Test: Symmetrical Components - 1 - Electrical Engineering (EE) MCQ

Concept:

The relation between the line currents in terms of the symmetrical components of currents is given below.

I_a0 = Zero Sequence Component of Current
I_a1 = Positive Sequence Component of Current
I_a2 = Negative Sequence Component of Current
a = 1∠120°, which represents the rotation of 120° in clockwise direction.
a² = 1∠-120° or 1∠240° in anticlockwise direction or in clockwise direction, respectively.

Calculation:
I_a = 100∠0°, I_b = 141.4∠225°, I_c = 100∠90°
Zero sequence component of current,

Positive sequence component of current,

⇒ I_a1 = 111∠15°
Negative sequence component of current,

I_a2 = 29.88 ∠ 105°

Given that, positive sequence impedance (Z₁) = (1 + j10) Ω

Zero sequence impedance (Z₀) = (4 + j31) Ω

We know that, Z₁ = Z_s – Z_m

and Z₀ = Z_s + 2Z_m

Z₁ = 1 + j10 = Z_s – Z_m → (1)

and Z₀ = 4 + j31 = Z_s + 2Z_m → (2)

from equations (1) and (2)

⇒ Z_m = 1 + j7

Imaginary part of Z_m = 7 Ω.

Concept:

Fortescue’s Theorem:

A unbalance set of ‘n’ phasors may be resolved into (n - 1) balance n-phase system of different phase sequence and one zero phase sequence system.

A zero-phase sequence system is one in which all phasors are of equal magnitude and angle.

Considered three phasors are represented by a, b, c in such a way that their phase sequence is (a b c).

The positive phase sequence will be (a b c) and the negative phase sequence will be (a c b).

Assumed that subscript 0, 1, 2 refer to zero sequences, positive sequence, negative sequence respectively.

Current I_a, I_b, I_c represented an unbalance set of current phasor as shown,

Each of the original unbalance phasor is the sum of its component and it can be written as,

I_a = I_a0 + I_a1 + I_a2

I_b = I_b0 + I_b1 + I_b2

I_c = I_c0 + I_c1 + I_c2

For a balance position phase sequence (a b c) we can write the following relation,

I_a0 = I_b0 = I_c0

I_b1 = α² I_a1

I_c1 = α I_a1

I_b2 = α I_a2

I_c2 = α² I_a2

From the above equation I_a, I_b, I_c can be written in terms of phase sequence component,

I_a = I_a0 + I_a1 + I_a2

I_b = I_b0 +  α² I_a1 + α I_a2

I_c = I_a0 + α I_a1 + α² I_a2

The above equation can be written in form of Matrix as shown,

Calculation:
I_a = 1∠-90° P.u
I_b2 = 4∠-150° P.u
I_c0 = 3∠90° P.u
I_a = I_a1 + I_a2 + I_a0 
I_b2 = α I_a2
⇒ 4∠-150° = (1∠120°) I_a2
⇒ I_a2 = 4∠-270° 
I_a0 = I_b0 = I_c0 = 3∠90° 
I_a1 = 8∠90° 
I_b1 = α² I_a1 = 8∠150° 
I_b = I_b0 + I_b1 + I_b2 
= 8∠150° + 4∠-150° + 3∠90° 
= 11.53∠154.4°  
Therefore the magnitude of phase current I_b in pu is 11.53∠154.4°  

Symmetrical components:
Any unbalanced system of n phasors can be resolved into the n-system of balanced phasors. These subsystems of balanced phasors are called symmetrical components. These are used in the analysis of unbalanced three-phase faults.

Positive sequence components:
Set of three phasors equal in magnitude, displaced from each other by 120° in phase, and having the same phase sequence as the original phasors constitute positive sequence components. They are denoted by suffix 1.

Negative sequence components:
Set of three phasors equal in magnitude, displaced from each other by 120° in phase, and having phase sequence opposite to that original phasors constitute the negative sequence component. They are denoted by suffix 2.

Zero sequence components:
The set of three phasors equal in magnitude and all phases (with no mutual phase displacement) constitute zero sequence components. They are denoted by suffix 0.

Sequence Impedance:

• The sequence impedance of the network describes the behavior of the system under asymmetrical fault conditions.

Positive Sequence Impedance

• The impedance offered by the network to the flow of positive sequence current is called the positive sequence impedance.
• The positive sequence means all the electrical quantities are numerically equal and displaces each other by 120º.

Negative Sequence Impedance

• The negative sequence impedance means the impedance offered by the network to the flows of the negative sequence current.

Zero Sequence Impedance

• The impedance offered to zero sequence current is called the zero sequence impedance.
• Generally, resistance offered by the power system is less when compared to reactance.
• So, the resistance of the system is neglected and reactance is considered.

Values of sequence reactance of some equipment are as follows:

Where,
X₀ is the zero sequence reactance
X₁ is the positive sequence reactance
X₂ is the negative sequence reactance
X_d” is the direct axis reactance

Using Thevenin's Theorem, V_th = V all the three impedances will be in parallel.

Single line diagram:

So, equivalent p.u. impedance of each generator is
Z_th = Z_n / 3 = 0.15 / 3 = 0.05 pu
Key Points

• All generators are connected in parallel to increase current going to the line because all generators are at equal voltages.
• Generators are shown with their impedances in a single line diagram.

Concept:
The relation between the line currents in terms of the symmetrical components of currents is given below.

I_a0 = Zero Sequence Component of Current

I_a1 = Positive Sequence Component of Current

I_a2 = Negative Sequence Component of Current

a = 1∠120°, which represents the rotation of 120° in the clockwise direction.

a² = 1∠-120° or 1∠240° in anticlockwise direction or clockwise direction, respectively.

1 + a + a² = 0

Calculation:

Given that

Ia = 4 + j6, I_b = 2 - j2, I_c = -3 + j2

Zero sequence component of current,

∴ I₀ = 1 + j2 A

Concept:
In a line to ground fault, the current through the neutral during fault is given by
I_f = 3I_a0

Where I_a0 is the zero sequence current of generator

Calculation:
Given that, the zero sequence current of generator for line to ground fault = j2.4 pu
The current through the neutral during the fault = 3 × j2.4 = j7.2 pu

Concept:
For the transformer,
Negative sequence component = positive sequence component = zero sequence component
X₀ = X₁ = X₂
For transmission line,
Negative sequence component = positive sequence component
Zero sequence component > positive sequence component
X₀ > X₁ = X₂
Application:
Given that, X₁ = 0.3 pu, X₂ = 0.3 pu, X₀ = 0.8 pu
⇒ X₀ > X₁ = X₂
Therefore, the element is a transmission line.

The contact in line C of the triple pole switch contactor fails to connect when switched on.

⇒ I_c = 0 A

And I_a + I_b = 0

⇒ I_a = -I_b

⇒ I_a = 25 ∠0°, I_b = 25 ∠-180°  
Positive sequence current,

 
= 1/3(25 ∠ 0^∘ + 1∠120 × 25∠ −180)
= 14.4 ∠ -30° 

Concept:
Fortescue’s Theorem: A unbalance set of ‘n’ phasors may be resolved into (n - 1) balance n-phase system of different phase sequence and one zero phase sequence system.
A zero-phase sequence system is one in which all phasors are of equal magnitude and angle.
Considered three phasors are represented by a, b, c in such a way that their phase sequence is (a b c).
The positive phase sequence will be (a b c) and the negative phase sequence will be (a c b).
Assumed that subscript 0, 1, 2 refer to zero sequences, positive sequence, negative sequence respectively.
Current I_a, I_b, I_c represented an unbalance set of current phasor as shown,

Each of the original unbalance phasor is the sum of its component and it can be written as,

I_a = I_a0 + I_a1 + I_a2

I_b = I_b0 + I_b1 + I_b2

I_c = I_c0 + I_c1 + I_c2

For a balance position phase sequence (a b c) we can write the following relation,

I_a0 = I_b0 = I_c0

I_b1 = α² I_a1

I_c1 = α I_a1

I_b2 = α I_a2

I_c2 = α² I_a2

From the above equation I_a, I_b, I_c can be written in terms of phase sequence component,

I_a = I_a0 + I_a1 + I_a2

I_b = I_b0 +  α² I_a1 + α I_a2

I_c = I_a0 + α I_a1 + α² I_a2

The above equation can be written in form of Matrix as shown,

Calculation:
Given,

I_B0 = 0.1 ∠0° p.u

I_A0 = I_B0 = I_C0 = 0.1 ∠0° p.u

I_A = 1.1 ∠0° p.u.

I_C = (1 ∠120° + 0.1) p.u.

From the above concept,

I_A = I_A0 + I_A1 + I_A2

I_A - I_A0 = I_A1 + I_A2 = 1.1 ∠0° + 0.1 ∠0° p.u = 1 ∠0° p.u

I_A1 + I_A2 = 1 ∠0° p.u .... (1)

I_C = I_C0 + I_C1 + I_C2 = I_A0 + α I_A1 +α² IA₂

1 ∠120° + 0.1 = 0.1 ∠0° + α I_A1 +α² I_A2

α I_A1 +α₂ I_A2 = 1 ∠120° .... (2)

Adding equation (1) and (2),

(I_A1 + I_A2) + (α I_A1 +α² I_A2) = 1 ∠0° + 1 ∠120°

I_A1(α + 1) + I_A2(α² + 1)  = 1 ∠0° + 1 ∠120° (Since, 1 + α + α² = 0 ⇒ 1 + α = -α²)

I_A1(-α²) + IA²(-α)  = 1 ∠0° + 1 ∠120°

α² I_A1 + α I_A2 = 1 ∠-120° .... (3)

From above concept,

I_B = I_A0 + α² I_A1 + α I_A2

I_C = 0.1 ∠0° + 1 ∠-120°

Purpose of Transposition:

• Transmission lines are transposed to prevent interference with neighbouring telephone lines.
• The transposition arrangement of high voltage lines helps to reduce the system power loss.
• We have developed transposition system for Single circuit tower using same tension tower with reduced deviation angle.
• Transposition arrangement of power line helps to reduce the effect of inductive coupling.
• It is proved more economical Solution, in comparison of the conventional transposition system.

Important:

Transposition arrangement

The transposition arrangement of the conductor can simply show in the following the figure. The conductor in Position 1, Position 2 and Position 3 changes in a specific arrangement to reduce the effect of capacitance and the electrostatic unbalanced voltages.

 

Z₁ = Z₂ = Z_S - Z_m
Z_o = Z_S + 2Z_m
Where,
Z₁ = positive sequence impedance
Z₂ = positive sequence impedance
Z_m = mutual impedance
Z_S = self impedance
Positive and negative sequence impedances are equal.