Question

Choosing system bases to be $\mathbf{240}\mathbf{/}\mathbf{24}\mathbf{}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{and}\mathbf{}\mathbf{100}\mathbf{}\mathbf{MVA}$, redraw the per unit equivalent circuit for Problem 3.39.

Short answer

The zero, positive and negative sequence diagram is drawn below:

Step-by-step solution

Determine the phase shift between star and delta connection of transformer and write the formula new leakage reactance of transformer.

It is given that, the primary side of transformer has star connection with solidly grounded neutral and secondary has delta connection.

There will be a $\mathbf{30}\mathbf{°}$phase shift between star and delta connection.

Write the formula of leakage reactance.

${\mathbf{X}}_{\mathbf{new}}\mathbf{=}{\mathbf{X}}_{\mathbf{old}}\left(\frac{S_{\mathrm{new}}}{S_{\mathrm{old}}}\right){\left(\frac{V_{\mathrm{old}}}{V_{\mathrm{new}}}\right)}^{\mathbf{2}}$ …… (1)

Determine the per-unit equivalent circuit.

Substitute $0.06\mathrm{pu}\mathrm{for}{\mathrm{X}}_{old},100\mathrm{MVA}\mathrm{for}{\mathrm{S}}_{new},300\mathrm{MVA}\mathrm{for}{\mathrm{S}}_{old},240\mathrm{kV}\mathrm{for}{\mathrm{V}}_{new}\mathrm{and}230\mathrm{kV}\mathrm{for}{\mathrm{V}}_{old}$in equation (1).

$$\begin{gathered} X_{new}=0.06\mathrm{pu}\times \frac{100\mathrm{MVA}}{300\mathrm{MVA}}\times {\left(\frac{230\mathrm{kV}}{240\mathrm{kV}}\right)}^2 \\ =0.01837\mathrm{pu} \end{gathered}$$

The zero sequence, positive sequence and negative sequence per-unit equivalent circuit is drawn below:

In the diagram, the transformer leakage reactance 0.01837 pu is shown.

Also, the phase shift is mentioned.

In zero sequence diagram. As the star connection is solidly grounded, it will be closed and delta will be open.

The positive and negative sequence diagram will be same.