Question

For a $\mathbf{\Delta }-\mathbf{Y}$ connected, 15-MVA ,33:11 KV transformer with differential relay protection and CT ratios shown in Figure 10.52, determine the relay currents at full load and calculate the minimum relay current setting to allow 125% overload.

Short answer

The value of relay current I_r at full load is 0.96 A and the minimum relay setting is 1.2 A .

Step-by-step solution

Write the given data from the question.

Consider the value of transformer power rating is 15 MVA .

Consider the value of transformer primary voltage is 33 kV .

Consider the value of transformer secondary voltage is 11 kV .

Consider percentage overload is 125% .

Determine the formula of relay current Ir at full load and minimum relay setting I'r.

Write the formula ofrelay current.

$I{}_{\mathbf{r}}={\mathbf{i}}_{\mathbf{p}}-{\mathbf{i}}_{\mathbf{s}}$ …… (1)

Here, i_p is CT primary current and i_s is CT secondary current.

Write the formula of minimum relay setting I_r.

${\mathbf{I}\mathbf{\text{'}}}_{\mathbf{r}}={\mathbf{I}}_{\mathbf{r}}\mathbf{\times }\mathbf{125}\mathbf{\%}$ …… (2)

Here, I_r is relay current.

 Determine the value of relay current Ir at full load and minimum relay setting I'r.

Draw the circuit diagram of connection of CTs and the currents in the instruments:

Connect CTs in the delta on the star side and CTs in a star on the 33 kV delta side (11 KV ).

Determine the transformer primary rated line current I_1rated as:

$$\begin{gathered} I_{\text{1rated}}=\frac{15\times {10}^6}{33\times {10}^3\times \sqrt{3}} \\ =262.43 \text{A} \end{gathered}$$

Choose a 300:5 standard CT ratio on the major side ( 33 kV). The following is the CT primary current I_p :

Similarly, determine the transformer secondary rated line current I_2rated as:

$$\begin{gathered} I_{2\text{rated}}=\frac{15\times {10}^6}{11\times {10}^3\times \sqrt{3}} \\ =787.3 \text{A} \end{gathered}$$

Choose a 2000:5 standard CT ratio on the major side (11 kV ). The following is the CT secondary current i_s :

$$\begin{gathered} I_S=\frac{\left(787.3\right)}{\left(\frac{2000}{5}\right)} \\ =1.96\text{A} \end{gathered}$$

Since CT is in the delta on the secondary side, the following current is flowing through the restraining winding i_s is:

$$\begin{gathered} i_s=I_s\times \sqrt{3} \\ =1.96\times \sqrt{3} \\ =3.41\text{A} \end{gathered}$$

The relay current I_r is as follows since the transformer is differentially protected:

Substitute 4.37 forlocalid="1656052958522" $i_P$ and 3.41 for i_s into equation (1).

$$\begin{gathered} I_{\text{r}}=\left(4.37-3.41\right) \\ =0.96\text{A} \end{gathered}$$

The overload ratio is specified as 125%. The relay setting I_r should be as follows:

Substitute 0.96 for I_r into equation (2).

$$\begin{gathered} I_{\text{r}}=0.96\times 1.25 \\ =1.2\text{A} \end{gathered}$$

Therefore, the value of relay current I_r at full load is 0.96 A and the minimum relay setting is 1.2 A.