Three-phase power and line-to-line voltage ratings of the system shown in the figure are given as follows:

\[ V_g \]

\[ T1 \]

\[ Bus 1 \]

\[ Bus 2 \]

\[ T2 \]

\[ Vm \]

\[ Line \]

\[ G \] : 60 MVA, 20 kV, $ Z = 9\% $

\[ T1 \] : 50 MVA, 20/200 kV, $ Z = 10\% $

\[ T2 \] : 80 MVA, 200/20 kV, $ Z = 12\% $

\[ Load \] : 32.4 MVA, 18 kV, $ pf = 0.8 $ (lag)

\[ Line \] : 200 kV, $ Z = 120 + j200 \Omega $

Draw the impedance diagram of the system in per unit, using $ S_{\text{base}} = 100 \text{ MVA} $ and $ V_{\text{base}} = 20 \text{ kV} $ (for the generator).

Note: Assume that generator and transformer resistances are negligible.

To draw the impedance diagram of the system in per unit, convert the given impedance values to per unit values using the formula: Z_perunit = (Z / S_base) * (V_base^2 / V^2).

What is the formula for calculating the apparent power in a three-phase system?

To draw the impedance diagram of the system in per unit, we need to convert the given impedance values to per unit values. Given that S_base = 100 MVA and V_base = 20 kV for the generator, we can calculate the per unit impedance values as follows:

Generator:

Zg = 9% of 60 MVA = 0.09 * 60 = 5.4 MVA

Zg_perunit = (Zg / S_base) * (V_base^2 / Vg^2) = (5.4 / 100) * (20^2 / 20^2) = 0.0027 pu

Transformer T1:

Zt1 = 10% of 50 MVA = 0.1 * 50 = 5 MVA

Zt1_perunit = (Zt1 / S_base) * (V_base^2 / Vt1^2) = (5 / 100) * (20^2 / 200^2) = 0.0005 pu

Transformer T2:

Zt2 = 12% of 80 MVA = 0.12 * 80 = 9.6 MVA

Zt2_perunit = (Zt2 / S_base) * (V_base^2 / Vt2^2) = (9.6 / 100) * (20^2 / 20^2) = 0.0048 pu

Load:

Zload = 120 + j200 Ω

Zload_perunit = (Zload / S_base) * (V_base^2 / S_base) = (120 + j200) / (100 * (20^2)) = 0.06 + j0.1 pu

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