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6. Multi-Voltage Level SystemA power system includes a generator (500 MVA, 13.8 kV, X'' = 0.2 pu), a transformer (500MVA, 13.8/138 kV, Z = 0.1 pu), and a transmission line (Z = 0.02 + j0.1 pu on 138 kV, 500MVA base). A load of 450 MW at 0.9 power factor lagging is connected at the end of thetransmission line.a. Convert all impedances to a common base of 500 MVA, 13.8 kV.b. Calculate the per-unit voltage at the load bus.

Question

  1. Multi-Voltage Level SystemA power system includes a generator (500 MVA, 13.8 kV, X'' = 0.2 pu), a transformer (500MVA, 13.8/138 kV, Z = 0.1 pu), and a transmission line (Z = 0.02 + j0.1 pu on 138 kV, 500MVA base). A load of 450 MW at 0.9 power factor lagging is connected at the end of thetransmission line.a. Convert all impedances to a common base of 500 MVA, 13.8 kV.b. Calculate the per-unit voltage at the load bus.
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Solution

a. Convert all impedances to a common base of 500 MVA, 13.8 kV.

The generator and transformer are already on the base of 500 MVA and 13.8 kV, so their per-unit impedances remain the same.

For the transmission line, we need to convert the base voltage from 138 kV to 13.8 kV. The new per-unit impedance can be calculated using the formula:

Z_new = Z_old * (V_new/V_old)^2 * (S_old/S_new)

Where: Z_old = 0.02 + j0.1 pu (old per-unit impedance) V_old = 138 kV (old voltage base) V_new = 13.8 kV (new voltage base) S_old = S_new = 500 MVA (power base remains the same)

Substituting the values, we get:

Z_new = (0.02 + j0.1) * (13.8/138)^2 * (500/500) Z_new = 0.02 + j0.1 pu

So, the per-unit impedance of the transmission line on the base of 500 MVA, 13.8 kV is also 0.02 + j0.1 pu.

b. Calculate the per-unit voltage at the load bus.

The per-unit voltage at the load bus can be calculated using the formula:

V_load = V_source - I_load * Z_total

Where: V_source = 1 pu (source voltage) I_load = P_load / (V_load * power factor) = 450 MW / (1 pu * 0.9) = 0.5 pu (load current) Z_total = Z_generator + Z_transformer + Z_transmission = 0.2 pu + 0.1 pu + 0.02 + j0.1 pu = 0.32 + j0.1 pu (total impedance)

Substituting the values, we get:

V_load = 1 pu - 0.5 pu * (0.32 + j0.1) V_load = 1 pu - 0.16 - j0.05 pu V_load = 0.84 - j0.05 pu

So, the per-unit voltage at the load bus is 0.84 - j0.05 pu.

This problem has been solved

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