Transmission Line Induct Ance

8/9/2019 Transmission Line Induct Ance

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TransmissionLine:

Inductanceand

CapacitanceCalculation

Prof S. A.Soman

SeriesImpedance of TransmissionLines

ShuntCapacitance of TransmissionLines

Transmission Line: Inductance and Capacitance

Calculation

Prof S. A. Soman

Visiting FacultyDepartment of Electrical and Computer Engineering

Virginia [email protected]

Department of Electrical EngineeringIIT-Bombay

October 5, 2012

Prof S. A. Soman (IIT-Bombay) Transmission Line: Inductance and Capacitance Calculation

October 5, 2012 1 / 33

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TransmissionLine:

Inductanceand


Prof S. A.

Soman



1 Series Impedance of Transmission Lines

2 Shunt Capacitance of Transmission Lines


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TransmissionLine:

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Soman



AC and DC Resistance

r DC = ρ

A

Ω

m

ρ of hard-drawn Cu at 20◦C = 1.77 × 10−8 Ωm.

ρ of Al at 20◦C = 2.83 × 10−8 Ωm.

ρ depends on temperature

ρT 2 = M + T 2M + T 1

ρT 1

where M = temperature constant

AC resistance is greater than DC resistance due to skineffect.


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Soman



Skin Effect

Skin effect - When AC current flows through a conductor, current density

near the surface is higher than current density inside.J = J s e

−d δ

where δ is called the skin depth.

Skin depth is given by

δ =

r 2ρ

ωµ

whereρ = resistivity of theconductor.ω = angular frequency of thecurrent = 2π× frequency.µ = absolute magneticpermeability of the

conductor.As the AC frequency increases, skin depth decreases. At 60 Hz in copper, itis about 8.5 mm. Over 98% of the current will flow within a layer 4 timesthe skin depth from the surface. Therefore, R AC > R DC .

Reference: http://en.wikipedia.org/wiki/Skin_effect

Prof S. A. Soman (IIT-Bombay) Transmission Line: Inductance and Capacitance Calculation October 5, 2012 4 / 33

http://en.wikipedia.org/wiki/Skin_effecthttp://en.wikipedia.org/wiki/Skin_effecthttp://find/


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TransmissionLine:

Inductanceand


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Soman



Series Inductance

Series Inductance of transmission line (per metre)It has two components

Internal inductance

External inductance

Recall that

Inductance, L

=

λ

I Henry;Energy stored, E =

1

2LI 2 Joules.

Ampere’s law states

¸ −→

H .−→

dl = I net = Ni

linesflux

current

I


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TransmissionLine:

Inductanceand


Prof S. A.

Soman



Flux density external to the conductor

xa

Case-I

Let radius of conductor be a.

Let x >a

H x .2πx = I

H x = I

2πx

B x = µ0.H x = µ0I

2πx Weber/m2

where

µ0 = 4π × 10−7 H/m




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Inductanceand


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Soman



Flux Density internal to the conductor I

x

xa

Let x ≤ aAssuming uniform current density,I encl (x ) =

πx 2

πa2 I =

x 2

a2 I

By Ampere’s law,

2πx .H x = I encl (x ) x

≤a

= x 2

a2 I

∴ H x = 1

2π

x

a2I A/m x ≤ a

B x = µ0

2π

x

a2I A/m


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Inductanceand


Prof S. A.

Soman



Overall flux density of a conductor

µ0 I

2 π a

Bx

xa

Figure: Plot of Flux density vs distance


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TransmissionLine:

Inductanceand


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Soman



Internal Inductance I

1m

a

1 Calculate the energy stored in magnetic field inside the conductor.

2 E int = 1

2

á 0

B x 2

µ0(2πx .dx )J/m (2)

∵ (dv = 2π×

dx ×

l where l is the length of the conductor)

3 B x = µ0

2π

x

a2I (3)

4 Substituting B x from (3) in (2)

E int = µ0

16π

I 2J/m (4)




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Soman


ShuntCapacitance of

TransmissionLines

Internal Inductance II

5 Energy stored in an inductor

E = 1

2LI 2J/m if L is H/m (5)

6 Equating (5) and (4)

Lint = µ0

8π=

4π × 10−78π

= 1

2 ×10−7H/m

Prof S. A. Soman (IIT-Bombay) Transmission Line: Inductance and Capacitance CalculationOctober 5, 2012 10 / 33

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Soman


ShuntCapacitance of

TransmissionLines

External Inductance

External Inductance in a cylindrical ring of inner diameter D 1 and outer diameterD 2

1 B x = µ0

2π

I

x x ≥ a

2 dv = 2πx .dx × 1m3 For a ≤ D 1 ≤ x ≤ D 2

E ext (D 1,D 2) = µ0

4πI 2

D 2ˆ

D 1

1

x dx

= µ0

4πln

D 2

D 1I 2J/m

4 E ext = 1

2Lext I

2 J/m

5 Equating (3) and (4)

Lext = µ0

2πln

D 2

D 1


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TransmissionLine:

Inductanceand


Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Inductance of a 1φ two-wire transmission line

D

Path 2

Path 1

a

a’

r

1 Path-1 encloses current I. Hence,¸

H .dl = I .

2 Path-2 encloses zero net current. As such, it does not contribute to netinductance of a 2-wire system.

3 Now,Lnet due to the above two-wire system is as follows.


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TransmissionLine:

Inductanceand


Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines


Similarly ,La′ = 2 × 10−7 ln D

r ′

2

H/m

Lnet = La + La′

= 4 × 10−7 ln( D q r ′

1 r ′

2

)

if (r ′

1 = r ′

2 = r ′)

Lnet = 4 × 10−7 ln(D

r ′) H/m


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TransmissionLine:

Inductanceand


Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines


Observation:

1 Greater separation between transmission line implies

greater inductance of the line.2 Greater the radius of the conductors in a transmission line,

the lower the inductance.

Inductive reactive,X L = j 2πfL Ω


T i i

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TransmissionLine:

Inductanceand


Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Inductance of a 3φ transmission line

We assume the following:

Three phase currents sum to zero i.e.−→

I a +−→

I b +−→

I c = 0.Lines are transposed i.e. a conductor in phase a sees thesame average topology e.g., conductor a sees config 1,2and 3 equal number of times. Hence, its inductance perkm is average of the inductance of the three configs.

a

b

c

c

a

b

b

c

a

Config−1 Config−2 Config−3

If the lines are not transposed then due to unsymmetricalgeometry, the phase inductances will not be equal. Thiswill lead to network unbalance.


T i i I d f 3φ i i li I

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TransmissionLine:

Inductanceand


Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Inductance of a 3φ transmission line I

c

a

b

1

3

2

D p1

D

D p2

p3

P

Cond.b

Cond.c

Cond.aCond.b

Cond.a

Cond.cCond.a

Cond.b

Cond.c

c1 c2 c3

We compute flux linkage of phase a conductor in config-1 upto point P,

λnet a,c 1 = L(0, D p 1)I a + L(D 12, D p 2)I b + L(D 13, D p 3)I c

λnet a,c 1 = 2 × 10−7

[lnD p 1

r ′I a + I b ln

D p 2

D 12+ I c ln

D p 3

D 13] H/m

Similarly,

λnet a,c 2 = 2 × 10

−7[ln

D p 2

r ′I a + I b ln

D p 3

D 23+ I c ln

D p 1

D 12] H/m

λnet a,c 3 = 2 × 10−7 [lnD

p 3r ′

]I a + I b lnD

p 1D 13

+ I c lnD

p 2D 23

] H/m

Due to transposition,

λnet a =

λnet a,c 1 + λnet a,c 2 + λ

net a,c 3

3

= 2 × 10−7

[ln 3p

D p 1D p 2D p 3

r ′I a + I b

ln 3p

D p 1D p 2D p 33p

D 12D 23D 13+ I c

ln 3p

D p 1D p 2D p 33p

D 12D 23D 13] H/m


T a s issio I d f 3φ i i li II

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TransmissionLine:

Inductanceand


Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Inductance of a 3φ transmission line II

We have used following identity,

ln(m, n) = ln m + ln n

ln(m

n

) = ln m

−ln n

ln mp = p ln m

ln(m

n)p = p ln(

m

n)

Let DP eq = 3p

D p 1D p 2D p 3.

Geometric Mean Distance, GMD = 3

√ D 12D 23D 13Then,

λnet a = 2 × 10−7[I a ln DP eq

GMR + I b ln

DP eq

GMD + I c ln

DP eq

GMD ]


Transmission I d f 3φ i i li III



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TransmissionLine:

Inductanceand


Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Inductance of a 3φ transmission line III

Since, I b + I c = −I a

λnet a = 2 × 10−7I a[ln DP eq

GMR − ln DP eq

GMD ]

= 2

×10−7 ln

GMD

GMR

∴ Lnet a = 2 × 10−7 ln GMD

GMR H/m

This is net inductance per phase of a transmission line. It considers mutualcoupling with other phases.

Transposition of lines is assumed.Ī a + Ī b + Ī c = 0 is assumed but not necessarily balanced currents.


Transmission C it f T i i Li I

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TransmissionLine:

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Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Capacitance of Transmission Lines I

Voltage and charge relation

+qE − q

Figure: Two transmission line conductors of a 1φ line.

Fig.2 shows that electric field exists between two conductors of atransmission line.

This implies the existence of shunt capacitance between the conductors.Recall the following concepts

a. C = Q (Charge)

V (Volts) Farads


Transmission C it f T i i Li II

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Inductanceand


Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Capacitance of Transmission Lines II

+ q

− q

EV

Figure: Electric field between two conductors.

b. Electric Flux Density −→

D (C /m2)−→

D = ǫ0ǫr −→

E

whereAbsolute permittivity in vacuum (free space),

ǫ0 = 8.85 × 10−12

Farad/meterRelative permittivity for air, ǫr = 1−→

E = Electric field intensity at a point.


Transmission Definition of Voltage and Electric field

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a s ss oLine:

Inductanceand


Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Definition of Voltage and Electric field

Defn-1: The voltage or potential difference, V measured involts between two points is the amount of work needed to

move a Coulomb of charge from one point to another one.Defn-2: The electric field intensity

−→

E (measured in N/Cor V/m)is the force exerted on a Coulomb of charge at agiven point in the electric field.


Transmission Gauss’s Law

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Line:Inductance

andCapacitanceCalculation

Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Gauss s Law

‚ −→D .−→

ds = q whereq = the charge inside the surface in coulombs.−→

ds = the unit vector normal to the surface in m2.


Transmission Electric field around a long straight conductor I

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Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Electric field around a long straight conductor I

Gauss law at a distance ’x ’ from

+ + + +

+++

+++

+

++

++

++++++

a b

D

D

A

B

+

Figure: Electric field around a charge of +q

ǫ0¸ ¸ −→

E .−→ds = q

where q = the charge in coulomb on a unit length of wire.

Since−→E and

−→ds are in parallel, cos θ = 1.

Remark: Inside of a conductor is an equipotential surface.


Transmission Electric field around a long straight conductor II

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Line:Inductance


Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Electric field around a long straight conductor II

Since−→E is constant in magnitude at a distance x , we get for a length ’l ’ of conductor

ǫ0E 2πx × l = q × l

E (x ) =q

2πǫ0 x

∴ V a − V b =

b ˆ

a

−→E −→dx

=q

2πǫ0

b ˆ

a

1

x

−→dx

=q

2πǫ0ln

D B

D A

∴ C ab =q

V a − V b =

2πǫ0

lnD B

D A

Farads/meter

Recall that electric field is a conservative field. Therefore, the results are independent of the path of integration.


TransmissionL Capacitance of 1φ two-wire transmission line I

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Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Capacitance of 1φ two-wire transmission line I

+q −q

DA B

r r

Figure: Capacitance of 1φ two-wire line.

Charge, q is in C/m.

V AB = ∆V +q AB − ∆V −q

AB

where ∆V +q AB

is the potential drop from A to B due to charge +q .

Similarly, we define ∆V −q AB

.

∆V +q AB

= q 2πǫ0

ln D r

Volts/metre

Note:

For points outside of the conductor, the charge can beimagined to be at the center of the conductor.


TransmissionLi Capacitance of 1φ two-wire transmission line II

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Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Capacitance of 1φ two-wire transmission line II

Surface of conductor is equipotential. So, −→

E is along theradius.

Similarly, ∆V −q BA

= −q 2πǫ0

ln D

r

i.e. ∆V −q

AB =

q

2πǫ0ln

D

r .

Hence, ∆V AB = q

πǫ0ln

D

r

∴ C AB = πǫ0

ln D

r

Farads/meter

Note that unlike inductance calculation, denominator does not have r ′.


TransmissionLi e Shunt capacitance to neutral I

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Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Shunt capacitance to neutral I

With respect to neutral plane, conductor A is on positive potential andconductor B is on negative potential.

By symmetry,

V an = V AB

2

∴ C n = C an = C bn = 2πǫ

ln

D

r

We state the following result (without proof)

For a 3φ transposed transmission line, the capacitance of phase to neutral

C = 2πǫ0

ln D eq r

F/m

where D eq =GMD between conductors.


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TransmissionLine: Bundled conductors

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Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Bundled conductors

When a big thick conductor is split into multiple conductors, it is called as a

bundled conductor. Bundling increases the effective GMR of the conductor.

dd

d

d d

d

d

d

Figure: Bundled conductor arrangements.

Advantages:

Reduced reactance. For e.g., for a four-strand bundle,

D b s = 16q

(D s × d × d × d × 212 )4

where D b s = GMR of bundled conductor, D s = GMR of individual

conductors composing the bundle.

Reduces the effect of corona. The stress on the conductors in a bundle iscomparatively less than what it is on a single conductor.


TransmissionLine: Review Questions

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Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Q

1 Why is shunt capacitance of cable much higher than thatof an overhead line at same kV level?

2 Why is series reactance of overhead line higher than thatof shunt capacitance of cable at same kV level?

3 What is a bundled conductor?


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Inductanceand


Prof S. A.

Soman


ShuntCapacitance of

TransmissionLines

Thank You

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Transmission Line Induct Ance

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