To solve this problem, we will calculate each of the required parameters step-by-step. ### Given Data: - Line length, \( l = 200 \, \text{km} \) - Resistance per phase, \( R = .2 \, \Omega/\text{km} \) - Reactance per phase, \( X = .4 \, \Omega/\text{km} \) - Shunt admittance per phase, \( Y = 4 \times 10^{-6} \, S/\text{km} \) - Load power, \( S_R = 100 \, \text{MVA} \) - Load voltage, \( V_R = 230 \, \text{kV} \) - Power factor at receiving end, \( \text{pf} = .9 \) (lagging) ### Calculations: #### 1. Calculate ABCD Constants: **Series Impedance \( Z \):** \[ Z = (R + jX) \times l = (.2 + j.4) \times 200 = (40 + j80) \, \Omega \] **Shunt Admittance \( Y \):** \[ Y = Y \times l = 4 \times 10^{-6} \times 200 = 8 \times 10^{-4} \, S \] **ABCD Constants:** - \( A = 1 + \frac{YZ}{2} \) \[ A = 1 + \frac{(8 \times 10^{-4})(40 + j80)}{2} = 1 + (.016 + j.032) = 1.016 + j.032 \] - \( B = Z \left(1 + \frac{Y}{2}\right) \) \[ B = (40 + j80)(1 + .0004) = (40 + j80)(1.0004) = 40.016 + j80.032 \] - \( C = Y = 8 \times 10^{-4} \) - \( D = A = 1.016 + j.032 \) #### 2. Calculate Receiving-End Power Components: - Real power \( P_R = S_R \times \text{pf} = 100 \times .9 = 90 \, \text{MW} \) - Reactive power \( Q_R = S_R \times \sin(\cos^{-1}(.9)) = 100 \times \sin(25.84^\circ) = 43.59 \, \text{MVAR} \) #### 3. Calculate Sending-End Voltage and Current: - Receiving-end current \( I_R \): \[ I_R = \frac{S_R}{\sqrt{3} \times V_R} = \frac{100 \times 10^6}{\sqrt{3} \times 230 \times 10^3} = 251.57 \, \text{A} \] - Sending-end voltage \( V_S \): \[ V_S = A \times V_R + B \times I_R \] Calculate \( B \times I_R \): \[ B \times I_R = (40.016 + j80.032) \times 251.57 = 10082.84 + j20165.68 \, \text{V} \] \[ V_S = (1.016 + j.032) \times 230000 + (10082.84 + j20165.68) \] \[ V_S = 233680 + j736 + 10082.84 + j20165.68 = 243762.84 + j27525.68 \, \text{V} \] Magnitude of \( V_S \): \[ |V_S| = \sqrt{(243762.84)^2 + (27525.68)^2} = 245321 \, \text{V} = 245.32 \, \text{kV} \] - Sending-end current \( I_S \): \[ I_S = \frac{P_R + jQ_R}{\sqrt{3} \times V_R} \] \[ I_S = \frac{90 \times 10^6 + j43.59 \times 10^6}{\sqrt{3} \times 230 \times 10^3} = 225.87 + j109.43 \, \text{A} \] #### 4. Calculate Sending-End Power: \[ S_S = 3V_SI_S^* \quad (\text{Conjugate of } I_S) \] Calculate the conjugate of \( I_S \): \[ I_S^* = 225.87 - j109.43 \] \[ S_S = 3 \times (243762.84 + j27525.68) \times (225.87 - j109.43) \] Calculate \( S_S \): \[ S_S = 3 \times (53256366.8 + j300000) = 160000000 + j900000 \, \text{VA} = 160 \, \text{MVA} \] #### 5. Transmission Efficiency: \[ \eta = \frac{P_R}{P_S} \times 100 = \frac{90}{160} \times 100 = 56.25\% \] ### Final Answers: 1. The sending-end voltage of the line: \( 245.32 \, \text{kV} \) 2. The sending-end current, power, and power factor: - Current: \( 251.57 \, \text{A} \) - Power: \( 160 \, \text{MVA} \) - Power factor: \( \cos \phi_s = \frac{P_S}{|S_S|} = \frac{90}{160} = .5625 \) 3. The efficiency of the line: \( 56.25\% \)