Question

Evaluate relay coordination for the minimum fault currents in Example 10.4. For the selected current tap settings and time dial settings, (a) determine the operating time of relays at B2 and B3 for the 700 A fault current. (b) Determine the operating time of relays at Bl and B2 for the 1500 A fault current. Are the fault-to-pickup current ratios $\mathbf{\ge }\mathbf{2}\mathbf{.}\mathbf{0}$(a requirement for reliable relay operation) in all cases? Are the coordination time intervals $\mathbf{\ge }\mathbf{0}\mathbf{.}\mathbf{3}$seconds in all cases?

Short answer

(a) The relay operating time for B3 and B2 are 0.10 sec and 1.3 sec respectively. The coordinate time between B3 and B2 relay is 1.12 sec.

(b) The relay operating time for B1 and B2 are 1.8 sec and 0.55 sec respectively. The coordinate time between B3 and B2 relay is 1.17 sec.

Step-by-step solution

Write the given data from the question.

(a) The fault current for the relay B2 and B3, I_F = 700 A.

(b) The fault current for the relay B1 and B2, I_F = 1500 A .

The circuit breaker time for 5 cycles, T_breaker= 0.083 sec.

Determine operating time of the relay.

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

$\frac{\mathbf{I}{\mathbf{\text{'}}}_{\mathbf{R}\mathbf{}\mathbf{f}\mathbf{a}\mathbf{u}\mathbf{l}\mathbf{t}}}{\mathbf{T}\mathbf{S}\mathbf{\left(}\mathbf{R}\mathbf{\right)}}\mathbf{=}\frac{{\displaystyle \frac{{\mathbf{I}}_{\mathbf{f}}}{\mathbf{n}}}}{\mathbf{T}\mathbf{S}\mathbf{\left(}\mathbf{R}\mathbf{\right)}}$ …… (1)

Here, represent the number of the relay.

Calculate the relay operating current of B2 and B3 for fault current 700 A.

From example 10.4

The current tap setting for B3, TS3 = 3 A

The CT ratio, n = 200 : 5

Calculate the fault to pick up current.

Substitute 200 : 5 for n, 700 A for I_F, 3 A for TS3 into equation (1).

$$\begin{gathered} \frac{I{\text{'}}_{3fault}}{TS3}=\frac{\left({\displaystyle \frac{{\displaystyle \frac{700}{200}}}{5}}\right)}{3} \\ \frac{I{\text{'}}_{3fault}}{TS3}=\frac{\left({\displaystyle \frac{700}{40}}\right)}{3} \\ \frac{I{\text{'}}_{3fault}}{TS3}=\frac{700}{120} \\ \frac{I{\text{'}}_{3fault}}{TS3}=5.83 \end{gathered}$$
The characteristics curve for relay CO-8.

The operating time of the relay B3 corresponding to ts = 3 and relay current input multiple of current tap setting 5.83 is 0.10 sec.

The clearance time by the primary protection can be calculated by adding the time for circuit breaker to operating time. i.e., 0.10 + 0.083 = 0183 sec.

From example 10.4

The CT ratio, n = 200 : 5

The current tap setting for B3, TS2 = 5 A

Calculate the fault to pick up current.

Substitute 200 : 5 for n, 700 A for I_F, 5 A for TS2 into equation (1).

$$\begin{gathered} \frac{I{\text{'}}_{2fault}}{TS2}=\frac{\left({\displaystyle \frac{{\displaystyle \frac{700}{200}}}{5}}\right)}{5} \\ \frac{I{\text{'}}_{2fault}}{TS2}=\frac{\left({\displaystyle \frac{700}{40}}\right)}{5} \\ \frac{I{\text{'}}_{3fault}}{TS2}=\frac{700}{200} \\ \frac{I{\text{'}}_{3fault}}{TS2}=3.5 \end{gathered}$$

The operating time of the relay B2 corresponding to TS = 5 and relay current input multiple of current tap setting 3.5 is 1.3 sec.

The coordinate time between the relay B3 and B2 is 1.3 - 0.183 = 1.12 sec.

Hence the relay operating time for B3 and B2 are 0.10 sec and 1.3 sec respectively. The coordinate time between B3 and B2 relay is 1.12 sec

Calculate the relay operating current of B1 and B2 for fault current 1500 A.

From example 10.4

The current tap setting for B2, TS2 = 5 A

The CT ratio, n = 200 : 5

Calculate the fault to pick up current.

Substitute 200 : 5 for n, 1500 A for I_F, 5 A for TS2 into equation (1).

$$\begin{gathered} \frac{I{\text{'}}_{2fault}}{TS2}=\frac{\left({\displaystyle \frac{{\displaystyle \frac{1500}{200}}}{5}}\right)}{5} \\ \frac{I{\text{'}}_{2fault}}{TS2}=\frac{\left({\displaystyle \frac{1500}{40}}\right)}{5} \\ \frac{I{\text{'}}_{3fault}}{TS2}=\frac{1500}{200} \\ \frac{I{\text{'}}_{3fault}}{TS2}=7.5 \end{gathered}$$

The operating time of the relay B2 corresponding to TS = 5 and relay current input multiple of current tap setting 7.5 is 0.55 sec.

The clearance time by the primary protection can be calculated by adding the time for circuit breaker to operating time. i.e., 0.55 + 0.083 = 0.633 sec.

From example 10.4

The current tap setting for B1, TS1 = 5 A

The CT ratio, n = 400 : 5

Calculate the fault to pick up current.

Substitute 400 : 5 for n, 1500 A for I_F, 5 A for TS1 into equation (1).

$$\begin{gathered} \frac{I{\text{'}}_{1fault}}{TS1}=\frac{\left({\displaystyle \frac{{\displaystyle \frac{1500}{400}}}{5}}\right)}{5} \\ \frac{I{\text{'}}_{1fault}}{TS1}=\frac{\left({\displaystyle \frac{1500}{80}}\right)}{5} \\ \frac{I{\text{'}}_{1fault}}{TS1}=\frac{1500}{400} \\ \frac{I{\text{'}}_{1fault}}{TS1}=3.75 \end{gathered}$$

The operating time of the relay B1 corresponding to TS = 5 and relay current input multiple of current tap setting 3.75 is 1.8 sec.

The coordinate time between the relay B3 and B2 is 1.8 - 0.633 = 1.17 sec.

Hence the relay operating time for B1 and B2 are 1.8 sec and 0.55 sec respectively. The coordinate time between B1 and B2 relay is 1.17 sec.