Question

Three single-phase transformers, each rated $\mathbf{10}\mathbf{}\mathbf{MW}\mathbf{,}\mathbf{}\mathbf{66}\mathbf{.}\mathbf{4}\mathbf{/}\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{kV}\mathbf{}\mathbf{,}\mathbf{}\mathbf{60}\mathbf{}\mathbf{Hz}$, with an equivalent series reactance of 0.1 per unit divided equally between primary and secondary, are connected in a three-phase bank. The highvoltage windings are Y-connected and their terminals are directly connected to a $\mathbf{115}\mathbf{}\mathbf{kV}$three-phase bus. The secondary terminals are all shorted together. Find the currents entering the high-voltage terminals and leaving the low-voltage terminals if the low-voltage windings are (a) Y-connected and (b) -connected.

Short answer

(a) The current entering a high voltage terminal and leaving a low voltage terminal in star connection at secondary is 1.506 kA and 8 kA .

(b) The current entering a high voltage terminal and leaving a low voltage terminal in delta connection at secondary l is 1.506 kA and 13.86 kA .

Step-by-step solution

Given data.

The rating of each transformer is 10MW , 66.4 kV/12.5 kV .

The leakage reactance of transformer is 0.1pu .

Determine the formulas of line-to-line voltage, per-unit current, base current and actual current.

Write the formula of line-to-line voltage.

${\mathbf{V}}_{\mathbf{L}\mathbf{-}\mathbf{L}}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{phase}}$ ……. (1)

Write the formulas of per-unit current.

${\mathbf{I}}_{\mathbf{pu}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{pu}}}{{\mathbf{X}}_{\mathbf{pu}}}$……. (2)

Write the formula of base current.

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

Write the formula of actual current.

${\mathbf{I}}_{\mathbf{actual}}\mathbf{=}{\mathbf{I}}_{\mathbf{base}}\mathbf{\times }{\mathbf{I}}_{\mathbf{pu}}$ ……. (4)

Determine the per-unit current.

The base power is 30 MVA.

Determine the base voltage at high voltage side.

Substitute $66.4\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{ph}}$in equation (1).

role="math" localid="1655197785151" $$\begin{gathered} V_{baseH}=66.4\mathrm{kV}\times \sqrt{3} \\ =115\mathrm{kV} \end{gathered}$$

Determine the base voltage at low voltage side.

Substitute $12.5\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{ph}}$in equation (1).

$$\begin{gathered} V_{baseL}=12.5\mathrm{kV}\times \sqrt{3} \\ =21.65\mathrm{kV} \end{gathered}$$

Assume that, the per-unit voltage at secondary terminal is $V_{pu}=1\angle 0°\mathrm{pu}$.

The leakage reactance given is $X_{pu}=0.1\mathrm{pu}$.

Determine the per-unit current in the below circuit:

Substitute $1\angle 0°\mathrm{pu}\mathrm{for}{\mathrm{V}}_{pu}\mathrm{and}\mathrm{j}0.1\mathrm{pu}\mathrm{for}{\mathrm{X}}_{pu}$ in equation (2).

$$\begin{gathered} I_{pu}=\frac{1\angle 0°\mathrm{pu}}{\mathrm{j}0.1\mathrm{pu}} \\ =10\angle -90°\mathrm{pu} \end{gathered}$$

Determine the current for star connection at secondary.

(a)

Determine thebase value of current at high-voltage side.

Substitute $30\mathrm{MVA}\mathrm{for}{\mathrm{S}}_{base}\mathrm{and}115\mathrm{kV}\mathrm{for}{\mathrm{V}}_{base}$in equation (3).

role="math" localid="1655198594925" $$\begin{gathered} I_{baseH}=\frac{30\mathrm{MVA}}{115\mathrm{kV}\times \sqrt{3}} \\ =0.1506\mathrm{kA} \end{gathered}$$

Determine thebase value of current at low-voltage side.

Substitute $30\mathrm{MVA}\mathrm{for}{\mathrm{S}}_{base}\mathrm{and}21.65\mathrm{kV}\mathrm{for}{\mathrm{V}}_{base}$in equation (3).

role="math" localid="1655198633960" $$\begin{gathered} I_{baseL}=\frac{30\mathrm{MVA}}{21.65\mathrm{kV}\times \sqrt{3}} \\ =0.8\mathrm{kA} \end{gathered}$$

Determine theactual current at high voltage side.

Substitute $0.1506\mathrm{kA}\mathrm{for}{\mathrm{I}}_{base\mathit{}H}\mathrm{and}10\mathrm{pu}\mathrm{for}{\mathrm{I}}_{pu}$ in equation (4).

$$\begin{gathered} I_H=0.1506\mathrm{kA}\times 10\mathrm{pu} \\ =1.506\mathrm{kA} \end{gathered}$$

Determine theactual current at low voltage side.

Substitute $0.8\mathrm{kA}\mathrm{for}{\mathrm{I}}_{baseH}\mathrm{and}10\mathrm{pu}\mathrm{for}{\mathrm{I}}_{pu}$in equation (4).

$$\begin{gathered} I_L=0.8\mathrm{kA}\times 10\mathrm{pu} \\ =8\mathrm{kA} \end{gathered}$$

Therefore, the current entering a high voltage terminal and leaving a low voltage terminal in star connection at secondary is 1.506 kA and 8 kA .

Determine the current for delta connection at secondary.

(b)

Determine thebase value of current at low-voltage side.

Substitute $30\mathrm{MVA}\mathrm{for}{\mathrm{S}}_{\mathrm{base}}\mathrm{and}12.5\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{base}}$in equation (3).

$$\begin{gathered} I_{baseL}=\frac{30\mathrm{MVA}}{12.5\mathrm{kV}\times \sqrt{3}} \\ =1.386\mathrm{kA} \end{gathered}$$

Theactual current at high voltage side will be same as star connection.

Determine theactual current at low voltage side.

Substitute $1.386\mathrm{kA}\mathrm{for}{\mathrm{I}}_{baseH}\mathrm{and}10\mathrm{pu}\mathrm{for}{\mathrm{I}}_{pu}$ in equation (4).

role="math" localid="1655199401445" $$\begin{gathered} I_L=1.386\mathrm{kA}\times 10\mathrm{pu} \\ =13.86\mathrm{kA} \end{gathered}$$

Therefore, the current entering a high voltage terminal and leaving a low voltage terminal in delta connection at secondary l is 1.506 kA and 13.86 kA .