ch5

CHAPTER 5 INSULATION DIAGNOSTIC & PARTIAL DISCHARGES (QUESTION AND ANSWER)

Question 1 a)

The dissipation factor or loss tangent is an indication to determine the performance of insulating property of insulators. The most common method used to determine the lost tangent is by using a.c bridges. i)

Sketch the circuit diagram of a high voltage Schering bridge for the



measurement measurement of loss tangent (tan ).

ii)

Derive the expression for tan



of the unknown series se ries model of tge tested

sample when the high voltage standard capacitor used in the Schering Bridge is a loss-free type.

Answer i)

 : C  : r 

Capacitance of the sample

: Resistance of the sample

     

 : Standard capacitor  ,

 and

 : Variable components of Schering bridge

ii)

                 

        [ ] [ ]                

At balance condition ;

;

By equating the real and imaginary parts of both side; Therefore;

                 

For series model, tan δ =

b)

Cr,

Breakdown in solid or liquid dielectrics dielectric s arise from the action of the electrical discharges in internal gaseous cavitties. Draw an equivalent circuit for such an arrangement and derive the expression for energy dissipated in the cavity in one discharge.

Answer

d

Distan ce

-

+

o

n0

Void in solid dielectric material

Equivalent circuit of void in the structure

   

    

   

Energy dissipated in the cavity in one discharge;

       Where

       Therefore

                 c)

A solid solid dielectric specimen of 5cm diameter and 10mm thickness has dielectric constant of 3. It has an embedded air filled void of 2.5mm diameter and 1mm thickness. It is subjected to a voltage of 100k Vrms. Find the voltage at which an internal discharge in the void can occur. The breakdown strength of air is 3kV/mm. Also determine the charge value each time there is discharge inside the void and what will be the value of charge as measured on the detector?

Answer The inception voltage of the stressed void Vi is given by:

  (    ⁄  )) Where



 

- breakdown stress of the void = 3 kV/mm - thickness of the specimen = 10mm - thickness of the void = 1mm - dielectric constant of the specimen = 3

when  = 1mm

 (  ⁄ )     =

Capacitance of void

    

⁶          ⁶     

When t = 1mm, area of void = π (2.5 x 10ˉ³)² / 4 = 491 x 10ˉ  m²

ˉ²

ˉ

ˉ

ˉ³

Capacitance of slab

                                 (   )                  ˉ²

²

ˉ²

5.79 x 10ˉ ² F

The measured change

ˉ

Therefore,

ˉ²

ˉ

ˉ²

Question 2 a)

Using the circuit for series and parallel models of an insulating sample, derive the loss tangent (tan sample.



) equation in terms of the capacitance (C) and resistance (R) of the

Answer  R4

r 

b1

C4

C2 c

R3

b2

C a Solid Insulator Model Series Model Rs = Cs =

Tan δ =

   

  



 =

 

 = ωCsRs

Va I Vc

Vr 

Cs

Vc

Ca

Rs t void

Solid Insulator Model Parallel Model

             b)

What are partial discharges? Describe some of the typical partial discharges.

Answer Some of the typical partial discharges are: i)

Corona or gas discharge. These occur due to non-uniform field in sharp edges of the conductor subjected to high voltage especially when the insulation provided is air or gas or liquid. [Fig. (a)]

ii)

Surface discharges and discharges in laminated materials on the interfaces of different dielectric material such as gas/solid interface. [Fig. (a) and (b)]

iii)

Cavity discharges: When cavities are formed in solid or liquid insulating materials the gas in the cavity is over stressed and discharges are formed. [Fig. (d)]

iv)

Treeing Channels: High intensity fields are produced in an insulating material at its sharp edges and this deteriorates the insulating material. The continuous partial discharges so produced are known as treeing channels. [Fig. (e)]

Cb

(a )

(b ) (c )

(d )

c)

(e )

A sample of impregnated paper (

Ԑ

=3.5) of thickness 2mm is placed between large

parallel-plate electrodes. A cubical air-filled cavity with 2x2 mm length of its edges and 0.05mm deep, with its axis at right angles to the electrodes is located in the centre of the dielectric. A sinusoidal alternating voltage is applied between the electrodes. The breakdown voltages of air at different pd level is shown in figure Q5 (i)

Determine the discharge inception stress in kV/mm in the dielectric for a pressure of 550mm Hg.

(ii)

Calculate the apparent discharge magnitude.

(iii)

If the pressure in the void is doubled, determine the new inception voltage.

Answer i)

       U     U

Discharge inception voltage, V ,

where U : breakdown voltage of the void

Ԑ     U [      ԐԐ ] U Ԑ 

[  ]    U U is taken from the graph  At pd = (500)(0.05) = 25; therefore U≈0.46kV

Vi = 12.14 (0.46) = 5.58kV Inception stress in kV/mm Ei = Vi/d = 5.58/2 = 2.79kV/mm

ii)

ₐ

Apparent discharge q ;

   ₐ  ₐ ₐ ₐ ₐ 

q  = δV [



since Cb t;

proven

In term of electric field strength

       and

Hence

 And

  



     and

Electric stress in the void is greater than the dielectric stress across the sample. Partial discharge occurs due to the very small gap of the void even at the normal service voltage.

Question 4 a)

The dissipation factor or loss tangent is an indication to determine the performance of insulating property of insulators. The most common method used to determine the loss tangent is by using a.c. bridges. i)

Sketch the circuit diagram of a high voltage Schering bridge for the measurement



of loss tangent (tan ) ii)



Derive the expression for tan  of the unknown series model of the tested sample when the high voltage standard capacitor used in the Schering bridge is a lossfree type.

Answer

C4 a

Sample Standard with losses

b2

r 

r2

 AC Supply

C: capacitance of the sample r: resistance of the sample

C2: standard capacitor R3, R4 and C4: variable components of Schering Bridge

ii) Given;

 At balance condition

                        

- equation (i)

By equating the real and imaginary parts of both side; Therefore:

; and

For series model,

b)

        ,

A solid insulating material of relative permittivity,

  comprises a cylindrical air-filled

cavity of depth t. The material has thickness T, where its value is much greater compared to the depth of the cavity. Show that the voltage across the sample (material),



is given by the expression.

Where



     is the voltage across the cavity.

From the above equation, explain why partial discharge can occur in the cavity even though only normal service voltage is applied across the insulating material.

Answer

T

Cb

t

bc1 void

Va

Va

Void in solid dielectric material

Void capacitance, Capacitance series with void, Remaining capacitance,

Equivalent circuit of void in the structure

          

(i)

 

(ii)

 

(iii)

    

Voltage across cavity,

Therefore

 

(iv)

                               

Since T >> t;

                          

Yields

proven

In term of electric field strength

       and

Hence

 

 And

Since

 





 

, the electrical stress in the void is greater than the electrical stress across

the sample. Thus, partial discharge occurs due to the very small gap of the void even at the normal service voltage.

Question 5

a)

Partial discharge detection and measurements were limited to the laboratory due to high levels of electrical noise at the switchyards. With the aid of proper diagrams describe the main sources of interferences or noise which hampered the partial discharge detection process and the techniques to suppress the interferences.

Answer Typical noise source : i)

Power supply

ii)

Voltage regulator

iii)

High voltage source

iv)

Filtering of the HV source

v)

Feeder line and electrodes

vi)

Coupling capacitor

vii)

Loose conductive objects in the vicinity of the last location

viii)

Pulse shaped interferences

ix)

Electromagnetic waves by radio transmitter (harmonic interferences)

x)

Interference currents in ground system of PD measuring instrument

The interferences can be reduced by;

b)

i)

Filter grounding

ii)

Shielded room

iii)

Separate source

iv)

Filter AC source (Harmonics)

A solid dielectric has a small cavity. Draw an equivalent circuit for such an arrangement and define all symbols. Derive the expression for electric field strength across the cavity.

Answer

R4

 AC Sup ply

r 

C

C2 C4

R3

c

Void in solid dielectric material

b

Equivalent circuit of void in the structure

In the absence of the void, the electric field within the insulation would everywhere be

ₐ ₐ

E  = V /d. Neglecting conductance, the insulation, which its void, can be represented by the three capacitances a, b, and c. The void capacitance is represented by c. If the void has length d 1 and the cross sectional area A 1 (perpendicular to d1) then;

C=

Ԑ₁₁

The void is connected to the conducting plates through two capacitance b 1 and b2 . Their series combination is represented by the single capacitance b. Clearly,

b=

Where

Ԑ

ԐԐ₁ ₁

  is the relative permittivity of the insulating material. The remaining

capacitor a has the value,

 ԐԐ Where A is the cross sectional area of the insulation, minus the (usually small) cross sectional area of the void.

The voltage across the void is;

  ₐ     And substituting for the capacitances gives,

Ԑ₁  ₐ Ԑ₁ ₁ In the term of electric field strength we have,

ₐ ₐ

V  =E .T

and

Vc = EC.t

Where Ec is the dielectric field strength in the void. Substituting in equation gives,

Ԑ₁  ₐ Ԑ₁ ₁ ₁ Ԑ ₁ Ԑ ₐ

Usually the void will be small, so that d