Question

A CO-8 relay with a current tap setting of $\mathbf{5}$ amperes is used with the $\mathbf{100}\mathbf{:}\mathbf{5}$ CT in Example 10.1. The CT secondary current${\mathbf{I}}^{\mathbf{\text{'}}}$ is the input to the relay operating coil. The CO-8 relay burden is shown in the following table for various relay input currents.

CO-8 relay Input current ${\mathbf{I}}^{\mathbf{\text{'}}}$,A	5	8	10	13	15
CO-8 relay burden ${\mathbf{Z}}_{\mathbf{B}}\mathbf{,}\mathbf{\Omega }$	0.5	0.8	1.0	1.3	1.5

Primary current and CT error are computed in Example 10.1 for the$\mathbf{5}\mathbf{-}\mathbf{,}\mathbf{8}\mathbf{-}\mathbf{,}$ and 1 relay input currents. Compute the primary current and CT error for (a)${\mathbf{I}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{10}\mathbf{A}$ and ${\mathbf{Z}}_{\mathbf{B}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\Omega }$and for (b) ${\mathbf{I}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{13}\mathbf{A}$and ${\mathbf{Z}}_{\mathbf{B}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\Omega }$. (c) Plot${\mathbf{I}}^{\mathbf{\text{'}}}$ versus$\mathit{I}$ for the above five values of ${\mathit{I}}^{\mathbf{\text{'}}}$. (d) For reliable relay operation, the fault-to-pickup current ratio with minimum fault current should be greater than two. Determine the minimum fault current for application of this CT and relay with$\mathbf{5}\mathbf{-}\mathbf{A}$ tap setting.

CT equivalent circuit.

Short answer

(a) The minimum primary fault current is$212\text{\hspace{0.17em}A}$ and CT error is$37.5\%$ .

(b) The minimum primary fault current is$296\text{\hspace{0.17em}A}$ and CT error is $12.16\%$.

(c) The graph between the input current and primary minimum fault current as,

(d) The relay will operate for fault current exceeding$212\text{\hspace{0.17em}A}$ .

Step-by-step solution

Write the given data from the question.

The CT ratio,$n=100:5$

Current tap setting of relays is$5\text{\hspace{0.17em}A}$ .

The secondary resistance corresponding to CT ratio from the figure 10.8,$Z^{\text{'}}=0.082\Omega$

The CO-8 relay burden is shown in the following table for various relay input currents.

CO-8 relay Input current $I^{\text{'}}$,A	5	8	10	13	15
CO-8 relay burden ${\mathrm{Z}}_{\mathrm{B}},\mathrm{\Omega }$	0.5	0.8	1.0	1.3	1.5

Determine the equations to calculate the minimum primary fault current.

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

${\mathbf{E}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{I}}^{\mathbf{\text{'}}}\mathbf{(}{\mathbf{Z}}^{\mathbf{\text{'}}}\mathbf{+}{\mathbf{Z}}_{\mathbf{B}}\mathbf{)}$ …… (1)

Here ${\mathbf{I}}^{\mathbf{\text{'}}}$ is the relay current, and${\mathbf{Z}}_{\mathbf{B}}$ is the relay impedance.

The equation to calculate the minimum primary fault current is given as follows.

role="math" localid="1656481532437" $\mathbf{I}\mathbf{=}\mathbf{n}\mathbf{(}{\mathbf{I}}^{\mathbf{\text{'}}}\mathbf{+}{\mathbf{I}}_{\mathbf{c}}\mathbf{)}$ …… (2)

Here, ${\mathbf{I}}_{\mathbf{c}}$ is the exciting current.

The equation to calculate the CT error is given as follows.

$\mathbf{CT}\mathbf{error}\mathbf{=}\frac{{\mathbf{I}}_{\mathbf{c}}}{{\mathbf{I}}^{\mathbf{\text{'}}}\mathbf{+}{\mathbf{I}}_{\mathbf{c}}}\mathbf{\times }\mathbf{100}$ …… (3)

Calculate the minimum primary fault current for I'=10 A, ZB=1 Ω .

(a)

The relay current, is $I^{\text{'}}=10\text{\hspace{0.17em}A}$.

Relay impedance,$Z_B=1\Omega$.

Calculate the secondary voltage of the CT.

Substitute $10\text{\hspace{0.17em}A}$for $I^{\text{'}}$, $0.082\Omega$for $Z^{\text{'}}$and $1\Omega$for $Z_B$into equation (1).

$\begin{array}{l}{E}^{\text{'}}=10(0.082+1)\\ E^{\text{'}}=10\times 1.082\\ E^{\text{'}}=10.82\text{\hspace{0.17em}V}\end{array}$

The exciting current corresponding to the secondary voltage of the CT $10.82\text{\hspace{0.17em}V}$ is $0.6\text{\hspace{0.17em}A}$.

$I_c=0.6\text{\hspace{0.17em}A}$

Calculate the minimum primary fault current.

Substitute$100:5$ for$n$ ,$10\text{\hspace{0.17em}A}$ for $I^{\text{'}}$and $0.6\text{\hspace{0.17em}A}$for$I_c$into equation (2).

$\begin{array}{l}I=\frac{100}{5}(10+0.6)\\ I=20\times 10.6\\ I=212\text{\hspace{0.17em}A}\end{array}$

Calculate the CT error.

Substitute$0.6\text{\hspace{0.17em}A}$ for$I_c$ and$10\text{\hspace{0.17em}A}$ for$I^{\text{'}}$ into equation (3).

$\begin{array}{c}\text{CT\hspace{0.17em}error}=\frac{0.6}{10+0.6}\times 100\\ =\frac{0.6}{1.6}\times 100\\ =37.5\%\end{array}$

Hence, the minimum primary fault current is$212\text{\hspace{0.17em}A}$ and CT error is $37.5\%$.

Calculate the minimum primary fault current for  I'=13 A, ZB=1.3 Ω.

(b)

The relay current is$I^{\text{'}}=13\text{\hspace{0.17em}A}$.

Relay impedance is $Z_B=1.3\Omega$.

Calculate the secondary voltage of the CT.

Substitute $13\text{\hspace{0.17em}A}$for $I^{\text{'}}$, $0.082\Omega$ for $Z^{\text{'}}$and $1.3\Omega$for $Z_B$into equation (1).

$\begin{array}{l}{E}^{\text{'}}=13(0.082+1.3)\\ E^{\text{'}}=13\times 1.382\\ E^{\text{'}}=18\text{\hspace{0.17em}V}\end{array}$

The exciting current corresponding to the secondary voltage of the CT$18\text{\hspace{0.17em}V}$ is$1.8\text{\hspace{0.17em}A}$.

$I_c=1.8\text{\hspace{0.17em}A}$

Calculate the minimum primary fault current.

Substitute$100:5$for$n$,$13\text{\hspace{0.17em}A}$for$I^{\text{'}}$and$1.8\text{\hspace{0.17em}A}$for$I_c$into equation (2).

$\begin{array}{l}I=\frac{100}{5}(13+1.8)\\ I=20\times 14.8\\ I=296\text{\hspace{0.17em}A}\end{array}$

Calculate the CT error.

Substitute$1.8\text{\hspace{0.17em}A}$ for$I_c$ and$13\text{\hspace{0.17em}A}$ for$I^{\text{'}}$ into equation (3).

$\begin{array}{l}\text{CT\hspace{0.17em}error\hspace{0.17em}=\hspace{0.17em}}\frac{1.8}{13+1.8}\times 100\\ \text{CT\hspace{0.17em}error\hspace{0.17em}=}\frac{1.8}{14.8}\times 100\\ \text{CT\hspace{0.17em}error\hspace{0.17em}=\hspace{0.17em}}12.16\%\end{array}$

Hence the minimum primary fault current is$296\text{\hspace{0.17em}A}$ and CT error is $12.16\%$.

Draw the curve between I' and I .

(c)

The primary current are$5\text{\hspace{0.17em}A,\hspace{0.17em}8\hspace{0.17em}A}$ and$15\text{\hspace{0.17em}A}$ .

Write the minimum fault current for the different value of input current from the example 1 and the problem 2 (a) and 2(b).

$I^{\text{'}}$	5	8	10	13	15
role="math" localid="1656482997659" $I$	105	168	212	296	700

Draw the graph between the input current and primary minimum fault current as,

Determine the minimum fault current for the application of the CT and relay 5 A tap setting.

(d)

Current tap setting is $5\mathrm{A}$.

The minimum fault picks up ratio is 2.

Calculate the minimum current that required to trip the coil for the reliable operation.

$\begin{array}{l}{I}_{\min }=2\times 5\\ I_{\min }=10\text{\hspace{0.17em}A}\end{array}$

The corresponding primary current from the graph shown in the part (c).

${I^{\text{'}}}_{\min }=212\text{\hspace{0.17em}A}$

Hence the relay will operate for fault current exceeding $212\text{\hspace{0.17em}A}$.