Question

An 11kV radial system is shown in Figure 10.42. Assuming a CO-7 relay with relay characteristic given in Figure 10.41 and the same power factor for all loads, select relay settings to protect the system.

Short answer

Answer

The time setting, and current dial setting are shown below.

Relay	TS	TDS
R1	6	0.5
R2	10	2
R3	10	2.5

Step-by-step solution

Write the given data from the question.

The transmission take place on voltage,$V_L=11kV$

The system data are given as,

Bus	Maximum Load MVA Lagging P. F	Fault current $\left(I_{SC}\right)$
1	4.0	2500
2	2.5	3000
3	6.75	3200

Breaker	CT ratio	Relay
R1	200:5	CO-7
R2	200:5	CO-7
R3	600:5	CO-7

The equation to calculate the current time setting and current dial setting for the relays.

The equation to calculate the CT primary current corresponding to maximum power at bus is given as follows.

${\mathit{I}}_{\mathbf{L}}\mathbf{=}\frac{\mathbf{S}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{L}}}$ …… (1)

Here,${\mathit{I}}_{\mathbf{L}}$is the primary CT current, S is the maximum power and ${\mathit{V}}_{\mathbf{L}}$is the line-to-line voltage.

The equation to calculate the CT secondary current is given as follows.

${\mathit{I}}_{\mathbf{L}}^{\mathbf{\text{'}}}\mathbf{=}\frac{{\mathbf{I}}_{\mathbf{L}}}{\mathbf{n}}$ …… (2)

The equation to calculate the fault to pick-up current ratio is given as follows.

$\frac{{\mathbf{I}}_{\mathbf{S}\mathbf{C}\mathbf{R}}^{\mathbf{\text{'}}}}{\mathbf{T}\mathbf{S}\left(R\right)}\mathbf{=}\frac{{\displaystyle \frac{{\mathbf{I}}_{\mathbf{t}}}{\mathbf{n}}}}{\mathbf{T}\mathbf{S}\left(R\right)}$ …… (3)

Here, R represent the number of the relay.

Calculate the current time setting and current dial setting for the relays.

For relay 1,

The CT ratio for relay R1, n = 200 : 5

Calculate the CT primary current.

Substitute 4MVA for Sand 11 kV for $V_L$ into equation (1).

$$\begin{gathered} I_L=\frac{4\times {10}^6}{\sqrt{3}\times 11\times {10}^3} \\ {I}_L=209.95A \end{gathered}$$

Calculate the CT secondary current.

Substitute 200 : 5 for n, 209.95 A for $I_L$into equation (2).

$$\begin{gathered} I_L^{\text{'}}=\frac{209.95}{{\displaystyle \frac{200}{}}} \\ {I}_L^{\text{'}}=\frac{209.95}{{\displaystyle 40}} \\ {I}_L^{\text{'}}=5.25A \end{gathered}$$

The CT secondary maximum relay current at the maximum load is 5.25 A. Take TS3 as 6 A because it is the lowest above 5.25 A.

For relay 2,

The CT ratio for relay R2, n = 200:5

Since the power factor of loads at both the buses are identical. So, the power added linearly.

$S_{12}=S_1+S_2$

Substitute 2MVA for $S_1$and 2.5 MVA for $S_2$into above equation.

$$\begin{gathered} S_{12}=4+2.5 \\ {S}_{12}=6.25MVA \end{gathered}$$

Calculate the CT primary current.

Substitute 6.25 MVA for S and 11kV for $V_L$ into equation (1).

$$\begin{gathered} I_L=\frac{6.25\times {10}^6}{\sqrt{3}\times 11\times {10}^3} \\ {I}_L=341.16A \end{gathered}$$

Calculate the CT secondary current.

Substitute 200 : 5 for n, 341.16 A for role="math" localid="1655903259721" $I_L$into equation (2).

$$\begin{gathered} I_L^{\text{'}}=\frac{341.16}{{\displaystyle \frac{200}{5}}} \\ {I}_L^{\text{'}}=\frac{341.16}{{\displaystyle 40}} \\ {I}_L^{\text{'}}=8.53A \end{gathered}$$

The CT secondary maximum relay current at the maximum load is 8.53 A. Take TS3 as 10 A because it is the lowest above 8.53 A.

For relay 3,

The CT ratio for relay R2, n = 400 : 5

Since the power factor of loads at both the buses are identical. So, the power added linearly.

$S_{123}=S_1+S_2+S_3$

Substitute 4 MVA for $S_1$, 2.5 MVA for $S_2$and 6.75 MVA for into above equation.

$$\begin{gathered} S_{23}=4+2.5+6.75 \\ {S}_{23}=13.25MVA \end{gathered}$$

Calculate the CT primary current.

Substitute 13.25 MVA for S and 11kVfor into equation (1).

Calculate the CT secondary current.

Substitute 400 : 5 for n, 69.44 Afor into equation (2).

The CT secondary maximum relay current at the maximum load is 8.69 A. Take TS3 as 10 A because it is the lowest above 8.69 A.

The largest fault current through R1 is 2500 A.

Calculate the fault to pick up current.

Substitute 200 : 5 for n, 2500 A for , 6 A for TS1 into equation (3).

Select the lowest TDS to clear the fault as soon as possible. So, select the curve corresponding to TDS 0.5 from the characteristic curve. Then the operating time for the relay R1 is.

Addbreaker to operating time andfor margin to obtain the R2 operating time.

For the same fault, the fault to pick up ratio at R2,

Substitute 200 : 5 for n , 2500 Afor, 10 A for TS2into equation (3).

The TDS value corresponding to operating time and fault to pick up ratio is 2

The largest fault current through the relay R2 is 3000 A.

Calculate the fault to pick up current.

Substitute 200 : 5 for n, 3000 Afor, 10 A for TS2into equation (3).

$$\begin{gathered} \frac{I_{SC3}}{TS3}=\frac{\left({\displaystyle \frac{{\displaystyle \frac{3000}{400}}}{5}}\right)}{10} \\ \frac{I_{SC3}}{TS3}=\frac{\left({\displaystyle \frac{3000}{40}}\right)}{10} \\ \frac{I_{SC3}}{TS3}=\frac{3000}{400} \\ \frac{I_{SC3}}{TS3}=7.5 \end{gathered}$$

Corresponding to the TDS curve 2, the operating time for the relay 2 is$T_{22}=0.5sec$.

Add breaker to operating time and 0.3 sec for margin to obtain the R3 operating time.

$$\begin{gathered} T_{32}=T_{22}+T_{breaker}+T_{coordinate} \\ {T}_{32}=0.5+0.1 \\ {T}_{32}=0.9sec \end{gathered}$$

For the same fault, the fault to pick up ratio at R3,

Substitute 400 : 5 for n, 300 A for $I_F$, 10 A for TS3 into equation (3).

$$\begin{gathered} \frac{I_{SC3}}{TS3}=\frac{\left({\displaystyle \frac{{\displaystyle \frac{3000}{400}}}{5}}\right)}{10} \\ \frac{I_{SC3}}{TS3}=\frac{37.5}{10} \\ \frac{I_{SC3}}{TS3}=3.75 \end{gathered}$$

The TDS value corresponding to operating time and fault to pick up ratio is 2.5 .

Therefore, the time setting, and current dial setting are shown below.

Relay	TS	TDS
R1	6	0.5
R2	10	2
R3	10	2.5