## Q5. Estimate the load angle of a swing curve over .2 seconds (sec) using the step-by-step method of Δt = .05 sec, considering a 50 Hz power system of the following parameters: - **Initial power (\(P_m\))** = 1 p.u. - **Maximum power (\(P_{max}\))** = 2 p.u. - **Inertia constant (\(H\))** = 5 MJ/MVA - **\(t_ = .1\) sec (fault starts)** - **\(t_f = .2\) sec (fault cleared)** --- ### **Step 1: Calculate the Swing Equation Constants and Initial Values** #### **a. Swing Equation Setup** The swing equation for a generator is: \[ \frac{d^2\delta}{dt^2} = \frac{\omega_s}{2H} (P_m - P_e) \] Where: - \(\delta\): Rotor angle (load angle) - \(\omega_s\): Synchronous speed (rad/sec) = \(2\pi f = 2\pi \times 50 = 314.16\) rad/sec - \(H\): Inertia constant (MJ/MVA) - \(P_m\): Mechanical power input (p.u.) - \(P_e\): Electrical power output (p.u.) - \(P_{max}\): Maximum power output (during fault, this may be reduced) #### **b. Per Unit System and Units** Since all powers are given in per unit and the inertia is in MJ/MVA, we use: \[ \frac{\omega_s}{2H} = \frac{314.16}{2 \times 5} = 31.416 \] #### **c. Initial Conditions** At \(t = \), before fault, assume \(\delta_ = \sin^{-1}(P_m / P_{max}) = \sin^{-1}(1/2) = 30^\circ = .5236\) rad. #### **d. Time Steps** - Total time = .2 sec - Step size = .05 sec - Number of steps = .2 / .05 = 4 steps (at , .05, .10, .15, .20 sec) **Explanation:** We have set up the swing equation and calculated all required constants for step-by-step solution. We have also determined the initial angle. --- ### **Step 2: Step-by-Step Solution Using the Step Method** #### **a. At Each Step** For each interval, calculate acceleration: \[ \alpha = \frac{\omega_s}{2H} (P_m - P_e) \] Update speed (first derivative of angle, \(\dot{\delta}\)) and then update angle (\(\delta\)) using: - \(\dot{\delta}_{n+1} = \dot{\delta}_n + \alpha_n \Delta t\) - \(\delta_{n+1} = \delta_n + \dot{\delta}_{n+1} \Delta t\) For the **fault period**, \(P_{max}\) may be less than 2 p.u., but since it's not specified to change, we use 2 p.u. #### **b. Step Calculations** | Step | t (sec) | \(\delta\) (rad) | \(\dot{\delta}\) (rad/s) | \(P_e = P_{max} \sin\delta\) | \(\alpha\) (rad/s\(^2\)) | |------|---------|------------------|--------------------------|-----------------------------|-------------------------| | | .00 | .5236 | | 1. | | | 1 | .05 | | | | | | 2 | .10 | | | | | | 3 | .15 | | | | | | 4 | .20 | | | | | **Let's do Step 1 (t = to t = .05 sec):** - \(P_e = 2 \sin(.5236) = 2 \times .5 = 1.\) - \(\alpha_ = 31.416 \times (1. - 1.) = \) - \(\dot{\delta}_1 = + \times .05 = \) - \(\delta_1 = .5236 + \times .05 = .5236\) **Now Step 2 (t = .05 to .10 sec):** - \(P_e = 2 \sin(.5236) = 1.\) - \(\alpha_1 = \) - \(\dot{\delta}_2 = \) - \(\delta_2 = .5236\) **Explanation:** After .1 seconds (at the start of the fault), no change in angle as \(P_m = P_e\). The system is initially at equilibrium. --- ### **Final Answer: Swing Curve Over .2 Seconds** After .1 seconds, **if a fault is applied and \(P_{max}\) is assumed unchanged (2 p.u.), there is no change in the load angle** as \(P_m = P_e\) at each step, so the rotor remains stable at the initial angle. #### **Summary Table** | Time (sec) | Load Angle (\(\delta\), rad) | |------------|------------------------------| | .00 | .5236 | | .05 | .5236 | | .10 | .5236 | | .15 | .5236 | | .20 | .5236 | **Final Answer:** **The load angle remains constant at \(.5236\) radians (30°) for each step over .2 seconds.** --- > **If you want to consider a reduced \(P_{max}\) during the fault, please provide the value, and I can recalculate!**