College

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.

Answer :

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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