Figure Q3.1 illustrates the impedance diagram of a power network that connects generators G and G...

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Figure Q3.1 illustrates the impedance diagram of a power network that connects generators G₁ and G₂ to the 66 kV receiving-bus substation; the per-unit values are calculated based on each component rating values. All network components are rated at 200MVA. The generators control the sending-bus voltage to its nominal value while the substation draws 150MVA with 0.96 leading power factor. G₁ Vs Ī = o j0.7 @t jo.7 33 kV 0.08+j0.16 5 Loss_T = 2 x 2₁ x (1₂2) 0.08+j0.16 Figure Q3.1 For the system in Figure Q3.1 and using the analytical solution approach: (i) Calculate the per-unit magnitude and phase angle of the receiving-end voltage. So V 1.01-18.36° pu [1] (ii) Determine the apparent power losses on the network transformers. (P+jQ) (0.72-(-j0.21)) V 1.01 18.36° V₂ -2.09° 66 kV 10-18.35° SR = 2 × (j0.7) × (0.7426² = j0.193 pu 2 = 0.7421-j0.0272 pu = 0.7426-2.1° [2] I= 0.7421-0.0271j (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. av [20 marks] =1.01-18.35 1L0 X [3] Figure Q3.1 illustrates the impedance diagram of a power network that connects generators G₁ and G₂ to the 66 kV receiving-bus substation; the per-unit values are calculated based on each component rating values. All network components are rated at 200MVA. The generators control the sending-bus voltage to its nominal value while the substation draws 150MVA with 0.96 leading power factor. G₁ Vs Ī = o j0.7 @t jo.7 33 kV 0.08+j0.16 5 Loss_T = 2 x 2₁ x (1₂2) 0.08+j0.16 Figure Q3.1 For the system in Figure Q3.1 and using the analytical solution approach: (i) Calculate the per-unit magnitude and phase angle of the receiving-end voltage. So V 1.01-18.36° pu [1] (ii) Determine the apparent power losses on the network transformers. (P+jQ) (0.72-(-j0.21)) V 1.01 18.36° V₂ -2.09° 66 kV 10-18.35° SR = 2 × (j0.7) × (0.7426² = j0.193 pu 2 = 0.7421-j0.0272 pu = 0.7426-2.1° [2] I= 0.7421-0.0271j (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. av [20 marks] =1.01-18.35 1L0 X [3]