Power System Analysis II — Chapter 5: Unsymmetrical Faults on Power Systems

Asked 5 times
2081 Bhadra · 8 marks
2078 Kartik · 8 marks
2073 Chaitra · 8 marks
2072 Kartik · 6 marks
2071 Shrawan · 6 marks

Starting from a suitable point, show that the three sequence networks are connected in series during a single line to ground fault in a 3-phase power system (unloaded generator), and develop the mathematical expression for the fault current.

Answer

In a single line to ground (L-G) fault, the three sequence currents of the faulted phase are equal and the sequence voltages add to $3Z_fI_{a1}$ , so the positive, negative and zero sequence networks are connected in series, and $I_f = 3E_a/(Z_1+Z_2+Z_0+3Z_f)$ .

Starting point: an unloaded star-connected generator with neutral grounded through $Z_n$ , generating balanced EMFs $E_a$ , $a^2E_a$ , $aE_a$ ; fault on phase a through fault impedance $Z_f$ .

Sequence networks of an unloaded generator

The generator EMFs are balanced (positive sequence only), so with $Z_0 = Z_{g0} + 3Z_n$ :

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

 Positive        Negative        Zero
  F1 o            F2 o            F0 o
     |Ia1 ^          |Ia2 ^          |Ia0 ^
    Z1              Z2              Z0
     |               |               |
   (Ea)              |               |
     |               |               |
  N1 o            N2 o            N0 o

Boundary conditions (fault on phase a through $Z_f$ )

  a o----------+
  b o---  (Ib=0)
  c o---  (Ic=0)
               |  Ia = If
              Zf
               |
            ground

I_b = 0, \qquad I_c = 0, \qquad V_a = Z_fI_a

Sequence currents

\begin{bmatrix} I_{a0}\\I_{a1}\\I_{a2} \end{bmatrix} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}\begin{bmatrix} I_a\\0\\0 \end{bmatrix} \quad\Rightarrow\quad I_{a0} = I_{a1} = I_{a2} = \frac{I_a}{3}

So the same current flows in all three sequence networks, which means they are in series.

Voltage condition

V_a = V_{a0} + V_{a1} + V_{a2} = Z_fI_a = 3Z_fI_{a1}

Substituting the network equations:

-Z_0I_{a1} + E_a - Z_1I_{a1} - Z_2I_{a1} = 3Z_fI_{a1}

I_{a1} = I_{a2} = I_{a0} = \frac{E_a}{Z_1 + Z_2 + Z_0 + 3Z_f}

Fault current

I_f = I_a = 3I_{a1} = \frac{3E_a}{Z_1 + Z_2 + Z_0 + 3Z_f}

For a bolted fault ( $Z_f = 0$ ): $I_f = \dfrac{3E_a}{Z_1+Z_2+Z_0}$ .

Interconnection of sequence networks

      Ia1 -->   F1         F2         F0
  +--(Ea)--Z1---o   +--Z2---o  +--Z0---o
  |             |   |       |  |       |
  N1            +---+  N2   +--+  N0   |
  |                                    |
  +---------------- 3Zf ---------------+

The positive, negative and zero sequence networks are connected in series with $3Z_f$ : F1 to the reference of the next network and so on, carrying the common current $I_{a1}$ . Both conditions ( $I_{a0} = I_{a1} = I_{a2}$ and $V_{a0}+V_{a1}+V_{a2} = 3Z_fI_{a1}$ ) are satisfied only by this series connection.

Remarks

The neutral current is $I_n = 3I_{a0} = I_f$ ; it flows only if the neutral is grounded. For an ungrounded neutral, $Z_0 = \infty$ and $I_f = 0$ .
If $Z_0$ is small (solidly grounded), the L-G fault current can exceed the three-phase fault current $E_a/Z_1$ .

Asked 4 times
2075 Asoj · 6 marks
2074 Asoj · 8 marks
2074 Chaitra · 9 marks
2068 Chaitra · 8 marks

Explain (derive) with necessary mathematical expressions and diagrams how fault current and bus voltages are calculated when a line-to-line fault occurs in a 3-phase power system network. Draw a diagram showing interconnection of sequence networks for this type of fault.

Answer

In a line-to-line (L-L) fault the zero sequence current is zero, $I_{a1} = -I_{a2}$ , and the positive and negative sequence networks are connected in parallel (opposition) through $Z_f$ at the fault point. Fault current is $I_f = -j\sqrt3\,E_a/(Z_1+Z_2+Z_f)$ .

Model: the network is represented at the fault bus k by its Thevenin sequence equivalents: prefault voltage $V_f = V_k(0)$ and impedances $Z_1 = Z_{kk}^{(1)}$ , $Z_2 = Z_{kk}^{(2)}$ , $Z_0 = Z_{kk}^{(0)}$ (from the sequence Z-bus matrices). For a generator terminal fault, $V_f = E_a$ and $Z$ are the machine's own values.

Boundary conditions (fault between phases b and c through $Z_f$ )

  a o-----------  Ia = 0
  b o-----+        Ib = If -->
          Zf
  c o-----+        Ic = -Ib

I_a = 0, \qquad I_b + I_c = 0, \qquad V_b - V_c = Z_fI_b

Sequence currents

\begin{aligned} I_{a0} &= \tfrac13(I_a + I_b + I_c) = \tfrac13(0 + I_b - I_b) = 0 \\ I_{a1} &= \tfrac13(I_a + aI_b + a^2I_c) = \tfrac13(a - a^2)I_b \\ I_{a2} &= \tfrac13(I_a + a^2I_b + aI_c) = \tfrac13(a^2 - a)I_b \end{aligned}

Hence

I_{a0} = 0, \qquad I_{a2} = -I_{a1}

Voltage condition

\begin{aligned} V_b - V_c &= (a^2 - a)V_{a1} + (a - a^2)V_{a2} = (a^2 - a)(V_{a1} - V_{a2}) \\ Z_fI_b &= Z_f\,(a^2I_{a1} + aI_{a2}) = Z_f(a^2 - a)I_{a1} \end{aligned}

so

V_{a1} - V_{a2} = Z_fI_{a1}

With $V_{a1} = V_f - Z_1I_{a1}$ and $V_{a2} = -Z_2I_{a2} = Z_2I_{a1}$ :

I_{a1} = -I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_f}

Fault current

\begin{aligned} I_b &= I_{a0} + a^2I_{a1} + aI_{a2} = (a^2 - a)I_{a1} = -j\sqrt3\,I_{a1} \\ I_f &= I_b = -I_c = \frac{-j\sqrt3\,V_f}{Z_1 + Z_2 + Z_f} \end{aligned}

Voltages at the fault bus

\begin{aligned} V_{a1} &= V_f - Z_1I_{a1}, \quad V_{a2} = Z_2I_{a1}, \quad V_{a0} = 0 \\ V_a &= V_{a1} + V_{a2}, \quad V_b = a^2V_{a1} + aV_{a2}, \quad V_c = aV_{a1} + a^2V_{a2} \end{aligned}

For a bolted fault ( $Z_f = 0$ ), $V_{a1} = V_{a2}$ , so $V_a = 2V_{a1}$ and $V_b = V_c = -V_{a1}$ .

Voltages at other buses (i) of a network

Using the sequence bus impedance matrices, with the fault at bus k:

V_i^{(1)} = V_i(0) - Z_{ik}^{(1)}I_{a1}, \qquad V_i^{(2)} = -Z_{ik}^{(2)}I_{a2}, \qquad V_i^{(0)} = 0

Phase voltages follow from $V_{abc} = AV_{012}$ (with the $\pm30^\circ$ phase shift of Δ-Y transformers applied to the positive and negative sequence values when crossing them). Line currents then follow from $I_{ij} = (V_i - V_j)/z_{ij}$ in each sequence network.

Interconnection of sequence networks

        Ia1 -->           <-- Ia2
  +--(Ea)--Z1---o F1--Zf--o F2---Z2--+
  |                                  |
  N1 ------------------------------- N2

  Zero sequence network: not connected (Ia0 = 0)

The positive and negative sequence networks are connected in parallel (in opposition) through $Z_f$ ; the zero sequence network is left open.

Steps in summary

Form $Z_1$ , $Z_2$ at the fault point (Thevenin), and take $V_f$ = prefault voltage.
$I_{a1} = -I_{a2} = V_f/(Z_1+Z_2+Z_f)$ , $I_{a0} = 0$ .
$I_f = -j\sqrt3 I_{a1}$ .
Find sequence voltages, then phase voltages at the fault and other buses.

Asked 3 times
2076 Chaitra · 6 marks
2075 Chaitra · 6 marks
2082 Bhadra (new course) · 5 marks

For a double line to ground fault (bolted, zero fault impedance) in an unloaded generator (at alternator terminal), obtain the sequence network and show that the positive sequence network is connected in series with the parallel connection of negative sequence and zero sequence networks.

Answer

In a double line to ground (L-L-G) fault on phases b and c, all three sequence voltages at the fault are equal and the sequence currents add to zero, so the positive sequence network is in series with the parallel combination of the negative and zero sequence networks.

Generator sequence equations (unloaded)

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

where $Z_0 = Z_{g0} + 3Z_n$ .

Boundary conditions (bolted fault, b and c to ground)

  a o-------------  Ia = 0
  b o-----+         Ib -->
  c o-----+         Ic -->
          |  If = Ib + Ic
        ground

I_a = 0, \qquad V_b = 0, \qquad V_c = 0

Sequence voltages

\begin{bmatrix} V_{a0}\\V_{a1}\\V_{a2} \end{bmatrix} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}\begin{bmatrix} V_a\\0\\0 \end{bmatrix} \quad\Rightarrow\quad V_{a0} = V_{a1} = V_{a2} = \frac{V_a}{3}

Equal voltages across the three networks at the fault point means they are connected in parallel at that point.

Sequence currents

I_a = I_{a0} + I_{a1} + I_{a2} = 0

So the positive sequence current $I_{a1}$ divides into $-I_{a2}$ and $-I_{a0}$ , i.e. it returns through the negative and zero sequence networks in parallel.

Solving

From $V_{a1} = V_{a2} = V_{a0}$ :

I_{a2} = -\frac{V_{a1}}{Z_2}, \qquad I_{a0} = -\frac{V_{a1}}{Z_0}

Substituting in $I_{a0} + I_{a1} + I_{a2} = 0$ with $V_{a1} = E_a - Z_1I_{a1}$ :

I_{a1} = (E_a - Z_1I_{a1})\left(\frac{1}{Z_2} + \frac{1}{Z_0}\right) \;\Rightarrow\; I_{a1} = \frac{E_a}{Z_1 + \dfrac{Z_2Z_0}{Z_2 + Z_0}}

This is exactly the current in a circuit with $E_a$ and $Z_1$ in series with ( $Z_2 \parallel Z_0$ ), which proves the connection.

I_{a2} = -I_{a1}\frac{Z_0}{Z_2 + Z_0}, \qquad I_{a0} = -I_{a1}\frac{Z_2}{Z_2 + Z_0}

Fault (ground) current: $I_f = I_b + I_c = 3I_{a0}$ .

Interconnection

        Ia1 -->        F
  +--(Ea)--Z1----------o------+---------+
  |                           |         |
  |                          Z2        Z0
  |                           | Ia2     | Ia0
  N1 -------------------------+---------+
                              N2        N0

Positive sequence network in series with the parallel combination of negative and zero sequence networks. With fault impedance $Z_f$ to ground, $3Z_f$ is added in series with $Z_0$ .

Asked 2 times
2082 Baishakh · 8 marks
2076 Asoj · 8 marks

From suitable diagram and mathematics show that during a line to line (double line) fault in a transmission line, zero sequence component is absent in the expression of fault current.

Answer

In a line-to-line fault there is no path to ground and the two fault currents are equal and opposite, so $I_a + I_b + I_c = 0$ and the zero sequence current $I_{a0} = 0$ ; the fault current depends only on $Z_1$ and $Z_2$ .

Model: at the fault point on the line, the system is replaced by its Thevenin sequence networks: prefault voltage $V_f$ and impedances $Z_1$ , $Z_2$ , $Z_0$ seen from the fault point.

Boundary conditions (fault between phases b and c through $Z_f$ )

  a o-----------  Ia = 0
  b o-----+        Ib = If -->
          Zf
  c o-----+        Ic = -Ib

I_a = 0, \qquad I_b + I_c = 0, \qquad V_b - V_c = Z_fI_b

Sequence currents

\begin{aligned} I_{a0} &= \tfrac13(I_a + I_b + I_c) = \tfrac13(0 + I_b - I_b) = 0 \\ I_{a1} &= \tfrac13(I_a + aI_b + a^2I_c) = \tfrac13(a - a^2)I_b \\ I_{a2} &= \tfrac13(I_a + a^2I_b + aI_c) = \tfrac13(a^2 - a)I_b \end{aligned}

Hence

I_{a0} = 0, \qquad I_{a2} = -I_{a1}

So the zero sequence current is absent: with no connection to ground, current that leaves through phase b returns through phase c, and the three line currents sum to zero.

Voltage condition

\begin{aligned} V_b - V_c &= (a^2 - a)V_{a1} + (a - a^2)V_{a2} = (a^2 - a)(V_{a1} - V_{a2}) \\ Z_fI_b &= Z_f\,(a^2I_{a1} + aI_{a2}) = Z_f(a^2 - a)I_{a1} \end{aligned}

so

V_{a1} - V_{a2} = Z_fI_{a1}

With $V_{a1} = V_f - Z_1I_{a1}$ and $V_{a2} = -Z_2I_{a2} = Z_2I_{a1}$ :

I_{a1} = -I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_f}

Fault current

\begin{aligned} I_f = I_b &= I_{a0} + a^2I_{a1} + aI_{a2} = (a^2 - a)I_{a1} \\ &= -j\sqrt3\,I_{a1} = \frac{-j\sqrt3\,V_f}{Z_1 + Z_2 + Z_f} \end{aligned}

$Z_0$ does not appear in the expression, confirming that the zero sequence network plays no part.

Interconnection of sequence networks

        Ia1 -->           <-- Ia2
  +--(Ea)--Z1---o F1--Zf--o F2---Z2--+
  |                                  |
  N1 ------------------------------- N2

  Zero sequence network: not connected (Ia0 = 0)

The positive and negative sequence networks are connected in parallel (in opposition) through $Z_f$ ; the zero sequence network is left open.

Consequences

The L-L fault current does not depend on the grounding of the neutral or on line zero sequence impedance.
With $Z_1 = Z_2$ and $Z_f = 0$ : $I_f = \frac{\sqrt3}{2}\cdot\frac{V_f}{Z_1} = 0.866 \times I_{3\phi}$ , i.e. smaller than the three-phase fault current.
Also $V_{a0} = 0$ at the fault, since the zero sequence network carries no current.

Asked 2 times
2070 Asar · 5+3+3+3 marks
2070 Chaitra · 5+3+3+3 marks

A three phase synchronous generator whose neutral is grounded through a reactance Xn has balanced emfs and sequence reactance as X1, X2 and X0 such that X1 = X2 >> X0. i) Derive the expression for fault current for solid line to ground fault on phase a. ii) Draw the sequence networks and their interconnection for above fault. iii) Show that if neutral is solidly grounded, L-G fault is more severe than 3-phase fault at the terminal of generator. iv) Find the limiting value of neutral grounding reactance so that fault current in L-G and 3 phase faults are equal.

Answer

For an L-G fault on a generator grounded through $X_n$ , $I_f = 3E_a/[j(X_1 + X_2 + X_0 + 3X_n)]$ ; with solid grounding and $X_0 < X_1$ this exceeds the 3-phase current $E_a/X_1$ , and the two are equal when $X_n = (X_1 - X_0)/3$ .

i) Fault current for solid L-G fault on phase a

Generator sequence equations (neutral reactance appears as $3X_n$ in the zero sequence network):

V_{a1} = E_a - jX_1I_{a1}, \quad V_{a2} = -jX_2I_{a2}, \quad V_{a0} = -j(X_0 + 3X_n)I_{a0}

Boundary conditions: $I_b = I_c = 0$ , $V_a = 0$ .

From $I_s = A^{-1}I_p$ : $I_{a0} = I_{a1} = I_{a2} = I_a/3$ .

From $V_a = V_{a0} + V_{a1} + V_{a2} = 0$ :

E_a - jI_{a1}(X_1 + X_2 + X_0 + 3X_n) = 0 \;\Rightarrow\; I_{a1} = \frac{E_a}{j(X_1 + X_2 + X_0 + 3X_n)}

I_f = I_a = 3I_{a1} = \frac{3E_a}{j(X_1 + X_2 + X_0 + 3X_n)}

ii) Sequence networks and interconnection

 Positive      Negative      Zero
 F1 o          F2 o          F0 o
    |jX1          |jX2          |jX0
  (Ea)            |             |j3Xn
    |             |             |
 N1 o          N2 o          N0 o

 L-G fault: networks in series
     Ia1-->
 +-(Ea)-jX1--F1--N2--jX2--F2--N0--j3Xn--jX0--F0-+
 |                                              |
 N1 --------------------------------------------+

(F1 joined to N2, F2 to N0, F0 back to N1: a single series loop carrying $I_{a1}$ .)

iii) Solidly grounded: L-G more severe than 3-phase fault

With $X_n = 0$ and $X_1 = X_2$ :

|I_{LG}| = \frac{3E_a}{2X_1 + X_0}, \qquad |I_{3\phi}| = \frac{E_a}{X_1} = \frac{3E_a}{3X_1}

Since $X_0 \ll X_1$ , $2X_1 + X_0 < 3X_1$ , so

|I_{LG}| > |I_{3\phi}|

For example, $X_1 = X_2 = 0.2$ , $X_0 = 0.05$ pu: $I_{LG} = 3/0.45 = 6.67$ pu while $I_{3\phi} = 1/0.2 = 5$ pu. Hence an L-G fault at the terminals of a solidly grounded generator is more severe, and the winding bracing must be designed for it, or the neutral grounded through a reactor.

iv) Limiting neutral reactance

For equal fault currents:

\frac{3E_a}{2X_1 + X_0 + 3X_n} = \frac{E_a}{X_1} \;\Rightarrow\; 3X_1 = 2X_1 + X_0 + 3X_n

X_n = \frac{X_1 - X_0}{3}

With $X_n$ larger than this value, the L-G fault current is less than the three-phase fault current. In the example above, $X_n = (0.2 - 0.05)/3 = 0.05$ pu.

Asked 2 times
2080 Bhadra · 8 marks
2075 Asoj · 8 marks

A 25 MVA, 13.2 kV alternator with solidly grounded neutral had a sub-transient reactance of 0.25 pu. The negative and zero sequence reactances are 0.35 and 0.1 pu respectively. A line to line fault occurs at the terminals of an alternator. Determine the line to line voltages under fault conditions. Neglect resistance.

Answer

For a line-to-line fault at the generator terminals, $I_{a1} = -I_{a2} = E_a/[j(X_1+X_2)]$ and $I_{a0} = 0$ , and for a bolted fault $V_{a1} = V_{a2}$ ; the phase voltages follow from these.

Data: $X_1 = 0.25$ , $X_2 = 0.35$ , $X_0 = 0.1$ pu; $E_a = 1\angle0^\circ$ pu (no-load prefault); fault between phases b and c, $Z_f = 0$ .

Sequence currents

I_{a1} = -I_{a2} = \frac{E_a}{j(X_1 + X_2)} = \frac{1}{j0.60} = -j1.667\ \text{pu}, \qquad I_{a0} = 0

Sequence voltages

\begin{aligned} V_{a1} &= E_a - jX_1I_{a1} = 1 - j0.25(-j1.667) = 1 - 0.4167 = 0.5833\ \text{pu} \\ V_{a2} &= -jX_2I_{a2} = -j0.35(j1.667) = 0.5833\ \text{pu} \\ V_{a0} &= 0 \end{aligned}

Phase voltages

\begin{aligned} V_a &= V_{a1} + V_{a2} = 1.1667\angle0^\circ\ \text{pu} \\ V_b &= a^2V_{a1} + aV_{a2} = (a^2 + a)(0.5833) = -0.5833\ \text{pu} \\ V_c &= aV_{a1} + a^2V_{a2} = -0.5833\ \text{pu} \end{aligned}

Line voltages

\begin{aligned} V_{ab} &= V_a - V_b = 1.1667 + 0.5833 = 1.75\angle0^\circ\ \text{pu} \\ V_{bc} &= V_b - V_c = 0 \\ V_{ca} &= V_c - V_a = -1.75 = 1.75\angle180^\circ\ \text{pu} \end{aligned}

These are in per unit of the base phase voltage $13.2/\sqrt3 = 7.621$ kV:

|V_{ab}| = |V_{ca}| = 1.75 \times 7.621 = 13.34\ \text{kV}, \qquad V_{bc} = 0

Quantity	pu (phase base)	Actual
$V_{ab}$	1.75∠0°	13.34∠0° kV
$V_{bc}$	0	0
$V_{ca}$	1.75∠180°	13.34∠180° kV

(Fault current for reference: $I_b = -j\sqrt3 I_{a1} = 2.887$ pu $= 2.887 \times 1093.5 = 3157$ A.)

Answer: $V_{ab} = 13.34\angle0^\circ$ kV, $V_{bc} = 0$ , $V_{ca} = 13.34\angle180^\circ$ kV.

2069 Chaitra · 10 marks

With the help of suitable mathematical aid verify that, for a line to line fault in a synchronous generator the zero sequence component of current is absent and positive-sequence of current is equal to the negative sequence component of the current.

Answer

For a line-to-line fault (phases b and c) on a synchronous generator, the line currents sum to zero, so $I_{a0} = 0$ ; and transforming the boundary conditions gives $I_{a2} = -I_{a1}$ , i.e. the positive and negative sequence currents are equal in magnitude (opposite in phase).

Generator and fault

An unloaded generator produces balanced EMFs; its sequence equations are

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

Boundary conditions (fault between phases b and c through $Z_f$ )

  a o-----------  Ia = 0
  b o-----+        Ib = If -->
          Zf
  c o-----+        Ic = -Ib

I_a = 0, \qquad I_b + I_c = 0, \qquad V_b - V_c = Z_fI_b

Sequence currents

\begin{aligned} I_{a0} &= \tfrac13(I_a + I_b + I_c) = \tfrac13(0 + I_b - I_b) = 0 \\ I_{a1} &= \tfrac13(I_a + aI_b + a^2I_c) = \tfrac13(a - a^2)I_b \\ I_{a2} &= \tfrac13(I_a + a^2I_b + aI_c) = \tfrac13(a^2 - a)I_b \end{aligned}

Hence

I_{a0} = 0, \qquad I_{a2} = -I_{a1}

Since $a - a^2 = j\sqrt3$ :

I_{a1} = \frac{j\sqrt3}{3}I_b = \frac{j}{\sqrt3}I_b, \qquad I_{a2} = -\frac{j}{\sqrt3}I_b

So $|I_{a1}| = |I_{a2}| = |I_b|/\sqrt3$ : the positive and negative sequence currents have equal magnitude; they are 180° apart. The zero sequence current is absent because there is no ground connection and no neutral current flows ( $I_n = 3I_{a0} = 0$ ).

Voltage condition

\begin{aligned} V_b - V_c &= (a^2 - a)V_{a1} + (a - a^2)V_{a2} = (a^2 - a)(V_{a1} - V_{a2}) \\ Z_fI_b &= Z_f\,(a^2I_{a1} + aI_{a2}) = Z_f(a^2 - a)I_{a1} \end{aligned}

so

V_{a1} - V_{a2} = Z_fI_{a1}

With $V_{a1} = E_a - Z_1I_{a1}$ and $V_{a2} = -Z_2I_{a2} = Z_2I_{a1}$ :

I_{a1} = -I_{a2} = \frac{E_a}{Z_1 + Z_2 + Z_f}

Fault current

I_f = I_b = -I_c = (a^2 - a)I_{a1} = -j\sqrt3\,I_{a1} = \frac{-j\sqrt3\,E_a}{Z_1 + Z_2 + Z_f}

Interconnection of sequence networks

        Ia1 -->           <-- Ia2
  +--(Ea)--Z1---o F1--Zf--o F2---Z2--+
  |                                  |
  N1 ------------------------------- N2

  Zero sequence network: not connected (Ia0 = 0)

The positive and negative sequence networks are connected in parallel (in opposition) through $Z_f$ ; the zero sequence network is left open.

Because the two networks form one series loop, the current entering the negative sequence network equals the current leaving the positive network, which is another way of seeing $I_{a1} = -I_{a2}$ in magnitude.

Fault-point voltages (bolted fault)

$V_{a1} = V_{a2} = E_a - Z_1I_{a1}$ , $V_{a0} = 0$ , so $V_a = 2V_{a1}$ and $V_b = V_c = -V_{a1}$ .

2071 Chaitra · 6 marks

Show that positive and negative sequence currents are equal in magnitude but out of phase by 180° in a line to line fault on a power system network. Draw a diagram showing inter-connection of sequence networks for this type of fault.

Answer

In a line-to-line fault, $I_{a2} = -I_{a1}$ : the positive and negative sequence currents have equal magnitude and are 180° out of phase, while $I_{a0} = 0$ .

Boundary conditions (fault between phases b and c through $Z_f$ )

  a o-----------  Ia = 0
  b o-----+        Ib = If -->
          Zf
  c o-----+        Ic = -Ib

I_a = 0, \qquad I_b + I_c = 0, \qquad V_b - V_c = Z_fI_b

Sequence currents

\begin{aligned} I_{a0} &= \tfrac13(I_a + I_b + I_c) = \tfrac13(0 + I_b - I_b) = 0 \\ I_{a1} &= \tfrac13(I_a + aI_b + a^2I_c) = \tfrac13(a - a^2)I_b \\ I_{a2} &= \tfrac13(I_a + a^2I_b + aI_c) = \tfrac13(a^2 - a)I_b \end{aligned}

Hence

I_{a0} = 0, \qquad I_{a2} = -I_{a1}

Now $a - a^2 = j\sqrt3$ and $a^2 - a = -j\sqrt3$ , so

I_{a1} = \frac{j\sqrt3}{3}I_b = \frac{I_b}{\sqrt3}\angle90^\circ, \qquad I_{a2} = \frac{-j\sqrt3}{3}I_b = \frac{I_b}{\sqrt3}\angle{-90^\circ}

Both have magnitude $|I_b|/\sqrt3$ and differ in angle by 180°: they are equal and opposite.

Voltage condition

\begin{aligned} V_b - V_c &= (a^2 - a)V_{a1} + (a - a^2)V_{a2} = (a^2 - a)(V_{a1} - V_{a2}) \\ Z_fI_b &= Z_f\,(a^2I_{a1} + aI_{a2}) = Z_f(a^2 - a)I_{a1} \end{aligned}

so

V_{a1} - V_{a2} = Z_fI_{a1}

With $V_{a1} = V_f - Z_1I_{a1}$ and $V_{a2} = -Z_2I_{a2} = Z_2I_{a1}$ :

I_{a1} = -I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_f}

Fault current: $I_f = I_b = (a^2 - a)I_{a1} = -j\sqrt3\,I_{a1} = \dfrac{-j\sqrt3\,V_f}{Z_1 + Z_2 + Z_f}$ .

Interconnection of sequence networks

        Ia1 -->           <-- Ia2
  +--(Ea)--Z1---o F1--Zf--o F2---Z2--+
  |                                  |
  N1 ------------------------------- N2

  Zero sequence network: not connected (Ia0 = 0)

The positive and negative sequence networks are connected in parallel (in opposition) through $Z_f$ ; the zero sequence network is left open.

The positive network drives $I_{a1}$ out of F1 and the same current enters F2 of the negative network, so $I_{a2} = -I_{a1}$ .

2080 Bhadra · 8 marks

Draw positive sequence, negative sequence and zero sequence network of synchronous generator. Show that all three sequence currents of faulty phase are equal in case of single line to ground fault of unloaded synchronous generator. Also find the expression of fault current.

Answer

For an L-G fault on phase a of an unloaded generator, $I_b = I_c = 0$ , which gives $I_{a0} = I_{a1} = I_{a2} = I_a/3$ ; the three sequence networks are in series and $I_f = 3E_a/(Z_1+Z_2+Z_0)$ (bolted) or $3E_a/(Z_1+Z_2+Z_0+3Z_f)$ with fault impedance.

Generator: star connected, balanced EMFs $E_a$ , $a^2E_a$ , $aE_a$ ; neutral grounded through $Z_n$ .

Sequence networks of an unloaded generator

The generator EMFs are balanced (positive sequence only), so with $Z_0 = Z_{g0} + 3Z_n$ :

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

 Positive        Negative        Zero
  F1 o            F2 o            F0 o
     |Ia1 ^          |Ia2 ^          |Ia0 ^
    Z1              Z2              Z0
     |               |               |
   (Ea)              |               |
     |               |               |
  N1 o            N2 o            N0 o

Positive sequence: contains the EMF $E_a$ and $Z_1$ (sub-transient reactance $X_d''$ for fault studies). No current in $Z_n$ .
Negative sequence: no EMF, impedance $Z_2$ , connected to the neutral directly.
Zero sequence: no EMF, impedance $Z_0 = Z_{g0} + 3Z_n$ ; the current through the neutral is $3I_{a0}$ .

Boundary conditions (fault on phase a through $Z_f$ )

  a o----------+
  b o---  (Ib=0)
  c o---  (Ic=0)
               |  Ia = If
              Zf
               |
            ground

I_b = 0, \qquad I_c = 0, \qquad V_a = Z_fI_a

Sequence currents are equal

\begin{bmatrix} I_{a0}\\I_{a1}\\I_{a2} \end{bmatrix} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}\begin{bmatrix} I_a\\0\\0 \end{bmatrix} \quad\Rightarrow\quad I_{a0} = I_{a1} = I_{a2} = \frac{I_a}{3}

So the same current flows in all three sequence networks, which means they are in series.

Voltage condition

V_a = V_{a0} + V_{a1} + V_{a2} = Z_fI_a = 3Z_fI_{a1}

Substituting the network equations:

-Z_0I_{a1} + E_a - Z_1I_{a1} - Z_2I_{a1} = 3Z_fI_{a1}

I_{a1} = I_{a2} = I_{a0} = \frac{E_a}{Z_1 + Z_2 + Z_0 + 3Z_f}

Fault current

I_f = I_a = 3I_{a1} = \frac{3E_a}{Z_1 + Z_2 + Z_0 + 3Z_f}

For a bolted fault ( $Z_f = 0$ ): $I_f = \dfrac{3E_a}{Z_1+Z_2+Z_0}$ .

Interconnection of sequence networks

      Ia1 -->   F1         F2         F0
  +--(Ea)--Z1---o   +--Z2---o  +--Z0---o
  |             |   |       |  |       |
  N1            +---+  N2   +--+  N0   |
  |                                    |
  +---------------- 3Zf ---------------+

The positive, negative and zero sequence networks are connected in series with $3Z_f$ : F1 to the reference of the next network and so on, carrying the common current $I_{a1}$ . Both conditions ( $I_{a0} = I_{a1} = I_{a2}$ and $V_{a0}+V_{a1}+V_{a2} = 3Z_fI_{a1}$ ) are satisfied only by this series connection.

2073 Shrawan · 8 marks

For an unloaded 3-phase, ABC phase sequence, synchronous generator, starting from the boundary condition draw and justify the interconnection of sequence networks if a L-G fault occurs in phase B.

Answer

For an L-G fault on phase B, the sequence currents of phase B are equal, $I_{b0} = I_{b1} = I_{b2} = I_b/3$ , so the three sequence networks are connected in series. In terms of phase-a reference components they are joined through phase shifts: $I_{a1} = aI_{a0}$ , $I_{a2} = a^2I_{a0}$ .

Generator sequence equations (phase a reference)

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

Boundary conditions (bolted fault on B)

I_a = 0, \qquad I_c = 0, \qquad V_b = 0

Sequence currents

\begin{aligned} I_{a0} &= \tfrac13(0 + I_b + 0) = \tfrac13I_b \\ I_{a1} &= \tfrac13(0 + aI_b + 0) = \tfrac13aI_b = aI_{a0} \\ I_{a2} &= \tfrac13(0 + a^2I_b + 0) = \tfrac13a^2I_b = a^2I_{a0} \end{aligned}

Phase-b components are $I_{b0} = I_{a0}$ , $I_{b1} = a^2I_{a1} = I_{a0}$ , $I_{b2} = aI_{a2} = I_{a0}$ . So

I_{b0} = I_{b1} = I_{b2} = \frac{I_b}{3}

The same current flows in all three networks when they are written for phase b, which means a series connection.

Voltage condition

V_b = V_{a0} + a^2V_{a1} + aV_{a2} = V_{b0} + V_{b1} + V_{b2} = 0

Substituting:

-Z_0I_{a0} + a^2(E_a - Z_1aI_{a0}) - aZ_2a^2I_{a0} = 0 \;\Rightarrow\; I_{a0} = \frac{a^2E_a}{Z_0 + Z_1 + Z_2} = \frac{E_b}{Z_0 + Z_1 + Z_2}

since $E_b = a^2E_a$ . Hence

I_f = I_b = 3I_{a0} = \frac{3E_b}{Z_1 + Z_2 + Z_0}

Interconnection of sequence networks

Drawn with phase b as reference (EMF $E_b$ in the positive network):

      Ib1 -->
  +--(Eb)--Z1--F1  N2--Z2--F2  N0--Z0--F0--+
  |             |__|        |__|           |
  N1 ---------------------------------------+
     Ib0 = Ib1 = Ib2 = Ib/3

With phase a as reference the same series loop is used, but the currents entering the networks are linked by phase shifts: $I_{a1} = aI_{a0}$ and $I_{a2} = a^2I_{a0}$ (ideal phase-shifting transformers of $+120^\circ$ and $+240^\circ$ in the positive and negative networks).

Justification: the series connection satisfies both conditions: equal phase-b sequence currents and zero total phase-b voltage. The result is the same as an L-G fault on phase a, only shifted by $-120^\circ$ , as expected for a symmetrical machine.

2070 Chaitra · 10 marks

What is the difference between symmetrical components of positive, negative and zero phase sequence? A 3-phase synchronous generator with its neutral solidly grounded and operating at no load develops an L-G fault in one of the phase have fault impedance Zf. Derive expressions for the fault currents and the line to ground voltage at the location of the fault at all the phases.

Answer

Difference between positive, negative and zero sequence components

Feature	Positive sequence	Negative sequence	Zero sequence
Magnitudes	Equal	Equal	Equal
Phase displacement	120°	120°	0° (all in phase)
Phase order	a-b-c (same as system)	a-c-b (reverse)	No rotation
Relation to phase a	$V_{b1} = a^2V_{a1}$ , $V_{c1} = aV_{a1}$	$V_{b2} = aV_{a2}$ , $V_{c2} = a^2V_{a2}$	$V_{b0} = V_{c0} = V_{a0}$
Source in generator	Yes (EMF)	No	No
Return path needed	No	No	Neutral/ground ( $I_n = 3I_0$ )
Effect	Useful power, torque	Reverse torque, rotor heating	Ground currents, interference

L-G fault through $Z_f$ on phase a (solidly grounded, no load)

Generator equations: $V_{a1} = E_a - Z_1I_{a1}$ , $V_{a2} = -Z_2I_{a2}$ , $V_{a0} = -Z_0I_{a0}$ (solid grounding: $Z_0 = Z_{g0}$ ).

Boundary conditions (fault on phase a through $Z_f$ )

  a o----------+
  b o---  (Ib=0)
  c o---  (Ic=0)
               |  Ia = If
              Zf
               |
            ground

I_b = 0, \qquad I_c = 0, \qquad V_a = Z_fI_a

Sequence currents

\begin{bmatrix} I_{a0}\\I_{a1}\\I_{a2} \end{bmatrix} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}\begin{bmatrix} I_a\\0\\0 \end{bmatrix} \quad\Rightarrow\quad I_{a0} = I_{a1} = I_{a2} = \frac{I_a}{3}

So the same current flows in all three sequence networks, which means they are in series.

Voltage condition

V_a = V_{a0} + V_{a1} + V_{a2} = Z_fI_a = 3Z_fI_{a1}

Substituting the network equations:

-Z_0I_{a1} + E_a - Z_1I_{a1} - Z_2I_{a1} = 3Z_fI_{a1}

I_{a1} = I_{a2} = I_{a0} = \frac{E_a}{Z_1 + Z_2 + Z_0 + 3Z_f}

Fault current

I_f = I_a = 3I_{a1} = \frac{3E_a}{Z_1 + Z_2 + Z_0 + 3Z_f}

For a bolted fault ( $Z_f = 0$ ): $I_f = \dfrac{3E_a}{Z_1+Z_2+Z_0}$ .

Interconnection of sequence networks

      Ia1 -->   F1         F2         F0
  +--(Ea)--Z1---o   +--Z2---o  +--Z0---o
  |             |   |       |  |       |
  N1            +---+  N2   +--+  N0   |
  |                                    |
  +---------------- 3Zf ---------------+

The positive, negative and zero sequence networks are connected in series with $3Z_f$ : F1 to the reference of the next network and so on, carrying the common current $I_{a1}$ . Both conditions ( $I_{a0} = I_{a1} = I_{a2}$ and $V_{a0}+V_{a1}+V_{a2} = 3Z_fI_{a1}$ ) are satisfied only by this series connection.

Line-to-ground voltages at the fault

Let $I_1 = I_{a1} = I_{a2} = I_{a0} = \dfrac{E_a}{Z_1+Z_2+Z_0+3Z_f}$ .

Phase a:

V_a = Z_fI_a = 3Z_fI_1 = \frac{3Z_fE_a}{Z_1 + Z_2 + Z_0 + 3Z_f}

Phase b:

\begin{aligned} V_b &= V_{a0} + a^2V_{a1} + aV_{a2} \\ &= -Z_0I_1 + a^2(E_a - Z_1I_1) - aZ_2I_1 \end{aligned}

Replacing $a^2E_a = a^2I_1(Z_1 + Z_2 + Z_0 + 3Z_f)$ :

V_b = I_1\left[(a^2 - a)Z_2 + (a^2 - 1)Z_0 + 3a^2Z_f\right]

Phase c: similarly, with $aE_a = aI_1(Z_1+Z_2+Z_0+3Z_f)$ :

V_c = I_1\left[(a - a^2)Z_2 + (a - 1)Z_0 + 3aZ_f\right]

For a bolted fault ( $Z_f = 0$ ): $V_a = 0$ and

V_b = \frac{E_a\left[(a^2 - a)Z_2 + (a^2 - 1)Z_0\right]}{Z_1 + Z_2 + Z_0}, \quad V_c = \frac{E_a\left[(a - a^2)Z_2 + (a - 1)Z_0\right]}{Z_1 + Z_2 + Z_0}

2083 Baishakh (new course) · 5 marks

Show that for the two-conductor open circuit fault in a 3-phase system, the sequence networks are to be connected in series to simulate the fault.

Answer

In a two-conductor open fault, two phases (say b and c) are broken between points $p$ and $q$ while phase a stays healthy. Writing the fault conditions in symmetrical components shows that the three sequence networks, seen between $p$ and $q$ , must be joined in series.

Fault conditions

  p                        q
 a ---------------------------  (healthy, Ia flows)
 b -----x            x--------  (open, Ib = 0)
 c -----x            x--------  (open, Ic = 0)

Let $V_{a}, V_{b}, V_{c}$ be the series voltage drops across the break (from $p$ to $q$ ) in each phase.

Phases b and c are open: $I_b = 0,\ I_c = 0$
Phase a is healthy (no break): $V_a = 0$ (drop across the break in phase a is zero)

Sequence currents

\begin{bmatrix} I_{a0} \\ I_{a1} \\ I_{a2} \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{bmatrix}\begin{bmatrix} I_a \\ 0 \\ 0 \end{bmatrix}

so

I_{a0} = I_{a1} = I_{a2} = \frac{I_a}{3}

The same current flows through all three sequence networks.

Sequence voltages

Since $V_a = V_{a0} + V_{a1} + V_{a2} = 0$ :

V_{a0} + V_{a1} + V_{a2} = 0

Sequence network equations

Seen from the break terminals $p$ - $q$ , each sequence network has a Thevenin form. Only the positive sequence network contains a source (the pre-fault open-circuit voltage $V_{pq}^{0}$ or the equivalent driving voltage $E$ ):

\begin{aligned} V_{a1} &= E - Z_1 I_{a1} \\ V_{a2} &= -Z_2 I_{a2} \\ V_{a0} &= -Z_0 I_{a0} \end{aligned}

Adding and using $V_{a0}+V_{a1}+V_{a2}=0$ and $I_{a0}=I_{a1}=I_{a2}$ :

I_{a1} = \frac{E}{Z_1 + Z_2 + Z_0}

Interpretation

Equal currents in all three networks and voltages that add to zero are exactly the conditions of three elements in series in a closed loop:

  p1 +--[ +ve seq N/W, Z1, E ]--+ q1
     |                          |
  p2 +--[ -ve seq N/W, Z2 ]-----+ q2   (series loop:
     |                          |      q1->p2, q2->p0,
  p0 +--[ zero seq N/W, Z0 ]----+ q0   q0->p1)

Hence the positive, negative and zero sequence networks are connected in series between the fault points to simulate a two-conductor open fault, and the healthy-phase current is $I_a = 3I_{a1} = \dfrac{3E}{Z_1+Z_2+Z_0}$ . This is the series-fault counterpart of the single line-to-ground shunt fault.

2071 Chaitra · 5 marks

How do the different vector groups of transformer affect the fault current in the power system network?

Answer

The vector group of a transformer (Yy, Yd, Dy, Dd, with neutral earthed or not) decides how the zero-sequence current can flow through it and what phase shift it gives. This changes the zero-sequence network and therefore the earth-fault current; positive and negative sequence impedances are the same for all groups.

Effect on sequence networks

Positive and negative sequence: every group behaves as a series leakage reactance $X_T$ . Only a phase shift of $\pm 30^\circ$ appears in Dy/Yd units (positive sequence shifted one way, negative the other). Magnitudes of balanced and L-L fault currents are not changed.
Zero sequence: zero-sequence currents are in phase in all three lines, so they need a return path through an earthed neutral or a circulating path inside a delta.

Vector group	Zero-sequence equivalent	Effect on earth fault current
Yg–Yg (both neutrals earthed)	Series $Z_0$ between both sides	Zero-seq current passes through; fault fed from both sides
Yg–Δ	Yg side connected to reference through $Z_0$ ; Δ side open	Acts as an earthing source on Yg side; blocks zero seq to Δ side
Y–Δ (Y not earthed)	Open on both sides	No zero-seq path; no earth fault current through it
Yg–Y (one neutral isolated)	Open circuit	Blocks zero-seq current
Δ–Δ	Open on both sides	No earth fault current through it
Neutral earthed through $Z_n$	$3Z_n$ in series in zero-seq path	Earth fault current reduced

How it affects fault current

L-G fault current $I_f = \dfrac{3E}{Z_1+Z_2+Z_0+3Z_f}$ depends strongly on $Z_0$ . A Δ-winding or an unearthed star makes $Z_0$ seen from that side very large (open), so the L-G fault current falls sharply; for an isolated system it becomes almost zero.
A Δ/Yg transformer feeding a fault on its star side provides a low-impedance zero-sequence path, so the L-G current can even exceed the three-phase fault current when $Z_0 < Z_1$ .
Yg–Yg units let the zero-sequence current pass to the other side, so earth faults are fed from both sides and are seen by relays on both sides.
Neutral impedance $Z_n$ appears as $3Z_n$ and is used deliberately to limit earth fault current.
Phase shift of Dy/Yd units shifts the positive and negative sequence currents by $\pm30^\circ$ . The current magnitudes on the other side are the same, but the phase currents are distributed differently, e.g. an L-G fault on the star side of a Dy transformer appears as current in two lines on the delta side.
L-L faults have no zero-sequence component, so the vector group does not change their magnitude.

Example

For a generator connected through a Δ/Yg step-up transformer, an L-G fault on the HV side is fed only through the transformer's own $Z_0$ (generator zero sequence is isolated by the delta), while an L-G fault on the delta side gives almost no fault current because that side has no earthed neutral.

2082 Baishakh · 6+2 marks

A 3-phase synchronous generator with isolated neutral at no-load is subjected to different types of faults at their terminal. Considering the line-to-line voltage rating of generator as 380V and positive, negative & zero sequence impedances of the generator are j2Ω, j0.5Ω & j0.25Ω respectively, determine the fault current for (i) L-G fault, (ii) L-L-G fault, and (iii) L-L-L-G fault. Comment on the results.

Answer

Data: $V_L = 380$ V, so phase emf at no load $E = 380/\sqrt3 = 219.39$ V. $Z_1 = j2\ \Omega$ , $Z_2 = j0.5\ \Omega$ , $Z_0 = j0.25\ \Omega$ . Neutral is isolated, so the neutral impedance $Z_n = \infty$ and the zero-sequence network is open: no zero-sequence current can flow.

(i) L-G fault (phase a to ground)

I_f = 3I_{a0} = \frac{3E}{Z_1 + Z_2 + Z_0 + 3Z_n}

With $Z_n \to \infty$ :

I_f = 0\ \text{A}

There is no return path for the current, so no fault current flows. Only the potential of the neutral shifts: the faulted phase goes to ground potential and the healthy phase voltages to ground rise to line voltage (380 V).

(ii) L-L-G fault (phases b, c to ground)

$I_{a0} = 0$ because the zero-sequence network is open. The fault behaves like an L-L fault:

\begin{aligned} I_{a1} &= \frac{E}{Z_1 + Z_2} = \frac{219.39}{j2.5} = -j87.76\ \text{A} \\ |I_b| = |I_c| &= \sqrt3\,|I_{a1}| = \frac{\sqrt3 \times 219.39}{2.5} = 152.0\ \text{A} \end{aligned}

Current to ground $= 3I_{a0} = 0$ . The line currents are 152.0 A each (flowing from b to c).

(iii) L-L-L-G fault

Balanced fault, only the positive sequence network acts:

I_f = \frac{E}{|Z_1|} = \frac{219.39}{2} = 109.70\ \text{A}

Results

Fault	Fault current
L-G	0 A
L-L-G	152.0 A (ground current 0)
L-L-L-G	109.7 A

Comments

With an isolated neutral, faults involving ground give no zero-sequence current; an L-G fault gives no fault current and an L-L-G fault reduces to an L-L fault.
The L-L (and L-L-G) current, 152 A, is larger than the three-phase current, 109.7 A, because $Z_2$ (0.5 Ω) is much smaller than $Z_1$ (2 Ω): $I_{LL}/I_{3\phi} = \sqrt3 Z_1/(Z_1+Z_2) = 1.386$ .
If the neutral were solidly grounded, the L-G current would be $3E/(Z_1+Z_2+Z_0) = 3\times219.39/2.75 = 239.3$ A, the highest of all. Isolating the neutral therefore removes the severe earth fault current but lets healthy phases rise to line voltage, which stresses insulation.

2081 Bhadra · 8 marks

For the power system network shown below, compute the fault current if a L-L fault occurs at point B. Assume pre-fault voltage at the point of fault is 1 p.u. [Figure: generator 20 MVA, 6 kV, Z1 = 10%, Z2 = 5%, Z0 = 2%, Zn = j2 Ω at bus A; transformer 50 MVA, 6/66 kV, Δ/Y, Y solidly grounded, X = 4%, between A and B; line X = 30 Ω between B and C; transformer 40 MVA, 66 kV/3.3 kV, Y/Δ, Y isolated, X = 4%, between C and D; machine 20 MVA, 3.3 kV, Z1 = 10%, Z2 = 5%, Z0 = 2%, Zn = j2 Ω at bus D]

Answer

Assumptions: base 20 MVA; base voltages 6 kV (generator side), 66 kV (line), 3.3 kV (motor side). The L-L fault needs only the positive and negative sequence networks, so zero-sequence data and the neutral impedances are not used. Pre-fault voltage $V_f = 1$ pu, no-load.

Per-unit reactances (20 MVA base)

\begin{aligned} X_{T1} &= 0.04 \times \frac{20}{50} = 0.016\ \text{pu} \\ Z_{base,66} &= \frac{66^2}{20} = 217.8\ \Omega,\quad X_{line} = \frac{30}{217.8} = 0.1377\ \text{pu} \\ X_{T2} &= 0.04 \times \frac{20}{40} = 0.02\ \text{pu} \\ \text{Generator: } X_1 &= 0.10,\ X_2 = 0.05;\quad \text{Machine at D: } X_1 = 0.10,\ X_2 = 0.05 \end{aligned}

Sequence networks seen from B

Positive:
 E-j0.1-A-j0.016-B-j0.1377-C-j0.02-D-j0.1-E
                |
           (fault point)
Negative:  same, with j0.05 for both machines, no emf

Thevenin impedances at B:

\begin{aligned} Z_1 &= j(0.10+0.016)\ \|\ j(0.1377+0.02+0.10) = j0.116\ \|\ j0.2577 = j0.0800\ \text{pu} \\ Z_2 &= j(0.05+0.016)\ \|\ j(0.1377+0.02+0.05) = j0.066\ \|\ j0.2077 = j0.0501\ \text{pu} \end{aligned}

L-L fault current (phases b and c)

\begin{aligned} I_{a1} &= -I_{a2} = \frac{V_f}{Z_1+Z_2} = \frac{1}{j0.1301} = -j7.687\ \text{pu} \\ |I_f| = |I_b| = |I_c| &= \sqrt3\,|I_{a1}| = \frac{\sqrt3}{0.1301} = 13.31\ \text{pu} \end{aligned}

Base current at B (66 kV):

I_{base} = \frac{20 \times 10^6}{\sqrt3 \times 66 \times 10^3} = 174.95\ \text{A}

I_f = 13.31 \times 174.95 = 2329.5\ \text{A}

Answer: L-L fault current at B ≈ 13.31 pu ≈ 2.33 kA (in phases b and c, opposite in direction).

2079 Bhadra · 6+2 marks

A 30 MVA, 11 kV solidly grounded generator has positive, negative and zero sequence impedance of j0.2 pu, j0.2 pu and j0.05 pu. Generator is under unloaded condition. i) Calculate fault current and line to line voltages during fault condition if L-G fault occurs at the generator terminals. ii) Find the line current for 3-phase fault.

Answer

Data: 30 MVA, 11 kV, $Z_1 = Z_2 = j0.2$ , $Z_0 = j0.05$ pu, solidly grounded, unloaded, so $E_a = 1\angle0^\circ$ pu.

Base current: $I_{base} = \dfrac{30\times10^6}{\sqrt3 \times 11\times10^3} = 1574.6$ A; base phase voltage $= 11/\sqrt3 = 6.351$ kV.

(i) L-G fault on phase a

Sequence networks in series:

\begin{aligned} I_{a1} = I_{a2} = I_{a0} &= \frac{E_a}{Z_1+Z_2+Z_0} = \frac{1}{j0.45} = -j2.222\ \text{pu} \\ I_f = I_a &= 3I_{a1} = -j6.667\ \text{pu} \\ |I_f| &= 6.667 \times 1574.6 = 10\,497\ \text{A} \end{aligned}

Sequence voltages

\begin{aligned} V_{a1} &= E_a - Z_1 I_{a1} = 1 - (j0.2)(-j2.222) = 0.5556 \\ V_{a2} &= -Z_2 I_{a2} = -0.4444 \\ V_{a0} &= -Z_0 I_{a0} = -0.1111 \end{aligned}

Phase voltages ( $V = V_{a0} + a^2 V_{a1} + aV_{a2}$ for phase b, etc.)

\begin{aligned} V_a &= 0.5556 - 0.4444 - 0.1111 = 0 \\ V_b &= V_{a0} + a^2V_{a1} + aV_{a2} = 0.8819\angle{-100.89^\circ}\ \text{pu} \\ V_c &= V_{a0} + aV_{a1} + a^2V_{a2} = 0.8819\angle{100.89^\circ}\ \text{pu} \end{aligned}

Line-to-line voltages

\begin{aligned} V_{ab} &= V_a - V_b = 0.8819\angle{79.11^\circ}\ \text{pu} \Rightarrow 0.8819 \times 6.351 = 5.60\ \text{kV} \\ V_{bc} &= V_b - V_c = 1.7321\angle{-90^\circ}\ \text{pu} \Rightarrow 1.7321 \times 6.351 = 11.0\ \text{kV} \\ V_{ca} &= V_c - V_a = 0.8819\angle{100.89^\circ}\ \text{pu} \Rightarrow 5.60\ \text{kV} \end{aligned}

(Line voltages are converted with the phase base 6.351 kV because the pu values are on a phase-voltage base.)

(ii) Three-phase fault

I_f = \frac{E_a}{Z_1} = \frac{1}{j0.2} = -j5\ \text{pu} \Rightarrow 5 \times 1574.6 = 7873\ \text{A}

Results

Quantity	Value
L-G fault current	6.667 pu = 10.50 kA
$V_{ab}$ , $V_{bc}$ , $V_{ca}$ during L-G	5.60 kV, 11.0 kV, 5.60 kV
3-phase fault line current	5 pu = 7.87 kA

The L-G current is larger than the 3-phase current because $Z_0$ (0.05) is much less than $Z_1$ .

2079 Bhadra · 8 marks

Figure shows a power system network. Draw positive, negative and zero sequence network. If LG fault occurs at bus 1, find fault current. Assume fault impedance Zf = 0.05 p.u.

Equipment MVA rating Voltage rating X1 (p.u.) X2 (p.u.) X0 (p.u.)
G1 50 11 KV 0.20 0.20 0.08
G2 30 11 KV 0.25 0.25 0.1
Transformer T1 50 11/132 KV 0.1 0.1 0.1
Transformer T1 (T2) 30 11/132 KV 0.09 0.09 0.09
Line L1 45 132 KV 0.1 0.1 0.25
Line L1 (L2) 45 132 KV 0.1 0.1 0.25
[Figure: G1 (Y grounded) - T1 (Δ on generator side, Y grounded on bus 1 side) - bus 1 - two parallel lines - bus 2 - T2 (Y grounded on bus 2 side, Y grounded through j0.03 on generator side) - G2 (Y grounded); fault F at bus 1]

Equipment	MVA rating	Voltage rating	X1 (p.u.)	X2 (p.u.)	X0 (p.u.)
G1	50	11 KV	0.20	0.20	0.08
G2	30	11 KV	0.25	0.25	0.1
Transformer T1	50	11/132 KV	0.1	0.1	0.1
Transformer T1 (T2)	30	11/132 KV	0.09	0.09	0.09
Line L1	45	132 KV	0.1	0.1	0.25
Line L1 (L2)	45	132 KV	0.1	0.1	0.25

Answer

Assumptions: base 50 MVA, 11 kV (generators) and 132 kV (lines). The neutral reactance j0.03 of T2 is taken as already on the 50 MVA base. Pre-fault voltage 1 pu, no load.

Per-unit values on 50 MVA

Element	$X_1 = X_2$	$X_0$
G1	0.20	0.08
G2 ( $\times 50/30$ )	0.4167	0.1667
T1	0.10	0.10
T2 ( $\times 50/30$ )	0.15	0.15
Each line ( $\times 50/45$ )	0.1111	0.2778
Two lines in parallel	0.0556	0.1389

Sequence networks

Positive (negative same, no emf):
 E1-j0.2-j0.1-(1)-j0.0556-(2)-j0.15-j0.4167-E2
               F
Zero:
 ref-j0.1-(1)-j0.1389-(2)-j0.15-[3x j0.03]-j0.1667-ref
 (G1 cut off by T1 delta; T2 is Yg-Yg, so G2
  zero seq path continues through T2)

Thevenin impedances at bus 1

\begin{aligned} Z_1 = Z_2 &= j0.30\ \|\ j(0.0556 + 0.15 + 0.4167) = j0.30\ \|\ j0.6222 = j0.2024\ \text{pu} \\ Z_0 &= j0.10\ \|\ j(0.1389 + 0.15 + 0.09 + 0.1667) = j0.10\ \|\ j0.5456 = j0.0845\ \text{pu} \end{aligned}

Fault current (L-G through $Z_f = 0.05$ pu)

I_{a0} = I_{a1} = I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_0 + 3Z_f}

Taking $Z_f$ as a reactance $j0.05$ (usual in these problems), $3Z_f = j0.15$ :

\begin{aligned} I_{a0} &= \frac{1}{j(0.2024+0.2024+0.0845+0.15)} = \frac{1}{j0.6393} = -j1.564\ \text{pu} \\ I_f &= 3I_{a0} = -j4.692\ \text{pu} \end{aligned}

Base current at 132 kV:

I_{base} = \frac{50\times10^6}{\sqrt3 \times 132\times10^3} = 218.7\ \text{A}

I_f = 4.692 \times 218.7 = 1026\ \text{A}

Answer: L-G fault current at bus 1 ≈ 4.69 pu ≈ 1.03 kA.

(If $Z_f$ is taken as a pure resistance 0.05 pu, $|I_f| = 3/|0.15 + j0.4893| = 5.86$ pu.)

2078 Bhadra · 16 marks

Determine the Fault current when a line to ground fault occurs at Bus 3 as shown in figure below. G1, G2: 100 MVA, 11 kV, X1 = X2 = 15%, X0 = 5%, Xn = 6%; T1, T2: 100 MVA, 11 kV/220 kV, Xleak = 10%; L1, L2: X1 = X2 = 10%, X0 = 10% on a base of 100 MVA. [Figure: G1 (Y grounded through Xn) - bus 1 - T1 (Y grounded/Y grounded) - bus 2 - parallel lines L1 and L2 - bus 3 - T2 (Y grounded/Y grounded) - bus 4 - G2 (Y grounded through Xn)]

Answer

Base: 100 MVA; 11 kV on generator side, 220 kV on line side. All data are already on this base. $X_n = 0.06$ pu for each generator. Pre-fault voltage $E = 1\angle0^\circ$ pu, system unloaded.

Single line diagram

 G1        T1                 L1            T2        G2
 (~)--(1)--)(--(2)====================(3)--)(--(4)--(~)
  |       Yg Yg     ====== L2 =======      Yg Yg     |
 Xn                                                  Xn
 |                                                   |
gnd                      F (L-G at bus 3)           gnd

Per-unit reactances

Element	$X_1$	$X_2$	$X_0$
G1, G2	0.15	0.15	0.05 + 3(0.06) = 0.23
T1, T2	0.10	0.10	0.10
L1 ∥ L2	0.05	0.05	0.05

Positive sequence network

 E1-j0.15-(1)-j0.1-(2)-j0.05-(3)-j0.1-(4)-j0.15-E2
                            |
                            F

Z_1 = j(0.15+0.1+0.05)\ \|\ j(0.1+0.15) = j0.30\ \|\ j0.25 = j0.1364\ \text{pu}

Negative sequence network

Same as positive, without the sources:

Z_2 = j0.1364\ \text{pu}

Zero sequence network

Both transformers are Yg–Yg, so zero-sequence current passes through them; generator neutrals are grounded through $X_n$ , which appears as $3X_n$ .

 ref-j0.23-(1)-j0.1-(2)-j0.05-(3)-j0.1-(4)-j0.23-ref
                             |
                             F

Z_0 = j(0.23+0.1+0.05)\ \|\ j(0.1+0.23) = j0.38\ \|\ j0.33 = j0.1766\ \text{pu}

Fault current (L-G at bus 3, $Z_f = 0$ )

Sequence networks in series:

\begin{aligned} I_{a1} = I_{a2} = I_{a0} &= \frac{E}{Z_1+Z_2+Z_0} = \frac{1}{j(0.1364+0.1364+0.1766)} = \frac{1}{j0.4494} = -j2.2255\ \text{pu} \\ I_f &= 3I_{a0} = -j6.676\ \text{pu} \end{aligned}

Base current at 220 kV:

I_{base} = \frac{100\times10^6}{\sqrt3 \times 220\times10^3} = 262.43\ \text{A}

I_f = 6.676 \times 262.43 = 1752\ \text{A}

Phase currents at the fault

$I_a = I_f = 6.676\angle{-90^\circ}$ pu $= 1.752$ kA
$I_b = I_c = 0$ (healthy phases, unloaded system)

Fault MVA (for reference)

Fault MVA $= |I_f| \times$ base MVA (with $V_f = 1$ ) $= 6.676 \times 100 = 667.6$ MVA.

Answer: L-G fault current at bus 3 ≈ 6.68 pu ≈ 1.75 kA.

Note: $Z_0$ (0.1766) is larger than $Z_1$ (0.1364) mainly because of the neutral reactances, so here the L-G current (6.68 pu) is less than the three-phase fault current ( $1/0.1364 = 7.33$ pu).

2078 Kartik · 8 marks

A 40 MVA, 11 kV generator has Z1 = Z2 = j0.3 pu, Z0 = j0.4 pu. A line to line fault occurs on the generator terminals. Find the fault current.

Answer

Data: 40 MVA, 11 kV, $Z_1 = Z_2 = j0.3$ pu, $Z_0 = j0.4$ pu. Generator unloaded, $E_a = 1\angle0^\circ$ pu. Fault between phases b and c.

Fault conditions

$I_a = 0,\ I_b = -I_c,\ V_b = V_c$ , which give $I_{a0} = 0$ and $I_{a2} = -I_{a1}$ : the positive and negative sequence networks are connected in parallel (opposing), and $Z_0$ does not appear.

Sequence currents

I_{a1} = -I_{a2} = \frac{E_a}{Z_1 + Z_2} = \frac{1}{j0.6} = -j1.667\ \text{pu}

Fault current

\begin{aligned} I_b &= (a^2 - a)I_{a1} = -j\sqrt3\, I_{a1} = -j\sqrt3(-j1.667) = -2.887\ \text{pu} \\ I_c &= -I_b = 2.887\ \text{pu} \end{aligned}

Base current:

I_{base} = \frac{40\times10^6}{\sqrt3 \times 11\times10^3} = 2099.5\ \text{A}

|I_f| = 2.887 \times 2099.5 = 6061\ \text{A}

Answer: L-L fault current = 2.887 pu ≈ 6.06 kA (equal and opposite in phases b and c; $I_a = 0$ ).

For comparison, a 3-phase fault would give $1/0.3 = 3.333$ pu, so $I_{LL} = \frac{\sqrt3}{2} I_{3\phi}$ when $Z_1 = Z_2$ .

2076 Chaitra · 10 marks

Compute the sequence currents for a LLG fault at bus 3 of the following network. The fault impedance is j0.1 pu. All the parameters are in pu. [Figure: G1 (Y grounded) - T1 (Y grounded/Y grounded) - bus 1; G2 (Y grounded) - T2 (Δ on generator side/Y grounded) - bus 2; L1 between buses 1 and 2, L2 between buses 1 and 3, L3 between buses 2 and 3]

Item Base MVA X1 X2 X0
G1 100 0.15 0.15 0.05
G2 100 0.15 0.15 0.05
T1 100 0.1 0.1 0.1
T2 100 0.1 0.1 0.1
L1 100 0.12 0.12 0.3
L2 100 0.15 0.15 0.35
L3 100 0.25 0.25 0.71

Item	Base MVA	X1	X2	X0
G1	100	0.15	0.15	0.05
G2	100	0.15	0.15	0.05
T1	100	0.1	0.1	0.1
T2	100	0.1	0.1	0.1
L1	100	0.12	0.12	0.3
L2	100	0.15	0.15	0.35
L3	100	0.25	0.25	0.71

Answer

Approach: find the Thevenin impedances at bus 3 from the bus impedance matrices of the positive and zero sequence networks ( $Z_2 = Z_1$ since all $X_1 = X_2$ ), then connect the networks for an LLG fault. Pre-fault voltage $V_f = 1\angle0^\circ$ pu, no load; $Z_f = j0.1$ pu is the impedance from the faulted phases' common point to ground.

Positive sequence network

Generator + transformer branches to reference: bus 1: $j(0.15+0.1) = j0.25$ ; bus 2: $j0.25$ . Lines: 1-2 $j0.12$ , 1-3 $j0.15$ , 2-3 $j0.25$ .

   E1          E2
   |           |
  j0.25       j0.25
   |           |
  (1)--j0.12--(2)
    \         /
   j0.15   j0.25
      \     /
       (3) F

Forming $Y_{bus}$ and inverting gives

Z_{bus}^{(1)} = j\begin{bmatrix} 0.1445 & 0.1055 & 0.1299 \\ 0.1055 & 0.1445 & 0.1201 \\ 0.1299 & 0.1201 & 0.2200 \end{bmatrix}

So $Z_1 = Z_2 = Z_{33}^{(1)} = j0.2200$ pu.

Zero sequence network

G1 (Yg) + T1 (Yg/Yg): zero-seq path passes, bus 1 to reference $j(0.05+0.1) = j0.15$
G2 behind T2 (Δ on generator side, Yg on bus side): bus 2 to reference through $j0.1$ only; G2 is cut off
Lines: 1-2 $j0.3$ , 1-3 $j0.35$ , 2-3 $j0.71$

Z_{bus}^{(0)} = j\begin{bmatrix} 0.1035 & 0.0310 & 0.0796 \\ 0.0310 & 0.0793 & 0.0470 \\ 0.0796 & 0.0470 & 0.3032 \end{bmatrix}

So $Z_0 = j0.3032$ pu.

Sequence currents for LLG fault (phases b, c to ground through $Z_f$ )

The negative network is in parallel with the zero network plus $3Z_f$ :

\begin{aligned} Z_0 + 3Z_f &= j0.3032 + j0.3 = j0.6032 \\ I_{a1} &= \frac{V_f}{Z_1 + \dfrac{Z_2(Z_0+3Z_f)}{Z_2 + Z_0 + 3Z_f}} = \frac{1}{j0.2200 + j\dfrac{0.2200 \times 0.6032}{0.8232}} = \frac{1}{j0.3812} = -j2.6236\ \text{pu} \\ I_{a2} &= -I_{a1}\frac{Z_0+3Z_f}{Z_2+Z_0+3Z_f} = j2.6236 \times \frac{0.6032}{0.8232} = j1.9225\ \text{pu} \\ I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0+3Z_f} = j2.6236 \times \frac{0.2200}{0.8232} = j0.7011\ \text{pu} \end{aligned}

Check: $I_{a1} + I_{a2} + I_{a0} = -j2.6236 + j1.9225 + j0.7011 = 0 = I_a$ .

Phase and fault currents

\begin{aligned} I_b &= I_{a0} + a^2 I_{a1} + a I_{a2} = 4.075\angle{165.05^\circ}\ \text{pu} \\ I_c &= I_{a0} + a I_{a1} + a^2 I_{a2} = 4.075\angle{14.95^\circ}\ \text{pu} \\ I_f &= I_b + I_c = 3I_{a0} = j2.103\ \text{pu} \end{aligned}

Answer: $I_{a1} = -j2.624$ pu, $I_{a2} = j1.923$ pu, $I_{a0} = j0.701$ pu; fault (ground) current $= 2.103$ pu; $|I_b| = |I_c| = 4.075$ pu.

2076 Asoj · 8 marks

A single line to ground fault occur at generator terminal of 20 MVA, 13.8 KV and having Z1 = j0.20 pu, Z2 = j0.3 pu. Find the fault current and line to line voltage under fault condition.

Answer

Data: 20 MVA, 13.8 kV, $Z_1 = j0.20$ pu, $Z_2 = j0.30$ pu. The zero-sequence impedance is not given; assume $Z_0 = j0.10$ pu (typical value for such a generator, about half of $Z_1$ ) and a solidly grounded neutral. Unloaded, $E_a = 1\angle0^\circ$ pu.

Base current: $I_{base} = \dfrac{20\times10^6}{\sqrt3 \times 13.8\times10^3} = 836.7$ A; base phase voltage $= 13.8/\sqrt3 = 7.967$ kV.

Fault current (phase a to ground)

Sequence networks in series:

\begin{aligned} I_{a1} = I_{a2} = I_{a0} &= \frac{E_a}{Z_1+Z_2+Z_0} = \frac{1}{j0.6} = -j1.667\ \text{pu} \\ I_f &= 3I_{a1} = -j5.0\ \text{pu} \\ |I_f| &= 5 \times 836.7 = 4184\ \text{A} \end{aligned}

Sequence voltages

\begin{aligned} V_{a1} &= 1 - (j0.2)(-j1.667) = 0.6667 \\ V_{a2} &= -(j0.3)(-j1.667) = -0.5 \\ V_{a0} &= -(j0.1)(-j1.667) = -0.1667 \end{aligned}

Phase voltages

\begin{aligned} V_a &= V_{a0}+V_{a1}+V_{a2} = 0 \\ V_b &= V_{a0} + a^2V_{a1} + aV_{a2} = 1.0408\angle{-103.90^\circ}\ \text{pu} \\ V_c &= V_{a0} + aV_{a1} + a^2V_{a2} = 1.0408\angle{103.90^\circ}\ \text{pu} \end{aligned}

Line-to-line voltages

\begin{aligned} V_{ab} &= V_a - V_b = 1.0408\angle{76.10^\circ}\ \text{pu} = 1.0408 \times 7.967 = 8.29\ \text{kV} \\ V_{bc} &= V_b - V_c = 2.0207\angle{-90^\circ}\ \text{pu} = 16.10\ \text{kV} \\ V_{ca} &= V_c - V_a = 1.0408\angle{103.90^\circ}\ \text{pu} = 8.29\ \text{kV} \end{aligned}

Answer (with $Z_0 = j0.1$ pu): $I_f = 5$ pu $= 4.18$ kA; $V_{ab} = V_{ca} = 8.29$ kV, $V_{bc} = 16.10$ kV.

General form for any $Z_0$ : $I_f = \dfrac{3}{|Z_1+Z_2+Z_0|}$ pu, so with the data given the fault current is $3/(0.5 + X_0)$ pu. $V_{bc}$ rises above the rated 13.8 kV because $Z_2 > Z_1$ .

2075 Chaitra · 10 marks

Two alternators are operating in parallel and supplying a synchronous motor which is receiving 60 MW power at 0.8 pf (lag) at 6 kV. Single line diagram for the system and its data are given below. Compute the fault current when a single line to ground fault occurs at the middle of the line through a fault resistance of 4.033 ohm. Data: G1 & G2: 11 kV, 100 MVA, xg1 = 0.20 pu, xg2 = xg0 = 0.10 pu; T1: 180 MVA, 11.5/115 kV, xT1 = 0.10 pu; T2: 170 MVA, 6.6/115 kV, xT2 = 0.10 pu; M: 6.3 kV, 160 MVA, xM1 = xM2 = 0.30 pu, xM0 = 0.10 pu; Line: xLINE1 = xLINE2 = 30.25 ohm, xLINE0 = 60.5 ohm. [Figure: G1 and G2 (star, grounded) in parallel on a bus - T1 - line - T2 - motor M (star, grounded)]

Answer

Assumptions: the figure does not mark transformer windings, so the usual arrangement is taken: T1 and T2 are Δ on the machine side and star-grounded on the 115 kV line side. Then the machine zero-sequence reactances are isolated by the deltas. Base: 100 MVA, 11 kV in the generator circuit.

Base voltages and per-unit values

\begin{aligned} V_{b,line} &= 11 \times \frac{115}{11.5} = 110\ \text{kV}, \quad Z_{b,line} = \frac{110^2}{100} = 121\ \Omega \\ V_{b,motor} &= 110 \times \frac{6.6}{115} = 6.313\ \text{kV} \end{aligned}

Element	Calculation	pu
G1 ∥ G2, $X_1$	$0.2 \\| 0.2$	0.10
G1 ∥ G2, $X_2$	$0.1 \\| 0.1$	0.05
T1	$0.1 \times \frac{100}{180}\times(\frac{11.5}{11})^2$	0.0607
Line $X_1 = X_2$	$30.25/121$	0.25
Line $X_0$	$60.5/121$	0.50
T2	$0.1 \times \frac{100}{170}\times(\frac{6.6}{6.313})^2$	0.0643
Motor $X_1 = X_2$	$0.3 \times \frac{100}{160}\times(\frac{6.3}{6.313})^2$	0.1867
Fault resistance $R_f$	$4.033/121$	0.0333

Pre-fault voltage at the middle of the line

Motor voltage $V_m = 6/6.313 = 0.9504\angle0^\circ$ pu (reference). Motor input $S = 60/0.8 = 75$ MVA $= 0.75$ pu.

\begin{aligned} I_L &= \frac{0.75}{0.9504}\angle{-36.87^\circ} = 0.7891\angle{-36.87^\circ}\ \text{pu} \\ V_f &= V_m + I_L \cdot j(X_{T2} + X_{line}/2) = 0.9504 + 0.7891\angle{-36.87^\circ} \times j0.3143 \\ &= 1.0469\angle{6.55^\circ}\ \text{pu} \end{aligned}

Thevenin impedances at the fault point (mid-line)

\begin{aligned} Z_1 &= j(0.10 + 0.0607 + 0.125)\ \|\ j(0.125 + 0.0643 + 0.1867) = j0.2857 \| j0.3760 = j0.1624\ \text{pu} \\ Z_2 &= j(0.05 + 0.0607 + 0.125)\ \|\ j0.3760 = j0.2357 \| j0.3760 = j0.1449\ \text{pu} \\ Z_0 &= j(0.0607 + 0.25)\ \|\ j(0.0643 + 0.25) = j0.3107 \| j0.3143 = j0.1562\ \text{pu} \end{aligned}

(Zero sequence: each half of the line, $X_0/2 = 0.25$ , goes to ground through the grounded star of its transformer.)

Fault current (sequence networks in series with $3R_f$ )

\begin{aligned} I_{a0} = I_{a1} = I_{a2} &= \frac{V_f}{Z_1+Z_2+Z_0+3R_f} = \frac{1.0469\angle{6.55^\circ}}{0.1 + j0.4635} \\ &= 2.208\angle{-71.27^\circ}\ \text{pu} \\ I_f &= 3I_{a0} = 6.624\angle{-71.27^\circ}\ \text{pu} \end{aligned}

Base current on the line:

I_{base} = \frac{100\times10^6}{\sqrt3 \times 110\times10^3} = 524.9\ \text{A}

|I_f| = 6.624 \times 524.9 = 3477\ \text{A}

Answer: SLG fault current ≈ 6.62 pu ≈ 3.48 kA (angle measured from the motor terminal voltage).

Note: $3R_f = 0.1$ pu adds resistance in series, so it both reduces the current and makes it lag less than 90°.

2075 Asoj · 12 marks

Three 6.6 kV, 3-phase, 10 MVA alternators are connected to a common bus. Each alternator has a positive sequence reactance of 0.15 pu. The negative and zero sequence reactances are 75% and 30% of positive sequence reactance. A single line-to-ground fault occurs on the bus. Find the fault current for the following cases: (i) All the alternator neutrals are solidly grounded. (ii) One alternator neutral is grounded through 0.3 ohm resistance and the other two neutrals are isolated.

Answer

Base: 10 MVA, 6.6 kV (rating of each alternator).

I_{base} = \frac{10\times10^6}{\sqrt3 \times 6.6\times10^3} = 874.8\ \text{A}, \quad Z_{base} = \frac{6.6^2}{10} = 4.356\ \Omega

Each alternator: $X_1 = 0.15$ , $X_2 = 0.75 \times 0.15 = 0.1125$ , $X_0 = 0.30 \times 0.15 = 0.045$ pu. Pre-fault voltage 1 pu, no load.

The positive and negative networks always have all three machines in parallel:

X_1 = \frac{0.15}{3} = 0.05\ \text{pu}, \quad X_2 = \frac{0.1125}{3} = 0.0375\ \text{pu}

(i) All neutrals solidly grounded

All three zero-sequence reactances are in parallel:

\begin{aligned} X_0 &= \frac{0.045}{3} = 0.015\ \text{pu} \\ I_f &= \frac{3E}{j(X_1+X_2+X_0)} = \frac{3}{j(0.05+0.0375+0.015)} = \frac{3}{j0.1025} = -j29.27\ \text{pu} \\ |I_f| &= 29.27 \times 874.8 = 25\,603\ \text{A} \approx 25.6\ \text{kA} \end{aligned}

(ii) One neutral grounded through 0.3 Ω, others isolated

Only the grounded machine provides a zero-sequence path; its neutral resistance appears as $3R_n$ :

\begin{aligned} R_n &= \frac{0.3}{4.356} = 0.06887\ \text{pu}, \quad 3R_n = 0.2066\ \text{pu} \\ Z_0 &= 3R_n + jX_0 = 0.2066 + j0.045\ \text{pu} \\ Z_1+Z_2+Z_0 &= 0.2066 + j(0.05 + 0.0375 + 0.045) = 0.2066 + j0.1325 \\ I_f &= \frac{3}{0.2066 + j0.1325} = 12.22\angle{-32.67^\circ}\ \text{pu} \\ |I_f| &= 12.22 \times 874.8 = 10\,692\ \text{A} \approx 10.7\ \text{kA} \end{aligned}

Results

Case	Fault current
(i) All neutrals solidly grounded	29.27 pu = 25.6 kA
(ii) One neutral through 0.3 Ω	12.22 pu = 10.7 kA

Grounding only one machine through a resistance cuts the earth-fault current to about 42 % and is why generator neutrals on a common bus are normally grounded through impedance, with only one neutral grounded.

2074 Asoj · 8 marks

A 30 MVA, 13.2 kV synchronous generator has a solidly grounded neutral. Its positive, negative and zero sequence impedances are 0.30, 0.40 and 0.05 pu respectively. Determine the following: i) The value of reactance that must be placed in the generator neutral so that the fault current for a line-to-ground fault of zero fault impedance shall not exceed the rated line current. ii) The value of resistance to be placed in the neutral that will serve the same purpose.

Answer

Data: 30 MVA, 13.2 kV; $X_1 = 0.30$ , $X_2 = 0.40$ , $X_0 = 0.05$ pu. Rated line current $= 1$ pu. Pre-fault voltage $E = 1$ pu.

Z_{base} = \frac{13.2^2}{30} = 5.808\ \Omega

With a neutral impedance $Z_n$ , the zero-sequence impedance becomes $Z_0 + 3Z_n$ :

I_f = \frac{3E}{|Z_1 + Z_2 + Z_0 + 3Z_n|}

Without $Z_n$ : $I_f = 3/0.75 = 4$ pu, i.e. four times rated current, so it must be limited to 1 pu.

(i) Neutral reactance $X_n$

\begin{aligned} \frac{3}{0.30 + 0.40 + 0.05 + 3X_n} &= 1 \\ 0.75 + 3X_n &= 3 \\ X_n &= 0.75\ \text{pu} = 0.75 \times 5.808 = 4.356\ \Omega \end{aligned}

(ii) Neutral resistance $R_n$

The resistance is in quadrature with the reactances:

\begin{aligned} \frac{3}{\sqrt{(3R_n)^2 + 0.75^2}} &= 1 \\ (3R_n)^2 &= 9 - 0.5625 = 8.4375 \\ 3R_n &= 2.9047, \quad R_n = 0.9682\ \text{pu} \\ R_n &= 0.9682 \times 5.808 = 5.624\ \Omega \end{aligned}

Answer: (i) neutral reactance $X_n = 0.75$ pu $= 4.36\ \Omega$ ; (ii) neutral resistance $R_n = 0.968$ pu $= 5.62\ \Omega$ .

A larger ohmic value is needed with resistance because it adds to the network reactance in quadrature, not directly.

2074 Chaitra · 9 marks

A double line to ground fault occur at generator terminal of 30 MVA, 11 kV and having Z1 = Z2 = j0.2 pu and Z0 = j0.05 pu. Find the line currents, fault current and line to neutral voltages under fault condition.

Answer

Data: 30 MVA, 11 kV, $Z_1 = Z_2 = j0.2$ , $Z_0 = j0.05$ pu, solidly grounded, unloaded ( $E_a = 1\angle0^\circ$ ). Fault: phases b and c to ground.

$I_{base} = \dfrac{30\times10^6}{\sqrt3\times11\times10^3} = 1574.6$ A; $V_{base,ph} = 11/\sqrt3 = 6.351$ kV.

Fault conditions

$I_a = 0$ , $V_b = V_c = 0$ , giving $V_{a0} = V_{a1} = V_{a2}$ : the three sequence networks are in parallel.

Sequence currents

\begin{aligned} Z_2 \| Z_0 &= \frac{(j0.2)(j0.05)}{j0.25} = j0.04 \\ I_{a1} &= \frac{E_a}{Z_1 + Z_2\|Z_0} = \frac{1}{j0.24} = -j4.1667\ \text{pu} \\ I_{a2} &= -I_{a1}\frac{Z_0}{Z_2+Z_0} = j4.1667 \times 0.2 = j0.8333\ \text{pu} \\ I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0} = j4.1667 \times 0.8 = j3.3333\ \text{pu} \end{aligned}

Line currents

\begin{aligned} I_a &= I_{a0} + I_{a1} + I_{a2} = 0 \\ I_b &= I_{a0} + a^2I_{a1} + aI_{a2} = 6.614\angle{130.89^\circ}\ \text{pu} \Rightarrow 6.614 \times 1574.6 = 10\,415\ \text{A} \\ I_c &= I_{a0} + aI_{a1} + a^2I_{a2} = 6.614\angle{49.11^\circ}\ \text{pu} \Rightarrow 10\,415\ \text{A} \end{aligned}

Fault current (to ground)

I_f = I_b + I_c = 3I_{a0} = j10.0\ \text{pu} \Rightarrow 10 \times 1574.6 = 15\,746\ \text{A}

Line-to-neutral voltages

\begin{aligned} V_{a1} = V_{a2} = V_{a0} &= E_a - Z_1I_{a1} = 1 - (j0.2)(-j4.1667) = 0.1667\ \text{pu} \\ V_a &= 3V_{a1} = 0.5\ \text{pu} = 0.5 \times 6.351 = 3.175\ \text{kV} \\ V_b &= V_c = 0 \end{aligned}

Results

Quantity	pu	Actual
$I_a$	0	0
$I_b$ , $I_c$	6.614	10.41 kA
Fault current $I_f$	10.0	15.75 kA
$V_a$	0.5	3.18 kV
$V_b$ , $V_c$	0	0

2073 Shrawan · 6 marks

A 3-phase generator rated 15 MVA, 13.2 kV has a solidly grounded neutral. Its positive, negative and zero sequence reactance are 40%, 30% and 5% respectively. Find the value of reactance to be connected in neutral circuit so that fault current for a single line to ground fault (of negligible fault impedance) at No-load does not exceed line current.

Answer

Data: 15 MVA, 13.2 kV, solidly grounded; $X_1 = 0.40$ , $X_2 = 0.30$ , $X_0 = 0.05$ pu. At no load $E = 1$ pu; rated line current $= 1$ pu.

Fault current without neutral reactance

I_f = \frac{3E}{X_1+X_2+X_0} = \frac{3}{0.75} = 4\ \text{pu}

This is four times rated current, so a neutral reactance $X_n$ is needed. $X_n$ carries $3I_{a0}$ , so it appears as $3X_n$ in the zero-sequence network.

Required neutral reactance

\begin{aligned} I_f &= \frac{3E}{X_1+X_2+X_0+3X_n} \le 1 \\ 0.40 + 0.30 + 0.05 + 3X_n &= 3 \\ X_n &= \frac{3 - 0.75}{3} = 0.75\ \text{pu} \end{aligned}

Ohmic value

Z_{base} = \frac{(13.2)^2}{15} = 11.616\ \Omega

X_n = 0.75 \times 11.616 = 8.712\ \Omega

Answer: neutral reactance $X_n = 0.75$ pu $= 8.71\ \Omega$ .

Check: with $X_n$ , $I_f = 3/(0.75 + 2.25) = 1$ pu = rated current $\left(\frac{15\times10^6}{\sqrt3\times13.2\times10^3} = 656\ \text{A}\right)$ .

2073 Shrawan · 10 marks

Figure below shows the power system network. i) Draw positive, negative and zero sequence networks ii) Determine fault current in kA if line to line fault occurs at Bus 3.
System Data:

Equipment MVA rating Voltage rating X1 pu X2 pu X0 pu
Generator, G1 50 11 kV 0.2 0.2 0.05
Generator, G2 50 11 kV 0.15 0.15 0.03
Transformer, T1 50 11/220 kV 0.1 0.1 0.1
Transformer, T2 50 11/220 kV 0.075 0.075 0.075
Line-1 50 220 kV 0.12 0.12 0.42
Line-2 50 220 kV 0.12 0.12 0.42
The pre-fault voltage at Bus-3 is 0.95 p.u at common base of G1. [Figure: G1 (Y, solidly grounded) - bus 1 - T1 (Δ/Y grounded) - bus 2 - Line 1 and Line 2 in parallel - bus 3 - T2 (Y grounded through j0.03 on bus 3 side / Δ) - bus 4 - G2 (Y, grounded through j0.024)]

Equipment	MVA rating	Voltage rating	X1 pu	X2 pu	X0 pu
Generator, G1	50	11 kV	0.2	0.2	0.05
Generator, G2	50	11 kV	0.15	0.15	0.03
Transformer, T1	50	11/220 kV	0.1	0.1	0.1
Transformer, T2	50	11/220 kV	0.075	0.075	0.075
Line-1	50	220 kV	0.12	0.12	0.42
Line-2	50	220 kV	0.12	0.12	0.42

Answer

Base: 50 MVA, 11 kV (generators), 220 kV (lines); all data are on this base. Pre-fault voltage at bus 3: $V_f = 0.95$ pu, no load.

Single line diagram

 G1      T1               Line-1              T2      G2
 (~)-(1)-)(-(2)======================(3)-)(-(4)-(~)
 Yg      D  Yg     Line-2 (parallel)     Yg   D     Yg
                                         |          |
                                       j0.03     j0.024

(i) Sequence networks

Lines in parallel: $X_1 = X_2 = 0.12/2 = 0.06$ , $X_0 = 0.42/2 = 0.21$ pu.

Positive sequence (emfs $E_1$ , $E_2$ ):

 E1-j0.2-(1)-j0.1-(2)-j0.06-(3)-j0.075-(4)-j0.15-E2
                             |
                             F

Negative sequence: same reactances (0.2, 0.1, 0.06, 0.075, 0.15), no emfs, all ends to reference.

Zero sequence:

G1 is behind the Δ of T1, so it is isolated; T1 connects bus 2 to reference through $j0.1$ .
T2 star side (bus 3) grounded through $j0.03$ : bus 3 to reference through $j(0.075 + 3\times0.03) = j0.165$ ; G2 is behind the Δ, so isolated.

 ref-j0.1-(2)-j0.21-(3)-j0.165-ref
                     |
                     F
 G1 zero seq: j0.05 isolated;  G2: j0.03+3(j0.024) isolated

$Z_0 = j0.31 \| j0.165 = j0.1077$ pu (not needed for L-L fault).

(ii) L-L fault at bus 3

Thevenin impedances:

\begin{aligned} Z_1 &= j(0.2+0.1+0.06)\ \|\ j(0.075+0.15) = j0.36\ \|\ j0.225 = j0.1385\ \text{pu} \\ Z_2 &= Z_1 = j0.1385\ \text{pu} \end{aligned}

Sequence currents:

I_{a1} = -I_{a2} = \frac{V_f}{Z_1+Z_2} = \frac{0.95}{j0.2769} = -j3.431\ \text{pu}

Fault current:

|I_f| = |I_b| = |I_c| = \sqrt3 \times 3.431 = 5.942\ \text{pu}

Base current at 220 kV:

I_{base} = \frac{50\times10^6}{\sqrt3 \times 220\times10^3} = 131.2\ \text{A}

I_f = 5.942 \times 131.2 = 779.7\ \text{A} = 0.780\ \text{kA}

Answer: L-L fault current at bus 3 ≈ 5.94 pu ≈ 0.78 kA.

2073 Chaitra · 8 marks

A 30 MVA, 11 kV generator has Z1 = Z2 = j0.2 pu, Z0 = j0.05 pu. A line to line fault occurs on the generator terminals. Find the line currents, fault currents and line to neutral voltage under fault conditions.

Answer

Data: 30 MVA, 11 kV, $Z_1 = Z_2 = j0.2$ , $Z_0 = j0.05$ pu. Unloaded, $E_a = 1\angle0^\circ$ pu. Fault between phases b and c (no ground).

$I_{base} = \dfrac{30\times10^6}{\sqrt3\times11\times10^3} = 1574.6$ A; $V_{base,ph} = 11/\sqrt3 = 6.351$ kV.

Fault conditions

$I_a = 0$ , $I_b = -I_c$ , $V_b = V_c$ → $I_{a0} = 0$ , $I_{a2} = -I_{a1}$ , $V_{a1} = V_{a2}$ . Positive and negative networks in parallel; zero network not involved.

Sequence currents

I_{a1} = -I_{a2} = \frac{E_a}{Z_1+Z_2} = \frac{1}{j0.4} = -j2.5\ \text{pu}, \quad I_{a0} = 0

Line currents and fault current

\begin{aligned} I_a &= 0 \\ I_b &= (a^2 - a)I_{a1} = (-j\sqrt3)(-j2.5) = -4.330\ \text{pu} \\ I_c &= -I_b = 4.330\ \text{pu} \\ |I_f| &= 4.330 \times 1574.6 = 6818\ \text{A} \end{aligned}

Line-to-neutral voltages

\begin{aligned} V_{a1} = V_{a2} &= E_a - Z_1I_{a1} = 1 - (j0.2)(-j2.5) = 0.5\ \text{pu}, \quad V_{a0} = 0 \\ V_a &= V_{a1}+V_{a2} = 1.0\ \text{pu} = 6.351\ \text{kV} \\ V_b &= (a^2 + a)V_{a1} = -V_{a1} = -0.5\ \text{pu} = 3.175\ \text{kV}\ (\angle180^\circ) \\ V_c &= V_b = -0.5\ \text{pu} = 3.175\ \text{kV}\ (\angle180^\circ) \end{aligned}

Results

Quantity	pu	Actual
$I_a$	0	0
$I_b = -I_c$ (fault current)	4.330	6.82 kA
$V_a$	1.0∠0°	6.35 kV
$V_b = V_c$	0.5∠180°	3.18 kV

The line voltage between the faulted phases is zero, and $V_{ab} = V_{ac} = 1.5$ pu $= 9.53$ kV.

2072 Kartik · 10 marks

For the given power system network, draw sequence networks for DLG fault at bus no 2 and determine the fault current, short circuit MVA. G1: 100 MVA, 15.75 kV, X1 = X2 = 0.15 pu, X0 = 0.05 pu; G2: 100 MVA, 15.75 kV, X1 = X2 = 0.2 pu, X0 = 0.1 pu; T1 = T2: 100 MVA, 15.75/138 kV, X1 = X2 = X0 = 0.1 pu; Line: X1 = X2 = 25 Ω, X0 = 70 Ω. [Figure: G1 (Y grounded) - bus 1 - T1 (Δ/Y grounded) - bus 2 - line - bus 3 - T2 (Y grounded/Δ) - bus 4 - G2 (Y grounded)]

Answer

Base: 100 MVA; 15.75 kV (generators), 138 kV (line). Pre-fault voltage 1 pu, no load, bolted fault ( $Z_f = 0$ ).

Z_{base,line} = \frac{138^2}{100} = 190.44\ \Omega, \quad X_{1,line} = \frac{25}{190.44} = 0.1313, \quad X_{0,line} = \frac{70}{190.44} = 0.3676\ \text{pu}

Sequence networks

Positive (emfs $E_1$ , $E_2$ ):

 E1-j0.15-(1)-j0.1-(2)-j0.1313-(3)-j0.1-(4)-j0.2-E2
                   |
                   F

Negative: same reactances ( $X_2 = X_1$ ), no emfs.

Zero:

T1 is Δ (bus 1 side)/Yg (bus 2 side): bus 2 to reference through $j0.1$ ; G1 isolated.
T2 is Yg (bus 3 side)/Δ (bus 4 side): bus 3 to reference through $j0.1$ ; G2 isolated.

 ref-j0.1-(2)-j0.3676-(3)-j0.1-ref
           |
           F      (G1 j0.05, G2 j0.1 isolated by deltas)

Thevenin impedances at bus 2

\begin{aligned} Z_1 = Z_2 &= j0.25\ \|\ j(0.1313 + 0.1 + 0.2) = j0.25\ \|\ j0.4313 = j0.1583\ \text{pu} \\ Z_0 &= j0.1\ \|\ j(0.3676 + 0.1) = j0.1\ \|\ j0.4676 = j0.0824\ \text{pu} \end{aligned}

DLG fault (b, c to ground): networks in parallel

\begin{aligned} Z_2\|Z_0 &= \frac{0.1583 \times 0.0824}{0.2406} = j0.0542 \\ I_{a1} &= \frac{1}{j(0.1583 + 0.0542)} = -j4.707\ \text{pu} \\ I_{a2} &= -I_{a1}\frac{Z_0}{Z_2+Z_0} = j1.612\ \text{pu} \\ I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0} = j3.096\ \text{pu} \end{aligned}

Phase currents:

I_b = 7.177\angle{139.68^\circ}\ \text{pu}, \quad I_c = 7.177\angle{40.32^\circ}\ \text{pu}

Fault (ground) current:

I_f = 3I_{a0} = j9.287\ \text{pu}

I_{base} = \frac{100\times10^6}{\sqrt3\times138\times10^3} = 418.4\ \text{A} \Rightarrow I_f = 9.287 \times 418.4 = 3886\ \text{A}

(Each faulted line carries $7.177 \times 418.4 = 3003$ A.)

Short-circuit MVA

\text{SC MVA} = |I_f|_{pu} \times V_f \times \text{MVA}_{base} = 9.287 \times 1 \times 100 = 928.7\ \text{MVA}

Answer: $I_f \approx 9.29$ pu $= 3.89$ kA; short-circuit level ≈ 929 MVA.

2072 Chaitra · 16 marks

Calculate the fault currents in each phase for the system shown in figure below if (i) L-G (ii) L-L (iii) L-L-G fault occurs at (q) bus. Neglect the fault impedances. Data for equipments are (in p.u.): G1: Xd'' = j0.16, X2 = j0.17, X0 = j0.06 (printed as X1); G2: Xd'' = j0.2, X2 = j0.22, X0 = j0.15 (printed as X1); T1: X1 = X2 = X0 = j0.1; T2: X1 = X2 = X0 = j0.1; Line: X1 = X2 = j0.11, X0 = j0.33. [Figure: G1 (Y grounded, E = 1∠0°) - bus p - T1 (Δ/Y grounded) - line - T2 (Y grounded/Δ) - bus q - G2 (Y grounded)]

Answer

Assumptions: $E_1 = E_2 = 1\angle0^\circ$ pu, no pre-fault load, all values on a common base, $Z_f = 0$ . The values printed as "X1" are the zero-sequence reactances ( $X_0$ ).

Sequence networks seen from bus q

Positive (negative similar, no emf):
 E1-j0.16-(p)-T1 j0.1-line j0.11-T2 j0.1-(q)-j0.2-E2
                                          |
                                          F
Zero:
 T1 delta on p side, T2 delta on q side
 -> line + transformers form a path that does not
    reach q.   q --j0.15-- ref  (G2 solidly grounded)

\begin{aligned} Z_1 &= j(0.16+0.1+0.11+0.1)\ \|\ j0.2 = j0.47 \| j0.2 = j0.1403\ \text{pu} \\ Z_2 &= j(0.17+0.1+0.11+0.1)\ \|\ j0.22 = j0.48 \| j0.22 = j0.1509\ \text{pu} \\ Z_0 &= j0.15\ \text{pu (only G2; the Δ of T2 blocks the rest)} \end{aligned}

(i) L-G fault (phase a)

\begin{aligned} I_{a1} = I_{a2} = I_{a0} &= \frac{1}{j(0.1403+0.1509+0.15)} = \frac{1}{j0.4412} = -j2.267\ \text{pu} \\ I_a &= 3I_{a1} = -j6.800\ \text{pu}, \quad I_b = I_c = 0 \end{aligned}

(ii) L-L fault (phases b, c)

\begin{aligned} I_{a1} = -I_{a2} &= \frac{1}{j(0.1403+0.1509)} = -j3.435\ \text{pu}, \quad I_{a0} = 0 \\ I_b &= -j\sqrt3\,I_{a1} = -5.949\ \text{pu} = 5.949\angle180^\circ \\ I_c &= 5.949\angle0^\circ\ \text{pu}, \quad I_a = 0 \end{aligned}

(iii) L-L-G fault (phases b, c to ground)

\begin{aligned} Z_2\|Z_0 &= \frac{0.1509\times0.15}{0.3009} = j0.0752 \\ I_{a1} &= \frac{1}{j(0.1403+0.0752)} = -j4.640\ \text{pu} \\ I_{a2} &= -I_{a1}\frac{Z_0}{Z_2+Z_0} = j2.313\ \text{pu} \\ I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0} = j2.327\ \text{pu} \\ I_b &= I_{a0}+a^2I_{a1}+aI_{a2} = 6.960\angle{149.91^\circ}\ \text{pu} \\ I_c &= I_{a0}+aI_{a1}+a^2I_{a2} = 6.960\angle{30.09^\circ}\ \text{pu} \\ I_n &= 3I_{a0} = j6.980\ \text{pu}, \quad I_a = 0 \end{aligned}

Summary

Fault	$I_a$ (pu)	$I_b$ (pu)	$I_c$ (pu)	Ground current (pu)
L-G	6.800∠−90°	0	0	6.800
L-L	0	5.949∠180°	5.949∠0°	0
L-L-G	0	6.960∠149.9°	6.960∠30.1°	6.980

The L-G current is high because only G2's small $X_0$ (0.15) is in the zero-sequence path.

2071 Shrawan · 3+7 marks

A single line diagram of a power system network is shown in figure below. The system data is given in the table below:

Element X1 (pu) X2 (pu) X0 (pu)
G 0.1 0.12 0.05
M1 0.05 0.06 0.025
M2 0.05 0.06 0.025
T1 0.07 0.07 0.07
T2 0.08 0.08 0.08
Line 0.1 0.1 0.3
i) Draw sequence networks ii) Find fault current for a line-to-line fault on phase b and c at point q. Assume 1.0 pu pre-fault voltage throughout. [Figure: G - bus p - T1 (Δ/Y grounded) - bus q - line - bus r - T2 (Y grounded/Δ) - bus s, which feeds motors M1 (Y, ungrounded) and M2 (Y grounded)]

Element	X1 (pu)	X2 (pu)	X0 (pu)
G	0.1	0.12	0.05
M1	0.05	0.06	0.025
M2	0.05	0.06	0.025
T1	0.07	0.07	0.07
T2	0.08	0.08	0.08
Line	0.1	0.1	0.3

Answer

Assumptions: all reactances on a common base; pre-fault voltage 1 pu everywhere (no load current); bolted fault between phases b and c at q.

(i) Sequence networks

Positive (emfs Eg, Em):
 Eg-j0.1-(p)-j0.07-(q)-j0.1-(r)-j0.08-(s)-+-j0.05-Em1
                    |                     +-j0.05-Em2
                    F
Negative: same layout, G j0.12, T1 j0.07,
 line j0.1, T2 j0.08, M1 j0.06, M2 j0.06, no emfs
Zero:
 ref-j0.07-(q)-j0.3-(r)-j0.08-ref
 G (j0.05) cut off by T1 delta; M1 ungrounded,
 M2 (j0.025) cut off by T2 delta

T1 is Δ/Yg with the grounded star at q, and T2 is Yg/Δ with the grounded star at r, so the zero network is the line between two transformer reactances to ground.

(ii) L-L fault at q

Motors in parallel: positive $0.05\|0.05 = 0.025$ , negative $0.06\|0.06 = 0.03$ .

\begin{aligned} Z_1 &= j(0.1+0.07)\ \|\ j(0.1+0.08+0.025) = j0.17 \| j0.205 = j0.0929\ \text{pu} \\ Z_2 &= j(0.12+0.07)\ \|\ j(0.1+0.08+0.03) = j0.19 \| j0.21 = j0.0998\ \text{pu} \end{aligned}

Sequence currents ( $I_{a0} = 0$ ):

I_{a1} = -I_{a2} = \frac{1}{j(0.0929+0.0998)} = \frac{1}{j0.1927} = -j5.190\ \text{pu}

Fault current:

\begin{aligned} I_b &= -j\sqrt3 I_{a1} = -8.989\ \text{pu} \\ I_c &= -I_b = 8.989\ \text{pu} \end{aligned}

Answer: fault current for L-L fault at q = 8.99 pu (in phases b and c, opposite in sign; $I_a = 0$ ). The zero-sequence network does not take part in an L-L fault.

2071 Chaitra · 10 marks

In a power system network shown in figure below, single line to ground (SLG) fault occurs at bus 3. i. Draw the positive, negative and zero sequence networks; ii. Determine phase currents in per units and amperes; iii. Phase voltages in per units and kilovolts. Specifications of the equipments are as under: G1: 100 MVA, 13.8 kV, X1 = X2 = 15% and X0 = 5%; G2: 100 MVA, 13.8 kV, X1 = X2 = 15% and X0 = 5%; G3: 100 MVA, 13.8 kV, X1 = X2 = 15% and X0 = 5%; T1: 100 MVA, 13.8/115 kV, X1 = X2 = X0 = 20%; T2: 100 MVA, 115/13.8 kV, X1 = X2 = X0 = 18%; Line: 100 MVA, 115 kV, X1 = X2 = 30% and X0 = 90%. [Figure: G1 (Y, ungrounded) and G2 (Y grounded) at bus 1 - T1 (Y grounded through j0.01 pu on bus 1 side / Δ) - bus 2 - line - bus 3 - T2 (Δ / Y grounded through 0.02 pu on bus 4 side) - bus 4 - G3 (Y grounded through j0.03 pu)]

Answer

Base: 100 MVA; 13.8 kV (generator buses 1 and 4), 115 kV (buses 2 and 3). Pre-fault voltage 1 pu, no load. From the figure: T1 is Yg (through j0.01) on the bus 1 side and Δ on the bus 2 side; T2 is Δ on the bus 3 side and Yg (through 0.02) on the bus 4 side; G1 is ungrounded, G2 solidly grounded, G3 grounded through j0.03.

(i) Sequence networks

Positive (emfs at G1, G2, G3):

 G1 j0.15 \
           (1)-j0.2-(2)-j0.3-(3)-j0.18-(4)-j0.15-G3
 G2 j0.15 /                   |
                              F

Negative: same reactances, no emfs.

Zero:

 G2: ref-j0.05-(1)        G1: open (ungrounded)
 T1: (1)-[3(j0.01)+j0.2]-ref  ;  Δ side (2): open
 line: (2)-j0.9-(3)  -> floating, no ground path
 T2: Δ side (3): open ; (4)-[0.06+j0.18]-ref
 G3: (4)-[j0.05+3(j0.03)]-ref
 -> bus 3 has NO connection to the reference

Because bus 3 sits between the Δ windings of T1 and T2, the zero-sequence impedance seen from bus 3 is infinite: $Z_0 = \infty$ .

Thevenin values: $Z_1 = Z_2 = j(0.075+0.2+0.3)\ \|\ j(0.18+0.15) = j0.575\|j0.33 = j0.2097$ pu.

(ii) Phase currents

I_{a0} = I_{a1} = I_{a2} = \frac{1}{Z_1+Z_2+Z_0} = \frac{1}{\infty} = 0

I_a = I_b = I_c = 0\ \text{pu} = 0\ \text{A}

No fault current flows: the 115 kV section between the two deltas is an ungrounded system (only a small capacitive current would flow in practice).

(iii) Phase voltages at bus 3

With no current, $V_{a1} = 1$ , $V_{a2} = 0$ , and the condition $V_a = 0$ forces $V_{a0} = -1$ pu (the whole neutral shifts):

\begin{aligned} V_a &= V_{a0}+V_{a1}+V_{a2} = 0 \\ V_b &= V_{a0} + a^2V_{a1} = -1 + 1\angle{-120^\circ} = 1.732\angle{-150^\circ}\ \text{pu} \\ V_c &= V_{a0} + aV_{a1} = -1 + 1\angle{120^\circ} = 1.732\angle{150^\circ}\ \text{pu} \end{aligned}

Base phase voltage at bus 3 $= 115/\sqrt3 = 66.40$ kV:

$V_a = 0$ kV
$|V_b| = |V_c| = 1.732 \times 66.40 = 115$ kV

Results

Quantity	pu	Actual
$I_a, I_b, I_c$	0	0 A
$V_a$	0	0 kV
$V_b$	1.732∠−150°	115 kV
$V_c$	1.732∠150°	115 kV

The healthy phases rise to full line voltage, showing why ungrounded sections need insulation for line voltage to ground.

2068 Chaitra · 8 marks

Draw positive, negative and zero sequence networks. If an unsymmetrical fault occurs at bus 3, determine equivalent Z0, Z1 and Z2. [Figure: G1 100 MVA, 11 kV, X1 = X2 = j0.2 pu, X0 = j0.05 pu (Y grounded) - T1 100 MVA, 11/220 kV, j0.1 pu (Δ / Y grounded) - parallel lines L1 (100 MVA, 220 kV, j0.1 pu) and L2 (100 MVA, 220 kV, 0.12 pu) - T2 50 MVA, 11/220 kV, j0.075 pu (Y grounded through j0.03 pu on line side / Δ) - G2 50 MVA, 11 kV, X1 = X2 = j0.15 pu, X0 = j0.03 pu (Y grounded through j0.024 pu)]

Answer

Assumptions: base 100 MVA; bus 1 = G1 terminal, bus 2 = T1 HV side, bus 3 = T2 HV side (end of the parallel lines), bus 4 = G2 terminal. Neutral reactances (j0.03 for T2, j0.024 for G2) are taken on their own 50 MVA rating. Line zero-sequence reactance is taken equal to the given value since no separate $X_0$ is given.

Per-unit values on 100 MVA

Element	$X_1 = X_2$	$X_0$
G1	0.20	0.05
T1	0.10	0.10
L1 ∥ L2	$0.1\\|0.12 = 0.0545$	0.0545
T2 ( $\times 2$ )	0.15	0.15
T2 neutral ( $\times 2$ )	–	$3\times0.06 = 0.18$
G2 ( $\times 2$ )	0.30	0.06
G2 neutral ( $\times 2$ )	–	$3\times0.048 = 0.144$

Positive sequence network

 E1-j0.2-(1)-j0.1-(2)-j0.0545-(3)-j0.15-(4)-j0.3-E2
                                |
                                F

Z_1 = j(0.2+0.1+0.0545)\ \|\ j(0.15+0.3) = j0.3545\ \|\ j0.45 = j0.1983\ \text{pu}

Negative sequence network

Same reactances, emfs shorted:

Z_2 = j0.1983\ \text{pu}

Zero sequence network

T1 Δ (G1 side)/Yg (line side): bus 2 to reference through $j0.1$ ; G1 isolated.
T2 Yg through j0.03 (line side)/Δ (G2 side): bus 3 to reference through $j(0.15+0.18) = j0.33$ ; G2 isolated.

 ref-j0.1-(2)-j0.0545-(3)-j0.33-ref
                       |
                       F
 G1 (j0.05) and G2 (j0.06+j0.144) isolated by deltas

Z_0 = j(0.1+0.0545)\ \|\ j0.33 = j0.1545\ \|\ j0.33 = j0.1053\ \text{pu}

Answer: $Z_1 = Z_2 = j0.198$ pu, $Z_0 = j0.105$ pu at bus 3.

(If the neutral reactance j0.03 is taken as already on the 100 MVA base, $Z_0 = j0.1545\|j0.24 = j0.094$ pu.)

2083 Baishakh (new course) · 7 marks

For the power system network shown below, if Line to Line fault occurs at point 'F' then find the fault current, take fault impedance Zf = 0.1 p.u. For Generator: X1 = X2 = 20%, X0 = 5%; For Both Motor: X1 = X2 = 25%, X0 = 10%; For Both Transformer: X1 = X2 = X0 = 10%. The neutral reactance of generator G1 and motor M2 are in p.u. [Figure: G1 25 MVA, 11 kV, Y grounded through Xn = 0.1 - T1 30 MVA, 10.8/121 kV, Δ/Y grounded - line X1 = X2 = 100 Ω, X0 = 250 Ω - T2 30 MVA, 121/10.8 kV, Y grounded/Δ - motor bus with M1 (15 MVA, 10 kV, Y ungrounded) and M2 (7.5 MVA, 10 kV, Y grounded through Xn = 0.036); fault F at the motor bus]

Answer

Assumptions: base 25 MVA, 11 kV in the generator circuit; pre-fault voltage at F is 1 pu with no load current; $Z_f = j0.1$ pu. An L-L fault involves only positive and negative sequence networks, so the neutral reactances and zero-sequence data are not needed.

Base voltages

V_{b,line} = 11 \times \frac{121}{10.8} = 123.24\ \text{kV}, \quad V_{b,motor} = 123.24 \times \frac{10.8}{121} = 11\ \text{kV}

Per-unit reactances (25 MVA)

Element	Calculation	$X_1 = X_2$ (pu)
G1	given	0.20
T1, T2	$0.1 \times \frac{25}{30} \times (\frac{10.8}{11})^2$	0.0803 each
Line	$100 / \frac{123.24^2}{25}$	0.1646
M1	$0.25 \times \frac{25}{15} \times (\frac{10}{11})^2$	0.3444
M2	$0.25 \times \frac{25}{7.5} \times (\frac{10}{11})^2$	0.6887

Positive (and negative) sequence network at F

 Eg-j0.2-T1 j0.0803-line j0.1646-T2 j0.0803-(F)
                                             |
                               +-------------+
                               |             |
                            j0.3444       j0.6887
                              Em1           Em2

\begin{aligned} X_{gen\ side} &= 0.2 + 0.0803 + 0.1646 + 0.0803 = 0.5253 \\ X_{motors} &= 0.3444 \| 0.6887 = 0.2296 \\ Z_1 = Z_2 &= j(0.5253 \| 0.2296) = j0.1597\ \text{pu} \end{aligned}

L-L fault through $Z_f$

\begin{aligned} I_{a1} = -I_{a2} &= \frac{V_f}{Z_1 + Z_2 + Z_f} = \frac{1}{j(0.1597 + 0.1597 + 0.1)} = \frac{1}{j0.4195} = -j2.384\ \text{pu} \\ |I_f| = |I_b| = |I_c| &= \sqrt3 \times 2.384 = 4.129\ \text{pu} \end{aligned}

Base current at the motor bus (11 kV):

I_{base} = \frac{25\times10^6}{\sqrt3 \times 11\times10^3} = 1312.2\ \text{A}

I_f = 4.129 \times 1312.2 = 5418\ \text{A}

Answer: L-L fault current at F ≈ 4.13 pu ≈ 5.42 kA.

2082 Bhadra (new course) · 7 marks

A 50 MVA, 12 kV, three-phase alternator were subjected to different types of faults. The fault currents were: (i) 1870 A for three-phase fault (ii) 2590 A for L-L fault (iii) 4130 A for L-G fault. The alternator neutral is solidly grounded. Find the three-sequence reactance of the alternator in per unit.

Answer

Assumptions: the faults are at the terminals of the unloaded alternator, so the emf $E = 1$ pu (rated voltage); resistance neglected; neutral solidly grounded ( $Z_n = 0$ ).

Base current

I_{base} = \frac{50\times10^6}{\sqrt3 \times 12\times10^3} = 2405.6\ \text{A}

Fault	Current (A)	Current (pu)
3-phase	1870	0.7773
L-L	2590	1.0766
L-G	4130	1.7168

Fault current formulas

I_{3\phi} = \frac{E}{X_1}, \qquad I_{LL} = \frac{\sqrt3 E}{X_1 + X_2}, \qquad I_{LG} = \frac{3E}{X_1 + X_2 + X_0}

Positive sequence reactance

X_1 = \frac{1}{0.7773} = 1.2864\ \text{pu}

Negative sequence reactance

\begin{aligned} X_1 + X_2 &= \frac{\sqrt3}{1.0766} = 1.6088 \\ X_2 &= 1.6088 - 1.2864 = 0.3223\ \text{pu} \end{aligned}

Zero sequence reactance

\begin{aligned} X_1 + X_2 + X_0 &= \frac{3}{1.7168} = 1.7474 \\ X_0 &= 1.7474 - 1.6088 = 0.1387\ \text{pu} \end{aligned}

Answer: $X_1 = 1.286$ pu, $X_2 = 0.322$ pu, $X_0 = 0.139$ pu.

The large $X_1$ (it is the steady-state synchronous reactance here, since sustained currents were measured) and small $X_0$ explain why the L-G current is the highest and the 3-phase current the lowest.

Past questions and answers

1.Starting from a suitable point, show that the three sequence networks are connected in series during a single line to ground fault in a 3-phase power system (unloaded generator), and develop the mathematical expression for the fault current.

Sequence networks of an unloaded generator

Boundary conditions (fault on phase a through ZfZ_fZf​)

Sequence currents

Voltage condition

Fault current

Interconnection of sequence networks

Remarks

2.Explain (derive) with necessary mathematical expressions and diagrams how fault current and bus voltages are calculated when a line-to-line fault occurs in a 3-phase power system network. Draw a diagram showing interconnection of sequence networks for this type of fault.

Boundary conditions (fault between phases b and c through ZfZ_fZf​)

Sequence currents

Voltage condition

Fault current

Voltages at the fault bus

Voltages at other buses (i) of a network

Interconnection of sequence networks

Steps in summary

3.For a double line to ground fault (bolted, zero fault impedance) in an unloaded generator (at alternator terminal), obtain the sequence network and show that the positive sequence network is connected in series with the parallel connection of negative sequence and zero sequence networks.

Generator sequence equations (unloaded)

Boundary conditions (bolted fault, b and c to ground)

Sequence voltages

Sequence currents

Solving

Interconnection

4.From suitable diagram and mathematics show that during a line to line (double line) fault in a transmission line, zero sequence component is absent in the expression of fault current.

Boundary conditions (fault between phases b and c through ZfZ_fZf​)

Sequence currents

Voltage condition

Fault current

Interconnection of sequence networks

Consequences

i) Fault current for solid L-G fault on phase a

ii) Sequence networks and interconnection

iii) Solidly grounded: L-G more severe than 3-phase fault

iv) Limiting neutral reactance

Sequence currents

Sequence voltages

Phase voltages

Line voltages

7.With the help of suitable mathematical aid verify that, for a line to line fault in a synchronous generator the zero sequence component of current is absent and positive-sequence of current is equal to the negative sequence component of the current.

Generator and fault

Boundary conditions (fault between phases b and c through ZfZ_fZf​)

Sequence currents

Voltage condition

Fault current

Interconnection of sequence networks

Fault-point voltages (bolted fault)

8.Show that positive and negative sequence currents are equal in magnitude but out of phase by 180° in a line to line fault on a power system network. Draw a diagram showing inter-connection of sequence networks for this type of fault.

Boundary conditions (fault between phases b and c through ZfZ_fZf​)

Sequence currents

Voltage condition

Interconnection of sequence networks

9.Draw positive sequence, negative sequence and zero sequence network of synchronous generator. Show that all three sequence currents of faulty phase are equal in case of single line to ground fault of unloaded synchronous generator. Also find the expression of fault current.

Sequence networks of an unloaded generator

Boundary conditions (fault on phase a through ZfZ_fZf​)

Sequence currents are equal

Voltage condition

Fault current

Interconnection of sequence networks

10.For an unloaded 3-phase, ABC phase sequence, synchronous generator, starting from the boundary condition draw and justify the interconnection of sequence networks if a L-G fault occurs in phase B.

Generator sequence equations (phase a reference)

Boundary conditions (bolted fault on B)

Sequence currents

Voltage condition

Interconnection of sequence networks

Difference between positive, negative and zero sequence components

L-G fault through ZfZ_fZf​ on phase a (solidly grounded, no load)

Boundary conditions (fault on phase a through ZfZ_fZf​)

Sequence currents

Voltage condition

Fault current

Interconnection of sequence networks

Line-to-ground voltages at the fault

12.Show that for the two-conductor open circuit fault in a 3-phase system, the sequence networks are to be connected in series to simulate the fault.

Fault conditions

Sequence currents

Sequence voltages

Sequence network equations

Interpretation

Starting from a suitable point, show that the three sequence networks are connected in series during a single line to ground fault in a 3-phase power system (unloaded generator), and develop the mathematical expression for the fault current.

Boundary conditions (fault on phase a through $Z_f$ )

Explain (derive) with necessary mathematical expressions and diagrams how fault current and bus voltages are calculated when a line-to-line fault occurs in a 3-phase power system network. Draw a diagram showing interconnection of sequence networks for this type of fault.

Boundary conditions (fault between phases b and c through $Z_f$ )

For a double line to ground fault (bolted, zero fault impedance) in an unloaded generator (at alternator terminal), obtain the sequence network and show that the positive sequence network is connected in series with the parallel connection of negative sequence and zero sequence networks.

From suitable diagram and mathematics show that during a line to line (double line) fault in a transmission line, zero sequence component is absent in the expression of fault current.

Boundary conditions (fault between phases b and c through $Z_f$ )

With the help of suitable mathematical aid verify that, for a line to line fault in a synchronous generator the zero sequence component of current is absent and positive-sequence of current is equal to the negative sequence component of the current.

Boundary conditions (fault between phases b and c through $Z_f$ )

Show that positive and negative sequence currents are equal in magnitude but out of phase by 180° in a line to line fault on a power system network. Draw a diagram showing inter-connection of sequence networks for this type of fault.

Boundary conditions (fault between phases b and c through $Z_f$ )

Draw positive sequence, negative sequence and zero sequence network of synchronous generator. Show that all three sequence currents of faulty phase are equal in case of single line to ground fault of unloaded synchronous generator. Also find the expression of fault current.

Boundary conditions (fault on phase a through $Z_f$ )

For an unloaded 3-phase, ABC phase sequence, synchronous generator, starting from the boundary condition draw and justify the interconnection of sequence networks if a L-G fault occurs in phase B.

L-G fault through $Z_f$ on phase a (solidly grounded, no load)

Boundary conditions (fault on phase a through $Z_f$ )

Show that for the two-conductor open circuit fault in a 3-phase system, the sequence networks are to be connected in series to simulate the fault.

How do the different vector groups of transformer affect the fault current in the power system network?

Fault current (L-G through $Z_f = 0.05$ pu)

Fault current (L-G at bus 3, $Z_f = 0$ )

A 40 MVA, 11 kV generator has Z1 = Z2 = j0.3 pu, Z0 = j0.4 pu. A line to line fault occurs on the generator terminals. Find the fault current.

Sequence currents for LLG fault (phases b, c to ground through $Z_f$ )

A single line to ground fault occur at generator terminal of 20 MVA, 13.8 KV and having Z1 = j0.20 pu, Z2 = j0.3 pu. Find the fault current and line to line voltage under fault condition.

Fault current (sequence networks in series with $3R_f$ )

(i) Neutral reactance $X_n$

(ii) Neutral resistance $R_n$

A double line to ground fault occur at generator terminal of 30 MVA, 11 kV and having Z1 = Z2 = j0.2 pu and Z0 = j0.05 pu. Find the line currents, fault current and line to neutral voltages under fault condition.