Question

The primary conductor in Figure 10.2 is one phase of a three-phase transmission line operating at$\mathbf{345}\mathbf{kV}$ , $\mathbf{700}\mathbf{MVA}$,$\mathbf{0}\mathbf{.}\mathbf{95}$ power factor lagging. The CT ratio is $\mathbf{1200}\mathbf{:}\mathbf{5}$, and the VT ratio is $\mathbf{3000}\mathbf{:}\mathbf{1}$. Determine the CT secondary current${\mathbf{I}}^{\mathbf{\text{'}}}$ and the VT secondary voltage${\mathbf{V}}^{\mathbf{\text{'}}}$ . Assume zero CT error.

Short answer

The CT secondary current$I^{\text{'}}$ is$4.88\text{\hspace{0.17em}A}$ and VT secondary voltage is$115\text{\hspace{0.17em}V}$ .

Step-by-step solution

Write the given data from the question:

The primary line to line voltage of transformer,$V_{LL}=345\text{\hspace{0.17em}kV}$

The three-phase power $S=700\text{\hspace{0.17em}MVA}$.

The power factor $pf=0.95$, leading.

Current transformer ratio$n_1=1200:5$ .

Voltage transformer ratio$n=3000:1$ .

Determine the equation to calculate the CT secondary current and VT secondary voltage.

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

${\mathbf{V}}^{\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{n}}{\mathbf{V}}_{\mathbf{LL}}$ …… (1)

The equation to calculate the current entering of current transformer is given as follows.

$\mathbf{I}\mathbf{=}\frac{\mathbf{S}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{LL}}}$ …… (2)

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

$\frac{\mathbf{1}}{{\mathbf{n}}_{\mathbf{1}}}\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{e}}\mathbf{+}{\mathbf{I}}^{\mathbf{\text{'}}}$ …… (3)

Here, ${\mathbf{I}}_{\mathbf{e}}$ is the exciting current and${\mathbf{I}}^{\mathbf{\text{'}}}$ is the secondary current of CT.

Calculate the secondary voltage of VT and secondary current of CT.

Calculate the secondary voltage of VT,

Substitute$345\text{\hspace{0.17em}kV}$ for$V_{LL}$ ,$3000:1$ for$n$ into equation (1).

$\begin{array}{l}{V}^{\text{'}}=\frac{1}{3000}\times 345\times {10}^3\\ V^{\text{'}}=115\text{\hspace{0.17em}V}\end{array}$

Calculate the current entering in the primary of CT.

Substitute$7000\text{\hspace{0.17em}MVA}$ for$S$ , and$345\text{\hspace{0.17em}kV}$ for$V_{LL}$ into equation (2).

$\begin{array}{l}I=\frac{7000}{\sqrt{3}\times 345}\\ I=1171.43\text{\hspace{0.17em}A}\end{array}$

Draw the equivalent circuit of the CT.

Calculate the secondary current of CT.

Substitute $1171.43\text{\hspace{0.17em}A}$for $I$, and $1200:5$for$n_1$ into equation (3).

$\begin{array}{l}\left(\frac{5}{1200}\right)1171.43=0+I^{\text{'}}\\ I^{\text{'}}=4.88\text{\hspace{0.17em}A}\end{array}$

Hence, the CT secondary current$I^{\text{'}}$ is$4.88\text{\hspace{0.17em}A}$ and VT secondary voltage is$115\text{\hspace{0.17em}V}$ .