Question

Question:A single-phase, 5 MVA 20/8.66 KV , transformer is protected by a differential relay with taps. Available relay tap settings are 5:5,5:55,5:66 , 5:73 , 5:8 ,5:9 ,and 5:10, giving tap ratios of 1,1:10,1:32,1:46.1:60,1:80, and 2:00. Select CT ratios and relay tap settings. Also, determine the percentage mismatch for the selected tap setting.

Short answer

Answer

The value of transformer primary rated current,$I_{\text{1rated}}$is 250 A .

The value of CT primary current $I{\text{'}}_1$, is 4.167 A.

The value of transformer secondary rated current $I_{\text{2rated}}$, is 577.4 A

The value of CT secondary current $I{\text{'}}_2$, is 4.811 A .

The value of percentage mismatch is 4.7% .

Step-by-step solution

 Step 1: Write the given data from the question.

Consider the value of transformer power rating = 5 MVA.

Consider the transformer primary voltage is 20 KV .

Consider the transformer secondary voltage is 8.66 KV .

Consider available CT tap settings are: 5:5,5:55,5:66 , 5:73 , 5:8 ,5:9 ,and 5:10,

Consider corresponding CT tap ratios are:1,1:10,1:32,1:46.1:60,1:80, and 2:00.

Determine the formula of transformer primary rated current, CT primary current and percentage mismatch.

Write the formula oftransformer primary rated current.

${\mathbf{I}}_{\mathbf{1}\mathbf{rated}}=\frac{\mathbf{CT}\left(\mathbf{Y}\right)}{\sqrt{\mathbf{3}}\mathbf{CT}\left(\mathbf{\Delta }\right)}$ …… (1)

Here, $\mathbf{CT}\left(Y\right)$ is current transformer in star side and $\mathbf{CT}\left(\mathbf{\Delta }\right)$ is current transformer in delta side.

Write the formula of CT primary current.

${\mathbf{I}\mathbf{\text{'}}}_{\mathbf{1}}=\frac{{\mathbf{I}}_{\mathbf{1}\mathbf{rated}}}{\mathbf{CT}\mathbf{ratio}}$ …… (2)

Here, ${\mathbf{I}}_{\mathbf{1}\mathbf{rated}}$ is transformer primary rated current.

Write the formula of transformer secondary rated current.

${\mathbf{I}}_{\mathbf{2}\mathbf{rated}}=\frac{\mathbf{CT}\left(\mathbf{\Delta }\right)}{\sqrt{\mathbf{3}}\mathbf{CT}\left(\mathbf{Y}\right)}$ …… (3)

Here, $\mathbf{CT}\left(\mathbf{\Delta }\right)$ is current transformer in delta side and is current transformer in star side.

Write the formula of CT secondary current.

${\mathbf{I}\mathbf{\text{'}}}_{\mathbf{2}}=\frac{{\mathbf{I}}_{\mathbf{2}\mathbf{rated}}}{\mathbf{CT} \mathbf{ratio}}$ …… (4)

Here,role="math" localid="1656151032235" ${\mathbf{I}}_{\mathbf{2}\mathbf{rated}}$ is transformer primary rated current.

Write the formula of percentage mismatch.

$\mathbf{M}=\frac{\left(\frac{{\mathbf{I}\mathbf{\text{'}}\mathbf{\text{'}}}_{\mathbf{2}}}{{\mathbf{T}\mathbf{\text{'}}}_{\mathbf{2}}}\right)-\left(\frac{{\mathbf{I}\mathbf{\text{'}}}_{\mathbf{1}}}{{\mathbf{T}\mathbf{\text{'}}}_{\mathbf{1}}}\right)}{\left(\frac{{\mathbf{I}\mathbf{\text{'}}}_{\mathbf{1}}}{{\mathbf{T}\mathbf{\text{'}}}_{\mathbf{1}}}\right)}\left(\mathbf{100}\right)$ …… (5)

Here, ${{\mathbf{I}}_{\mathbf{2}}}^{\mathbf{\text{'}\text{'}}}$ is restraining winding current, ${{\mathbf{I}}_{\mathbf{1}}}^{\mathbf{\text{'}}}$ is CT primary current, ${\mathbf{T}}_{\mathbf{1}}\text{'}$ and ${\mathbf{T}}_{\mathbf{2}}\text{'}$ is taps of current transformer.

 Determine the value of transformer primary rated current, CT primary current and percentage mismatch.

Choose the CT ratio first.

Determine the transformer primary rated current.

Substitute $5\times {10}^6$ for and $20\times {10}^3$for $I_{\text{1rated}}$into equation (1).

$$\begin{gathered} I_{\text{1rated}}=\frac{5\times {10}^6}{20\times {10}^3} \\ =250\text{A} \end{gathered}$$

Therefore, the value of transformer primary rated current , is .

Choose a 300:5 standard CT ratio on the primary side(20 KV) .

Determine the CT primary current$I{\text{'}}_1$.

Substitute 250 A for $I_{\text{1rated}}$and 300/5 for CT into equation (2).

$$\begin{gathered} {I\text{'}}_1=\frac{250 \text{A}}{\left(\frac{300}{5}\right)} \\ =4.167\text{A} \end{gathered}$$

Therefore, the value of CT primary current $I{\text{'}}_1$, is .

Similarly, determine the transformer secondary rated current.

Substitute $5\times {10}^6\text{VA}$for CT($∆$) and $8.66\times {10}^3\text{V}$ for CT(Y) .

$$\begin{gathered} I_{\text{2rated}}=\frac{5\times {10}^6}{8.66\times {10}^3\text{V}} \\ =577.4\text{A} \end{gathered}$$

Therefore, the value of transformer secondary rated current$I_{\text{2rated}}$, is 577.4 A .

Choose a 600:5 standard CT ratio on the secondary side (8.66 KV).

Determine the CT secondary current $I{\text{'}}_2$.

Substitute 577.4 A for I_2RATED and 600/5 for CT into equation (3).

$$\begin{gathered} {I\text{'}}_2=\frac{577.4 \text{A}}{\frac{600}{5}} \\ =4.811\text{A} \end{gathered}$$

Therefore, the value of CT secondary current ,I₂ ^'is 4.811 .

To balance currents in the restraining windings, use certain relay taps.

Determine the ratio of the currents in the restraining winding is:

$$\begin{gathered} \frac{{I\text{'}\text{'}}_2}{{I\text{'}}_1}=\frac{4.811\text{A}}{4.167\text{A}} \\ =1.155 \end{gathered}$$

The closest relay tap ratio is 1:10 , so,

$\frac{{T\text{'}}_2}{{T\text{'}}_1}=1.10$

Here, $T{\text{'}}_1$ and $T{\text{'}}_2$ are the taps of the current transformer.

Therefore, the relay tap ratio is 1:10 and CT ratio is5:55 .

Determine the percentage mismatch for the tap setting ratio .

Substitute 4.167 A for $I{\text{'}}_1$, 4.811 A for $I\text{'}{\text{'}}_2$, 5 for T₁ ^'and 5:5 for T₂^' into equation (5).

$$\begin{gathered} M=\frac{\left(\frac{4.167\text{A}}{5}\right)-\left(\frac{4.811\text{A}}{5.5}\right)}{\left(\frac{4.811}{5.5}\right)}\left(100\right) \\ =\left(\frac{0.8334-0.8747}{0.8747}\right)\left(100\right) \\ =\left(-0.0472\right)\left(100\right) \\ =-4.7\% \end{gathered}$$

Therefore, the value of percentage mismatch is -4.7% .