Question

Question:A three-phase, 500 MVA ,345 KV$∆$/500 KV transformer is protected by differential relays with taps. Select CT ratios, CT connections, and relay tap settings. Determine the currents in the transformer and in the CTs at rated conditions. Also determine the percentage mismatch for the selected relay tap settings. Available relay tap settings are given in Problem 10.26.

Short answer

Answer

The value of transformer primary rated current $I_{\text{1rated}}$, is 834.74 a .

The value of CT primary current l₁' , is 4.65 A .

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

The value of CT secondary current I₂,^' is 40811 A.

The value of percentage mismatch is -0.43 .

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 current transformer (CT) in star to delta side is 345 K V .

Consider the value of current transformer (CT) in delta to star side is 500 KV .

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, CT(Y) is current transformer in star side and CT($∆$) 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, CT($∆$), is current transformer in delta side and CT(Y) 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,${\mathbf{I}}_{\mathbf{2}\mathbf{rated}}$ 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, I₂^' is restraining winding current, I₁^' is CT primary current, T₁' and T₂' 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 $500\times {10}^6$for CT(Y) and $345\times {10}^3$for CT($∆$) into equation (1).

$$\begin{gathered} I_{\text{1rated}}=\frac{500\times {10}^6}{\sqrt{3}\left(345\times {10}^3\right)} \\ =\frac{500\times {10}^6}{597.56\times {10}^3} \\ =836.74\text{A} \end{gathered}$$

Therefore, the value of transformer primary rated current $I_{\text{1rated}}$, is 836.74

Choose a 900:5 standard CT ratio on the primary side (345 KV ).

Determine the CT primary current I₁^'.

Substitute for and for into equation (2).

$$\begin{gathered} {I\text{'}}_1=\frac{836.74 \text{A}}{\left(\frac{900}{5}\right)} \\ =4.65\text{A} \end{gathered}$$

Therefore, the value of CT primary current, is 4.65 .

Similarly, determine the transformer secondary rated current.

Substitute $5\times {10}^6\text{VA}$for and $8.66\times {10}^3$ 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 , is577.4 A .

Choose a 600:5 standard CT ratio on the secondary side .

Determine the CT secondary current .

Substitute 577..4 A forrole="math" localid="1656148850815" $I_{\text{2rated}}$ and role="math" localid="1656148784314" $\frac{600}{5}$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,$I_{\text{2rated}}$iS 4.811 A .

Due to CT's delta position on the secondary side. As a result, the current in the restraint winding ${I\text{'}}_2$ is:

$$\begin{gathered} {I\text{'}\text{'}}_2={I\text{'}}_2\times \sqrt{3} \\ =4.811\text{A}\times \sqrt{3} \\ =8.333\text{A} \end{gathered}$$

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

Determine the restraining windings' currents are distributed as follows:

$$\begin{gathered} \frac{{I\text{'}\text{'}}_2}{{I\text{'}}_1}=\frac{8.333\text{A}}{4.65\text{A}} \\ =1.792 \end{gathered}$$

The closest relay tap ratio is ,1.8 so,

role="math" localid="1656149151996" $$\begin{gathered} \frac{{T\text{'}}_2}{{T\text{'}}_1}=1.8 \\ =\frac{9}{5} \end{gathered}$$

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

Determine the percentage mismatch for the tap setting ratio .

Substitute 4.65 A for $I{\text{'}}_1$, 8.333 A for $I{\text{'}}_2$ and $1.8T{\text{'}}_1$ for T₂^' into equation (5).

$$\begin{gathered} M=\frac{\left(\frac{8.333\text{A}}{1.8{T\text{'}}_1}\right)-\left(\frac{4.65\text{A}}{{T\text{'}}_1}\right)}{\left(\frac{4.65\text{A}}{{T\text{'}}_1}\right)}\left(100\right) \\ =\left(\frac{4.63\text{A}-4.65\text{A}}{4.65\text{A}}\right)\left(100\right) \\ =\left(-0.0043\right)\left(100\right) \\ =-0.43\% \end{gathered}$$

Therefore, the value of percentage mismatch is -0.43%.