Power System Analysis II — Chapter 4: Unbalanced System Analysis

Asked 3 times
2078 Kartik · 8 marks
2076 Asoj · 4 marks
2073 Chaitra · 6 marks

Derive an expression to determine 3-phase (complex) power of an unbalanced system in terms of symmetrical components of voltages and currents.

Answer

The total complex power in a three-phase (balanced or unbalanced) system equals three times the sum of the complex powers of the zero, positive and negative sequence components: $S = 3(V_{a0}I_{a0}^* + V_{a1}I_{a1}^* + V_{a2}I_{a2}^*)$ .

Phase-domain power

With phase-to-neutral voltages $V_a, V_b, V_c$ and line currents $I_a, I_b, I_c$ :

S = P + jQ = V_a I_a^* + V_b I_b^* + V_c I_c^*

In matrix form, with $V_p = [V_a\ V_b\ V_c]^T$ and $I_p = [I_a\ I_b\ I_c]^T$ :

S = V_p^T I_p^*

Symmetrical component relations

V_p = A V_s, \qquad I_p = A I_s, \qquad A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & a^2 & a \\ 1 & a & a^2 \end{bmatrix}, \quad a = 1\angle 120^\circ

where $V_s = [V_{a0}\ V_{a1}\ V_{a2}]^T$ and $I_s = [I_{a0}\ I_{a1}\ I_{a2}]^T$ .

Substitution

\begin{aligned} S &= (A V_s)^T (A I_s)^* \\ &= V_s^T A^T A^* I_s^* \end{aligned}

Since $A$ is symmetric, $A^T = A$ . Also $a^* = a^2$ and $(a^2)^* = a$ , so

A^* = \begin{bmatrix} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{bmatrix}

Multiplying, using $1 + a + a^2 = 0$ and $a^3 = 1$ :

A^T A^* = \begin{bmatrix} 1+1+1 & 1+a+a^2 & 1+a^2+a \\ 1+a^2+a & 1+a^3+a^3 & 1+a^4+a^2 \\ 1+a+a^2 & 1+a^2+a^4 & 1+a^3+a^3 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} = 3U

Therefore

\begin{aligned} S &= 3\,V_s^T I_s^* \\ &= 3\left(V_{a0} I_{a0}^* + V_{a1} I_{a1}^* + V_{a2} I_{a2}^*\right) \end{aligned}

Remarks

The transformation is power invariant except for the factor 3; if the transformation is defined with $\frac{1}{\sqrt3}A$ (unitary form), the factor 3 disappears.
There are no cross terms such as $V_{a1}I_{a2}^*$ : a voltage of one sequence and a current of another sequence produce no net average power.
For a balanced system $V_{a0}=V_{a2}=0$ and $I_{a0}=I_{a2}=0$ , so $S = 3V_{a1}I_{a1}^*$ , the familiar $3V_{ph}I_{ph}^*$ .

Asked 2 times
2078 Bhadra · 8 marks
2070 Asar · 8 marks

Starting from suitable point, derive an expression to determine sequence impedances for a balanced three-phase star-connected load with impedance Zs in each phase and grounded neutral with impedance Zn, and show that Z0 = Zs + 3Zn, Z1 = Zs and Z2 = Zs.

Answer

Sequence impedance is the impedance offered by a circuit to currents of one sequence only: $Z_0 = V_{a0}/I_{a0}$ , $Z_1 = V_{a1}/I_{a1}$ , $Z_2 = V_{a2}/I_{a2}$ . We start from the phase-domain voltage equations of the load and transform them with the symmetrical component matrix $A$ .

Circuit and phase equations

   Ia -->  Zs
 a o-----/\/\---+
   Ib -->  Zs   |
 b o-----/\/\---+ n
   Ic -->  Zs   |
 c o-----/\/\---+
                |  In = Ia+Ib+Ic
               Zn
                |
              ground

Voltage of each phase to ground = drop in $Z_s$ + drop in $Z_n$ :

\begin{aligned} V_a &= Z_s I_a + Z_n (I_a + I_b + I_c) \\ V_b &= Z_s I_b + Z_n (I_a + I_b + I_c) \\ V_c &= Z_s I_c + Z_n (I_a + I_b + I_c) \end{aligned}

In matrix form $V_p = Z_p I_p$ with

Z_p = \begin{bmatrix} Z_s+Z_n & Z_n & Z_n \\ Z_n & Z_s+Z_n & Z_n \\ Z_n & Z_n & Z_s+Z_n \end{bmatrix}

Transformation to sequence quantities

Put $V_p = A V_s$ and $I_p = A I_s$ :

A V_s = Z_p A I_s \;\Rightarrow\; V_s = A^{-1} Z_p A\, I_s = Z_s^{seq} I_s

with $A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}$ , $A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$ .

Write $Z_p = Z_s U + Z_n J$ , where $U$ is the unit matrix and $J$ is the matrix of all ones. Then $A^{-1}(Z_sU)A = Z_sU$ . For $J$ : every row of $J A$ is $[3\ \ 0\ \ 0]$ (since column sums of $A$ are $3, 1+a^2+a=0, 1+a+a^2=0$ ), and $A^{-1}$ times that gives

A^{-1} J A = \begin{bmatrix} 3&0&0\\0&0&0\\0&0&0 \end{bmatrix}

Hence

Z_{seq} = A^{-1} Z_p A = \begin{bmatrix} Z_s+3Z_n & 0 & 0 \\ 0 & Z_s & 0 \\ 0 & 0 & Z_s \end{bmatrix}

Result

Writing the three rows separately:

\begin{aligned} V_{a0} &= (Z_s + 3Z_n) I_{a0} &\Rightarrow\; Z_0 &= Z_s + 3Z_n \\ V_{a1} &= Z_s I_{a1} &\Rightarrow\; Z_1 &= Z_s \\ V_{a2} &= Z_s I_{a2} &\Rightarrow\; Z_2 &= Z_s \end{aligned}

Physical meaning

Positive and negative sequence currents are balanced sets; their sum is zero, so no current flows in $Z_n$ and the impedance is just $Z_s$ .
Zero sequence currents are equal and in phase in all three phases, so the neutral carries $3I_{a0}$ . The drop across $Z_n$ is $3I_{a0}Z_n$ , which appears as an extra $3Z_n$ in the zero sequence circuit of each phase.
If the neutral is solidly grounded ( $Z_n = 0$ ), $Z_0 = Z_s$ ; if it is isolated ( $Z_n \to \infty$ ), $Z_0 = \infty$ and no zero sequence current can flow.
Since the sequence impedance matrix is diagonal, the three sequence networks are independent (decoupled).

Asked 2 times
2073 Shrawan · 8 marks
2071 Shrawan · 10 marks

Deduce the sequence networks for balanced Y connected load with neutral grounded through impedance Zn and show also that current through neutral impedance is three times the zero sequence current.

Answer

For a balanced star load with neutral grounded through $Z_n$ , the sequence networks are three separate circuits with impedances $Z_1 = Z_2 = Z_s$ and $Z_0 = Z_s + 3Z_n$ , and the neutral carries $I_n = 3I_{a0}$ .

Neutral current is three times zero sequence current

By the symmetrical component definitions:

\begin{aligned} I_a &= I_{a0} + I_{a1} + I_{a2} \\ I_b &= I_{a0} + a^2 I_{a1} + a I_{a2} \\ I_c &= I_{a0} + a I_{a1} + a^2 I_{a2} \end{aligned}

Adding and using $1 + a + a^2 = 0$ :

\begin{aligned} I_n &= I_a + I_b + I_c \\ &= 3I_{a0} + (1+a^2+a)I_{a1} + (1+a+a^2)I_{a2} \\ &= 3I_{a0} \end{aligned}

So only zero sequence currents flow through the neutral, and the neutral current is $3I_{a0}$ . Positive and negative sequence currents form balanced sets that cancel at the star point.

Circuit and phase equations

   Ia -->  Zs
 a o-----/\/\---+
   Ib -->  Zs   |
 b o-----/\/\---+ n
   Ic -->  Zs   |
 c o-----/\/\---+
                |  In = Ia+Ib+Ic
               Zn
                |
              ground

Voltage of each phase to ground = drop in $Z_s$ + drop in $Z_n$ :

\begin{aligned} V_a &= Z_s I_a + Z_n (I_a + I_b + I_c) \\ V_b &= Z_s I_b + Z_n (I_a + I_b + I_c) \\ V_c &= Z_s I_c + Z_n (I_a + I_b + I_c) \end{aligned}

In matrix form $V_p = Z_p I_p$ with

Z_p = \begin{bmatrix} Z_s+Z_n & Z_n & Z_n \\ Z_n & Z_s+Z_n & Z_n \\ Z_n & Z_n & Z_s+Z_n \end{bmatrix}

Transformation to sequence quantities

Put $V_p = A V_s$ and $I_p = A I_s$ :

A V_s = Z_p A I_s \;\Rightarrow\; V_s = A^{-1} Z_p A\, I_s = Z_s^{seq} I_s

with $A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}$ , $A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$ .

Write $Z_p = Z_s U + Z_n J$ , where $U$ is the unit matrix and $J$ is the matrix of all ones. Then $A^{-1}(Z_sU)A = Z_sU$ . For $J$ : every row of $J A$ is $[3\ \ 0\ \ 0]$ (since column sums of $A$ are $3, 1+a^2+a=0, 1+a+a^2=0$ ), and $A^{-1}$ times that gives

A^{-1} J A = \begin{bmatrix} 3&0&0\\0&0&0\\0&0&0 \end{bmatrix}

Hence

Z_{seq} = A^{-1} Z_p A = \begin{bmatrix} Z_s+3Z_n & 0 & 0 \\ 0 & Z_s & 0 \\ 0 & 0 & Z_s \end{bmatrix}

Sequence networks

 Positive         Negative         Zero
 a1 --Ia1-->      a2 --Ia2-->      a0 --Ia0-->
   |                |                |
  Zs               Zs               Zs
   |                |                |
   n1 (ref)         n2 (ref)        3Zn
                                     |
                                    n0 (ref)

$V_{a1} = Z_s I_{a1}$ , $V_{a2} = Z_s I_{a2}$ , $V_{a0} = (Z_s + 3Z_n) I_{a0}$ .
The impedance $Z_n$ appears as $3Z_n$ in the zero sequence network only because the voltage drop across it is $Z_n \times 3I_{a0}$ , while the network is drawn for one phase carrying $I_{a0}$ .
Solidly grounded neutral: $Z_n = 0$ ; ungrounded neutral: zero sequence network open between $Z_s$ and the reference.

Asked 2 times
2072 Chaitra · 6 marks
2083 Baishakh (new course) · 5 marks

With suitable mathematical aid, show (justify) that in a delta connected system, zero sequence components of current are absent in the line currents.

Answer

In a delta connected system there is no neutral or ground return path, so the three line currents always add to zero; hence their zero sequence component $I_{a0} = \frac13(I_a+I_b+I_c)$ is zero.

Line currents of a delta

       Ia -->  a
   o----------o
             / \
         Iab/   \Ica
           /     \
   Ib --> b-------c <-- Ic
              Ibc

Applying KCL at the three corners:

\begin{aligned} I_a &= I_{ab} - I_{ca} \\ I_b &= I_{bc} - I_{ab} \\ I_c &= I_{ca} - I_{bc} \end{aligned}

Zero sequence line current

\begin{aligned} I_{a0} &= \frac13 (I_a + I_b + I_c) \\ &= \frac13\left[(I_{ab}-I_{ca}) + (I_{bc}-I_{ab}) + (I_{ca}-I_{bc})\right] \\ &= 0 \end{aligned}

This holds for any values of the delta currents, balanced or unbalanced. So line currents into a delta never contain zero sequence components.

Phase (delta) currents in sequence form

Let the delta currents have components $I_{ab0}, I_{ab1}, I_{ab2}$ . Then

I_a = I_{ab} - I_{ca}

For the zero sequence part, $I_{ab0} = I_{bc0} = I_{ca0}$ , so it cancels in every line current. For positive and negative sequence:

I_{a1} = (1 - a)\,I_{ab1} = \sqrt3\,\angle -30^\circ\, I_{ab1}, \qquad I_{a2} = (1 - a^2)\,I_{ab2} = \sqrt3\,\angle 30^\circ\, I_{ab2}

Consequences

A zero sequence current can circulate inside the delta (for example, from induced zero sequence voltages or third harmonics) but it cannot leave the delta into the lines.
In the zero sequence network, a delta connected load or winding is open from the line side; for a transformer, the delta winding is shown as a short to the reference on its own side (current circulates) but open to the line.
Similarly, the line-to-line voltages of any system have no zero sequence component because $V_{ab}+V_{bc}+V_{ca} = 0$ .

2080 Bhadra · 2+6 marks

What do you understand by symmetrical components? Derive an expression of 3-phase complex power in terms of symmetrical components of voltages and currents.

Answer

Symmetrical components

Symmetrical components (Fortescue's theorem, 1918) state that any set of three unbalanced phasors can be replaced by the sum of three balanced sets:

Positive sequence: three equal phasors 120° apart with the same phase sequence as the original (a-b-c).
Negative sequence: three equal phasors 120° apart with opposite sequence (a-c-b).
Zero sequence: three equal phasors in phase with each other.

With operator $a = 1\angle120^\circ$ :

\begin{bmatrix} V_a \\ V_b \\ V_c \end{bmatrix} = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix} \begin{bmatrix} V_{a0} \\ V_{a1} \\ V_{a2} \end{bmatrix}, \quad\text{i.e. } V_p = A V_s

The same relation holds for currents, $I_p = A I_s$ . This lets an unbalanced problem be solved as three balanced, independent problems.

Complex power in terms of symmetrical components

Total complex power in phase quantities:

S = V_a I_a^* + V_b I_b^* + V_c I_c^* = V_p^T I_p^*

Substituting $V_p = AV_s$ , $I_p = AI_s$ :

S = (AV_s)^T (AI_s)^* = V_s^T A^T A^* I_s^*

$A$ is symmetric ( $A^T = A$ ), and since $a^* = a^2$ :

A^* = \begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}

A^T A^* = \begin{bmatrix} 3 & 1+a+a^2 & 1+a^2+a \\ 1+a^2+a & 1+a^3+a^3 & 1+a^4+a^2 \\ 1+a+a^2 & 1+a^2+a^4 & 1+a^3+a^3 \end{bmatrix} = \begin{bmatrix} 3&0&0\\0&3&0\\0&0&3 \end{bmatrix}

using $1+a+a^2 = 0$ , $a^3 = 1$ , $a^4 = a$ . Hence

S = 3V_s^T I_s^* = 3\left(V_{a0}I_{a0}^* + V_{a1}I_{a1}^* + V_{a2}I_{a2}^*\right)

So total power is three times the sum of the sequence powers of phase a, with no cross products between different sequences. For a balanced system only the positive sequence term remains: $S = 3V_{a1}I_{a1}^*$ .

2075 Chaitra · 8 marks

Define Sequence impedances. For the star connected load show that impedance matrix has non-zero elements on its diagonal matrix.

Answer

Sequence impedance of an element is the impedance it offers to the flow of currents of one particular sequence: $Z_1 = V_{a1}/I_{a1}$ (positive), $Z_2 = V_{a2}/I_{a2}$ (negative) and $Z_0 = V_{a0}/I_{a0}$ (zero sequence), when only that sequence is present.

Star connected load

Consider a balanced star load with impedance $Z_s$ per phase and neutral grounded through $Z_n$ (solid grounding: $Z_n = 0$ ).

Circuit and phase equations

   Ia -->  Zs
 a o-----/\/\---+
   Ib -->  Zs   |
 b o-----/\/\---+ n
   Ic -->  Zs   |
 c o-----/\/\---+
                |  In = Ia+Ib+Ic
               Zn
                |
              ground

Voltage of each phase to ground = drop in $Z_s$ + drop in $Z_n$ :

\begin{aligned} V_a &= Z_s I_a + Z_n (I_a + I_b + I_c) \\ V_b &= Z_s I_b + Z_n (I_a + I_b + I_c) \\ V_c &= Z_s I_c + Z_n (I_a + I_b + I_c) \end{aligned}

In matrix form $V_p = Z_p I_p$ with

Z_p = \begin{bmatrix} Z_s+Z_n & Z_n & Z_n \\ Z_n & Z_s+Z_n & Z_n \\ Z_n & Z_n & Z_s+Z_n \end{bmatrix}

Transformation to sequence quantities

Put $V_p = A V_s$ and $I_p = A I_s$ :

A V_s = Z_p A I_s \;\Rightarrow\; V_s = A^{-1} Z_p A\, I_s = Z_s^{seq} I_s

with $A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}$ , $A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$ .

Write $Z_p = Z_s U + Z_n J$ , where $U$ is the unit matrix and $J$ is the matrix of all ones. Then $A^{-1}(Z_sU)A = Z_sU$ . For $J$ : every row of $J A$ is $[3\ \ 0\ \ 0]$ (since column sums of $A$ are $3, 1+a^2+a=0, 1+a+a^2=0$ ), and $A^{-1}$ times that gives

A^{-1} J A = \begin{bmatrix} 3&0&0\\0&0&0\\0&0&0 \end{bmatrix}

Hence

Z_{seq} = A^{-1} Z_p A = \begin{bmatrix} Z_s+3Z_n & 0 & 0 \\ 0 & Z_s & 0 \\ 0 & 0 & Z_s \end{bmatrix}

The sequence impedance matrix has non-zero elements only on its diagonal:

Z_0 = Z_s + 3Z_n, \qquad Z_1 = Z_s, \qquad Z_2 = Z_s

All off-diagonal elements are zero, which means a current of one sequence produces a voltage drop of the same sequence only. So the three sequence networks are decoupled and can be solved separately, which is the main advantage of symmetrical components.

The same holds if the load also has equal mutual impedances $Z_m$ between phases: then $Z_1 = Z_2 = Z_s - Z_m$ and $Z_0 = Z_s + 2Z_m + 3Z_n$ , still diagonal.

2080 Bhadra · 8 marks

Define sequence impedances. For the star connected load, verify that impedance matrix has non-zero elements on its diagonal matrix. Comment on zero sequence impedance.

Answer

Sequence impedance of an element is the impedance it offers to the flow of currents of one particular sequence: $Z_1 = V_{a1}/I_{a1}$ (positive), $Z_2 = V_{a2}/I_{a2}$ (negative) and $Z_0 = V_{a0}/I_{a0}$ (zero sequence), when only that sequence is present.

Star connected load

Consider a balanced star load with impedance $Z_s$ per phase and neutral grounded through $Z_n$ (solid grounding: $Z_n = 0$ ).

Circuit and phase equations

   Ia -->  Zs
 a o-----/\/\---+
   Ib -->  Zs   |
 b o-----/\/\---+ n
   Ic -->  Zs   |
 c o-----/\/\---+
                |  In = Ia+Ib+Ic
               Zn
                |
              ground

Voltage of each phase to ground = drop in $Z_s$ + drop in $Z_n$ :

\begin{aligned} V_a &= Z_s I_a + Z_n (I_a + I_b + I_c) \\ V_b &= Z_s I_b + Z_n (I_a + I_b + I_c) \\ V_c &= Z_s I_c + Z_n (I_a + I_b + I_c) \end{aligned}

In matrix form $V_p = Z_p I_p$ with

Z_p = \begin{bmatrix} Z_s+Z_n & Z_n & Z_n \\ Z_n & Z_s+Z_n & Z_n \\ Z_n & Z_n & Z_s+Z_n \end{bmatrix}

Transformation to sequence quantities

Put $V_p = A V_s$ and $I_p = A I_s$ :

A V_s = Z_p A I_s \;\Rightarrow\; V_s = A^{-1} Z_p A\, I_s = Z_s^{seq} I_s

with $A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}$ , $A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$ .

Write $Z_p = Z_s U + Z_n J$ , where $U$ is the unit matrix and $J$ is the matrix of all ones. Then $A^{-1}(Z_sU)A = Z_sU$ . For $J$ : every row of $J A$ is $[3\ \ 0\ \ 0]$ (since column sums of $A$ are $3, 1+a^2+a=0, 1+a+a^2=0$ ), and $A^{-1}$ times that gives

A^{-1} J A = \begin{bmatrix} 3&0&0\\0&0&0\\0&0&0 \end{bmatrix}

Hence

Z_{seq} = A^{-1} Z_p A = \begin{bmatrix} Z_s+3Z_n & 0 & 0 \\ 0 & Z_s & 0 \\ 0 & 0 & Z_s \end{bmatrix}

The sequence impedance matrix has non-zero elements only on its diagonal:

Z_0 = Z_s + 3Z_n, \qquad Z_1 = Z_s, \qquad Z_2 = Z_s

All off-diagonal elements are zero, which means a current of one sequence produces a voltage drop of the same sequence only. So the three sequence networks are decoupled and can be solved separately, which is the main advantage of symmetrical components.

The same holds if the load also has equal mutual impedances $Z_m$ between phases: then $Z_1 = Z_2 = Z_s - Z_m$ and $Z_0 = Z_s + 2Z_m + 3Z_n$ , still diagonal.

Comment on zero sequence impedance

$Z_0 = Z_s + 3Z_n$ is larger than $Z_1$ and $Z_2$ , because zero sequence currents of all three phases are in phase and add up to $3I_{a0}$ in the neutral; the drop $3I_{a0}Z_n$ is felt by each phase.
Solidly grounded star ( $Z_n = 0$ ): $Z_0 = Z_s$ .
Ungrounded star ( $Z_n = \infty$ ): $Z_0 = \infty$ ; zero sequence current cannot flow, and the zero sequence network is open.
Delta connected load: no return path, so $Z_0$ seen from the lines is infinite.
In a fault study, the method of neutral grounding therefore mainly changes the zero sequence network and hence the single line to ground fault current; this is why resistance or reactance grounding is used to limit earth fault current.

2082 Baishakh · 3+5 marks

Write down the importance of symmetrical components transformation for unbalance system analysis. Also derive the expression to calculate the components of 3-phase unbalance currents.

Answer

Symmetrical components transform an unbalanced set of three phasors into three balanced sets (zero, positive and negative sequence), so an unbalanced three-phase problem can be solved as three simple single-phase problems.

Importance for unbalanced system analysis

Decoupling: for symmetrical (balanced) equipment such as lines, transformers and machines, the sequence impedance matrix is diagonal. So the zero, positive and negative sequence networks are independent and can be solved separately, instead of solving three coupled phase equations.
Unsymmetrical fault calculation: L-G, L-L and L-L-G faults are solved by connecting the sequence networks at the fault point in a simple way (series, parallel). This is the standard method for fault level and relay setting.
Per phase analysis is kept: each sequence network is a single-phase circuit, so per unit methods and Thevenin equivalents still apply.
Physical meaning: negative sequence current shows rotor heating in machines; zero sequence current shows earth faults and flows only through neutral/ground paths. Relays (earth fault, negative sequence) are built on these quantities.
Measurable: sequence components can be measured with sequence filters, useful for protection.

Derivation for currents

Let $I_a, I_b, I_c$ be an unbalanced set of line currents. By Fortescue's theorem each is the sum of zero, positive and negative sequence components:

\begin{aligned} I_a &= I_{a0} + I_{a1} + I_{a2} \\ I_b &= I_{b0} + I_{b1} + I_{b2} \\ I_c &= I_{c0} + I_{c1} + I_{c2} \end{aligned}

Using operator $a = 1\angle120^\circ$ (so $a^2 = 1\angle240^\circ$ , $1 + a + a^2 = 0$ , $a^3 = 1$ ), all components are written in terms of phase a:

Zero sequence: $I_{b0} = I_{c0} = I_{a0}$
Positive sequence (a-b-c): $I_{b1} = a^2 I_{a1}$ , $I_{c1} = a I_{a1}$
Negative sequence (a-c-b): $I_{b2} = a I_{a2}$ , $I_{c2} = a^2 I_{a2}$

Therefore

\begin{bmatrix} I_a \\ I_b \\ I_c \end{bmatrix} = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix} \begin{bmatrix} I_{a0} \\ I_{a1} \\ I_{a2} \end{bmatrix} \quad\Rightarrow\quad I_p = A I_s

Solving for the components:

Add the three equations: $I_a + I_b + I_c = 3I_{a0} + (1+a^2+a)I_{a1} + (1+a+a^2)I_{a2} = 3I_{a0}$
Multiply the second by $a$ and the third by $a^2$ , then add: $I_a + aI_b + a^2I_c = (1+a+a^2)I_{a0} + (1+a^3+a^3)I_{a1} + (1+a^2+a^4)I_{a2} = 3I_{a1}$
Multiply the second by $a^2$ and the third by $a$ , then add: $I_a + a^2I_b + aI_c = 3I_{a2}$

\begin{aligned} I_{a0} &= \tfrac13\left(I_a + I_b + I_c\right) \\ I_{a1} &= \tfrac13\left(I_a + aI_b + a^2I_c\right) \\ I_{a2} &= \tfrac13\left(I_a + a^2I_b + aI_c\right) \end{aligned}

or in matrix form

I_s = A^{-1}I_p, \qquad A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}

Components of phases b and c follow from those of phase a using the relations above.

Notes: $I_n = I_a + I_b + I_c = 3I_{a0}$ , so zero sequence current exists only if there is a neutral or ground return. In a 3-wire or delta system $I_{a0} = 0$ .

2072 Kartik · 5 marks

Derive the expression to calculate symmetrical components of 3-phase un-balance currents.

Answer

Any unbalanced set of three-phase currents can be expressed as the sum of a zero sequence, a positive sequence and a negative sequence balanced set; the components of phase a are $I_{a0} = \frac13(I_a+I_b+I_c)$ , $I_{a1} = \frac13(I_a+aI_b+a^2I_c)$ and $I_{a2} = \frac13(I_a+a^2I_b+aI_c)$ .

Derivation for currents

Let $I_a, I_b, I_c$ be an unbalanced set of line currents. By Fortescue's theorem each is the sum of zero, positive and negative sequence components:

\begin{aligned} I_a &= I_{a0} + I_{a1} + I_{a2} \\ I_b &= I_{b0} + I_{b1} + I_{b2} \\ I_c &= I_{c0} + I_{c1} + I_{c2} \end{aligned}

Using operator $a = 1\angle120^\circ$ (so $a^2 = 1\angle240^\circ$ , $1 + a + a^2 = 0$ , $a^3 = 1$ ), all components are written in terms of phase a:

Zero sequence: $I_{b0} = I_{c0} = I_{a0}$
Positive sequence (a-b-c): $I_{b1} = a^2 I_{a1}$ , $I_{c1} = a I_{a1}$
Negative sequence (a-c-b): $I_{b2} = a I_{a2}$ , $I_{c2} = a^2 I_{a2}$

Therefore

\begin{bmatrix} I_a \\ I_b \\ I_c \end{bmatrix} = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix} \begin{bmatrix} I_{a0} \\ I_{a1} \\ I_{a2} \end{bmatrix} \quad\Rightarrow\quad I_p = A I_s

Solving for the components:

Add the three equations: $I_a + I_b + I_c = 3I_{a0} + (1+a^2+a)I_{a1} + (1+a+a^2)I_{a2} = 3I_{a0}$
Multiply the second by $a$ and the third by $a^2$ , then add: $I_a + aI_b + a^2I_c = (1+a+a^2)I_{a0} + (1+a^3+a^3)I_{a1} + (1+a^2+a^4)I_{a2} = 3I_{a1}$
Multiply the second by $a^2$ and the third by $a$ , then add: $I_a + a^2I_b + aI_c = 3I_{a2}$

\begin{aligned} I_{a0} &= \tfrac13\left(I_a + I_b + I_c\right) \\ I_{a1} &= \tfrac13\left(I_a + aI_b + a^2I_c\right) \\ I_{a2} &= \tfrac13\left(I_a + a^2I_b + aI_c\right) \end{aligned}

or in matrix form

I_s = A^{-1}I_p, \qquad A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}

Components of phases b and c follow from those of phase a using the relations above.

Notes: $I_n = I_a + I_b + I_c = 3I_{a0}$ , so zero sequence current exists only if there is a neutral or ground return. In a 3-wire or delta system $I_{a0} = 0$ .

2076 Asoj · 6 marks

Starting from a suitable point show that a 3-phase unbalanced system of voltages can be represented by a symmetrical components.

Answer

Fortescue's theorem states that three unbalanced phasors can be represented by three balanced sets of phasors: positive, negative and zero sequence. We show this by writing the phase voltages as sums of sequence components and proving the components can always be found uniquely.

Starting point: operator a

a = 1\angle120^\circ = -0.5 + j0.866, \quad a^2 = 1\angle240^\circ, \quad a^3 = 1, \quad 1 + a + a^2 = 0

Multiplying a phasor by $a$ rotates it by +120°.

Three balanced sets

 Positive (a-b-c)    Negative (a-c-b)    Zero
     Va1                 Va2           Va0 Vb0 Vc0
      ^                   ^             ^   ^   ^
     / \                 / \            |   |   |
  Vc1   Vb1          Vb2   Vc2        (all in phase)

Positive: $V_{a1}$ , $V_{b1} = a^2V_{a1}$ , $V_{c1} = aV_{a1}$
Negative: $V_{a2}$ , $V_{b2} = aV_{a2}$ , $V_{c2} = a^2V_{a2}$
Zero: $V_{a0} = V_{b0} = V_{c0}$

Phase voltages as sums

\begin{aligned} V_a &= V_{a0} + V_{a1} + V_{a2} \\ V_b &= V_{a0} + a^2V_{a1} + aV_{a2} \\ V_c &= V_{a0} + aV_{a1} + a^2V_{a2} \end{aligned} \qquad\Longleftrightarrow\qquad V_p = A V_s,\ \ A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}

The components always exist

The determinant of $A$ is $3(a - a^2) = j3\sqrt3 \neq 0$ , so $A$ is invertible, and for any given $V_a, V_b, V_c$ there is one unique set of components:

\begin{aligned} V_{a0} &= \tfrac13\left(V_a + V_b + V_c\right) \\ V_{a1} &= \tfrac13\left(V_a + aV_b + a^2V_c\right) \\ V_{a2} &= \tfrac13\left(V_a + a^2V_b + aV_c\right) \end{aligned} \qquad\Longleftrightarrow\qquad V_s = A^{-1}V_p

(obtained by adding the three equations, and by adding them after multiplying by $a, a^2$ or $a^2, a$ , using $1+a+a^2=0$ ).

Hence a 3-phase unbalanced voltage system is exactly represented by its symmetrical components. A balanced positive sequence system gives $V_{a0} = V_{a2} = 0$ ; line-to-line voltages always have $V_0 = 0$ because they sum to zero.

2082 Bhadra (new course) · 2+3 marks

Explain the significance of symmetrical components in power system analysis. Also derive the expression for phasor voltages in terms of the symmetrical components of voltages.

Answer

Significance of symmetrical components

Symmetrical components replace an unbalanced set of three phasors by three balanced sets (zero, positive and negative sequence).

For symmetrical equipment the three sequence networks are independent, so unbalanced faults and loads are solved with three single-phase circuits instead of coupled three-phase equations.
Unsymmetrical faults (L-G, L-L, L-L-G) are solved by simple interconnection of sequence networks.
Each sequence has a physical meaning: negative sequence causes rotor heating of machines; zero sequence shows ground currents. Many protective relays work on these quantities.

Phase voltages in terms of symmetrical components

With $a = 1\angle120^\circ$ ( $a^2 = 1\angle240^\circ$ , $1+a+a^2 = 0$ ), the components of phases b and c are expressed through phase a:

Zero sequence: $V_{b0} = V_{c0} = V_{a0}$
Positive sequence: $V_{b1} = a^2V_{a1}$ , $V_{c1} = aV_{a1}$
Negative sequence: $V_{b2} = aV_{a2}$ , $V_{c2} = a^2V_{a2}$

Each phase voltage is the sum of its components:

\begin{aligned} V_a &= V_{a0} + V_{a1} + V_{a2} \\ V_b &= V_{a0} + a^2V_{a1} + aV_{a2} \\ V_c &= V_{a0} + aV_{a1} + a^2V_{a2} \end{aligned}

\begin{bmatrix} V_a\\V_b\\V_c \end{bmatrix} = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}\begin{bmatrix} V_{a0}\\V_{a1}\\V_{a2} \end{bmatrix}

2076 Chaitra · 4 marks

How do the symmetrical components help us in analyzing unbalance electric power system? Explain briefly.

Answer

Symmetrical components convert one coupled unbalanced three-phase problem into three uncoupled balanced single-phase problems, which are easy to solve.

Decomposition: any unbalanced voltages or currents are split into positive, negative and zero sequence sets: $V_p = AV_s$ , $V_s = A^{-1}V_p$ .
Decoupled networks: for balanced (symmetrical) equipment, the impedance matrix becomes diagonal in sequence form. A positive sequence current causes only positive sequence drops, and so on. So each sequence network is a separate single-phase circuit with its own impedance $Z_1, Z_2, Z_0$ .
Unbalance only at one point: in a real fault, the network is balanced everywhere except at the fault. Only the fault point conditions couple the sequence networks, e.g. for an L-G fault the three networks are connected in series: $I_{a1} = I_{a2} = I_{a0} = \dfrac{E_a}{Z_1+Z_2+Z_0+3Z_f}$ .
Known sequence data: manufacturers and textbooks give $Z_1, Z_2, Z_0$ of generators, transformers and lines, so the networks are easy to build.
Recombination: after solving, phase quantities are found back with $I_p = AI_s$ .
Physical insight: negative sequence shows unbalance and rotor heating; zero sequence shows earth paths. This helps in protection design and choice of neutral grounding.

2079 Bhadra · 2+6 marks

What is significance of sequence components? Deduce the sequence network for a loaded synchronous generator grounded through impedance Zn.

Answer

Significance of sequence components

Sequence components let an unbalanced three-phase system be analysed as three independent balanced single-phase networks (positive, negative, zero). For balanced equipment the sequence impedances are uncoupled, so unbalanced faults are solved by connecting the sequence networks only at the fault point.

Sequence networks of a synchronous generator grounded through $Z_n$

A generator produces only balanced positive sequence EMFs $E_a$ , $E_b = a^2E_a$ , $E_c = aE_a$ . Let $Z_1$ , $Z_2$ be its positive and negative sequence impedances and $Z_{g0}$ its zero sequence impedance per phase; the neutral is grounded through $Z_n$ .

            Ia -->
    +-(Ea)--Zs--o a
    |
 n  +-(Eb)--Zs--o b     Va, Vb, Vc measured
    |                   to ground
    +-(Ec)--Zs--o c
    |
   Zn  In = Ia+Ib+Ic
    |
  ground

Positive sequence: the EMF is purely positive sequence; positive sequence currents are balanced so no current flows in $Z_n$ :

V_{a1} = E_a - I_{a1}Z_1

Negative sequence: no negative sequence EMF; balanced currents, no current in $Z_n$ :

V_{a2} = -I_{a2}Z_2

Zero sequence: no zero sequence EMF; the neutral current is $I_n = 3I_{a0}$ , so the drop in $Z_n$ is $3I_{a0}Z_n$ :

V_{a0} = -I_{a0}Z_{g0} - 3I_{a0}Z_n = -I_{a0}(Z_{g0} + 3Z_n) = -I_{a0}Z_0

In matrix form:

\begin{bmatrix} V_{a0}\\V_{a1}\\V_{a2} \end{bmatrix} = \begin{bmatrix} 0\\E_a\\0 \end{bmatrix} - \begin{bmatrix} Z_0&0&0\\0&Z_1&0\\0&0&Z_2 \end{bmatrix}\begin{bmatrix} I_{a0}\\I_{a1}\\I_{a2} \end{bmatrix}

 Positive          Negative          Zero
 F1 o              F2 o              F0 o
    | Ia1 ^           | Ia2 ^           | Ia0 ^
   Z1                Z2               Zg0
    |                 |                 |
  (Ea)                |                3Zn
    |                 |                 |
 N1 o (ref)        N2 o (ref)        N0 o (ground)

Loaded generator: when the generator supplies a balanced load before the fault, the load appears in the positive sequence network (and in the negative/zero networks with its own sequence impedances). Using Thevenin's theorem at the fault point, $E_a$ is replaced by the prefault voltage $V_f$ at that point and $Z_1$ by the Thevenin impedance of generator in parallel with load; the zero sequence network still contains $3Z_n$ .

2081 Bhadra · 6+2 marks

With proper mathematical interpretations, justify that for a symmetrical static circuit, positive and negative sequence impedances are equal. Also discuss why the sequence impedances for a generator are not equal.

Answer

A symmetrical static circuit (transmission line, transformer, static load) has equal self impedances $Z_s$ in all phases and equal mutual impedances $Z_m$ between all pairs of phases; for such a circuit $Z_1 = Z_2 = Z_s - Z_m$ .

Mathematical proof

Voltage drops across the three phases:

\begin{aligned} V_a &= Z_s I_a + Z_m I_b + Z_m I_c \\ V_b &= Z_m I_a + Z_s I_b + Z_m I_c \\ V_c &= Z_m I_a + Z_m I_b + Z_s I_c \end{aligned} \qquad\Rightarrow\qquad Z_p = \begin{bmatrix} Z_s&Z_m&Z_m\\Z_m&Z_s&Z_m\\Z_m&Z_m&Z_s \end{bmatrix}

Transform with $V_p = AV_s$ , $I_p = AI_s$ : $V_s = A^{-1}Z_pA\,I_s$ .

Write $Z_p = (Z_s - Z_m)U + Z_mJ$ ( $U$ = unit matrix, $J$ = matrix of ones). Since $A^{-1}UA = U$ and $A^{-1}JA = \text{diag}(3, 0, 0)$ (column sums of $A$ are $3$ , $1+a+a^2 = 0$ , $0$ ):

Z_{seq} = A^{-1}Z_pA = \begin{bmatrix} Z_s+2Z_m & 0 & 0 \\ 0 & Z_s-Z_m & 0 \\ 0 & 0 & Z_s-Z_m \end{bmatrix}

Hence

Z_0 = Z_s + 2Z_m, \qquad Z_1 = Z_2 = Z_s - Z_m

Physically, in a static circuit the impedance does not depend on the order in which the phase currents reach their peaks (a-b-c or a-c-b); reversing the phase sequence only interchanges b and c, which gives the same circuit. So $Z_1 = Z_2$ . Zero sequence currents are in phase, so the mutual effects add ( $+2Z_m$ ) and $Z_0$ is different, usually larger.

Why generator sequence impedances are not equal

A synchronous generator is a rotating machine, so the sequence of the current changes the field it produces relative to the rotor:

Positive sequence: the armature MMF rotates at synchronous speed in the same direction as the rotor, stationary relative to it. The reactance is $X_d''$ , $X_d'$ or $X_d$ depending on the time after the fault.
Negative sequence: the MMF rotates opposite to the rotor at relative speed $2\omega_s$ . It induces double-frequency currents in the field and damper windings, which oppose the flux, so the reactance is low, about $X_2 \approx \frac12(X_d'' + X_q'')$ .
Zero sequence: the three in-phase currents produce almost no air-gap MMF (they cancel in space), so only leakage flux remains and $X_0$ is the smallest.

So for a generator $X_1 > X_2 > X_0$ in general, and $Z_1 \neq Z_2$ .

2068 Chaitra · 8 marks

For a 3-phase symmetrical static circuit, justify that sequence networks are decoupled.

Answer

For a three-phase symmetrical static circuit (line, transformer, static load), the impedance matrix in sequence quantities is diagonal, so a current of one sequence produces voltage drops of that sequence only; this means the three sequence networks are decoupled.

Phase equations

Each phase has self impedance $Z_s$ and equal mutual impedance $Z_m$ to each other phase:

\begin{bmatrix} V_a\\V_b\\V_c \end{bmatrix} = \begin{bmatrix} Z_s&Z_m&Z_m\\Z_m&Z_s&Z_m\\Z_m&Z_m&Z_s \end{bmatrix}\begin{bmatrix} I_a\\I_b\\I_c \end{bmatrix}, \qquad V_p = Z_pI_p

Transformation

With $V_p = AV_s$ , $I_p = AI_s$ :

V_s = A^{-1}Z_pA\,I_s = Z_{seq}I_s

where

A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}, \qquad A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}

First find $Z_pA$ , column by column (using $1 + a + a^2 = 0$ ):

Column 1: each row $= Z_s + 2Z_m$ , so column $= (Z_s+2Z_m)[1\ 1\ 1]^T$
Column 2: rows $= Z_s + Z_m(a^2+a) = Z_s - Z_m$ times $[1\ a^2\ a]^T$
Column 3: rows $= (Z_s - Z_m)$ times $[1\ a\ a^2]^T$

So $Z_pA = A\,\text{diag}(Z_s+2Z_m,\ Z_s-Z_m,\ Z_s-Z_m)$ ; the columns of $A$ are eigenvectors of $Z_p$ . Therefore

Z_{seq} = A^{-1}Z_pA = \begin{bmatrix} Z_s+2Z_m & 0 & 0 \\ 0 & Z_s-Z_m & 0 \\ 0 & 0 & Z_s-Z_m \end{bmatrix}

Result

V_{a0} = Z_0I_{a0}, \qquad V_{a1} = Z_1I_{a1}, \qquad V_{a2} = Z_2I_{a2}

with $Z_0 = Z_s + 2Z_m$ and $Z_1 = Z_2 = Z_s - Z_m$ .

All off-diagonal (mutual) terms between sequences are zero, so:

Positive sequence current causes only positive sequence voltage drop, and similarly for negative and zero.
Each sequence network can be drawn and solved independently as a single-phase circuit.
The networks are connected to each other only at points of unbalance (fault points), which is the basis of unsymmetrical fault analysis.

If the circuit is not symmetrical (e.g. unequal mutual impedances in an untransposed line), off-diagonal terms appear and the sequence networks become coupled.

2075 Chaitra · 8 marks

Show that the sequence component are decoupled for complex power flow in a power system.

Answer

The complex power of a three-phase system splits into three separate terms, one for each sequence, with no cross terms between different sequences. So each sequence network carries its own power independently; this is what "decoupled for power flow" means.

Proof

Complex power in phase quantities:

S = V_aI_a^* + V_bI_b^* + V_cI_c^* = V_p^TI_p^*

With $V_p = AV_s$ , $I_p = AI_s$ , where $A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}$ :

S = (AV_s)^T(AI_s)^* = V_s^T\,A^TA^*\,I_s^*

$A$ is symmetric, so $A^T = A$ ; and $a^* = a^2$ , so $A^* = \begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$ . Using $1+a+a^2 = 0$ , $a^3 = 1$ :

A^TA^* = \begin{bmatrix} 3 & 1+a+a^2 & 1+a^2+a \\ 1+a^2+a & 1+a^3+a^3 & 1+a^4+a^2 \\ 1+a+a^2 & 1+a^2+a^4 & 1+a^3+a^3 \end{bmatrix} = \begin{bmatrix} 3&0&0\\0&3&0\\0&0&3 \end{bmatrix}

Hence

S = 3V_s^TI_s^* = 3V_{a0}I_{a0}^* + 3V_{a1}I_{a1}^* + 3V_{a2}I_{a2}^*

Why this shows decoupling

Each term contains a voltage and a current of the same sequence only. Terms like $V_{a1}I_{a2}^*$ or $V_{a0}I_{a1}^*$ do not appear, because the off-diagonal elements of $A^TA^*$ are zero.
Physically, a positive sequence voltage acting with a negative (or zero) sequence current produces only an oscillating power whose sum over the three phases is zero; it gives no net power.
Therefore the power delivered by the positive sequence network ( $S_1 = 3V_{a1}I_{a1}^*$ ), negative sequence network ( $S_2$ ) and zero sequence network ( $S_0$ ) can be calculated separately and simply added:

S = S_0 + S_1 + S_2

Together with the diagonal sequence impedance matrix of symmetrical equipment (voltage of one sequence depends only on current of the same sequence), this shows the three sequence networks are fully decoupled for both voltage drop and power.

Example

A balanced system has only positive sequence: $S = 3V_{a1}I_{a1}^*$ . A generator feeding an unbalanced load delivers useful power mainly through the positive sequence network; power in the negative sequence network appears as losses (for example, extra rotor heating).

The factor 3 appears because $A$ is not unitary. With the power-invariant form $\frac{1}{\sqrt3}A$ , $S = V_s^TI_s^*$ exactly.

2075 Chaitra · 6+2 marks

Prove that the symmetrical component transformation is power invariant. Discuss also the significance of zero sequence circuit.

Answer

A transformation is power invariant if the power calculated from the transformed (sequence) quantities equals the power calculated from the original (phase) quantities. The symmetrical component transformation gives $S = 3(V_{a0}I_{a0}^* + V_{a1}I_{a1}^* + V_{a2}I_{a2}^*)$ , which is power invariant apart from a constant factor 3, and exactly invariant in its unitary form.

Proof

Complex power in phase quantities:

S = V_aI_a^* + V_bI_b^* + V_cI_c^* = V_p^TI_p^*

With $V_p = AV_s$ , $I_p = AI_s$ , where $A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}$ :

S = (AV_s)^T(AI_s)^* = V_s^T\,A^TA^*\,I_s^*

$A$ is symmetric, so $A^T = A$ ; and $a^* = a^2$ , so $A^* = \begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$ . Using $1+a+a^2 = 0$ , $a^3 = 1$ :

A^TA^* = \begin{bmatrix} 3 & 1+a+a^2 & 1+a^2+a \\ 1+a^2+a & 1+a^3+a^3 & 1+a^4+a^2 \\ 1+a+a^2 & 1+a^2+a^4 & 1+a^3+a^3 \end{bmatrix} = \begin{bmatrix} 3&0&0\\0&3&0\\0&0&3 \end{bmatrix}

Hence

S = 3V_s^TI_s^* = 3V_{a0}I_{a0}^* + 3V_{a1}I_{a1}^* + 3V_{a2}I_{a2}^*

Exact power invariance

If the transformation matrix is taken as $T = \frac{1}{\sqrt3}A$ , then $T^TT^* = \frac13\cdot3U = U$ (unit matrix) and

S = V_p^TI_p^* = V_s^TI_s^* = V_{0}I_{0}^* + V_{1}I_{1}^* + V_{2}I_{2}^*

So the power is the same in both reference frames: the transformation is power invariant. With the usual matrix $A$ the sequence power of phase a is multiplied by 3 (three phases), and there are still no cross terms.

Significance of the zero sequence circuit

Zero sequence currents are equal and in phase in all three phases, so they need a return path through neutral or ground: $I_n = 3I_{a0}$ . The zero sequence network therefore depends on transformer winding connections (star grounded, star ungrounded, delta) and on neutral grounding.
It decides the current in ground faults (L-G, L-L-G). Since about 70–80% of faults are L-G, the zero sequence network is essential for finding earth fault current, choosing neutral grounding impedance and setting earth fault relays.
A neutral impedance $Z_n$ appears as $3Z_n$ only in the zero sequence circuit, so grounding methods change only this network.
Delta windings block zero sequence current from the lines but let it circulate inside; this is used to trap third harmonics and to isolate ground faults between voltage levels.
Zero sequence currents in lines induce voltages in nearby telephone lines (interference) and are used for ground-fault detection.

2076 Asoj · 6 marks

A three phase balanced Y-connected load with self and mutual elements is shown in figure below. The load neutral is grounded with Zn = 0.0. Determine the sequence impedance. [Figure: phases R, Y, B carrying IR, IY, IB through self impedances ZR, ZY, ZB, with mutual impedances ZRY, ZYB and ZRB between phases; star point returned to ground carrying In; phase voltages VR, VY, VB]

Answer

For a balanced star load with self impedance $Z_s$ in each phase and equal mutual impedance $Z_m$ between phases, solidly grounded ( $Z_n = 0$ ), the sequence impedances are $Z_0 = Z_s + 2Z_m$ and $Z_1 = Z_2 = Z_s - Z_m$ .

Since the load is balanced: $Z_R = Z_Y = Z_B = Z_s$ and $Z_{RY} = Z_{YB} = Z_{RB} = Z_m$ .

Phase equations

With $Z_n = 0$ the star point is at ground potential:

\begin{aligned} V_R &= Z_sI_R + Z_mI_Y + Z_mI_B \\ V_Y &= Z_mI_R + Z_sI_Y + Z_mI_B \\ V_B &= Z_mI_R + Z_mI_Y + Z_sI_B \end{aligned} \qquad Z_p = \begin{bmatrix} Z_s&Z_m&Z_m\\Z_m&Z_s&Z_m\\Z_m&Z_m&Z_s \end{bmatrix}

Transformation

$V_p = AV_s$ and $I_p = AI_s$ give $V_s = A^{-1}Z_pA\,I_s$ .

Multiply $Z_p$ by each column of $A$ , using $1 + a + a^2 = 0$ :

Column $[1\ 1\ 1]^T$ : each row gives $Z_s + 2Z_m$
Column $[1\ a^2\ a]^T$ : row 1 gives $Z_s + Z_m(a^2+a) = Z_s - Z_m$ , and the other rows give the same factor times $a^2$ , $a$
Column $[1\ a\ a^2]^T$ : factor $Z_s - Z_m$

So

Z_{seq} = A^{-1}Z_pA = \begin{bmatrix} Z_s+2Z_m & 0 & 0 \\ 0 & Z_s-Z_m & 0 \\ 0 & 0 & Z_s-Z_m \end{bmatrix}

Result

Z_0 = Z_s + 2Z_m, \qquad Z_1 = Z_s - Z_m, \qquad Z_2 = Z_s - Z_m

The matrix is diagonal, so the sequence networks are decoupled.
If the neutral were grounded through $Z_n$ , every element of $Z_p$ would gain $Z_n$ and only $Z_0$ would change: $Z_0 = Z_s + 2Z_m + 3Z_n$ .

2076 Chaitra · 8 marks

A 3-phase unbalanced source with voltage given by Vabc = [200∠25° 100∠−155° 80∠100°]ᵀ V is feeding a 3-phase star connected load with series impedance per phase of Zs = 8 + j24 Ohms and load neutral impedance of Zn = j1.5 Ohms. Determine the: (i) Symmetrical components of voltage (ii) Symmetrical components of currents

Answer

Find the sequence voltages with $V_s = A^{-1}V_p$ , then divide each by its own sequence impedance of the load, since a balanced star load has a diagonal sequence impedance matrix.

Assumption: the given voltages are the phase-to-ground voltages applied to the load terminals (source neutral grounded), $a = 1\angle120^\circ$ .

Phase voltages in rectangular form

\begin{aligned} V_a &= 200\angle25^\circ = 181.26 + j84.52\ \text{V} \\ V_b &= 100\angle{-155^\circ} = -90.63 - j42.26\ \text{V} \\ V_c &= 80\angle100^\circ = -13.89 + j78.78\ \text{V} \end{aligned}

(i) Symmetrical components of voltage

\begin{aligned} V_{a0} &= \tfrac13(V_a + V_b + V_c) = 25.58 + j40.35 = 47.77\angle57.63^\circ\ \text{V} \\ V_{a1} &= \tfrac13(V_a + aV_b + a^2V_c) = 112.78 - j0.07 = 112.78\angle{-0.03^\circ}\ \text{V} \\ V_{a2} &= \tfrac13(V_a + a^2V_b + aV_c) = 42.90 + j44.24 = 61.62\angle45.88^\circ\ \text{V} \end{aligned}

Check: $V_{a0} + V_{a1} + V_{a2} = 181.26 + j84.52 = V_a$ .

Sequence impedances of the load

\begin{aligned} Z_1 = Z_2 &= Z_s = 8 + j24 = 25.30\angle71.57^\circ\ \Omega \\ Z_0 &= Z_s + 3Z_n = 8 + j(24 + 4.5) = 8 + j28.5 = 29.60\angle74.32^\circ\ \Omega \end{aligned}

(ii) Symmetrical components of current

\begin{aligned} I_{a0} &= \frac{V_{a0}}{Z_0} = \frac{47.77\angle57.63^\circ}{29.60\angle74.32^\circ} = 1.614\angle{-16.69^\circ}\ \text{A} \\ I_{a1} &= \frac{V_{a1}}{Z_1} = \frac{112.78\angle{-0.03^\circ}}{25.30\angle71.57^\circ} = 4.458\angle{-71.60^\circ}\ \text{A} \\ I_{a2} &= \frac{V_{a2}}{Z_2} = \frac{61.62\angle45.88^\circ}{25.30\angle71.57^\circ} = 2.436\angle{-25.68^\circ}\ \text{A} \end{aligned}

For reference, the line currents are $I_a = 7.72\angle{-48.16^\circ}$ A, $I_b = 4.15\angle136.40^\circ$ A, $I_c = 2.91\angle30.98^\circ$ A, and the neutral current $I_n = 3I_{a0} = 4.84\angle{-16.69^\circ}$ A.

Answer: $V_0 = 47.77\angle57.63^\circ$ , $V_1 = 112.78\angle{-0.03^\circ}$ , $V_2 = 61.62\angle45.88^\circ$ V; $I_0 = 1.614\angle{-16.69^\circ}$ , $I_1 = 4.458\angle{-71.60^\circ}$ , $I_2 = 2.436\angle{-25.68^\circ}$ A.

2074 Asoj · 6 marks

The phase voltages across a certain load are given as: Va = (176 − j132) Volts, Vb = (−[?]2[?] − j96) Volts, Vc = (−160 + j100) Volts. Compute positive, negative and zero sequence component voltages. (The real part of Vb is illegible in the scan.)

Answer

The sequence components are found with $V_{a0} = \frac13(V_a+V_b+V_c)$ , $V_{a1} = \frac13(V_a+aV_b+a^2V_c)$ , $V_{a2} = \frac13(V_a+a^2V_b+aV_c)$ , where $a = 1\angle120^\circ = -0.5 + j0.866$ .

Data: $V_a = 176 - j132$ V, $V_b = -128 - j96$ V (the real part is unclear in the scan; $-128$ is the value in the standard textbook version of this problem), $V_c = -160 + j100$ V.

Zero sequence

\begin{aligned} V_{a0} &= \tfrac13\left[(176 - 128 - 160) + j(-132 - 96 + 100)\right] \\ &= \tfrac13(-112 - j128) = -37.33 - j42.67 = 56.69\angle{-131.19^\circ}\ \text{V} \end{aligned}

Positive sequence

\begin{aligned} aV_b &= (1\angle120^\circ)(160\angle{-143.13^\circ}) = 160\angle{-23.13^\circ} = 147.14 - j62.85 \\ a^2V_c &= (1\angle240^\circ)(188.68\angle148.0^\circ) = 188.68\angle28.0^\circ = 166.60 + j88.56 \\ V_{a1} &= \tfrac13\left[(176 + 147.14 + 166.60) + j(-132 - 62.85 + 88.56)\right] \\ &= \tfrac13(489.74 - j106.29) = 163.25 - j35.43 = 167.05\angle{-12.24^\circ}\ \text{V} \end{aligned}

Negative sequence

\begin{aligned} a^2V_b &= 160\angle96.87^\circ = -19.14 + j158.85 \\ aV_c &= 188.68\angle268.0^\circ = -6.60 - j188.56 \\ V_{a2} &= \tfrac13\left[(176 - 19.14 - 6.60) + j(-132 + 158.85 - 188.56)\right] \\ &= \tfrac13(150.26 - j161.71) = 50.09 - j53.90 = 73.58\angle{-47.10^\circ}\ \text{V} \end{aligned}

Check

$V_{a0} + V_{a1} + V_{a2} = (-37.33 + 163.25 + 50.09) + j(-42.67 - 35.43 - 53.90) = 176.0 - j132.0 = V_a$ ✓

Components of phases b and c: $V_{b1} = a^2V_{a1}$ , $V_{c1} = aV_{a1}$ , $V_{b2} = aV_{a2}$ , $V_{c2} = a^2V_{a2}$ , $V_{b0} = V_{c0} = V_{a0}$ .

Answer: $V_1 = 167.05\angle{-12.24^\circ}$ V, $V_2 = 73.58\angle{-47.10^\circ}$ V, $V_0 = 56.69\angle{-131.19^\circ}$ V (phase a reference).

2072 Kartik · 5 marks

A 3 phase system has following voltage and currents: Va = 220V∠0°, Vb = 120V∠−120°, Vc = 300V∠−240°; Ia = 10A∠−10°, Ib = 5A∠−100°, Ic = 15A∠−200°. Calculate sequence voltages, currents and powers.

Answer

Sequence components are found with $X_0 = \frac13(X_a+X_b+X_c)$ , $X_1 = \frac13(X_a+aX_b+a^2X_c)$ , $X_2 = \frac13(X_a+a^2X_b+aX_c)$ ( $a = 1\angle120^\circ$ ), and sequence powers with $S_k = 3V_kI_k^*$ .

Sequence voltages

Note $V_c = 300\angle{-240^\circ} = 300\angle120^\circ$ ; $aV_b = 120\angle0^\circ$ , $a^2V_c = 300\angle0^\circ$ ; $a^2V_b = 120\angle120^\circ$ , $aV_c = 300\angle240^\circ$ .

\begin{aligned} V_0 &= \tfrac13(220\angle0^\circ + 120\angle{-120^\circ} + 300\angle120^\circ) = 3.33 + j51.96 = 52.07\angle86.33^\circ\ \text{V} \\ V_1 &= \tfrac13(220 + 120 + 300) = 213.33\angle0^\circ\ \text{V} \\ V_2 &= \tfrac13(220\angle0^\circ + 120\angle120^\circ + 300\angle240^\circ) = 3.33 - j51.96 = 52.07\angle{-86.33^\circ}\ \text{V} \end{aligned}

Sequence currents

\begin{aligned} I_0 &= \tfrac13(10\angle{-10^\circ} + 5\angle{-100^\circ} + 15\angle{-200^\circ}) = -1.705 - j0.510 = 1.780\angle{-163.35^\circ}\ \text{A} \\ I_1 &= \tfrac13(10\angle{-10^\circ} + 5\angle20^\circ + 15\angle40^\circ) = 8.679 + j3.205 = 9.252\angle20.27^\circ\ \text{A} \\ I_2 &= \tfrac13(10\angle{-10^\circ} + 5\angle140^\circ + 15\angle{-80^\circ}) = 2.874 - j4.432 = 5.282\angle{-57.03^\circ}\ \text{A} \end{aligned}

Sequence powers ( $S_k = 3V_kI_k^*$ )

\begin{aligned} S_0 &= 3(52.07\angle86.33^\circ)(1.780\angle163.35^\circ) = -96.6 - j260.7\ \text{VA} \\ S_1 &= 3(213.33\angle0^\circ)(9.252\angle{-20.27^\circ}) = 5554.6 - j2051.3\ \text{VA} \\ S_2 &= 3(52.07\angle{-86.33^\circ})(5.282\angle57.03^\circ) = 719.6 - j403.7\ \text{VA} \end{aligned}

Sequence	V (V)	I (A)	P (W)	Q (var)
Zero	52.07∠86.33°	1.780∠−163.35°	−96.6	−260.7
Positive	213.33∠0°	9.252∠20.27°	5554.6	−2051.3
Negative	52.07∠−86.33°	5.282∠−57.03°	719.6	−403.7
Total			6177.6	−2715.7

Check with phase quantities

$S = V_aI_a^* + V_bI_b^* + V_cI_c^* = 2200\angle10^\circ + 600\angle{-20^\circ} + 4500\angle{-40^\circ} = 6177.6 - j2715.7$ VA ✓

Answer: total $S = 6177.6 - j2715.7$ VA (P ≈ 6.18 kW, Q ≈ −2.72 kvar), equal to the sum of the three sequence powers.

2071 Chaitra · 5 marks

A 3-phase supply system has following voltages VA = 220∠120° V, VB = 120∠30° V and VC = 240∠210° V and the currents are IA = 10∠150° A, IB = 5∠0° A and IC = 15∠180° A. Determine sequence currents, voltages and powers.

Answer

Sequence components (phase A reference, sequence A-B-C, $a = 1\angle120^\circ$ ): $X_0 = \frac13(X_A+X_B+X_C)$ , $X_1 = \frac13(X_A+aX_B+a^2X_C)$ , $X_2 = \frac13(X_A+a^2X_B+aX_C)$ ; sequence power $S_k = 3V_kI_k^*$ .

Sequence voltages

\begin{aligned} V_0 &= \tfrac13(220\angle120^\circ + 120\angle30^\circ + 240\angle210^\circ) = -71.31 + j43.51 = 83.53\angle148.61^\circ\ \text{V} \\ V_1 &= \tfrac13(220\angle120^\circ + 120\angle150^\circ + 240\angle90^\circ) = -71.31 + j163.51 = 178.38\angle113.56^\circ\ \text{V} \\ V_2 &= \tfrac13(220\angle120^\circ + 120\angle270^\circ + 240\angle330^\circ) = 32.62 - j16.49 = 36.55\angle{-26.82^\circ}\ \text{V} \end{aligned}

Sequence currents

\begin{aligned} I_0 &= \tfrac13(10\angle150^\circ + 5\angle0^\circ + 15\angle180^\circ) = -6.220 + j1.667 = 6.440\angle165.0^\circ\ \text{A} \\ I_1 &= \tfrac13(10\angle150^\circ + 5\angle120^\circ + 15\angle60^\circ) = -1.220 + j7.440 = 7.540\angle99.31^\circ\ \text{A} \\ I_2 &= \tfrac13(10\angle150^\circ + 5\angle240^\circ + 15\angle300^\circ) = -1.220 - j4.107 = 4.284\angle{-106.55^\circ}\ \text{A} \end{aligned}

Sequence powers

\begin{aligned} S_0 &= 3V_0I_0^* = 3(83.53\angle148.61^\circ)(6.440\angle{-165.0^\circ}) = 1548.2 - j455.3\ \text{VA} \\ S_1 &= 3V_1I_1^* = 3(178.38\angle113.56^\circ)(7.540\angle{-99.31^\circ}) = 3910.6 + j993.1\ \text{VA} \\ S_2 &= 3V_2I_2^* = 3(36.55\angle{-26.82^\circ})(4.284\angle106.55^\circ) = 83.8 + j462.2\ \text{VA} \end{aligned}

Sequence	V (V)	I (A)	P (W)	Q (var)
Zero	83.53∠148.61°	6.440∠165.0°	1548.2	−455.3
Positive	178.38∠113.56°	7.540∠99.31°	3910.6	993.1
Negative	36.55∠−26.82°	4.284∠−106.55°	83.8	462.2
Total			5542.6	1000.0

Check

$S = V_AI_A^* + V_BI_B^* + V_CI_C^* = 2200\angle{-30^\circ} + 600\angle30^\circ + 3600\angle30^\circ = 5542.6 + j1000$ VA ✓

Answer: $S_0 = 1548 - j455$ VA, $S_1 = 3911 + j993$ VA, $S_2 = 84 + j462$ VA; total $S = 5542.6 + j1000$ VA.

2070 Chaitra · 4 marks

Determine the symmetrical components of three unbalanced voltage Va = 200∠0°, Vb = 200∠245°, Vc = 200∠105° V.

Answer

Using $a = 1\angle120^\circ$ : $V_0 = \frac13(V_a+V_b+V_c)$ , $V_1 = \frac13(V_a+aV_b+a^2V_c)$ , $V_2 = \frac13(V_a+a^2V_b+aV_c)$ .

Rectangular form

\begin{aligned} V_a &= 200\angle0^\circ = 200 + j0 \\ V_b &= 200\angle245^\circ = -84.52 - j181.26 \\ V_c &= 200\angle105^\circ = -51.76 + j193.19 \end{aligned}

Zero sequence

V_0 = \tfrac13(63.71 + j11.92) = 21.24 + j3.97 = 21.61\angle10.60^\circ\ \text{V}

Positive sequence

$aV_b = 200\angle365^\circ = 200\angle5^\circ$ , $a^2V_c = 200\angle345^\circ = 200\angle{-15^\circ}$ :

\begin{aligned} V_1 &= \tfrac13(200\angle0^\circ + 200\angle5^\circ + 200\angle{-15^\circ}) \\ &= \tfrac13(592.42 - j34.33) = 197.47 - j11.44 = 197.81\angle{-3.32^\circ}\ \text{V} \end{aligned}

Negative sequence

$a^2V_b = 200\angle485^\circ = 200\angle125^\circ$ , $aV_c = 200\angle225^\circ$ :

\begin{aligned} V_2 &= \tfrac13(200\angle0^\circ + 200\angle125^\circ + 200\angle225^\circ) \\ &= \tfrac13(-56.14 + j22.41) = -18.71 + j7.47 = 20.15\angle158.24^\circ\ \text{V} \end{aligned}

Check: $V_0 + V_1 + V_2 = 200.0 + j0 = V_a$ ✓

Answer: $V_0 = 21.61\angle10.60^\circ$ V, $V_1 = 197.81\angle{-3.32^\circ}$ V, $V_2 = 20.15\angle158.24^\circ$ V. The positive sequence dominates because the set is nearly balanced (only the angles are unbalanced).

2069 Chaitra · 12 marks

Across a symmetrical star connected impedance system with Z = 10 ohms in each phase, a three phase unbalanced system of Voltages with Va = 220∠0°, Vb = 200∠−110° & Vc = 180∠+110° volts is applied. Determine the line currents if the system neutral is (a) isolated (b) solidly grounded.

Answer

For a symmetrical star load, $Z_1 = Z_2 = Z$ and $Z_0 = Z + 3Z_n$ . With an isolated neutral $Z_n = \infty$ , so no zero sequence current flows; with a solidly grounded neutral each phase current is simply $V/Z$ .

Assumption: the applied voltages are phase-to-ground (supply neutral grounded) voltages, $Z = 10\ \Omega$ resistive.

Sequence components of the applied voltages

\begin{aligned} V_a &= 220\angle0^\circ = 220 + j0 \\ V_b &= 200\angle{-110^\circ} = -68.40 - j187.94 \\ V_c &= 180\angle110^\circ = -61.56 + j169.14 \end{aligned}

\begin{aligned} V_0 &= \tfrac13(V_a+V_b+V_c) = 30.01 - j6.26 = 30.66\angle{-11.79^\circ}\ \text{V} \\ V_1 &= \tfrac13(V_a+aV_b+a^2V_c) = 198.08 + j1.16 = 198.08\angle0.33^\circ\ \text{V} \\ V_2 &= \tfrac13(V_a+a^2V_b+aV_c) = -8.09 + j5.11 = 9.56\angle147.73^\circ\ \text{V} \end{aligned}

(a) Neutral isolated

Zero sequence current cannot flow ( $I_0 = 0$ ); the zero sequence voltage appears as the neutral shift $V_{nN} = V_0 = 30.66\angle{-11.79^\circ}$ V.

\begin{aligned} I_1 &= \frac{V_1}{Z} = 19.81 + j0.12\ \text{A}, \qquad I_2 = \frac{V_2}{Z} = -0.81 + j0.51\ \text{A} \\ I_a &= I_1 + I_2 = 19.00 + j0.63 = 19.01\angle1.89^\circ\ \text{A} \\ I_b &= a^2I_1 + aI_2 = -9.84 - j18.17 = 20.66\angle{-118.44^\circ}\ \text{A} \\ I_c &= aI_1 + a^2I_2 = -9.16 + j17.54 = 19.79\angle117.57^\circ\ \text{A} \end{aligned}

Check: $I_a + I_b + I_c = 0$ ✓

(b) Neutral solidly grounded

$Z_0 = Z_1 = Z_2 = 10\ \Omega$ , so each phase sees its own voltage:

\begin{aligned} I_a &= \frac{220\angle0^\circ}{10} = 22.0\angle0^\circ\ \text{A} \\ I_b &= \frac{200\angle{-110^\circ}}{10} = 20.0\angle{-110^\circ}\ \text{A} \\ I_c &= \frac{180\angle110^\circ}{10} = 18.0\angle110^\circ\ \text{A} \end{aligned}

Neutral current: $I_n = I_a + I_b + I_c = 3V_0/Z = 9.00 - j1.88 = 9.20\angle{-11.79^\circ}$ A.

Line current	Isolated neutral	Solidly grounded
$I_a$	19.01∠1.89° A	22.0∠0° A
$I_b$	20.66∠−118.44° A	20.0∠−110° A
$I_c$	19.79∠117.57° A	18.0∠110° A
$I_n$	0	9.20∠−11.79° A

2079 Bhadra · 8 marks

Draw the positive, negative and zero sequence impedance networks for the power system shown in figure below. Choose a base of 50 MVA, 220 kV in 50 Ω transmission lines and mark all reactances in pu. The ratings of the generators and transformers are: Generator 1: 25 MVA, 11 kV, X'' = 20%; Generator 2: 25 MVA, 11 kV, X'' = 20%; Three-phase transformer (each): 20 MVA, 11 Y/220 kV, x = 15%. The negative sequence reactance of each synchronous machine is equal to its subtransient reactance. The zero sequence reactance of each machine is 8%. Assume that the zero sequence reactances of lines are 25% of their positive sequence reactances. [Figure: machine 1 (star, neutral grounded through X = 5% on machine 1 rating) at bus 1; from bus 1 two parallel paths, each consisting of a Y-grounded/Y-grounded transformer, a j50 Ω line and another Y-grounded/Y-grounded transformer, to bus 2; machine 2 (star, neutral grounded through X = 5% on machine 2 rating) at bus 2]

Answer

Sequence networks are drawn by replacing each element with its positive, negative or zero sequence reactance on a common base. Only the generators have EMFs (positive network); neutral reactances appear as $3X_n$ in the zero sequence network only.

Base values

$S_b = 50$ MVA; $V_b = 220$ kV on the lines and 11 kV at the machines. $Z_{b,line} = 220^2/50 = 968\ \Omega$ .

Per unit reactances

Element	Positive	Negative	Zero
G1, G2: $0.20 \times 50/25$	0.40	0.40	$0.08 \times 2 = 0.16$
Neutral reactance: $0.05 \times 50/25 = 0.10$	–	–	$3X_n = 0.30$
Each transformer: $0.15 \times 50/20$	0.375	0.375	0.375
Each line: $50/968$	0.0517	0.0517	$0.25 \times 0.0517 = 0.0129$

Each parallel path (transformer + line + transformer):

Positive/negative: $0.375 + 0.0517 + 0.375 = 0.8017$ pu; two paths in parallel $= 0.4008$ pu
Zero: $0.375 + 0.0129 + 0.375 = 0.7629$ pu; two paths in parallel $= 0.3815$ pu

All transformers are star-grounded/star-grounded, so zero sequence current passes through them (series path, as in the positive network).

Positive sequence network

  1                                     2
  o---j0.375--j0.0517--j0.375---o
  |                             |
  +---j0.375--j0.0517--j0.375---+
  |                             |
 j0.40                        j0.40
  |                             |
 (E1)                          (E2)
  |                             |
  +======= reference bus =======+

Negative sequence network

Same as positive but without EMFs: G1 and G2 are j0.40 each, connected straight to the reference; the two paths of 0.8017 pu each between buses 1 and 2.

Zero sequence network

  1                                     2
  o---j0.375--j0.0129--j0.375---o
  |                             |
  +---j0.375--j0.0129--j0.375---+
  |                             |
 j0.16                        j0.16
  |                             |
 j0.30 (3Xn)                  j0.30 (3Xn)
  |                             |
  +===== reference (ground) ====+

Each machine branch in the zero sequence network: $0.16 + 0.30 = 0.46$ pu.

Summary: Positive: buses 1-2 joined by $j0.4008$ (equivalent), machines $j0.40$ with EMFs. Negative: same, no EMFs. Zero: buses joined by $j0.3815$ (equivalent), machines $j0.46$ to ground.

2070 Asar · 4 marks

For the power system shown in figure below, draw the zero sequence network hence determine the Thevenin's equivalent zero sequence impedance as view from bus-2. [Figure: generator G 100 MVA, 13.8 kV, X0 = 0.05 pu (Y grounded) - T1 100 MVA, 13.8/138 kV, Δ/Y grounded, x = 0.10 pu - line X0 = 60 Ω - T2 100 MVA, 138/13.8 kV, Y grounded/Δ, x = 0.1 pu - bus 2 - motor M 100 MVA, 13.8 kV, X0 = 0.1 pu, neutral grounded through Xn = 0.05 pu]

Answer

The zero sequence network is drawn from the winding connections: a delta winding blocks zero sequence current from its line side, and a neutral reactance appears as $3X_n$ . Bus 2 lies between T2 (delta on the 13.8 kV side) and the motor.

Per unit values (100 MVA, 138 kV line base)

Generator: $X_0 = 0.05$ pu (Y grounded)
T1 (Δ / Y-grounded): $0.10$ pu
Line: $Z_b = 138^2/100 = 190.44\ \Omega$ , $X_0 = 60/190.44 = 0.315$ pu
T2 (Y-grounded / Δ): $0.10$ pu
Motor: $X_0 = 0.1$ pu, neutral $3X_n = 3 \times 0.05 = 0.15$ pu

Zero sequence connections

T1 (Δ on generator side, Yg on line side): the line side is connected to the reference through 0.10; the generator side is open. So the generator is isolated from the rest.
T2 (Yg on line side, Δ on bus-2 side): the line side is connected to the reference through 0.10; bus 2 side is open.
Motor: star grounded through $X_n$ , so its branch is $0.1 + 0.15 = 0.25$ pu to reference.

   G         T1          Line         T2      bus 2   M
   o    o  o----j0.315----o  o         o-------+
   |    |  |              |  |                 |
 j0.05 open j0.10      j0.10 open           j0.10
   |       |              |                    |
   |       |              |                 j0.15 (3Xn)
   |       |              |                    |
 ==+=======+==== reference (ground) ===========+==

Thevenin zero sequence impedance at bus 2

Looking into bus 2, the path through T2 is open (delta on the bus-2 side), so only the motor branch is seen:

Z_{0,th} = j(X_{0M} + 3X_n) = j(0.10 + 0.15) = j0.25\ \text{pu}

Answer: $Z_{0,th}$ at bus 2 = j0.25 pu (100 MVA base). The generator, T1, line and T2 do not contribute because of the delta windings.

Past questions and answers

1.Derive an expression to determine 3-phase (complex) power of an unbalanced system in terms of symmetrical components of voltages and currents.

Phase-domain power

Symmetrical component relations

Substitution

Remarks

2.Starting from suitable point, derive an expression to determine sequence impedances for a balanced three-phase star-connected load with impedance Zs in each phase and grounded neutral with impedance Zn, and show that Z0 = Zs + 3Zn, Z1 = Zs and Z2 = Zs.

Circuit and phase equations

Transformation to sequence quantities

Result

Physical meaning

3.Deduce the sequence networks for balanced Y connected load with neutral grounded through impedance Zn and show also that current through neutral impedance is three times the zero sequence current.

Neutral current is three times zero sequence current

Circuit and phase equations

Transformation to sequence quantities

Sequence networks

4.With suitable mathematical aid, show (justify) that in a delta connected system, zero sequence components of current are absent in the line currents.

Line currents of a delta

Zero sequence line current

Phase (delta) currents in sequence form

Consequences

5.What do you understand by symmetrical components? Derive an expression of 3-phase complex power in terms of symmetrical components of voltages and currents.

Symmetrical components

Complex power in terms of symmetrical components

6.Define Sequence impedances. For the star connected load show that impedance matrix has non-zero elements on its diagonal matrix.

Star connected load

Circuit and phase equations

Transformation to sequence quantities

7.Define sequence impedances. For the star connected load, verify that impedance matrix has non-zero elements on its diagonal matrix. Comment on zero sequence impedance.

Star connected load

Circuit and phase equations

Transformation to sequence quantities

Comment on zero sequence impedance

8.Write down the importance of symmetrical components transformation for unbalance system analysis. Also derive the expression to calculate the components of 3-phase unbalance currents.

Importance for unbalanced system analysis

Derivation for currents

9.Derive the expression to calculate symmetrical components of 3-phase un-balance currents.

Derivation for currents

10.Starting from a suitable point show that a 3-phase unbalanced system of voltages can be represented by a symmetrical components.

Starting point: operator a

Three balanced sets

Phase voltages as sums

The components always exist

11.Explain the significance of symmetrical components in power system analysis. Also derive the expression for phasor voltages in terms of the symmetrical components of voltages.

Significance of symmetrical components

Phase voltages in terms of symmetrical components

12.How do the symmetrical components help us in analyzing unbalance electric power system? Explain briefly.

13.What is significance of sequence components? Deduce the sequence network for a loaded synchronous generator grounded through impedance Zn.

Significance of sequence components

Sequence networks of a synchronous generator grounded through ZnZ_nZn​

14.With proper mathematical interpretations, justify that for a symmetrical static circuit, positive and negative sequence impedances are equal. Also discuss why the sequence impedances for a generator are not equal.

Mathematical proof

Why generator sequence impedances are not equal

15.For a 3-phase symmetrical static circuit, justify that sequence networks are decoupled.

Phase equations

Transformation

Result

16.Show that the sequence component are decoupled for complex power flow in a power system.

Proof

Why this shows decoupling

Example

17.Prove that the symmetrical component transformation is power invariant. Discuss also the significance of zero sequence circuit.

Proof

Exact power invariance

Significance of the zero sequence circuit

Phase equations

Transformation

Result

Phase voltages in rectangular form

(i) Symmetrical components of voltage

Sequence impedances of the load

(ii) Symmetrical components of current

20.The phase voltages across a certain load are given as: Va = (176 − j132) Volts, Vb = (−[?]2[?] − j96) Volts, Vc = (−160 + j100) Volts. Compute positive, negative and zero sequence component voltages. (The real part of Vb is illegible in the scan.)

Zero sequence

Positive sequence

Negative sequence

Check

21.A 3 phase system has following voltage and currents: Va = 220V∠0°, Vb = 120V∠−120°, Vc = 300V∠−240°; Ia = 10A∠−10°, Ib = 5A∠−100°, Ic = 15A∠−200°. Calculate sequence voltages, currents and powers.

Sequence voltages

Sequence currents

Derive an expression to determine 3-phase (complex) power of an unbalanced system in terms of symmetrical components of voltages and currents.

Starting from suitable point, derive an expression to determine sequence impedances for a balanced three-phase star-connected load with impedance Zs in each phase and grounded neutral with impedance Zn, and show that Z0 = Zs + 3Zn, Z1 = Zs and Z2 = Zs.

Deduce the sequence networks for balanced Y connected load with neutral grounded through impedance Zn and show also that current through neutral impedance is three times the zero sequence current.

With suitable mathematical aid, show (justify) that in a delta connected system, zero sequence components of current are absent in the line currents.

What do you understand by symmetrical components? Derive an expression of 3-phase complex power in terms of symmetrical components of voltages and currents.

Define Sequence impedances. For the star connected load show that impedance matrix has non-zero elements on its diagonal matrix.

Define sequence impedances. For the star connected load, verify that impedance matrix has non-zero elements on its diagonal matrix. Comment on zero sequence impedance.

Write down the importance of symmetrical components transformation for unbalance system analysis. Also derive the expression to calculate the components of 3-phase unbalance currents.

Derive the expression to calculate symmetrical components of 3-phase un-balance currents.

Starting from a suitable point show that a 3-phase unbalanced system of voltages can be represented by a symmetrical components.

Explain the significance of symmetrical components in power system analysis. Also derive the expression for phasor voltages in terms of the symmetrical components of voltages.

How do the symmetrical components help us in analyzing unbalance electric power system? Explain briefly.

What is significance of sequence components? Deduce the sequence network for a loaded synchronous generator grounded through impedance Zn.

Sequence networks of a synchronous generator grounded through $Z_n$

With proper mathematical interpretations, justify that for a symmetrical static circuit, positive and negative sequence impedances are equal. Also discuss why the sequence impedances for a generator are not equal.

For a 3-phase symmetrical static circuit, justify that sequence networks are decoupled.

Show that the sequence component are decoupled for complex power flow in a power system.

Prove that the symmetrical component transformation is power invariant. Discuss also the significance of zero sequence circuit.

The phase voltages across a certain load are given as: Va = (176 − j132) Volts, Vb = (−[?]2[?] − j96) Volts, Vc = (−160 + j100) Volts. Compute positive, negative and zero sequence component voltages. (The real part of Vb is illegible in the scan.)

A 3 phase system has following voltage and currents: Va = 220V∠0°, Vb = 120V∠−120°, Vc = 300V∠−240°; Ia = 10A∠−10°, Ib = 5A∠−100°, Ic = 15A∠−200°. Calculate sequence voltages, currents and powers.

Sequence powers ( $S_k = 3V_kI_k^*$ )

A 3-phase supply system has following voltages VA = 220∠120° V, VB = 120∠30° V and VC = 240∠210° V and the currents are IA = 10∠150° A, IB = 5∠0° A and IC = 15∠180° A. Determine sequence currents, voltages and powers.

Determine the symmetrical components of three unbalanced voltage Va = 200∠0°, Vb = 200∠245°, Vc = 200∠105° V.

Across a symmetrical star connected impedance system with Z = 10 ohms in each phase, a three phase unbalanced system of Voltages with Va = 220∠0°, Vb = 200∠−110° & Vc = 180∠+110° volts is applied. Determine the line currents if the system neutral is (a) isolated (b) solidly grounded.