# Transmission Line Calculation (Nominal π Method) Given: - **Length of line, \( l \)**: 32 km - **Resistance per km, \( R \)**: .075 Ω/km - **Outside diameter of conductor, \( d \)**: 2.4 cm = .024 m - **Spacing between conductors, \( D \)**: 2 m - **Frequency, \( f \)**: 50 Hz - **Load delivered, \( P \)**: 26 MW - **Line-to-line voltage at receiving end, \( V_R \)**: 41 kV - **Power factor, \( pf \)**: .82 lagging - **System**: 3-phase --- ## Step 1: Calculate Line Parameters ### 1. Total Resistance (\( R \)) \[ R_{total} = R \times l = .075 \times 32 = 2.4\ \Omega \] ### 2. Inductance per Phase per km (\( L \)) \[ L = 2 \times 10^{-7} \ln\left(\frac{D}{r'}\right) \quad \text{(H/m)} \] where \( r' = .7788r \), and \( r = \frac{d}{2} = .012 \) m. \[ r' = .7788 \times .012 = .0093456\ \text{m} \] \[ L = 2 \times 10^{-7} \ln\left(\frac{2}{.0093456}\right) = 2 \times 10^{-7} \ln(214.14) \] \[ \ln(214.14) \approx 5.366 \] \[ L = 2 \times 10^{-7} \times 5.366 = 1.073 \times 10^{-6}\ \text{H/m} \] \[ L_\text{per km} = 1.073 \times 10^{-6} \times 100 = 1.073 \times 10^{-3}\ \text{H/km} \] \[ L_{total} = 1.073 \times 10^{-3} \times 32 = .03434\ \text{H} \] ### 3. Capacitance per Phase per km (\( C \)) \[ C = \frac{2\pi \varepsilon}{\ln(D/r')} \] where \( \varepsilon = 8.854 \times 10^{-12}\ \text{F/m} \). \[ C = \frac{2\pi \times 8.854 \times 10^{-12}}{5.366} = \frac{5.564 \times 10^{-11}}{5.366} = 1.037 \times 10^{-11}\ \text{F/m} \] \[ C_\text{per km} = 1.037 \times 10^{-11} \times 100 = 1.037 \times 10^{-8} \ \text{F/km} \] \[ C_{total} = 1.037 \times 10^{-8} \times 32 = 3.318 \times 10^{-7} \ \text{F} \] --- ## Step 2: Load Current \[ \text{Apparent Power } S = \frac{P}{pf} = \frac{26 \times 10^6}{.82} = 31.71 \times 10^6 \text{ VA} \] \[ I_R = \frac{S}{\sqrt{3} V_R} = \frac{31.71 \times 10^6}{\sqrt{3} \times 41 \times 10^3} \] \[ I_R = \frac{31.71 \times 10^6}{71,009.4} = 446.4\ \text{A} \] --- ## Step 3: Nominal π Model Parameters ### Series Impedance per phase \[ Z = R_{total} + j \omega L_{total} \] \[ \omega = 2\pi f = 314.16 \text{ rad/s} \] \[ j \omega L_{total} = j \times 314.16 \times .03434 = j 10.79 \] \[ Z = 2.4 + j 10.79\ \Omega \] ### Shunt Admittance per phase \[ Y = j \omega C_{total} \] \[ Y = j \times 314.16 \times 3.318 \times 10^{-7} = j .0001043 \] --- ## Step 4: Nominal π ABCD Parameters \[ A = D = 1 + \frac{YZ}{2} \] \[ B = Z \] \[ C = Y \left[1 + \frac{YZ}{4}\right] \] First, calculate \( YZ \): \[ YZ = (j .0001043) \times (2.4 + j10.79) = j .0001043 \times 2.4 + j .0001043 \times j 10.79 = j .0002503 - .001126 \] \[ A = 1 + \frac{YZ}{2} = 1 + \frac{-.001126 + j .0002503}{2} = 1 - .000563 + j .000125 = .99944 + j .000125 \] --- ## Step 5: Receiving End Per Phase Values \[ V_{R,\text{phase}} = \frac{41,000}{\sqrt{3}} = 23,677\ \text{V} \] \[ I_R = 446.4\ \text{A} \] \[ \text{Phase angle} = \cos^{-1}(.82) = 34.9^\circ \] \[ I_R = 446.4 \angle -34.9^\circ\ \text{A} \] --- ## Step 6: Sending End Voltage (Per Phase) \[ V_S = A V_R + B I_R \] With - \( V_R = 23,677 \angle ^\circ \) V - \( I_R = 446.4 \angle -34.9^\circ \) A - \( B = 2.4 + j10.79 = 11.06 \angle 77.45^\circ \) Ω Calculate \( B I_R \): \[ |B I_R| = 11.06 \times 446.4 = 4,936.8 \] \[ \text{Angle} = 77.45^\circ + (-34.9^\circ) = 42.55^\circ \] \[ B I_R = 4,936.8 \angle 42.55^\circ \] Now, \( A V_R \approx .99944 \times 23,677 = 23,663 \) (Due to small imaginary part, phase shift is negligible.) Sum using rectangular form: - \( A V_R \approx 23,663 + j 2.96 \) V - \( B I_R = 4,936.8 \cos 42.55^\circ + j 4,936.8 \sin 42.55^\circ \) - \( = 4,936.8 \times .7368 = 3,637.6 \) - \( = 4,936.8 \times .6761 = 3,340.7 \) - So, \( B I_R \approx 3,637.6 + j 3,340.7 \) V Add: \[ V_S = (23,663 + 3,637.6) + j (2.96 + 3,340.7) = 27,300.6 + j 3,343.7 \] \[ |V_S| = \sqrt{27,300.6^2 + 3,343.7^2} = \sqrt{745,312,000 + 11,185,300} = \sqrt{756,497,300} = 27,509\ \text{V} \] Per phase, so total line-to-line: \[ V_{S,LL} = 27,509 \times \sqrt{3} = 47,646\ \text{V} \] **(a) Sending end voltage = \(\boxed{47.65\ \text{kV}}\) (line-to-line)** --- ## Step 7: Voltage Regulation \[ \text{Regulation} = \frac{V_{S,LL} - V_{R,LL}}{V_{R,LL}} \times 100\% = \frac{47,646 - 41,000}{41,000} \times 100 = 16.2\% \] **(b) Voltage Regulation = \(\boxed{16.2\%}\)** --- ## Step 8: Efficiency \[ \text{Sending end apparent power} = \sqrt{3} \times V_{S,LL} \times I_S \] First, compute sending end current (\( I_S \)): \[ I_S = A I_R + C V_R \] But since shunt admittance is small, \( I_S \approx I_R \), so: \[ \text{Sending end real power} = \sqrt{3} \times V_{S,LL} \times I_R \times \text{pf}_S \] Estimate \( pf_S \approx .82 \) (since losses are relatively small): \[ P_S \approx \sqrt{3} \times 47,646 \times 446.4 \times .82 = 28,374,000\ \text{W} = 28.37\ \text{MW} \] \[ \text{Efficiency} = \frac{P_{load}}{P_S} \times 100 = \frac{26}{28.37} \times 100 = 91.7\% \] **(c) Efficiency = \(\boxed{91.7\%}\)** --- # **Summary Table** | Quantity | Value | |-------------------------|-----------| | Sending End Voltage | 47.65 kV (L-L) | | Voltage Regulation | 16.2% | | Efficiency | 91.7% | --- ## **Assumptions and Approximations** - Calculations are per phase, then converted to line-to-line where needed. - Some small effects (e.g., shunt admittance at sending end) are neglected for simplicity. - If you need three-phase phasor diagrams or more detailed steps, let me know.