Question

A simple system with circuit breaker-relay locations is shown in Figure 10.49. The six transmission-line circuit breakers are controlled by zone distance and directional relays, as shown in Figure 10.50. The three transmission lines have the same positive-sequence impedance of j0.1 per unit. The reaches for zones 1, 2 and 3 are 80, 120 and 250% , respectively. Consider only three-phase faults. (a) Find the settings ${\mathit{Z}}_{\mathbf{r}}$in per unit for all distance relays. (b) Convert the settings in V if the VTs are rated 133 kV : 115 V and the CTs are rated 400 : 5 A. (c) For a fault at location X, which is 10% down line TL31 from bus 3, discuss relay operations.

Figure 10.49

Figure 10.50

Short answer

Answer

(a)

The value of zone $\left(Z_r\right)$in per unit for zone 1 is 0.08 per unit.

The value of zone $\left(Z_r\right)$in per unit for zone 2 is 0.12 per unit.

The value of zone $\left(Z_r\right)$in per unit for zone 3 is 0.25 per unit.

(b)

The value of zone $\left(Z_r\right)$in terms of the ohms for zone 1 is $2.93\Omega$.

The value of zone $\left(Z_r\right)$in terms of the ohms for zone 2 is$4.39\Omega$

.

The value of zone $\left(Z_r\right)$in terms of the ohms for zone 3 is$9.15\Omega$.

(c) The issue is located 10% of the way along line TL31 from bus 3. In below Figure depicts the working region for a three-zone distance relay with directional constraint according to the setup.

Step-by-step solution

Write the given data from the question.

Refer to figure 10.49 from the textbook.

Consider the three transmission lines have the same positive-sequence impedance of j0.1 per unit.

Consider the value of zone 1, 2 and 3 reaches at 80%, 120% and 250%.

Consider the value of VTs rated at 133 kV: 115 V.

Consider the value of CTs rated at 400:5 A.

Determine the formula of zones settings in per unit, zones settings in ohm.

Write the formula of zone settings $\left(Z_r\right)$in per unit.

…… (1)

Here, Z represent the per unit impedance of the transmission line.

Write the formula of zone settings $\left(Z_r\right)$in ohms.

role="math" localid="1656327151654" $\mathbf{0}\mathbf{\left(}{\mathbf{Z}}_{\mathbf{r}}\mathbf{\right)}{\mathit{Z}}_{\mathbf{r}}\mathbf{=}{\mathit{Z}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}$ …… (2)

Here, ${\mathit{Z}}_{\mathbf{r}}$is a zone setting in per unit and ${\mathit{Z}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}$is base impedance.

(a) Determine the zone settings (Zr) in per unit.

Determine the value of zone 1 reaches at 80%.

Substitute j0.1 for Z and 0.8 for percentage of reach into equation (1).

$$\begin{gathered} Z_r=\left(0.1\right)\left(0.8\right) \\ =0.08perunit \end{gathered}$$

Therefore, the zone $\left(Z_r\right)$in per unit for zone 1 is 0.08 per unit

Determine the value of zone 2 reaches at 120%.

Substitute j0.1for Z and 1.2 for percentage of reach into equation (1).

$$\begin{gathered} Z_r=\left(0.1\right)\left(1.20\right) \\ =0.12perunit \end{gathered}$$

Therefore, the zone $\left(Z_r\right)$in per unit for zone 2 is 0.12 per unit

Determine the value of zone 3 reaches at 250%.

Substitute j0.1for Z and 2.5for percentage of reach into equation (1).

$$\begin{gathered} Z_r=\left(0.1\right)\left(2.5\right) \\ =0.25perunit \end{gathered}$$

Therefore, the zone in per unit for zone 3 is 0.25 per unit

(b) Construct the zone settings (Zr) in ohms.

The voltage turn’s ratio of the transformer is 133 kV:115 V and the current ratio of the transformer is 400 : 5 A.

Let's pretend that the system in Figure 1 has a base MVA of $MV{A}_{base}=100MVA$and a base line-to-line voltage of$V_{LL.base}=230kV$.

Determine the expression for the base line-to-neutral voltage of the system.

$V_{LNbase}=\frac{V_{LLbase}}{\sqrt{3}}$

Here, $V_{LLbase}$is base line-to-line voltage.

Substitute 230 kVfor $V_{LLbase}$into above equation.

$$\begin{gathered} V_{LNbase}=\frac{230kV}{\sqrt{3}} \\ =132.79kV \end{gathered}$$

Therefore, the base line-to-neutral voltage $V_{LNbase}$is 132.79 kV.

Determine the equivalent transformer secondary voltage for the system.

$$\begin{gathered} V_{base}=\left(132.79kV\right)\left(\frac{115V}{133kV}\right) \\ =\left(132.79kV\right)\left(0.86\times {10}^{-3}\right) \\ =114.82V \end{gathered}$$

Therefore, the equivalent transformer secondary voltage $V_{base}$is114.82 V.

Determine the base line current flowing in the system.

$I_{L.base}=\frac{MV{A}_{base}}{\sqrt{3}{V}_{LLbase}}$

Here, $MV{A}_{base}$is base MVA and $V_{LLbase}$is base line-to-line voltage.

Substitute 230 kV for $V_{LLbase}$and 100 MVA for $MV{A}_{base}$into above equation.

$$\begin{gathered} I_{Lbase}=\frac{100\times {10}^6}{\sqrt{3}\left(230\times {10}^3\right)} \\ =\frac{100\times {10}^6}{398.37\times {10}^3} \\ =251.02A \end{gathered}$$

Therefore, the line base line current, $I_{base}$is 251.02 A.

Determine the equivalent transformer secondary current for the given system.

$$\begin{gathered} I_{base}=\left(251.02\right)\left(\frac{5}{400}\right) \\ =\left(251.02\right)\left(0.0125\right) \\ =3.138A \end{gathered}$$

Therefore, the equivalent transformer secondary current, $I_{base}$is 3.138 A.

Determine the equivalent base impedance of the system.

$Z_{base}=\frac{V_{base}}{I_{base}}$

Here, $V_{base}$is equivalent transformer secondary voltage and $I_{base}$is equivalent transformer secondary current.

Substitute 114.82 V for$V_{base}$and 3.138 A for $I_{base}$into above equation.

$$\begin{gathered} Z_{base}=\frac{114.82}{3.138} \\ =36.59\Omega \end{gathered}$$

Therefore, the base impedance, $Z_{base}$of the system is$36.59\Omega$.

Determine the value of zone $\left(Z_r\right)$in terms of the ohms for zone.

Substitute 0.08 for $Z_r$and $36.59\Omega$for $Z_{base}$into equation (2).

$$\begin{gathered} Z_r\left(\Omega \right)=\left(0.08\right)\left(36.59\right) \\ =2.93\Omega \end{gathered}$$

Therefore, the settings for the zone 1, $Z_r\left(\Omega \right)$in terms of the ohms is$2.93\Omega$.

Determine the value of zone $\left(Z_r\right)$in terms of the ohms for zone 2.

Substitute 0.12 for $Z_r$and $36.59\Omega$for $Z_{base}$into equation (2).

$$\begin{gathered} Z_r\left(\Omega \right)=\left(0.12\right)\left(36.59\right) \\ =4.39\Omega \end{gathered}$$

Therefore, the settings for the zone 2, $Z_r\left(\Omega \right)$in terms of the ohms is$4.39\Omega$.

Determine the value of zone $\left(Z_r\right)$in terms of the ohms for zone 3.

Substitute 0.25 for $Z_r$and $36.59\Omega$for $Z_{base}$into equation (2).

$$\begin{gathered} Z_r\left(\Omega \right)=\left(0.25\right)\left(36.59\right) \\ =9.15\Omega \end{gathered}$$

Therefore, the settings for the zone 3, $Z_r\left(\Omega \right)$in terms of the ohms is$9.15\Omega$.

(c) Discuss relay operations.

The issue is located 10% of the way along line TL31 from bus 3. In below Figure depicts the working region for a three-zone distance relay with directional constraint according to the setup.

Load currents are often less than fault currents during normal operation. As a result, as illustrated in Figure 3, the system's impedance will be greater beyond the circle.

Figure 1

Determine line breaker operations:

B31: When a failure occurs in zone 1, immediate action is taken.

B32: Directional unit should block operation.

B23: For a fault in Zone 2; delayed operation takes place.

B31: Should trip first, preventing B23from tripping.

B21: Fault duty is light.

B12: Directional unit should block operation.

B13: Fault in Zone 2, just outside of Zone 1, Delayed operation takes place.

The fault is cleared as requested using line breakers B13 and B31. Breakers B1 and B4 must work along with Breaker B13. From quickest to slowest, the trip sequence should be B13, B1, and B4; similarly, B23, B31, and B23 should be quicker than B2 and B5.