(b) Figure Q2.1 illustrates the impedance diagram of a power network that connects generators G1 and G2 to the 66 kV substation; all per-unit values are calculated based on each component rating values. All network components are rated at 200 MVA. The generators maintain the sending-bus voltage, V??, at its nominal value while the substation draws 150 MVA with 0.96 leading power factor. V?? V?? G1 j0.7 0.08+j0.16 G2 j0.7 0.08+j0.16 S? 33 kV 66 kV Figure Q2.1 For the system in Figure Q2.1 and using the analytical solution approach: (i) Calculate the per-unit magnitude and phase angle of the receiving-end voltage. (ii) Determine the apparent power losses on the network transformers. (iii) Draw the vector diagram of the sending-end voltage, V??, and receiving-end voltage, V??. Also, indicate on your diagram the vectors of the load current and the voltage drop across the network.
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Given that the substation draws 150 MVA with a 0.96 leading power factor, the complex power S can be calculated as S = P + jQ, where P is the real power and Q is the reactive power. P = 150 MVA * 0.96 = 144 MW, and Q = 150 MVA * sqrt(1 - 0.96^2) = 36 MVAR. Show more…
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The one-line diagram of a three-phase power system is shown in Figure 3.29. Select a common base of 100 MVA and 15 kV on the motor side. Draw an impedance diagram with all impedances, including the load impedance, marked in per unit. The manufacturer's data for each device is given as follows: G: 90 MVA, 22 kV, X=18% T1: 50 MVA, 22/220 kV, X=10% T2: 40 MVA, 220/11 kV, X=6.0% T3: 40 MVA, 22/110 kV, X=6.4% T4: 40 MVA, 110/11 kV, X=8.0% M: 66.5 MVA, 10.45 kV, X=18.5% The three-phase load at bus 4 absorbs 57 MVA, 0.6 power factor lagging at 10.45 kV. Line 1 and line 2 have reactances of 48.4 Ω and 65.43 Ω, respectively.
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