Let's solve the problem **step-by-step** using the **Bergeron method** for the two cases. --- ## **Step 1: Find Cable Parameters** ### **Given:** - \( L_c = 0.38 \, \mu H/m = 0.38 \times 10^{-6} \, H/m \) - \( C_c = 100 \, pF/m = 100 \times 10^{-12} \, F/m \) - Length of cable, \( l = 150 \, m \) - Overhead line surge impedance \( Z_L = 250 \, \Omega \) - Initial voltage step \( U_0 = 300 \, kV \) ### **Cable Surge Impedance (\( Z_c \)):** \[ Z_c = \sqrt{\frac{L_c}{C_c}} \] \[ Z_c = \sqrt{\frac{0.38 \times 10^{-6}}{100 \times 10^{-12}}} = \sqrt{\frac{0.38}{0.0001}} = \sqrt{3800} \approx 61.64 \, \Omega \] ### **Propagation Speed (\( v \)):** \[ v = \frac{1}{\sqrt{L_c C_c}} \] \[ v = \frac{1}{\sqrt{0.38 \times 10^{-6} \times 100 \times 10^{-12}}} = \frac{1}{\sqrt{0.38 \times 10^{-6} \times 10^{-10}}} = \frac{1}{\sqrt{0.38 \times 10^{-16}}} = \frac{1}{0.616 \times 10^{-8}} \approx 1.623 \times 10^{8} \, m/s \] ### **Time to Traverse Cable (\( t_{cable} \)):** \[ t_{cable} = \frac{l}{v} = \frac{150}{1.623 \times 10^8} \approx 0.924 \, \mu s \] --- ## **Step 2: Bergeron Method (Successive Reflections)** ### **Initial Wave Arrives at A** - The step \( U_0 \) travels down the overhead line and reaches point A (junction). - Reflection and transmission occur at junction A. #### **Voltage and Current at Point A (Junction):** Let: - \( Z_L \) = Surge impedance of overhead line (\( 250 \Omega \)) - \( Z_c \) = Surge impedance of cable (\( 61.64 \Omega \)) #### **Transmission Coefficient (\( T \)) from Line to Cable:** \[ T = \frac{2Z_c}{Z_L + Z_c} = \frac{2 \times 61.64}{250 + 61.64} = \frac{123.28}{311.64} \approx 0.396 \] #### **Reflection Coefficient (\( \Gamma \)) at Junction (from Line):** \[ \Gamma = \frac{Z_c - Z_L}{Z_c + Z_L} = \frac{61.64 - 250}{61.64 + 250} = \frac{-188.36}{311.64} \approx -0.605 \] ### **Step 3: Find Initial Voltages and Currents** #### **Transmitted Voltage into Cable (\( V_{A1} \)):** \[ V_{A1} = U_0 \cdot T = 300\,kV \times 0.396 \approx 118.8\,kV \] #### **Reflected Voltage Back into Line at A (\( V_{ref} \)):** \[ V_{ref} = U_0 \cdot \Gamma = 300\,kV \times (-0.605) \approx -181.5\,kV \] ### **Step 4: Propagation to Point B** - The wave travels down the cable to point B (takes \( t_{cable} \)). - **At B**, consider two cases: --- ## **Case 1: Point B Open-Circuited** **Open-circuit reflection coefficient (\( \Gamma_B \)) = +1 (full reflection, no current out).** ### **First Reflection at B:** - The incident voltage at B is \( V_{A1} \). - The reflected voltage is also \( V_{A1} \). #### **At B:** \[ \text{Total voltage at B after first arrival:} \quad V_{B1} = V_{A1} + V_{A1} = 2 \times 118.8 = 237.6 \, kV \] - This reflected wave travels back to A (takes \( t_{cable} \)), gets partially reflected and transmitted at A, and so on. ### **Summary Table for Voltages at A and B:** | Event | At Point | Time (\( \mu s \)) | Voltage (kV) | Comments | |-------|----------|--------------------|--------------|----------| | Initial | A | 0 | 118.8 | Transmission at A | | Arrival | B | 0.924 | 118.8 | First arrival | | Reflection | B | 0.924 | 237.6 | After open-circuit reflection | | Back to A | A | 1.848 | \( 237.6 \times \frac{2Z_L}{Z_L+Z_c} \) | Transmission back to line | | ... | ... | ... | ... | Continue for each round trip | ### **Volatile at B:** - At \( t = 0.924 \mu s \): \( V_B = 118.8 \) kV (first arrival) - At \( t = 0.924 \mu s^+ \): \( V_B = 237.6 \) kV (after reflection) - At \( t = 2 \times 0.924 = 1.848 \mu s \): next reflected wave arrives, etc. --- ## **Case 2: Surge Arrester at Point B** Given V-I characteristic: \[ V(i) = 200000 \cdot \sqrt[3]{i} \quad [V] \] - Unlike the open circuit, the reflection coefficient at B is **nonlinear** and must be computed for each step using the V-I curve. ### **Iterative Bergeron Approach:** 1. At each arrival, calculate the **incident current** (\( i_{inc} = V_{inc}/Z_c \)). 2. Use the V-I characteristic to find the **voltage at B** for the total current \( i = i_{inc} + i_{refl} \). 3. The reflected voltage is then \( V_{refl} = V_B - V_{inc} \). 4. Continue the iteration as above. --- ## **Step 5: Drawing the Diagrams** - **Voltage at A:** Step up to 118.8 kV at \( t = 0 \). Subsequent jumps at each round trip as reflected waves return. - **Voltage at B (Case 1):** Steps up to 118.8 kV at \( t = 0.924 \mu s \), then to 237.6 kV immediately after, and so on. - **Voltage at B (Case 2):** Rises to a value determined by current through the arrester (use V-I curve, iterative as described). --- ## **Summary of Key Results** ### **Cable Parameters:** - \( Z_c = 61.64 \, \Omega \) - \( v = 1.623 \times 10^8 \, m/s \) - \( t_{cable} = 0.924 \, \mu s \) ### **First Transmission at A:** - \( V_{A1} = 118.8 \, kV \) ### **First Arrival at B (Case 1):** - \( V_{B1} = 118.8 \, kV \) - After reflection, \( V_{B2} = 237.6 \, kV \) ### **First Arrival at B (Case 2):** - Solve \( V_B = 200000 \cdot \sqrt[3]{i} \), where \( i = \frac{V_{B1}}{Z_c} \) --- ### **To Draw the Variations:** - Plot voltage at A and B as step functions, with each step occurring at multiples of \( t_{cable} \). - For Case 2, use the arrester's V-I curve for each arrival. --- ## **If you need the full table of all reflections and more detailed diagrams, let me know!**