Power System‑II (EX-601) - Unit 5 Short Notes

UNIT 5: Power System-II - Comprehensive Exam-Focused Notes

I. Load Flow Analysis (Power Flow Studies)

Significance and Applications

• Purpose: Determines the steady-state operating condition (voltage magnitudes, angles, real/reactive power flows) of a power system for a given load and generation schedule.

• Applications:

  • Planning: System expansion, feasibility studies.

  • Operation: Real-time monitoring, contingency analysis (N-1), optimal power flow.

  • Optimization: Loss minimization, voltage profile improvement.

Bus Classification

Buses are classified based on which two quantities are specified (known) and which two are to be solved (unknown). For each bus, the knowns are: $$\displaystyle P_{spec} $$, $$\displaystyle Q_{spec} $$, $$\displaystyle |V|_{spec} $$, $$\displaystyle \delta_{spec} $$.

[!TIP] Justification: The classification ensures a well-posed mathematical problem. The slack bus provides a reference for angles and absorbs losses. PV buses model generators with AVRs. Most load buses are PQ type.

Formation of Admittance Matrix ($$\displaystyle Y_{bus} $$)

$$\displaystyle Y_{bus} $$ is a sparse, symmetric matrix where diagonal element $$\displaystyle Y_{ii} $$ is the sum of all admittances connected to bus $i$, and off-diagonal $$\displaystyle Y_{ij} = -y_{ij} $$ (negative of line admittance between $i$ and $j$).

Step-by-Step Procedure (Neglecting Shunt Capacitance):

1. Initialize all $$\displaystyle Y_{ii} = 0 $$, $$\displaystyle Y_{ij} = 0 $$.

2. For each line between bus $i$ and $j$ with impedance $$\displaystyle z_{ij} = r_{ij} + jx_{ij} $$:

   • $$\displaystyle y_{ij} = 1 / z_{ij} $$

   • $$\displaystyle Y_{ii} \leftarrow Y_{ii} + y_{ij} $$

   • $$\displaystyle Y_{jj} \leftarrow Y_{jj} + y_{ij} $$

   • $$\displaystyle Y_{ij} \leftarrow Y_{ij} - y_{ij} $$

   • $$\displaystyle Y_{ji} \leftarrow Y_{ji} - y_{ij} $$ (ensures symmetry)

3. If transformer tap ratios exist, incorporate them into the off-diagonal elements.

Example (4-Bus System from Jun 2025):

Given line impedances (p.u.), neglecting shunt capacitance.

• $$\displaystyle Y_{11} = 1/(0.25+j0.1) + 1/(0.20+j0.8) + 1/(0.30+j1.2) $$

• $$\displaystyle Y_{12} = -1/(0.25+j0.1) $$

• $$\displaystyle Y_{13} = -1/(0.20+j0.8) $$

• $$\displaystyle Y_{14} = -1/(0.30+j1.2) $$

• $$\displaystyle Y_{22} = Y_{11} $$'s contribution from line 1-2 + $1/(0.20+j0.8)$ (line 2-3)

• ... and so on.

Solution Methods

1. Gauss-Seidel (GS) Method

• Principle: Iterative solution of nodal equations $$\displaystyle I_i = Y_{ii}V_i + \sum_{j \ne i} Y_{ij}V_j $$.

• Iteration for PQ Bus $i$:

$$V_i^{(k+1)} = \frac{1}{Y_{ii}} \left( \frac{P_i - jQ_i}{(V_i^{(k)})^*} - \sum_{j \ne i} Y_{ij} V_j^{(k+1)} \right)$$

(Uses latest voltages for already-updated buses in the same iteration).

• Modification for PV Bus:

  1. Calculate $$\displaystyle V_i^{(k+1)} $$ using above equation.

  2. Compute resulting reactive power: $$\displaystyle Q_i^{(k+1)} = -\text{Im}\left[ (V_i^{(k+1)})^* \left( Y_{ii}V_i^{(k+1)} + \sum_{j \ne i} Y_{ij} V_j \right) \right] $$.

  3. If $$\displaystyle Q_i^{(k+1)} $$ violates limits, convert bus to PQ type with $$\displaystyle Q_i = Q_{limit} $$ and fix $$\displaystyle |V_i| $$ at specified value for subsequent iterations.

• Flow Chart (PQ Buses only):

  1. Initialize all bus voltages (flat start: $$\displaystyle |V_i|=1.0 $$, $$\displaystyle \delta_i=0 $$).

  2. For each PQ bus $i$, compute new $$\displaystyle V_i $$ using GS equation.

  3. Check convergence: $$\displaystyle | \Delta V_i | < \epsilon $$ for all buses.

  4. If not converged, go to step 2.

• Advantages: Simple, low memory per iteration.

• Disadvantages: Slow convergence (linear), may diverge for large systems.

2. Newton-Raphson (NR) Method

• Principle: Solves nonlinear power equations using Taylor series and Jacobian matrix. Quadratic convergence near solution.

• Rectangular Coordinates (Voltage $$\displaystyle V_i = e_i + jf_i $$):

  Mismatches: $$\displaystyle \Delta P_i $$, $$\displaystyle \Delta Q_i $$.

  Unknowns: $$\displaystyle \Delta e_i $$, $$\displaystyle \Delta f_i $$.

  Jacobian: $$\displaystyle J = \begin{bmatrix} \frac{\partial P}{\partial e} & \frac{\partial P}{\partial f} \\ \frac{\partial Q}{\partial e} & \frac{\partial Q}{\partial f} \end{bmatrix} $$

  Solve: $$\displaystyle \begin{bmatrix} \Delta P \\ \Delta Q \end{bmatrix} = J \begin{bmatrix} \Delta e \\ \Delta f \end{bmatrix} $$

• Polar Coordinates (Voltage $$\displaystyle V_i = |V_i| \angle \delta_i $$): More common.

  Mismatches: $$\displaystyle \Delta P_i $$, $$\displaystyle \Delta Q_i $$.

  Unknowns: $$\displaystyle \Delta \delta_i $$, $$\displaystyle \Delta |V_i| $$.

  Jacobian: $$\displaystyle J = \begin{bmatrix} \frac{\partial P}{\partial \delta} & \frac{\partial P}{\partial |V|} \\ \frac{\partial Q}{\partial \delta} & \frac{\partial Q}{\partial |V|} \end{bmatrix} $$

  • $$\displaystyle \frac{\partial P_i}{\partial \delta_i} = |V_i| \sum_{j=1}^{n} |V_j| Y_{ij} \sin(\delta_i - \delta_j) $$ (diagonal)

  • $$\displaystyle \frac{\partial P_i}{\partial \delta_j} = -|V_i||V_j| Y_{ij} \sin(\delta_i - \delta_j) $$ (off-diagonal, $j \ne i$)

  • $$\displaystyle \frac{\partial P_i}{\partial |V_i|} = 2|V_i|Y_{ii}\cos\phi_{ii} + \sum_{j \ne i} |V_j| Y_{ij} \cos(\delta_i - \delta_j) $$

  • $$\displaystyle \frac{\partial Q_i}{\partial \delta_i} = |V_i| \sum_{j=1}^{n} |V_j| Y_{ij} \cos(\delta_i - \delta_j) $$

  • $$\displaystyle \frac{\partial Q_i}{\partial |V_i|} = 2|V_i|Y_{ii}\sin\phi_{ii} + \sum_{j \ne i} |V_j| Y_{ij} \sin(\delta_i - \delta_j) $$

  where $$\displaystyle \phi_{ii} = \arg(Y_{ii}) $$.

• PV Bus Handling: For a PV bus, $$\displaystyle |V_i| $$ is fixed. The $$\displaystyle \Delta Q_i $$ equation is replaced by $$\displaystyle \Delta |V_i| = 0 $$. The Jacobian is reduced by removing the row/column corresponding to $$\displaystyle \Delta |V_i| $$.

• Flow Chart:

  1. Initialize voltages.

  2. Calculate $$\displaystyle P_i $$, $$\displaystyle Q_i $$ from power equations.

  3. Form mismatch vector $\Delta P, \Delta Q$.

  4. Check convergence: $$\displaystyle \max |\Delta P, \Delta Q| < \epsilon $$.

  5. Form Jacobian $J$.

  6. Solve $$\displaystyle J \cdot \Delta X = -\Delta M $$ for $\Delta X$ ($\Delta \delta, \Delta |V|$).

  7. Update voltages: $$\displaystyle |V_i|^{new} = |V_i|^{old} + \Delta |V_i| $$, $$\displaystyle \delta_i^{new} = \delta_i^{old} + \Delta \delta_i $$.

  8. Go to step 2.

• Advantages: Quadratic convergence, robust for large systems.

• Disadvantages: High memory (Jacobian is $2n \times 2n$), complex programming, each iteration computationally heavy.

Comparison of Methods

[!TIP] Exam Focus: Be prepared to write the GS iteration formula for a PQ bus and explain the PV bus modification. For NR, know the structure of the Jacobian in polar coordinates and how PV buses are handled. Fast Decoupled (assumes weak coupling between $\Delta P/\Delta \delta$ and $\Delta Q/\Delta |V|$) and DC Load Flow (neglects reactive power, $$\displaystyle |V|=1.0 $$, $\cos\phi \approx 1$) are faster approximations used for large systems.

Data Requirements and Convergence Criteria

• Data: Line/cable parameters ($r, x, b/2$), transformer tap ratios & phase shifts, generator $P$, $|V|$ (PV buses), load $P, Q$ (PQ buses), slack bus specs.

• Convergence Criteria:

  1. Mismatch Criterion: $$\displaystyle \max |\Delta P_i, \Delta Q_i| < \epsilon_P, \epsilon_Q $$ (e.g., 0.001 p.u.).

  2. Voltage Change Criterion: $$\displaystyle \max |\Delta |V_i|| < \epsilon_V $$, $$\displaystyle \max |\Delta \delta_i| < \epsilon_\delta $$.

  3. Typically, mismatch in active power is the primary criterion.

II. Power System Stability

Introduction to Stability

• Definition: Ability of a power system to regain a state of operating equilibrium after being subjected to a disturbance, with system variables bounded.

• Steady-State (Small-Signal) Stability: Ability to maintain synchronism under small, slow disturbances (e.g., normal load fluctuations). Analyzed by linearizing equations around an equilibrium.

• Transient Stability: Ability to maintain synchronism after a large, sudden disturbance (e.g., short circuit, line outage). Involves nonlinear, large-swing rotor angle behavior.

• Factors Affecting:

  • Steady-State: Transfer reactance ($X$), generator excitation, system voltage levels.

  • Transient: Fault clearing time, type/nearness of fault, system strength (inertia, $X$), initial operating point.

Swing Equation

Derivation: Based on rotor dynamics. Net accelerating torque $$\displaystyle T_a = T_m - T_e $$. Using $$\displaystyle T = J \frac{d\omega}{dt} $$ and $$\displaystyle P = T\omega_s $$ (synchronous speed):

$$M \frac{d^2\delta}{dt^2} = P_m - P_e = P_a$$

where:

• $$\displaystyle M = 2H / \omega_s $$ is the angular momentum constant (MJ-sec/rad).

• $H$ is the inertia constant (MJ/MVA or kW-sec/kVA).

• $\delta$ is the rotor angle relative to a synchronously rotating reference (rad or deg).

• $$\displaystyle P_m, P_e $$ are mechanical and electrical powers (p.u. or MW).

• $$\displaystyle P_a $$ is the accelerating power.

Physical Significance: It's the fundamental equation describing the rotor angle stability of a synchronous machine. The difference between mechanical input and electrical output causes the rotor to accelerate or decelerate, changing its angle $\delta$. Stability requires $\delta$ to remain bounded.

Linearized Swing Equation (for small $\Delta\delta$):

$$M \frac{d^2(\Delta\delta)}{dt^2} + D \frac{d(\Delta\delta)}{dt} + \omega_s^2 \Delta\delta \cdot \frac{\partial P_e}{\partial \delta} \bigg|_{\delta_0} = 0$$

where $D$ is damping coefficient. The term $$\displaystyle \frac{\partial P_e}{\partial \delta} \big|_{\delta_0} $$ is the synchronizing torque coefficient. Steady-state stability limit is when this coefficient becomes zero.

Equal Area Criterion (EAC)

• Statement: For a single machine connected to an infinite bus (SMIB), the system is stable if the accelerating area ($$\displaystyle A_{acc} $$) equals the decelerating area ($$\displaystyle A_{dec} $$) during the swing. If $$\displaystyle A_{acc} > A_{dec} $$, the machine loses synchronism.

• Graphical Interpretation: Plot $$\displaystyle P_e(\delta) $$ (power-angle curve).

  • Pre-fault: $$\displaystyle P_{e0}(\delta) $$.

  • During-fault: $$\displaystyle P_{ef}(\delta) $$ (lower curve, higher reactance).

  • Post-fault: $$\displaystyle P_{e1}(\delta) $$ (after fault clearance).

  • At fault inception ($$\displaystyle \delta = \delta_0 $$), $$\displaystyle P_m $$ is constant. Accelerating area is area between $$\displaystyle P_m $$ and $$\displaystyle P_{ef}(\delta) $$ from $$\displaystyle \delta_0 $$ to $$\displaystyle \delta_c $$ (clearing angle). Decelerating area is area between $$\displaystyle P_{e1}(\delta) $$ and $$\displaystyle P_m $$ from $$\displaystyle \delta_c $$ to $$\displaystyle \delta_{max} $$ (where $$\displaystyle \frac{d\delta}{dt}=0 $$ again).

  Condition for Stability: $$\displaystyle A_{acc} = A_{dec} $$.

  Condition for Instability: $$\displaystyle A_{acc} > A_{dec} $$.

• Application: Used to determine Critical Clearing Angle ($$\displaystyle \delta_{cr} $$) and Critical Clearing Time ($$\displaystyle t_{cr} $$) for a given fault.

Critical Clearing Angle and Time

• Concept: Maximum allowable fault duration (clearing time) for which the system remains transiently stable. Corresponds to the point where $$\displaystyle A_{acc} = A_{dec} $$.

• Calculation Procedure:

  1. Determine $$\displaystyle \delta_0 $$ from pre-fault power $$\displaystyle P_{m0} = P_{e0}(\delta_0) $$.

  2. Determine $$\displaystyle \delta_{cr} $$ by solving: $$\displaystyle \int_{\delta_0}^{\delta_{cr}} (P_m - P_{ef}(\delta)) d\delta = \int_{\delta_{cr}}^{\delta_{max}} (P_{e1}(\delta) - P_m) d\delta $$, where $$\displaystyle \delta_{max} $$ is the angle where $$\displaystyle P_{e1}(\delta_{max}) = P_m $$ (the other intersection point of $$\displaystyle P_{e1} $$ and $$\displaystyle P_m $$).

  3. Critical clearing time $$\displaystyle t_{cr} $$ is the time taken for the rotor to swing from $$\displaystyle \delta_0 $$ to $$\displaystyle \delta_{cr} $$ during the fault, found by integrating the swing equation numerically (e.g., step-by-step method) or graphically.

• Example (Jun 2025): Given pre-fault max power $$\displaystyle P_{max0} $$, during-fault max power $$\displaystyle P_{maxf} $$, post-fault max power $$\displaystyle P_{max1} $$, and initial operating point (50% of $$\displaystyle P_{max0} $$). Find $$\displaystyle \delta_{cr} $$.

  • $$\displaystyle \delta_0 $$ from $$\displaystyle P_{m} = 0.5 P_{max0} = P_{max0} \sin\delta_0 \Rightarrow \delta_0 = \sin^{-1}(0.5) = 30^\circ $$.

  • During fault: $$\displaystyle P_{ef} = P_{maxf} \sin\delta = 0.4 P_{max0} \sin\delta $$.

  • Post-fault: $$\displaystyle P_{e1} = P_{max1} \sin\delta = 0.75 P_{max0} \sin\delta $$.

  • $$\displaystyle \delta_{max} $$ from $$\displaystyle P_m = P_{e1}(\delta_{max}) \Rightarrow 0.5 P_{max0} = 0.75 P_{max0} \sin\delta_{max} \Rightarrow \delta_{max} = \sin^{-1}(2/3) \approx 41.8^\circ $$.

  • Solve $$\displaystyle \int_{30^\circ}^{\delta_{cr}} (0.5P_{max0} - 0.4P_{max0}\sin\delta) d\delta = \int_{\delta_{cr}}^{41.8^\circ} (0.75P_{max0}\sin\delta - 0.5P_{max0}) d\delta $$ for $$\displaystyle \delta_{cr} $$.

Methods to Improve Stability

• Steady-State Stability:

  • Reduce transfer reactance (shorter lines, parallel circuits, series compensation).

  • Increase system voltage levels.

  • Use high-speed excitation systems (improves voltage support).

  • Employ power system stabilizers (PSS) to add damping.

• Transient Stability:

  • Fast Fault Clearing: Reduce fault duration (fast circuit breakers, protection).

  • Auto-Reclosing: Restore line quickly after transient faults.

  • Braking Resistors: Insert resistors at generator terminals during fault to reduce acceleration.

  • Fast Valving: Quickly reduce steam input to turbine during fault.

  • Advanced Excitation Control: High-initial-response (HIR) excitation, PSS.

  • FACTS Devices: SVC, STATCOM for dynamic reactive power support; TCSC for series compensation control.

III. Load Frequency Control (LFC) and Governor Systems

Importance of Frequency Regulation

• Frequency is a direct indicator of active power balance ($$\displaystyle P_m = P_e + \frac{dE_k}{dt} $$).

• Effects of Deviation:

  • Motors: Speed changes, affecting industrial processes.

  • Turbines: Blade stress, vibration.

  • Clocks & Instruments: Inaccurate timekeeping, measurement errors.

  • System Operation: Can trigger under-frequency load shedding, affect parallel operation.

• Permissible Limits: Typically $\pm 0.5$ Hz in interconnected systems, $\pm 1$ Hz in isolated.

Inertia Constant (H) and Kinetic Energy

• Definition: $$\displaystyle H = \frac{\text{Kinetic Energy stored in rotor at synchronous speed } (E_k)}{\text{Generator MVA base rating}} $$

$$H = \frac{E_k}{S_{base}} \quad \text{(units: MJ/MVA or kW-sec/kVA)}$$

• Kinetic Energy: $$\displaystyle E_k = H \times S_{base} $$.

• Angular Momentum: $$\displaystyle M = \frac{2H}{\omega_s} $$ (MJ-sec/rad), where $$\displaystyle \omega_s = 2\pi f $$ (rad/sec).

• Equivalent Inertia for Multiple Machines:

$$H_{eq} = \frac{\sum_{i=1}^{n} H_i S_{i}}{\sum_{i=1}^{n} S_{i}}$$

(All $$\displaystyle H_i $$ and $$\displaystyle S_i $$ on a common base).

[!TIP] Numerical Example (May 2024): 50 Hz, 4-pole, 20 MVA turbo generator, $$\displaystyle H=9 $$ kW-sec/kVA.

• $$\displaystyle S_{base} = 20 \times 10^3 $$ kVA.

• $$\displaystyle E_k = H \times S_{base} = 9 \times 20 \times 10^3 = 180,000 $$ kW-sec = 180 MJ.

Turbine Speed Governing System

• Purpose: To adjust mechanical power input $$\displaystyle P_m $$ in response to speed/frequency changes.

• Functional Block Diagram:

  Speed Sensor → Governor Amplifier → Servomotor → Control Valves → Turbine

• Governor Time Constant ($$\displaystyle T_g $$): Time delay from speed change to final valve position change (typically 0.2-0.5 sec). Represents governor deadband and transient response.

• Speed Droop (Regulation): $$\displaystyle R = \frac{\Delta f / f_s}{\Delta P_m / P_{m,rated}} $$ (fraction or %). Defines steady-state frequency change per unit change in load. A 4% droop means full load change causes 2% frequency change (since $R$ is usually defined from no-load to full-load).

• Governor Time Delay: The inherent delay between a load change and the initiation of steam valve movement (e.g., 0.6 sec in Jun 2025 problem). During this delay, frequency drops solely due to stored kinetic energy dissipation.

Load Frequency Control (LFC) / ALFC

• Primary Control (Governor Action - Free Governor Operation):

  • Action: Each generator's governor responds instantaneously to local frequency deviation $\Delta f$ via its droop characteristic.

  • Characteristic: $$\displaystyle \Delta P_m = -\frac{1}{R} \Delta f $$ (in p.u.). Negative sign: frequency drop → increase in $$\displaystyle P_m $$.

  • Steady-State Error: Results in a non-zero frequency deviation after a load change because all generators share load based on their $1/R$ (inverse droop). Frequency does not return to nominal.

• Secondary Control (Automatic Load Frequency Control - ALFC):

  • Action: A centralized controller (for each control area) adjusts the set-point of one or more governors (the regulating units) to:

    1. Restore system frequency to nominal ($$\displaystyle \Delta f = 0 $$).

    2. Restore tie-line power flows to their scheduled values ($$\displaystyle \Delta P_{tie} = 0 $$).

  • Area Control Error (ACE): The key feedback signal.

$$ACE = \Delta P_{tie} + B \Delta f$$

    where $B$ is the **frequency bias constant** (MW/Hz) for the area. $B$ is chosen so that ACE=0 when $$\displaystyle \Delta f=0 $$ and $$\displaystyle \Delta P_{tie}=0 $$.

*   **Controller:** Typically an **integral controller**: $$\displaystyle \Delta P_{set} = -K_I \int ACE \, dt $$. This ensures steady-state elimination of ACE.

• Control Area Concept & Tie-Line Modeling:

  • A control area is a group of generators and loads with coherent frequency (tied together by AC lines).

  • Tie-line with adjacent area is modeled as a DC line for LFC: $$\displaystyle \Delta P_{tie} = T \Delta \delta $$, where $T$ is the tie-line synchronizing coefficient (p.u. power/rad).

  • In steady-state, $$\displaystyle \Delta P_{tie} = \frac{2\pi}{s} T \Delta f $$ in Laplace domain.

Numerical Problems

1. Frequency Drop with Governor Delay (Jun/Dec 2025):

   • Given: $$\displaystyle S_{base} $$, $H$, sudden load $$\displaystyle \Delta P_L $$, governor delay $$\displaystyle T_d $$.

   • Assumption: During delay, $$\displaystyle P_m $$ constant = initial $$\displaystyle P_{m0} $$. $$\displaystyle P_e = P_{m0} $$ initially, then $$\displaystyle P_e = P_{m0} - \Delta P_L $$ after load change.

   • Swing Equation (p.u.): $$\displaystyle 2H \frac{d^2\delta}{dt^2} = P_m - P_e $$.

   • At $$\displaystyle t=0 $$, $$\displaystyle \delta=0 $$, $$\displaystyle \frac{d\delta}{dt}=0 $$. For $$\displaystyle 0 < t < T_d $$, $$\displaystyle P_m - P_e = \Delta P_L $$ (constant accelerating power).

   • Integrate: $$\displaystyle \frac{d\delta}{dt} = \frac{\Delta P_L}{2H} t $$.

   • Frequency: $$\displaystyle f = f_s + \frac{1}{2\pi} \frac{d\delta}{dt} $$. So $$\displaystyle \Delta f = \frac{1}{2\pi} \frac{\Delta P_L}{2H} T_d $$.

   Formula: $$\displaystyle \boxed{\Delta f = \frac{\Delta P_L \cdot T_d}{4\pi H}} $$ (in Hz, with $$\displaystyle \Delta P_L $$ in p.u. on $$\displaystyle S_{base} $$).

   • Example (Jun 2025): $$\displaystyle \Delta P_L = 25 $$ MW, $$\displaystyle S_{base}=100 $$ MVA $$\displaystyle \Rightarrow \Delta P_L(p.u.)=0.25 $$. $$\displaystyle H=4.5 $$ kW-sec/kVA = 4.5 sec (since 1 kW-sec/kVA = 1 sec). $$\displaystyle T_d=0.6 $$ sec.

     $$\displaystyle \Delta f = \frac{0.25 \times 0.6}{4\pi \times 4.5} \approx 0.00266 $$ p.u. $$\displaystyle \Rightarrow \Delta f \approx 0.00266 \times 50 = 0.133 $$ Hz. Final frequency ≈ 49.867 Hz.

2. Rotor Acceleration (May 2024):

   • Given $$\displaystyle P_m $$ (new), $$\displaystyle P_e $$ (load), find initial acceleration.

   • $$\displaystyle P_a = P_m - P_e $$ (MW or p.u.).

   • $$\displaystyle \frac{d^2\delta}{dt^2} = \frac{P_a \cdot \omega_s}{2E_k} = \frac{P_a}{M} $$.

   • Example: $$\displaystyle P_m=80 $$ MW, $$\displaystyle P_e=50 $$ MW, $$\displaystyle S_{base}=100 $$ MVA, $$\displaystyle H=8 $$ MJ/MVA.

     $$\displaystyle E_k = H \times S_{base} = 8 \times 100 = 800 $$ MJ. $$\displaystyle M = 2E_k / \omega_s = 1600 / (2\pi \times 50) \approx 5.09 $$ MJ-sec/rad.

     $$\displaystyle P_a = 30 $$ MW = 0.3 p.u. on 100 MVA.

     $$\displaystyle \frac{d^2\delta}{dt^2} = \frac{0.3 \times 2\pi \times 50}{2 \times 8} \approx 5.89 $$ rad/sec².

3. Load Sharing in Interconnected Plants (May 2024):

   • Two plants A, B with capacities $$\displaystyle C_A, C_B $$, speed regulations $$\displaystyle R_A, R_B $$ (from no-load to full-load).

   • Total load $$\displaystyle P_L $$ on the tie-line bus. Each plant's load $$\displaystyle P_A, P_B $$.

   • Principle: In steady-state, frequency deviation $\Delta f$ is same for both. Load change shares inversely with regulation.

$$\frac{\Delta P_A}{\Delta P_B} = \frac{1/R_A}{1/R_B} = \frac{R_B}{R_A}$$

    and $$\displaystyle \Delta P_A + \Delta P_B = \Delta P_L $$ (assuming no net tie-line flow change).

*   **Example:** $$\displaystyle C_A=200 $$ MW, $$\displaystyle R_A=1.5\%=0.015 $$; $$\displaystyle C_B=100 $$ MW, $$\displaystyle R_B=3\%=0.03 $$; Bus load $$\displaystyle P_L=100 $$ MW. Assume both initially at some load, but total load on bus is 100 MW. Need to find generation and tie-line flow.

    *   Let initial generation be $$\displaystyle P_{A0}, P_{B0} $$ with $$\displaystyle P_{A0}+P_{B0}=100 $$ MW. But we need more info (like initial frequency or scheduled tie-flow). Typically, such problems assume the two plants are connected by a line and supply a common load. The load sharing is determined by their droop settings.

    *   If both are on the same bus (no tie-line), then $$\displaystyle P_A + P_B = 100 $$ MW and $$\displaystyle \frac{P_A - P_{A,no-load}}{C_A} / \frac{P_B - P_{B,no-load}}{C_B} = \frac{R_B}{R_A} $$. Usually, no-load is 0 for simplicity. Then $$\displaystyle P_A / C_A \cdot R_A = P_B / C_B \cdot R_B $$ (since regulation is % change from no-load to full-load, so $$\displaystyle R = \frac{\Delta f / f_s}{\Delta P / C} $$). So $$\displaystyle \frac{P_A}{200} \times 0.015 = \frac{P_B}{100} \times 0.03 \Rightarrow P_A = 4 P_B $$. With $$\displaystyle P_A+P_B=100 $$, $$\displaystyle P_A=80 $$ MW, $$\displaystyle P_B=20 $$ MW. Tie-line flow depends on how they are interconnected. If they are simply parallel on the same bus, tie-line flow is 0. If they are connected by a line and supply separate loads plus exchange power, the problem statement must specify. The May 2024 question says "interconnected by a short line. Capacity of A is 200 MW and that of B is 100 MW. Their speed regulations... are 1.5% and 3% respectively. The load on bus of each station is 100 MW." This implies each station has its own local load of 100 MW, and they are connected by a line. So total generation must cover local load plus any export/import.

    *   Let $$\displaystyle P_A $$ = total gen at A, $$\displaystyle P_B $$ = total gen at B.

    *   Local load at A = 100 MW, at B = 100 MW.

    *   Tie-line flow from A to B = $$\displaystyle P_A - 100 = 100 - P_B $$.

    *   Frequency deviation $\Delta f$ causes generation change: $$\displaystyle \Delta P_A = \frac{C_A}{R_A} (-\Delta f) $$, $$\displaystyle \Delta P_B = \frac{C_B}{R_B} (-\Delta f) $$ (assuming linear droop around operating point).

    *   In steady-state, net interchange is scheduled. If no schedule, then $$\displaystyle \Delta P_A = \Delta P_B $$? Not necessarily. The system will reach a frequency where the total generation equals total load (200 MW). So $$\displaystyle P_A + P_B = 200 $$ MW.

    *   Also, from droop: $$\displaystyle \frac{P_A - P_{A,ref}}{C_A / R_A} = \frac{P_B - P_{B,ref}}{C_B / R_B} = -\Delta f $$. But we don't know references. Typically, references are set so that at nominal frequency, each generates its local load? That would mean $$\displaystyle P_{A,ref}=100 $$ MW, $$\displaystyle P_{B,ref}=100 $$ MW. Then:

        $$\displaystyle P_A = 100 + \frac{C_A}{R_A} (-\Delta f) = 100 + \frac{200}{0.015} (-\Delta f) $$

        $$\displaystyle P_B = 100 + \frac{C_B}{R_B} (-\Delta f) = 100 + \frac{100}{0.03} (-\Delta f) $$

        But $$\displaystyle P_A + P_B = 200 $$ MW (since total load 200 MW). So:

        $$\displaystyle 200 + \left( \frac{200}{0.015} + \frac{100}{0.03} \right) (-\Delta f) = 200 \Rightarrow -\Delta f = 0 $$. So $$\displaystyle \Delta f=0 $$, $$\displaystyle P_A=100 $$, $$\displaystyle P_B=100 $$, tie-flow=0. That seems trivial. Maybe the "load on bus of each station is 100 MW" means the total load on the system is 100 MW, shared between the two buses? The wording is ambiguous. In many textbook problems, two stations connected by a line supply a common load. Let's assume total load $$\displaystyle P_L=100 $$ MW on the tie-line bus. Then:

        $$\displaystyle P_A + P_B = 100 $$ MW (total generation).

        Droop sharing: $$\displaystyle \frac{P_A}{C_A / R_A} = \frac{P_B}{C_B / R_B} $$ (since both experience same $\Delta f$ from a common reference).

        $$\displaystyle \frac{P_A}{200/0.015} = \frac{P_B}{100/0.03} \Rightarrow \frac{P_A}{13333.3} = \frac{P_B}{3333.3} \Rightarrow P_A = 4 P_B $$.

        Then $$\displaystyle 4P_B + P_B = 100 \Rightarrow P_B=20 $$ MW, $$\displaystyle P_A=80 $$ MW.

        Tie-line flow from A to B = $$\displaystyle P_A - \text{load at A} $$. But load at A? Not specified. If the 100 MW load is at the tie-line bus (between them), then each plant's generation minus its local load? Not given. Possibly both plants are at their own buses with no local load, and the 100 MW is at a third bus? The problem likely expects: $$\displaystyle P_A=80 $$ MW, $$\displaystyle P_B=20 $$ MW, and since they are interconnected, the power from A to B is $$\displaystyle P_A - P_B = 60 $$ MW? That doesn't make sense because net flow must be zero if no other load. Actually, if they are connected by a line and supply a common load of 100 MW at one of the buses, the generation at each minus its own local load (if any) equals the flow on the line. Without local load data, we assume the 100 MW is the total system load, and the generation at each plant is as calculated. The tie-line power is then $$\displaystyle P_A - P_{A,local} $$ but $$\displaystyle P_{A,local} $$ unknown. Perhaps the "bus of each station" means the bus where the station is located has a load of 100 MW. So station A bus has 100 MW load, station B bus has 100 MW load. Total load 200 MW. Then as above, with references set to 100 MW each, we get $$\displaystyle P_A=100 $$, $$\displaystyle P_B=100 $$, tie-flow=0. That seems too simple. I'll check the exact wording: "Capacity of A is 200 MW and that of B is 100 MW. Their speed regulations... are 1.5% and 3% respectively. The load on bus of each station is 100 MW." So each station's bus has a 100 MW load. So total load 200 MW. Each station must generate at least 100 MW to supply its local load. Any excess/deficit is exchanged via the interconnecting line. In steady-state, frequency will be such that total generation = total load = 200 MW. Let $$\displaystyle P_A $$ and $$\displaystyle P_B $$ be total generation. Then $$\displaystyle P_A + P_B = 200 $$. The load sharing is determined by droop: the change in generation from no-load is proportional to $1/R$. But we don't know no-load points. Usually, we assume no-load generation is 0. Then $$\displaystyle P_A / (200/0.015) = P_B / (100/0.03) = -\Delta f $$. So $$\displaystyle P_A / 13333.3 = P_B / 3333.3 \Rightarrow P_A = 4 P_B $$. Then $$\displaystyle 4P_B + P_B = 200 \Rightarrow P_B=40 $$ MW, $$\displaystyle P_A=160 $$ MW. But station A's local load is 100 MW, so it exports $$\displaystyle 160-100=60 $$ MW to station B. Station B generates 40 MW but has 100 MW load, so it imports 60 MW from A. That makes sense. So answer: Gen A=160 MW, Gen B=40 MW, tie-line flow A→B = 60 MW.

> **General Formula:** If total load $$\displaystyle P_L $$, and each plant has capacity $$\displaystyle C_i $$ and regulation $$\displaystyle R_i $$ (fraction), and no-load generation is 0, then:

> 

$$P_i = \frac{C_i / R_i}{\sum (C_j / R_j)} P_L$$

> and tie-line flow depends on local loads.

IV. Voltage Control and Excitation Systems

Reactive Power and Voltage Relationship

• Phasor Diagram (Simple Short Line): $$\displaystyle V_r = V_s - I (R \cos\phi + jX \sin\phi) - I (jR \sin\phi - X \cos\phi) $$.

  For a short line ($R \approx 0$): $$\displaystyle V_r \approx V_s - j I X $$.

$$|V_r| \approx |V_s - j I X| \approx |V_s| - \frac{Q X}{|V_s|}$$

(for small angle, $$\displaystyle Q \approx |V_s||I|\sin\phi $$).

**Key Equation:** $$\displaystyle \Delta V \approx \frac{Q X}{V_s} $$ (for inductive line, $Q$ positive for lagging load).

• Correlation: Voltage drop along a transmission line is highly sensitive to reactive power flow. Injecting capacitive VARs (leading) raises voltage; absorbing inductive VARs (lagging) lowers voltage.

Generation and Absorption of Reactive Power

[!TIP] Protection & Absorption Methods (Dec 2024 - 14m):

• Generation: Over-excited generators, synchronous condensers, SVC/STATCOM, shunt capacitors.

• Absorption: Under-excited generators (within limits), shunt reactors, SVC/STATCOM in inductive mode, series reactors (for capacitive lines), load.

• Why needed? To maintain voltage profile, prevent over-voltages (light load) or under-voltages (heavy load), and ensure system stability.

Voltage Control Methods

Excitation Systems

• Need & Objectives:

  1. Voltage Regulation: Maintain terminal voltage constant under varying load.

  2. Stability Improvement: Increase steady-state and transient stability limits.

  3. Reactive Power Control: Control VAR output of generator.

  4. Damping: Provide damping to power oscillations (with PSS).

• Types & Functional Block Diagrams:

  1. DC Excitation System (with Amplidyne):

     • Components: DC exciter (shunt/series), amplidyne (high-gain rotating amplifier), voltage regulator (sensing terminal voltage, comparing to reference, controlling amplidyne field), power rectifier (if needed), main generator field.

     • Operation: Voltage regulator controls amplidyne field, which amplifies the signal and supplies DC to generator field. Fast response due to amplidyne.

  2. AC Static Excitation System (Brushless):

     • Principle: Uses an AC exciter (on same shaft) and a rotating diode rectifier (mounted on shaft) to supply DC directly to main generator field. No brushes/slip rings.

     • Components: AC exciter (stator field, rotor armature), rotating diode bridge, main generator.

     • Voltage regulator controls AC exciter field. Response slower than DC type but more reliable.

  3. Brushless Excitation System: Often used interchangeably with AC static. Detailed block: Voltage Regulator → AC Exciter Field → AC Exciter Armature (3-phase) → Rotating Diode Rectifier → Main Generator Field.

Limiting Features (Protection)

• Over-Excitation Limiter (OEL): Prevents field current/voltage from exceeding thermal limits of rotor winding. Acts to reduce $$\displaystyle E_f $$ if exceeded.

• Under-Excitation Limiter (UEL): Prevents field current from dropping below a minimum, avoiding loss of synchronism (underexcited operation) and stator end-iron heating. May increase $$\displaystyle E_f $$ or signal LFC to reduce output.

• Volts/Hz (V/Hz) Limiter: Prevents excessive core flux density ($\phi \propto V/f$) during low-frequency operation (e.g., islanding). Reduces voltage if $V/f$ exceeds limit.

Automatic Voltage Regulator (AVR)

• Functional Block Diagram (Turbo-Generator):

  Terminal Voltage (VT) → Measuring Element (PT, CT) → Comparator (with $$\displaystyle V_{ref} $$) → Error Signal → Amplifier (gain $$\displaystyle K_a $$) → Exciter (gain $$\displaystyle K_e $$, time constant $$\displaystyle T_e $$) → Generator Field (time constant $$\displaystyle T_f $$) → $$\displaystyle E_f $$ → Terminal Voltage $$\displaystyle V_t $$

• Explanation:

  • Measuring Element: Steps down voltage/current to proportional DC signal.

  • Comparator: Subtracts measured $$\displaystyle V_t $$ from reference $$\displaystyle V_{ref} $$ to get error.

  • Amplifier: Increases signal strength (may be static or rotating).

  • Exciter: Supplies DC field current to generator. Modeled as $$\displaystyle K_e/(1+sT_e) $$.

  • Generator Field: Winding time constant $$\displaystyle T_f $$.

• Transfer Function (Simplified): $$\displaystyle \frac{\Delta V_t(s)}{\Delta V_{ref}(s)} = \frac{K_a K_e / (1+sT_e)}{1 + K_a K_e T_f s + ...} $$ (more complex with load effects).

V. Economic Operation and Power System Economics

Economic Dispatch Problem (EDP)

• Definition: Determine the optimal generation schedule of online plants to minimize total fuel cost while meeting the load demand and satisfying generator operating limits.

• Objective Function: Minimize $$\displaystyle F_{total} = \sum_{i=1}^{n} F_i(P_i) $$, where $$\displaystyle F_i(P_i) = a_i P_i^2 + b_i P_i + c_i $$ ($/hr$ or appropriate unit).

• Constraints:

  1. Power Balance: $$\displaystyle \sum_{i=1}^{n} P_i = P_D + P_{loss} $$ (Demand + Losses).

  2. Generator Limits: $$\displaystyle P_{i,\min} \le P_i \le P_{i,\max} $$.

Economic Dispatch Solution (Without Losses)

• Lagrangian Multiplier Method:

  Lagrangian: $$\displaystyle L = \sum_{i=1}^{n} F_i(P_i) + \lambda \left( P_D - \sum_{i=1}^{n} P_i \right) $$.

  For optimum: $$\displaystyle \frac{\partial L}{\partial P_i} = 0 \Rightarrow \frac{dF_i}{dP_i} = \lambda \quad \forall i $$.

  • $\lambda$ is the incremental fuel cost ($\partial F / \partial P$) in $/MWh$.

  • Condition: All online units must operate at the same incremental cost $\lambda$.

• Procedure:

  1. Arrange units in increasing order of their incremental cost curves ($$\displaystyle \frac{dF_i}{dP_i} $$).

  2. Start with the cheapest unit, increase its output until either $\lambda$ equals that of next unit or it hits its max.

  3. Bring in next unit, adjust outputs to equalize $\lambda$, and so on.

  4. Continue until total demand is met.

• With Piecewise Linear Heat Rate: Use successive $\lambda$ values where unit switches between blocks.

Transmission Loss Considerations

• Loss Formula (B Coefficients):

$$P_{loss} = \sum_{i=1}^{n} \sum_{j=1}^{n} P_i B_{ij} P_j$$

where $$\displaystyle B_{ij} $$ are **loss coefficients** (symmetric, $$\displaystyle B_{ii} \ge 0 $$, $$\displaystyle B_{ij} \le 0 $$). Derived from DC load flow assumptions ($$\displaystyle |V|=1 $$, $\delta$ small, $$\displaystyle P_{ij} \approx (\delta_i - \delta_j)/X_{ij} $$).

• Penalty Factor ($$\displaystyle L_i $$):

$$L_i = \frac{1}{1 - \frac{\partial P_{loss}}{\partial P_i}}$$

where $$\displaystyle \frac{\partial P_{loss}}{\partial P_i} = 2 \sum_{j=1}^{n} B_{ij} P_j $$.

*   **Economic Dispatch with Losses:** $$\displaystyle \frac{dF_i}{dP_i} \cdot L_i = \lambda $$ for all $i$.

*   **Interpretation:** Loss factor accounts for the **marginal cost of delivering power** from plant $i$ to the load, including its contribution to losses.

• Procedure:

  1. Guess initial $$\displaystyle P_i $$ (e.g., without losses).

  2. Compute $$\displaystyle P_{loss} $$ and $$\displaystyle \frac{\partial P_{loss}}{\partial P_i} $$.

  3. Calculate $$\displaystyle L_i $$ for each unit.

  4. Adjust $\lambda$ such that $$\displaystyle \sum P_i = P_D + P_{loss} $$ and $$\displaystyle \frac{dF_i}{dP_i} L_i = \lambda $$.

  5. Iterate until convergence.

Pricing of Energy and Transmission Service

• Cost Components:

  • Generation Cost: Capital, fuel, O&M.

  • Transmission Cost: Lines, substations, losses, congestion.

  • Distribution Cost: Lines, transformers, customer service.

• Pricing Methodologies:

  • Embedded Cost: Traditional, based on average historical costs (rate-of-return regulation).

  • Incremental Cost (Marginal Cost): Based on cost of producing the next MWh. Economically efficient but may not cover fixed costs.

  • Locational Marginal Pricing (LMP): Price at each node = cost of supplying next MW at that location, including generation cost, losses, and congestion. Used in ISOs (e.g., PJM, NYISO). Components: Energy ($\lambda$), Congestion, Losses.

Deregulation, Restructuring, and Distributed Generation

• Deregulation: Introduction of competition in generation and sometimes retail, while transmission/distribution remain regulated.

  • Motivations: Lower prices, innovation, customer choice.

  • Effects: Separation of generation, transmission, distribution; creation of markets; need for Independent System Operator (ISO).

• Restructuring: Reorganization of the power industry from vertically integrated utilities to separate entities.

  • Models:

    1. POOL (Single Buyer): All generation sold to a central pool (ISO/PX).

    2. Bilateral: Direct contracts between generators and customers/retailers.

    3. Hybrid: Combination of pool and bilateral.

  • Role of ISO: Independent operator of transmission grid, ensures reliability, runs spot market, manages congestion.

  • Role of PX (Power Exchange): Marketplace for trading energy (may be same as ISO).

• Distributed Generation (DG): Small-scale generation (e.g., solar PV, wind, microturbines, fuel cells) located close to the load.

  • Technologies: Renewables, IC engines, gas turbines.

  • Impact:

    • Positive: Reduced transmission losses, deferred T&D upgrades, improved reliability (if islanded), environmental benefits.

    • Challenges: Reverse power flow, voltage regulation issues, protection coordination, intermittent output, grid integration standards.

VI. System Interconnection and Modern Power System Issues

Need for Interconnected Power Systems

• Reliability: Shared reserves, mutual assistance during emergencies.

• Economy: Economies of scale (large efficient plants), load diversity reduces spinning reserve requirement, access to cheap remote energy.

• Operational Flexibility: Easier to schedule maintenance, handle contingencies.

• Resource Utilization: Enables sharing of hydro/thermal resources across regions.

Problems Associated with Interconnection

• Control Complexity: Coordinated frequency and voltage control across multiple control areas. Need for ALFC and AVR coordination.

• Stability Challenges: Synchronizing torque issues, increased fault levels, risk of cascading outages, need for stability analysis of multi-machine systems.

• Protection Coordination: More complex fault currents, need for adaptive/communication-assisted protection.

• Regulatory & Market Challenges: Settlement of interchange, congestion management, transmission pricing, jurisdictional issues.

Real and Reactive Power Control in Interconnected Systems

• Real Power Control:

  • Governor Set-Points: Primary control (droop) handles instantaneous imbalances.

  • Tie-Line Power Scheduling: Secondary control (ALFC) adjusts set-points to maintain scheduled interchange ($$\displaystyle P_{tie,scheduled} $$) and system frequency.

  • ACE is the key: $$\displaystyle ACE = (P_{tie,actual} - P_{tie,scheduled}) + B \Delta f $$.

• Reactive Power & Voltage Control:

  • Local Control: Each area controls its own voltage via generator AVRs, OLTCs, shunt devices.

  • Coordination: Interconnected operation requires coordination to avoid excessive VAR flows on tie-lines. Often, voltage is controlled at "pilot" buses or via coordinated voltage control schemes.

  • Role of AVR & ALFC: AVRs maintain local voltage, indirectly influencing reactive power flows. ALFC primarily for real power/frequency, but can be extended for voltage control in some schemes (e.g., using voltage as an additional input to ACE).

[!TIP] Exam Focus: Be ready to differentiate steady-state vs. transient stability, derive/state swing equation, apply equal area criterion, calculate critical clearing angle, and explain LFC (primary vs. secondary). For economic dispatch, know the Lagrangian condition and penalty factor. For voltage control, correlate reactive power with voltage and list devices. For interconnection, list needs and problems.