Let's break the problem into steps and solve it using the **equal area criterion**. ### **1. Problem Summary** - **Power transferred:** \(P = 500\, \text{MW}\) - **Voltage:** \(V = 275\, \text{kV}\) (line-to-line) - **Reactance per circuit:** \(X = 30\, \Omega\) - **Number of circuits:** 2 (operating in parallel) - **Event:** 3-phase short circuit on one circuit, then that circuit is isolated and all load is transferred to the healthy circuit. - **Load angle acceleration during fault:** \(\frac{d^2\delta}{dt^2}\) corresponds to \(\frac{d\delta}{dt} = 0.5^\circ/\text{ms} = 500^\circ/\text{s}\) (rate of change of \(\delta\)) - **Required:** Maximum duration of fault (\(t_{max}\)) before instability using the equal area criterion. --- ### **2. Calculate Pre-fault Power Transfer Capability (\(P_{max,pre}\))** Each circuit has \(X = 30\, \Omega\), so two in parallel: \[ X_{total,pre} = \frac{30}{2} = 15\, \Omega \] The **maximum power transfer** for a lossless line: \[ P_{max} = \frac{EV}{X} \] Assuming both ends at same voltage (\(E = V\)), for a per-phase calculation (since all values are per phase): \[ P_{max,pre} = \frac{V^2}{X_{total,pre}} \] But since the data is three-phase, and \(V\) is line-to-line RMS, the max three-phase power is: \[ P_{max,pre} = \frac{3 V^2}{X_{total,pre}} \] Plug in values (\(V = 275 \times 10^3\) V): \[ P_{max,pre} = \frac{3 \times (275 \times 10^3)^2}{15} \] \[ = \frac{3 \times 75.625 \times 10^{9}}{15} \] \[ = \frac{226.875 \times 10^9}{15} \] \[ = 15.125 \times 10^9\, \text{W} = 15,125\, \text{MW} \] But this is an **unrealistically high value**. Let's check the formula. #### **Correct formula for maximum three-phase power transfer:** \[ P_{max,3\phi} = \frac{3 V^2}{X} \] Where \(V\) is **phase-to-neutral** voltage: \[ V_{phase} = \frac{V_{LL}}{\sqrt{3}} = \frac{275,000}{\sqrt{3}} = 158,113\, \text{V} \] So, \[ P_{max,pre} = \frac{3 \times (158,113)^2}{15} \] \[ = \frac{3 \times 25 \times 10^9}{15} \] \[ = \frac{75 \times 10^9}{15} \] \[ = 5 \times 10^9\, \text{W} = 5,000\, \text{MW} \] Alternatively, since this is a common confusion, the general power equation is: \[ P = \frac{EV}{X}\sin\delta \] Assume \(E = V = 275\,\text{kV}\), per phase. \[ P_{max,pre} = \frac{275,000 \times 275,000}{15} = \frac{75,625,000,000}{15} = 5,041,666,667\, \text{W} = 5,042\, \text{MW} \] This matches our earlier result. --- ### **3. Calculate Initial Power Angle (\(\delta_0\))** Given the actual power transfer is \(P = 500\, \text{MW}\): \[ P = P_{max,pre} \sin\delta_0 = 5,042 \sin\delta_0 \] \[ \sin\delta_0 = \frac{500}{5,042} = 0.0992 \] \[ \delta_0 = \arcsin(0.0992) \approx 5.7^\circ \] --- ### **4. Power Transfer After Fault is Cleared (Single Circuit)** After fault clearing, only **one circuit** is in operation: \[ X_{total,post} = 30\, \Omega \] \[ P_{max,post} = \frac{275,000 \times 275,000}{30} = \frac{75,625,000,000}{30} = 2,520,833,333\, \text{W} = 2,521\, \text{MW} \] --- ### **5. Power Transfer During Fault (\(P_{max,fault}\))** During a **three-phase short circuit** on one circuit, the healthy circuit is still in operation, but the total power transferred to the load is only through the healthy circuit. However, the generator end sees the shorted circuit as a zero-impedance path, so effectively, the power transfer capability is **zero** during the fault. \[ P_{max,fault} = 0 \] --- ### **6. Equal Area Criterion Setup** #### **Areas:** - **A1:** Area "acceleration" (from \(\delta_0\) to \(\delta_c\)), when power output is zero (during fault), so the accelerating power is simply the mechanical input (\(P_m\)). - **A2:** Area "deceleration" (from \(\delta_c\) to \(\delta_{max}\)), when power output is via only one circuit. **For stability:** \(A_1 = A_2\) #### **Mechanical Power:** \[ P_m = 500\, \text{MW} \] #### **Power-Angle Curves:** Pre-fault: \(P_{max,pre} \sin\delta\) (not needed for area calculation) During fault: \(P_{max,fault} = 0\) Post-fault: \(P_{max,post} \sin\delta\) --- #### **Area \(A_1\) (Acceleration, from \(\delta_0\) to \(\delta_c\)):** \[ A_1 = \int_{\delta_0}^{\delta_c} [P_m - 0]\, d\delta = P_m (\delta_c - \delta_0) \] --- #### **Area \(A_2\) (Deceleration, from \(\delta_c\) to \(\delta_{max}\)):** \[ A_2 = \int_{\delta_c}^{\delta_{max}} [P_{max,post} \sin\delta - P_m]\, d\delta \] \[ = P_{max,post} [\cos\delta_c - \cos\delta_{max}] - P_m (\delta_{max} - \delta_c) \] --- #### **Equal Area Condition:** \[ A_1 = A_2 \] \[ P_m (\delta_c - \delta_0) = P_{max,post} [\cos\delta_c - \cos\delta_{max}] - P_m (\delta_{max} - \delta_c) \] \[ P_m (\delta_c - \delta_0) + P_m (\delta_{max} - \delta_c) = P_{max,post} [\cos\delta_c - \cos\delta_{max}] \] \[ P_m (\delta_{max} - \delta_0) = P_{max,post} [\cos\delta_c - \cos\delta_{max}] \] --- ### **7. Find \(\delta_{max}\) (Maximum Allowable Rotor Angle)** After fault clearing, the system is operating with one circuit: \[ P_{max,post} = 2,521\, \text{MW} \] \[ P_m = 500\, \text{MW} \] \[ P_m = P_{max,post} \sin\delta_{max} \] \[ \sin\delta_{max} = \frac{500}{2,521} = 0.1983 \] \[ \delta_{max} = \arcsin(0.1983) \approx 11.45^\circ \] But this is not the actual \(\delta_{max}\) for the equal area criterion; this is the new steady-state value. For stability, the actual maximum angle that can be tolerated is when the decelerating area equals the accelerating area. But at \(\delta_{max}\), the **post-fault power equals mechanical power**, but the system could swing up to a much larger angle before returning, so we need to find \(\delta_{max}\) such that: \[ P_{max,post} \sin\delta_{max} = P_m \] \[ \sin\delta_{max} = \frac{500}{2521} = 0.1983 \] \[ \delta_{max,ss} = 11.45^\circ \] But the actual maximum swing is up to \(180^\circ - \delta_{max,ss}\). However, **for a post-fault power curve**, the maximum possible swing is up to the first peak of the curve, i.e., up to \(\delta = 180^\circ\). --- ### **8. Substitute Known Values and Solve for \(\delta_c\)** Recall: \[ P_m (\delta_{max} - \delta_0) = P_{max,post} [\cos\delta_c - \cos\delta_{max}] \] We know: - \(P_m = 500\) - \(P_{max,post} = 2,521\) - \(\delta_0 = 5.7^\circ = 0.0995\, \text{rad}\) - \(\delta_{max,ss} = 11.45^\circ = 0.1998\, \text{rad}\) - For a one-machine-infinite-bus system, the critical clearing angle \(\delta_c\) is less than \(\delta_{max,ss}\). But we need to solve for \(\delta_c\). \[ 500 (\delta_{max} - 0.0995) = 2521 [\cos\delta_c - \cos\delta_{max}] \] Let's try \(\delta_{max} = \pi - \arcsin(P_m / P_{max,post})\): \[ \delta_{max} = \pi - \arcsin(0.1983) = 180^\circ - 11.45^\circ = 168.55^\circ = 2.943\, \text{rad} \] Now calculate: \[ 500 (2.943 - 0.0995) = 2521 [\cos\delta_c - \cos(2.943)] \] \[ 500 \times 2.8435 = 2521 [\cos\delta_c - \cos(2.943)] \] \[ 1421.75 = 2521 [\cos\delta_c - \cos(2.943)] \] \[ \cos(2.943) = -0.9785 \] \[ \cos\delta_c - (-0.9785) = \frac{1421.75}{2521} \] \[ \cos\delta_c = \frac{1421.75}{2521} - (-0.9785) = 0.5642 + 0.9785 = 1.5427 \] This is impossible since \(\cos\delta_c\) cannot be greater than 1. Clearly, we need to use the correct value for \(\delta_{max}\): **\(\delta_{max}\) is not \(\pi - \arcsin(P_m/P_{max,post})\), but rather the angle where the area for deceleration is maximized, which is at \(\pi\) (180°), i.e., the first peak of the power-angle curve.** So, \[ \delta_{max} = \pi = 3.142\, \text{rad} \] \[ \cos(\pi) = -1 \] Now, \[ 500 (\pi - 0.0995) = 2521 [\cos\delta_c - (-1)] \] \[ 500 \times (3.142 - 0.0995) = 2521 (\cos\delta_c + 1) \] \[ 500 \times 3.0425 = 2521 (\cos\delta_c + 1) \] \[ 1521.25 = 2521 (\cos\delta_c + 1) \] \[ \cos\delta_c + 1 = \frac{1521.25}{2521} = 0.6036 \] \[ \cos\delta_c = 0.6036 - 1 = -0.3964 \] \[ \delta_c = \arccos(-0.3964) = 113.4^\circ = 1.98\, \text{rad} \] --- ### **9. Find Maximum Duration of Fault (\(t_{max}\))** The load angle increases from \(\delta_0\) to \(\delta_c\) during the fault. The rate of increase is \(0.5^\circ/\text{ms}\). \[ \Delta\delta = \delta_c -

Let's carefully solve this problem step-by-step using the equal area criterion. --- ## **Problem Summary** - Power transfer, \(P = 500\, \text{MW}\) - Line-to-line voltage, \(V_{LL} = 275\, \text{kV}\) - Per-phase voltage, \(V_{ph} = \frac{V_{LL}}{\sqrt{3}} \approx 158.7\, \text{kV}\) - Reactance per circuit, \(X = 30\, \Omega\) - Number of circuits: 2 (parallel) - During a 3-phase short circuit on one circuit: - That circuit trips and is isolated. - All load is transferred to the remaining circuit. - The rotor angle increases at a rate of **0.5°/ms** during the short circuit. - Objective: Find the **maximum duration** of the short circuit before system instability occurs. --- ## **Step 1: Determine pre-fault power transfer capability** ### **a) Total reactance after parallel operation:** \[ X_{pre} = \frac{X}{2} = \frac{30\, \Omega}{2} = 15\, \Omega \] ### **b) Max power transfer (per phase):** \[ P_{max} = \frac{3 V_{ph}^2}{X} \] Calculate: \[ V_{ph} = \frac{275,000}{\sqrt{3}} \approx 158,700\, \text{V} \] \[ P_{max} = \frac{3 \times (158,700)^2}{15} \] \[ = \frac{3 \times 25.17 \times 10^{9}}{15} = \frac{75.51 \times 10^{9}}{15} \approx 5.03 \times 10^{9}\, \text{W} \] or approximately **5030 MW**. --- ## **Step 2: Determine initial load angle \(\delta_0\)** Given the operating power \(P = 500\, \text{MW}\): \[ P = P_{max} \sin \delta_0 \] \[ \sin \delta_0 = \frac{P}{P_{max}} = \frac{500}{5030} \approx 0.0994 \] \[ \delta_0 \approx \arcsin(0.0994) \approx 5.7^\circ \] --- ## **Step 3: Power capability after one circuit trips** After the short circuit clears and only **one circuit** operates: \[ X_{post} = 30\, \Omega \] \[ P_{max,post} = \frac{3 V_{ph}^2}{X} = \frac{75.51 \times 10^{9}}{30} \approx 2.52 \times 10^9\, \text{W} \] or **2520 MW**. The **steady-state load angle** after the system stabilizes: \[ \sin \delta_{ss} = \frac{P}{P_{max,post}} = \frac{500}{2520} \approx 0.1984 \] \[ \delta_{ss} \approx \arcsin(0.1984) \approx 11.45^\circ \] --- ## **Step 4: Behavior during the fault** - **Before fault:** \(\delta_0 \approx 5.7^\circ\). - **During fault:** No power is transferred because one circuit is shorted; the generator accelerates freely at the mechanical power \(P = 500\, \text{MW}\). - The rotor accelerates, increasing \(\delta\) at a rate of **0.5°/ms**. --- ## **Step 5: Equal area criterion** The system remains stable if the rotor angle does **not** reach the **unstable swing** point, which corresponds roughly to the **critical clearing angle** \(\delta_c\). The stability condition: \[ \text{Accelerating area} = \text{Decelerating area} \] - **Accelerating area (A1):** from \(\delta_0\) to \(\delta_c\): \[ A_1 = P_m (\delta_c - \delta_0) \] where \(P_m = 500\, \text{MW}\). - **Decelerating area (A2):** from \(\delta_c\) to \(\delta_{max}\): \[ A_2 = P_{max,post} \left[\cos \delta_c - \cos \delta_{max}\right] - P_m (\delta_{max} - \delta_c) \] The maximum swing angle \(\delta_{max}\) where the rotor could go if the system is marginally stable is approximately where the power angle curve peaks, i.e., \(\delta \approx 180^\circ\). --- ## **Step 6: Find \(\delta_{max}\)** The **critical swing angle**: \[ \delta_{max} \approx 180^\circ = \pi\, \text{rad} \] Calculate: \[ \cos \pi = -1 \] To find the **critical clearing angle** \(\delta_c\), set: \[ P_m (\delta_c - \delta_0) = P_{max,post} [\cos \delta_c - \cos \delta_{max}] - P_m (\delta_{max} - \delta_c) \] Rearranged: \[ P_m (\delta_{max} - \delta_0) = P_{max,post} [\cos \delta_c - \cos \delta_{max}] \] Plugging values: \[ P_m = 500\, \text{MW} \] \[ P_{max,post} = 2520\, \text{MW} \] \[ \delta_0 = 5.7^\circ = 0.0995\, \text{rad} \] \[ \delta_{max} \approx \pi\, \text{rad} = 3.1416\, \text{rad} \] Compute: \[ \boxed{ 500 \times (3.1416 - 0.0995) = 2520 [\cos \delta_c - (-1)] } \] \[ 500 \times 3.0421 \approx 2520 (\cos \delta_c + 1) \] \[ 1521.05 = 2520 (\cos \delta_c + 1) \] Solve for \(\cos \delta_c\): \[ \cos \delta_c + 1 = \frac{1521.05}{2520} \approx 0.6036 \] \[ \cos \delta_c = 0.6036 - 1 = -0.3964 \] Find: \[ \delta_c = \arccos(-0.3964) \approx 113.4^\circ \approx 1.98\, \text{rad} \] --- ## **Step 7: Calculate the maximum fault clearing time** Since the rotor accelerates at **0.5°/ms**, the **time to reach \(\delta_c\)**: \[ \Delta \delta = \delta_c - \delta_0 = 113.4^\circ - 5.7^\circ = 107.7^\circ \] Convert to radians: \[ 107.7^\circ \times \frac{\pi}{180} \approx 1.88\, \text{rad} \] The rate of increase of rotor angle: \[ \frac{d\delta}{dt} = 0.5^\circ/\text{ms} \approx 0.0087\, \text{rad/ms} \] Time: \[ t_{max} = \frac{\Delta \delta}{d\delta/dt} = \frac{1.88\, \text{rad}}{0.0087\, \text{rad/ms}} \approx 216\, \text{ms} \] --- ## **## **Final Answer:** ### **The maximum duration of the three-phase short circuit before system instability occurs is approximately:** \[ \boxed{ \boxed{ \textbf{216 milliseconds} } } \] --- ## **Summary:** - The rotor angle can increase by about 107.7° before instability. - At an acceleration rate of 0.5°/ms, this corresponds to roughly **216 ms**. - Beyond this duration, the system becomes unstable per the equal area criterion. --- **Note:** In real systems, additional factors such as damping, system inertia, and protections influence actual stability margins, but this provides a theoretical estimate based on the equal area criterion.