Let's solve the problem step by step using the **Bergeron Method** for each case. --- ### **Given:** - Cable parameters: \( L_c = 0.38 \,\mu\text{H/m} \), \( C_c = 100 \,\text{pF/m} \), \( R_c = G_c = 0 \) - Length of cable, \( l = 150 \,\text{m} \) - Overhead line surge impedance, \( Z_L = 250 \,\Omega \) - Step voltage, \( U_0 = 300\,\text{kV} \) - Time interval: 0 to 5 μs - **Point A**: Connection between overhead line and cable - **Point B**: End of cable --- ## **Step 1: Calculate Cable Surge Impedance and Velocity** ### **1.1 Surge Impedance of the Cable, \( Z_c \):** \[ Z_c = \sqrt{\frac{L_c}{C_c}} \] Substitute: - \( L_c = 0.38 \times 10^{-6} \,\text{H/m} \) - \( C_c = 100 \times 10^{-12} \,\text{F/m} \) \[ Z_c = \sqrt{\frac{0.38 \times 10^{-6}}{100 \times 10^{-12}}} = \sqrt{\frac{0.38}{0.0001}} = \sqrt{3800} \approx 61.6\,\Omega \] --- ### **1.2 Propagation Velocity in the Cable, \( v \):** \[ v = \frac{1}{\sqrt{L_c C_c}} \] \[ v = \frac{1}{\sqrt{0.38 \times 10^{-6} \times 100 \times 10^{-12}}} = \frac{1}{\sqrt{0.38 \times 10^{-16}}} = \frac{1}{\sqrt{3.8 \times 10^{-17}}} \] \[ \sqrt{3.8 \times 10^{-17}} \approx 6.16 \times 10^{-9} \] \[ v \approx \frac{1}{6.16 \times 10^{-9}} \approx 1.62 \times 10^8\,\text{m/s} \] --- ### **1.3 Time for Wave to Travel Cable Length (\(\tau\)):** \[ t = \frac{l}{v} = \frac{150}{1.62 \times 10^8} \approx 0.926 \,\mu\text{s} \] --- ## **Step 2: Bergeron Method for Reflections** ### **2.1 Initial Step at \( t = 0 \) (\*incoming at A\*)** - Incident voltage at A: \( U_0 = 300\,\text{kV} \) - Overhead line to cable transition. #### **Reflection and Transmission Coefficients at A:** \[ \text{Reflection at A:} \quad \Gamma_A = \frac{Z_c - Z_L}{Z_c + Z_L} \] \[ \Gamma_A = \frac{61.6 - 250}{61.6 + 250} = \frac{-188.4}{311.6} \approx -0.605 \] \[ \text{Transmission at A:} \quad T_A = 1 + \Gamma_A = 0.395 \] #### **Transmitted Voltage into the Cable:** \[ V_{\text{trans}} = U_0 \cdot T_A = 300\,\text{kV} \times 0.395 \approx 118.5\,\text{kV} \] --- ### **2.2 Wave Travels to B (\( t = 0.926\,\mu\text{s} \))** #### **Case 1: B is open-circuited (\( Z_B = \infty \))** - Reflection coefficient at B: \[ \Gamma_B = \frac{Z_B - Z_c}{Z_B + Z_c} = \frac{\infty - Z_c}{\infty + Z_c} = 1 \] - Voltage doubles at B upon reflection (since open end). - Reflected voltage at B: \( V_{\text{ref, B}} = V_{\text{incident, B}} \times \Gamma_B = 118.5\,\text{kV} \times 1 = 118.5\,\text{kV} \) - **Voltage at B just after arrival:** \[ V_B = V_{\text{incident, B}} + V_{\text{ref, B}} = 118.5 + 118.5 = 237\,\text{kV} \] --- ### **2.3 Wave Returns to A (\( t = 2 \times 0.926 = 1.85\,\mu\text{s} \))** - Back at A, reflected again (use \( \Gamma_A \) for returning wave): \[ V_{\text{ref, A}} = V_{\text{incident, A}} \times \Gamma_A = 118.5\,\text{kV} \times (-0.605) \approx -71.7\,\text{kV} \] - This negative wave travels towards B again. --- ### **2.4 Continue the Process (Bergeron Table)** You repeat the process for each round-trip, summing incident and reflected voltages at each end at each time step, until you reach 5 μs. --- ### **Step 3: Summary Table for Case 1 (Open End B)** | Event | Time (μs) | At A (kV) | At B (kV) | Comments | |---------------------------------|------------|-----------|-----------|------------------| | Initial at A | 0 | 300 | | Step applied | | After transmission to cable | 0 | 118.5 | | Through T_A | | Arrival at B | 0.93 | | 237 | Open end doubles | | Reflection returns to A | 1.85 | 46.8 | | -71.7 kV added | | 2nd reflection at B | 2.78 | | 94.2 | | | 3rd at A | 3.7 | 74.1 | | | | ... | ... | ... | ... | Continue cycle | --- ## **Case 2: Surge Arrester at B** The surge arrester at B provides a nonlinear boundary. The voltage-current relationship: \[ V(i) = 200,000 \cdot \sqrt[30]{i} \] The reflection at B is not full; you would need to **match the incident and reflected waves to the nonlinear characteristic** at each time step. Typically, you'd proceed iteratively: 1. At each arrival of a wave at B, compute the total voltage as the sum of incident and reflected waves. 2. The total current through the arrester is \( i = \frac{V_{\text{B}}}{Z_c} \). 3. Use the V-I characteristic to match the reflected wave voltage such that the sum matches the non-linear V-I relation. --- ### **Step-by-Step Reflection Process at B (with Arrester):** - At each time a wave arrives at B: 1. Let \( V_{\text{incident}} \) be the incoming voltage. 2. Assume reflected voltage \( V_{\text{ref}} \). 3. Total voltage at B: \( V_B = V_{\text{incident}} + V_{\text{ref}} \). 4. Current into arrester: \( i = \frac{V_{\text{incident}} - V_{\text{ref}}}{Z_c} \). 5. Must satisfy \( V_B = 200,000 \cdot \sqrt[30]{i} \). 6. Solve for \( V_{\text{ref}} \) iteratively. --- ## **Final Answers and Plots** ### **Case 1: Open-Circuit at B** - Voltage at A drops stepwise with each reflection. - Voltage at B steps up and approaches a steady value with each reflection. **Key values:** - At \( t = 0 \): \( V_A = 118.5\,\text{kV} \), \( V_B = 0 \) - At \( t = 0.93 \,\mu\text{s} \): \( V_B = 237\,\text{kV} \) - At \( t = 1.85 \,\mu\text{s} \): \( V_A = 46.8\,\text{kV} \) - At \( t = 2.78 \,\mu\text{s} \): \( V_B = 94.2\,\text{kV} \) - Continue until 5 μs. ### **Case 2: Arrester at B** - At each arrival, solve the nonlinear equation for \( V_{\text{ref}} \) at B using the Bergeron method. - The voltage at B will be **clamped** by the arrester, so it will not exceed a certain value (much less than open-circuit case). ### **Plots** - \( V_A(t) \) and \( V_B(t) \) will be stepwise for case 1, and more rounded/limited for case 2 due to the arrester. - Current at A and B can be found by \( I = V/Z \) at each point. --- ## **Summary Table** | Time (μs) | \( V_A \) (kV) | \( V_B \) (kV) | Notes (Case 1) | |-----------|----------------|----------------|----------------| | 0 | 118.5 | 0 | Initial | | 0.93 | | 237 | 1st at B | | 1.85 | 46.8 | | 2nd at A | | 2.78 | | 94.2 | 2nd at B | | 3.7 | 74.1 | | 3rd at A | | ... | ... | ... | ... | For **Case 2**, you would similarly fill in the table, but use the nonlinear boundary condition at B. --- ## **Conclusion** - **Case 1:** Voltage at B approaches twice the transmitted voltage due to full reflection at open end. - **Case 2:** Voltage at B is limited by the arrester, and you must solve the nonlinear equation at each reflection. If you want to see the full step-by-step **Bergeron Table (with iterative calculations for Case 2)** or the **plots**, let me know!

Certainly! Let's proceed with a detailed, step-by-step solution for both cases using the Bergeron method, including all calculations and reasoning. --- ## **Given Data Recap** | Parameter | Value | |--------------|---------| | Line inductance per meter, \(L_c\) | 0.38 μH/m = \(0.38 \times 10^{-6}\) H/m | | Line capacitance per meter, \(C_c\) | 100 pF/m = \(100 \times 10^{-12}\) F/m | | Line length, \(l\) | 150 m | | Overhead line impedance, \(Z_L\) | 250 Ω | | Voltage source, \(U_s\) | 300 kV | | V-I characteristic of arrester: \(V(i) = 200,000 \times \sqrt[30]{i}\) (V) | | | Time interval for analysis | 0 – 5 μs | --- ## **Step 1: Calculate the line parameters** ### 1.1 Surge impedance, \( Z_c \): \[ Z_c = \sqrt{\frac{L_c}{C_c}} = \sqrt{\frac{0.38 \times 10^{-6}}{100 \times 10^{-12}}} = \sqrt{3800} \approx 61.6\, \Omega \] ### 1.2 Propagation velocity in the cable: \[ v = \frac{1}{\sqrt{L_c C_c}} = \frac{1}{\sqrt{0.38 \times 10^{-6} \times 100 \times 10^{-12}}} \] Calculate denominator: \[ \sqrt{0.38 \times 10^{-6} \times 100 \times 10^{-12}} = \sqrt{38 \times 10^{-18}} \approx 6.16 \times 10^{-9} \] Thus, \[ v \approx \frac{1}{6.16 \times 10^{-9}} \approx 1.62 \times 10^{8} \text{ m/s} \] ### 1.3 Time for wave to travel length \(l\): \[ t_{prop} = \frac{l}{v} = \frac{150}{1.62 \times 10^{8}} \approx 0.926\, \mu s \] --- ## **Step 2: Initial incident wave** - The initial voltage wave incident at point A: \[ U_{incident} = 300\,kV \] - Transmission coefficient from the overhead line to the cable: \[ T_A = \frac{2 Z_c}{Z_c + Z_L} \] But in the Bergeron method, the voltage transmitted into the cable (assuming incident wave) is: \[ V_{trans} = U_s \times T = U_s \times \frac{2 Z_c}{Z_c + Z_L} \] Alternatively, since the incident voltage source is 300 kV and the line is being excited with a step, the initial incident wave arriving at A is: \[ U_{incident} = 300\,kV \] The *reflection coefficient at A* (overhead line to cable): \[ \Gamma_A = \frac{Z_c - Z_L}{Z_c + Z_L} = \frac{61.6 - 250}{61.6 + 250} \approx \frac{-188.4}{311.6} \approx -0.605 \] The *transmission coefficient* into the cable: \[ T_A = 1 + \Gamma_A = 0.395 \] The *initial transmitted voltage* into the cable: \[ V_{0} = U_s \times T_A = 300\,kV \times 0.395 \approx 118.5\,kV \] --- ## **Step 3: Reflection at the end B** ### Case 1: B is **open-circuited** \[ \Gamma_B = 1 \] The wave reaching B reflects fully with same polarity: \[ V_{reflected at B} = V_{incident} \times \Gamma_B = 118.5\,kV \times 1 = 118.5\,kV \] ### Wave at B after wave arrives: \[ V_B(t_{arrival}) = V_{incident} + V_{reflected} = 118.5\,kV + 118.5\,kV = 237\,kV \] ### Wave returning to A: Time for wave to go from A to B and back: \[ t_{roundtrip} = 2 \times 0.926\, \mu s \approx 1.852\, \mu s \] ### Reflection at A: \[ \Gamma_A = -0.605 \] Reflected wave at A: \[ V_{ref, A} = V_{incident} \times \Gamma_A = 118.5\,kV \times -0.605 \approx -71.7\,kV \] This negative wave travels toward B, arriving at: \[ t \approx 1.852\, \mu s \] At B, the total voltage: \[ V_B = V_{incident} + V_{ref, B} \quad \text{(from previous wave)} \] But now, the incident wave at B is: \[ V_{incident, B} = -71.7\,kV \] Since the wave is traveling back, the total at B: \[ V_B(t_{2nd}) = V_{incident, B} + V_{ref, B} = -71.7\,kV + V_{ref, B} \] But the boundary condition at B (open circuit): \[ \Gamma_B = 1 \] which means the wave reflects with same polarity, so: \[ V_{ref, B} = V_{incident, B} \times 1 = -71.7\,kV \] Total voltage at B: \[ V_B = -71.7\,kV + (-71.7\,kV) = -143.4\,kV \] This indicates the voltage oscillates with decreasing amplitude if losses were included, but since R=G=0, the reflections continue with the same magnitude, leading to a standing wave pattern. --- ## **Summary of Case 1: Voltage and Current at A and B** | Time (μs) | Event | \(V_A\) (kV) | \(V_B\) (kV) | Comments | |------------|--------|--------------|--------------|----------| | 0 | Initial incident | 118.5 | 0 | Incident wave enters cable | | 0.93 | Arrival at B | 118.5 | 237 | Full reflection at open end | | 1.85 | Reflection at B reaches A | -71.7 | | Wave reflected back to A | | 2.78 | Reflection at A reaches B | | -143.4 | Another reflection at B | | 3.7 | Next cycle | ... | ... | Continue similarly | **Currents**: \[ I = \frac{V}{Z_c} \] At each point, for example, when \(V_B = 237\,kV\): \[ I_B = \frac{237 \times 10^3}{61.6} \approx 3841\,mA \] --- ## **Case 2: Surge arrester at B** The main difference is the nonlinear V–I characteristic: \[ V(i) = 200,000 \times \sqrt[30]{i} \] ### **Method to handle nonlinear boundary:** At each wave arrival at B: 1. **Estimate the total voltage \(V_B\)** as the sum of incident and reflected waves. 2. **Calculate the current:** \[ i = \frac{V_{incident} - V_{reflected}}{Z_c} \] 3. **Apply the arrester characteristic:** \[ V_B = 200,000 \times \sqrt[30]{i} \] 4. **Solve for \(V_{reflected}\)**: \[ V_{reflected} = V_{incident} - i Z_c \] But since \(V_B\) depends on \(i\), and \(i\) depends on \(V_{reflected}\), **we need to solve this iteratively**. --- ## **Step-by-step calculation at first wave arrival** - Incident wave at B: \[ V_{incident} = 118.5\,kV \] - Assume initial \(V_{reflected} = V_{ref, initial} = a\) - Total voltage at B: \[ V_B = V_{incident} + V_{reflected} \] - Current: \[ i = \frac{V_{incident} - V_{reflected}}{Z_c} \] - From the arrester characteristic: \[ V_B = 200,000 \times \sqrt[30]{i} \] - Combining: \[ V_{incident} + V_{reflected} = 200,000 \times \sqrt[30]{i} \] - Also, \[ i = \frac{V_{incident} - V_{reflected}}{61.6} \] - Substitute: \[ V_{incident} + V_{reflected} = 200,000 \times \left( \frac{V_{incident} - V_{reflected}}{61.6} \right)^{1/30} \] This is a nonlinear equation in \(V_{reflected}\), which can be solved iteratively or numerically. --- ## **Numerical solution idea (approximate)** - For small initial \(V_{reflected}\), initial guess: \[ V_{reflected} \approx 0 \] - Then, \[ V_B \approx 118.5\,kV \] - Current: \[ i \approx \frac{118.5\,kV - 0}{61.6} \approx 1924\,A \] - Using the arrester: \[ V_{arrest} = 200,000 \times \sqrt[30]{1924} \] Calculate: \[ \sqrt[30]{1924} = e^{\frac{1}{30} \ln 1924} \approx e^{\frac{1}{30} \times 7.56} \approx e^{0.252} \approx 1.286 \] Hence, \[ V_{arrest} \approx 200,000 \times 1.286 \approx 257,200\,V \approx 257\,kV \] Compare with initial \(V_B=118.5\,kV\). Since \(V_{arrest} > V_B\), the actual voltage at B will increase until the nonlinear relation is satisfied. **Iterate:** - New \(V_B \approx 257\,kV\) - New current: \[ i = \frac{118.5\,kV - V_{reflected}}{61.6} \] - Since \(V_B = V_{incident} + V_{reflected}\): \[ V_{reflected} = V_B - V_{incident} = 257\,kV - 118.5\,kV = 138.5\,kV \] - Now, \[ i = \frac{118.5\,kV - 138.5\,kV}{61.6} = \frac{-20\,kV}{61.6} \approx -324\,A \] Negative current indicates the wave is reversing direction (possible in the nonlinear case). The magnitude is small, so the process converges. In reality, the arrester's V–I characteristic limits the voltage at B to approximately 200–250 kV, preventing the voltage from rising indefinitely. --- ## **Final approximation for Case 2 (at first wave):** - Voltage at B: approximately **200–250 kV**, limited by the arrester. - Reflection coefficient is effectively **less than 1**, because part of the wave energy is dissipated or absorbed by the arrester. --- ## **Summary** | Case | Reflection coefficient at B | Voltage at B (approx.) | Notes | |---------|----------------------------|------------------------|--------| | **Open circuit** | \(+1\) | approaches ~300 kV (initial step), oscillates | Full reflection, standing wave | | **Arrester** | \(< 1\), nonlinear | limited around 200–250 kV | Nonlinear boundary absorbs part of energy | --- ## **Final Plots** - **Case 1:** - Voltage at A: stepwise decreasing with reflections, oscillations with amplitude diminishing if losses exist. - Voltage at B: oscillates between 0 and 237 kV, with reflections doubling the voltage at each bounce. - **Case 2:** - Voltage at B is capped at ~200–250 kV. - Voltage at A shows similar stepwise pattern but with diminished oscillations due to energy absorption. --- ## **Conclusion** - The **Bergeron method** involves tracking incident and reflected waves, applying boundary conditions at each interface. - For **Case 1**, reflections are full, leading to standing waves. - For **Case 2**, the nonlinear arrester limits the voltage, damping reflections, and preventing voltage from reaching dangerous levels. --- Let me know if you'd like the detailed iterative calculations for the nonlinear boundary or specific plots!