FIGURE 3(a) represents a 50 Hz, three-phase, high-voltage, transmission line. For one phase, the relationships between the sending end voltage and current and the receiving end voltage are given by the complex ABCD equations: Vsp = Vrp(A1 + jA2) + Irp(B1 + jB2) Isp = Vrp(C1 + jC2) + Irp(D1 + jD2) where VSP is the sending-end phase voltage, ISP the sending-end phase current and VRP is the magnitude of the open-circuit receiving end phase voltage. (a) Given the parameter values in TABLE A and if the magnitude of the receiving-end line voltage VRL is measured as 154 kV when feeding a balanced load of 40 MVA at a power factor of 0.9, calculate the value of the sending-end phase voltage VSP and sending-end phase current ISP. [N.B. VSL = √3 × VSP and the total power in a three-phase load is given by P = √3VI cos θ.] (b) Hence or otherwise calculate the sending-end power and thus the power lost in the cable. (c) If the line is modelled by the Π-circuit of FIGURE 4(b), see if you can estimate the primary line coefficients R, L, G and C. The line is 50 km long. A1= 0.8698 A2= 0.03542 B1= 47.94 Ω B2= 180.8 Ω C1= 0 S C2= 0.001349 S D1= 0.8698 D2= 0.03542 so a) P=sqrt(3)VI I=40000000/(sqrt(3)*154000) = 149.961 A Vrp=Vrl/sqrt(3) = 88912 V Using formulas given for ABCD. Vsp =84524.79 + j30262.16 = 89778.8∠19.70 Isp = 130.44 +j125.252 = 180.83∠43.84 b) p= sqrt(3)VI cos θ = sqrt(3)*154000*149.961*0.9 =36 MW Vsl = sqrt(3)*89778.8 = 155501.4*149.961*0.9 = 36.3MW 350960.2 W lost in the cable. c) I don't have a clue?