r1

• Change in Umax with respect to r1

1. Change in 𝐸𝑚𝑎𝑥 with respect to 𝑟1

• 𝑟2 → 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 and 𝑈 → 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

• Radius of outer sphere is constant, radius of inner sphere is variable.

• 𝐸𝑚𝑎𝑥 =𝑈

𝑟1𝑟2(𝑟2−𝑟1)

= 𝐸𝑚𝑎𝑥 𝑟1

• Q: How do we find the minimum value for Emax?

• A: derivative is zero!!!

← 𝐸𝑚𝑎𝑥 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 𝑟1

𝑑𝐸𝑚𝑎𝑥

𝑑𝑟1= 0 →

𝐸𝑚𝑎𝑥 =𝑟2. 𝑈

𝑟1. (𝑟2 − 𝑟1)=

𝑟2. 𝑈

𝑟1. 𝑟2 − 𝑟12

𝑑𝐸𝑚𝑎𝑥

𝑑𝑟1=− 𝑟2 − 2𝑟1 . 𝑟2. 𝑈

(𝑟1. 𝑟2 − 𝑟12)2

= 0

(−𝑟2+2𝑟1) . 𝑟2. 𝑈 = 0

𝑟2 − 2𝑟1 = 0 → 𝑟2 = 2𝑟1 →𝑟2𝑟1= 2

Optimum geometric characteristic

𝑟2𝑟1= 𝑝 = 2

𝑝 = 𝑝𝑏

fittest, optimum geometric char. with regard to breakdown in concentric spheres

Breakdown behavior

𝐸𝑚𝑎𝑥(𝑟1)

𝐸𝑚𝑎𝑥(𝑟1)

A B

𝐸𝑏 Breakdown strength of the insulator

𝑟𝐴 𝑟𝐵 𝑟1 =𝑟22

Breakdown region

𝑟𝐵 ≤ 𝑟1 ≤ 𝑟2

𝐸𝑚𝑎𝑥 ≥ 𝐸𝑏

No discharge region

No-Breakdown

𝑟𝐴 ≤ 𝑟1 ≤ 𝑟𝐵

𝐸𝑚𝑎𝑥 < 𝐸𝑏

Pre-Breakdown region

Partial discharge region

𝑟1 ≤ 𝑟𝐴

𝐸𝑚𝑎𝑥 ≥ 𝐸𝑏

𝑟1

1. Change in 𝑼𝑚𝑎𝑥 with respect to 𝑟1

• 𝑟2 → 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝐸𝑏 → 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

• 𝐸𝑚𝑎𝑥 =𝑈

𝑟1𝑟2(𝑟2−𝑟1)

=𝑈

𝛼

• 𝑈 = 𝛼. 𝐸𝑚𝑎𝑥

• 𝛼 = 𝛼𝑚𝑎𝑥, 𝑈 = 𝑈𝑚𝑎𝑥, 𝐸𝑚𝑎𝑥 = 𝐸𝑏

𝑈𝑚𝑎𝑥 = 𝛼𝑚𝑎𝑥 . 𝐸𝑏

𝛼 =𝑟1𝑟2

𝑟2 − 𝑟1 = 𝑓(𝑟1)

Maximum Voltage will be Maximum

Maximum value for

• Q: how do we find maximum value of ?

• A: 𝑑𝛼

𝑑𝑟1= 0; 𝛼 → 𝛼𝑚𝑎𝑥

𝛼 = 𝑟1 −𝑟1

2

𝑟2

𝑑𝛼

𝑑𝑟1= 1 −

2𝑟1𝑟2

= 0

→2𝑟1𝑟2

= 1 → 𝑟2 = 2𝑟1

𝑟2𝑟1= 2 = 𝑃𝑏 →

Optimum characteristic with regard to breakdown in spherical system.

𝑟1 =𝑟22

𝑟𝐴 𝑟2 𝑟1

𝑈𝑏

𝑈𝑚𝑎𝑥

Breakdown voltage of the electrode system

Breakdown voltage of the insulator

Breakdown Partial

discharge

No discharge

A B

Breakdown behavior

𝑈𝑏𝑠𝑦𝑠 > 𝑈𝑏𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑜𝑟 → 𝑁𝑜 𝑑𝑖𝑐ℎ𝑎𝑟𝑔𝑒

𝑈𝑏𝑠𝑦𝑠 ≤ 𝑈𝑏𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑜𝑟 → 𝐷𝑖𝑐ℎ𝑎𝑟𝑔𝑒

Example

In a concentric spherical electrode system; radius of the outer sphere is 𝑟2 = 250 𝑐𝑚 (constant), the insulator in between is air with a breakdown strength of 𝐸𝑏 = 30 𝑘𝑣/𝑐𝑚.

In case of optimum values with regard to breakdown, what is the maximum voltage that can be applied to the system?

What is the actual separation, effective electrode separation and system capacitance?

Solution

Optimum case with regard to breakdown:

• 𝑝 =𝑟2

𝑟1= 2

• 𝑟1 =𝑟2

2=

250

2= 125𝑐𝑚

Solution (Maximum voltage)

• 𝐸𝑚𝑎𝑥 =𝑈

𝑟1𝑟2(𝑟2−𝑟1)

= 𝐸𝑏

30 𝑘𝑣

𝑐𝑚=

𝑈𝑚𝑎𝑥

125250

(250 − 125)→

𝑈𝑚𝑎𝑥 = 30𝑘𝑣

𝑐𝑚.1

2250 − 125 = 1875 kV

If I apply 𝑼𝒎𝒂𝒙, I will end up with 𝑬𝒃!

Maximum voltage!!!

Solution (System’s parameters)

𝑑 = 𝑟2 − 𝑟1 = 125 𝑐𝑚

𝛼 =𝑈

𝐸𝑚𝑎𝑥=𝑟1𝑟2

𝑟2 − 𝑟1 =1

2. 125 = 62.5 𝑐𝑚

Solution (Capacitance)

𝐶 =𝑄

𝑈=𝐷. 𝑆

𝑈=𝜀. 𝐸. 𝑆

𝑈=

𝜀 ∙𝑈

𝑟1𝑟2

𝑟2 − 𝑟1∙1𝑟2

∙ 4𝜋 ∙ 𝑟2

𝑈

𝐶 = 4𝜋𝜀0 ∙ 𝜀𝑟 ∙𝑟1. 𝑟2𝑟2 − 𝑟1

= 4𝜋. 8.854. 10−12.1

100.125.250

125

𝐶 = 2 ∙ 781. 10−10 𝐹 𝐶 = 278.1. 10−12 𝐹

𝐶 = 278.1 𝑝𝐹

𝐹

𝑚 𝑢𝑛𝑖𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 [𝑚 → 𝑐𝑚]