Question

Question:Consider a three-phase $\mathbf{\Delta }--\mathbf{Y}$ connected, 30-MVA,33:11KV , transformer with differential relay protection. If the CT ratios are 500:5A on the primary side and 2000:5A on the secondary side, compute the relay current setting for faults drawing up to 200% of rated transformer current.

Short answer

Answer

The value of relay current I at differential protected is 1.57 A and the minimum relay setting is 3.14 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 200% .

Determine the formula of relay current   and relay setting  for faults drawing up to 200%.

Write the formula ofrelay current.

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

Here, I_s is CT secondary current.

Write the formula of minimum relay setting.

${\mathbf{I}\mathbf{\text{'}}}_{\mathbf{r}}={\mathbf{I}}_{\mathbf{r}}\mathbf{\times }\mathbf{200}\mathbf{\%}$ …… (2)

Here, I_r is relay current.

 Determine the value of relay current   and relay setting  for faults drawing up to 200%.

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

Determine the transformer primary rated line current $I_{\text{1rated}}$ as:

$$\begin{gathered} I_{\text{1rated}}=\frac{30\times {10}^6}{33\times {10}^3\times \sqrt{3}} \\ =524.88\text{A} \end{gathered}$$

Choose a 500:5 standard CT ratio on the primary side. The following is the CT secondary current$i_P$ :

$i_P=\frac{I_{\text{1rated}}}{\text{CT}\text{ratio}}$

Here, is transformer primary rated line current.

Substitute 524.88 for $I_{\text{1rated}}$ and $\frac{500}{5}$for into above equation.

$$\begin{gathered} i_{\text{P}}=\frac{\left(524.88\right)}{\left(\frac{500}{5}\right)} \\ =5.25\text{A} \end{gathered}$$

Determine the transformer secondary rated line current is calculated similarly as follows:

$$\begin{gathered} I_{\text{2rated}}=\frac{30\times {10}^6}{11\times {10}^3\times \sqrt{3}} \\ =1574.64\text{A} \end{gathered}$$

Choose a 2000:5 standard CT ratio on the secondary side. The following is the CT secondary current :

${\text{I}}_{\text{S}}=\frac{I_{\text{2rated}}}{\text{CT}\text{ratio}}$

Here, is transformer secondary rated line current.

Substitute 1574.64 for $I_{\text{2rated}}$ and $\frac{2000}{5}$for into above equation.

$$\begin{gathered} {\text{I}}_{\text{S}}=\frac{\left(1574.64\right)}{\frac{2000}{5}} \\ =3.92\text{A} \end{gathered}$$

Due to CT's delta position on the secondary side. As a result, the current in the restraint winding is:

$$\begin{gathered} i_{\text{S}}=I_{\text{S}}\times \sqrt{3} \\ =3.92\times \sqrt{3} \\ =6.82\text{A} \end{gathered}$$

Determine the relay current for differentially protected transformer.

Substitute 6.82 for I_s and 5.25 for I_p into equation (1).

$$\begin{gathered} I_r=\left(6.82-5.25\right) \\ =1.57A \end{gathered}$$

Therefore, the value of relay current I_r is 1.57 .

Determine the overload ratio is given as. So, the relay setting should be:

Substitute 1.57 for I_r into equation (2).

$$\begin{gathered} {I\text{'}}_{\text{r}}=1.57\times 2\text{A} \\ =\text{3.14}\text{A} \end{gathered}$$

Therefore, the minimum relay setting is 3.14 A .