Economic Load Dispatch and Compensation GATE EE Exam Quick Notes

Economic Load Dispatch

Economic Load Dispatch (ELD) Basics

Key Concepts for GATE

  • Objective: Minimize total generation cost while meeting demand

    \[\text{Minimize } C_T = \sum_{i=1}^n C_i(P_{Gi})\]
  • Cost Function: Typically quadratic

    \[C_i(P_{Gi}) = a_i + b_i P_{Gi} + c_i P_{Gi}^2 \text{ Rs/hr}\]
  • Incremental Cost: Slope of cost curve

    \[IC_i = \dfrac{dC_i}{dP_{Gi}} = b_i + 2c_i P_{Gi} \text{ Rs/MWh}\]
  • Constraints:

    • Power balance: \(\sum P_{Gi} = P_D + P_L\)

    • Generator limits: \(P_{Gi,min} \leq P_{Gi} \leq P_{Gi,max}\)

ELD Without Transmission Losses

Problem Formulation

Minimize \(C_T = \sum_{i=1}^n (a_i + b_i P_{Gi} + c_i P_{Gi}^2)\)
Subject to: \(\sum_{i=1}^n P_{Gi} = P_D\)

  • Optimal Condition: All units operate at same incremental cost

    \[\dfrac{dC_1}{dP_{G1}} = \dfrac{dC_2}{dP_{G2}} = \cdots = \lambda\]
  • Solution Steps:

    1. Calculate: \(\lambda = \dfrac{\sum b_i + 2\sum c_i P_D}{\sum \dfrac{1}{2c_i}}\)

    2. Find: \(P_{Gi} = \dfrac{\lambda - b_i}{2c_i}\)

    3. Verify: \(\sum P_{Gi} = P_D\)

ELD Without Losses - Numerical Example

Example

Two generators with costs:

\[\begin{aligned} C_1 &= 0.1 P_{G1}^2 + 40 P_{G1} + 100 \text{ Rs/hr} \\ C_2 &= 0.2 P_{G2}^2 + 30 P_{G2} + 80 \text{ Rs/hr} \\ \text{Demand} &= 200 \text{ MW} \end{aligned}\]

Solution:

  • \(IC_1 = 0.2 P_{G1} + 40\), \(IC_2 = 0.4 P_{G2} + 30\)

  • At optimum: \(0.2 P_{G1} + 40 = 0.4 P_{G2} + 30\)

  • With constraint: \(P_{G1} + P_{G2} = 200\)

  • Result: \(P_{G1} = 116.67\) MW, \(P_{G2} = 83.33\) MW

  • \(\lambda = 63.33\) Rs/MWh

ELD With Transmission Losses

  • Transmission Loss Formula (B-coefficients):

    \[P_L = \sum_{i=1}^n \sum_{j=1}^n P_{Gi} B_{ij} P_{Gj}\]
  • Modified Optimality Condition:

    \[\dfrac{dC_i}{dP_{Gi}} \times \dfrac{1}{1 - \dfrac{\partial P_L}{\partial P_{Gi}}} = \lambda\]
  • Penalty Factor:

    \[L_i = \dfrac{1}{1 - \dfrac{\partial P_L}{\partial P_{Gi}}}\]
  • Loss Sensitivity:

    \[\dfrac{\partial P_L}{\partial P_{Gi}} = 2 \sum_{j=1}^n B_{ij} P_{Gj}\]

ELD With Losses - Solution Algorithm

Iterative Solution Method

  1. Assume initial \(\lambda\) (start with loss-free solution)

  2. Calculate penalty factors: \(L_i = \dfrac{1}{1 - \dfrac{\partial P_L}{\partial P_{Gi}}}\)

  3. Find generations: \(\dfrac{dC_i}{dP_{Gi}} = \dfrac{\lambda}{L_i}\)

  4. Calculate total loss: \(P_L = \sum_{i=1}^n \sum_{j=1}^n P_{Gi} B_{ij} P_{Gj}\)

  5. Check power balance: \(\sum P_{Gi} = P_D + P_L\)

  6. If not balanced, adjust \(\lambda\) and repeat

Important GATE Formula

For simplified loss model: \(P_L = B_{00} + \sum B_{0i} P_{Gi} + \sum \sum B_{ij} P_{Gi} P_{Gj}\)

Generator Limits and Lagrange Multipliers

  • When generator hits limits:

    • If \(P_{Gi} < P_{Gi,min}\): Set \(P_{Gi} = P_{Gi,min}\)

    • If \(P_{Gi} > P_{Gi,max}\): Set \(P_{Gi} = P_{Gi,max}\)

    • Remove from optimization and redistribute load

  • Lagrangian Method:

    \[L = \sum C_i(P_{Gi}) + \lambda(\sum P_{Gi} - P_D - P_L)\]
  • KKT Conditions:

    \[\dfrac{\partial L}{\partial P_{Gi}} = \dfrac{dC_i}{dP_{Gi}} + \lambda(1 - \dfrac{\partial P_L}{\partial P_{Gi}}) = 0\]
  • Physical Interpretation: \(\lambda\) represents system marginal cost

Compensation Techniques

Series Compensation

  • Purpose: Improve power transfer capability and stability

  • Implementation: Series capacitor in transmission line

  • Effect on Line Reactance:

    \[X_{eff} = X_L - X_C\]
  • Power Transfer Enhancement:

    \[P_{max} = \dfrac{V_S V_R}{X_L - X_C} \sin \delta\]
  • Compensation Degree:

    \[K = \dfrac{X_C}{X_L} \times 100\%\]
  • Typical values: 25-70%

Equivalent Circuit:

\(V_S\)
\(\downarrow\)
\(X_L\)
\(\downarrow\)
\(-jX_C\)
\(\downarrow\)
\(V_R\)

Series Compensation - Benefits and Drawbacks

Benefits

  • Increases power transfer capability

  • Improves transient stability

  • Reduces voltage drop across line

  • Better voltage regulation

  • Reduces line losses

Drawbacks

  • Sub-synchronous resonance (SSR) problems

  • High installation and maintenance cost

  • Protection complexity

  • Fault current may increase

GATE Important

Maximum compensation limited to 70% due to SSR concerns

Shunt Compensation

  • Types:

    • Shunt capacitors (lagging PF loads)

    • Shunt reactors (leading PF, long lines)

  • Applications:

    • Voltage regulation

    • Power factor correction

    • Reactive power compensation

  • Location: Load centers, substations

  • Voltage Effect:

    \[\Delta V = \dfrac{Q_C X}{V}\]

Shunt Capacitor:

\(I_L\)
\(\rightarrow\)
Load
\(\uparrow\)
\(I_C\)
\(\uparrow\)
\(C\)

Key Formula

Reactive power from capacitor: \(Q_C = V^2 \omega C\)

Shunt Compensation - Detailed Analysis

  • Capacitive Compensation:

    • Provides leading reactive power

    • Improves lagging power factor

    • Reduces line current: \(I_{new} = \sqrt{P^2 + (Q_L - Q_C)^2}/V\)

  • Inductive Compensation:

    • Absorbs leading reactive power

    • Used for long transmission lines

    • Prevents voltage rise during light load

  • Optimal Location:

    • Capacitors: At load centers (2/3 distance from source)

    • Reactors: At line terminals

Power Factor Correction

  • Objective: Improve power factor from \(\cos \theta_1\) to \(\cos \theta_2\)

  • Capacitor Size Calculation:

    \[Q_C = P (\tan \theta_1 - \tan \theta_2)\]
    where:
    • \(\theta_1 = \cos^{-1}(\text{PF}_1)\) (original)

    • \(\theta_2 = \cos^{-1}(\text{PF}_2)\) (desired)

  • Alternative Form:

    \[Q_C = P \left(\dfrac{\sin \theta_1}{\cos \theta_1} - \dfrac{\sin \theta_2}{\cos \theta_2}\right)\]
  • Capacitor Rating:

    \[C = \dfrac{Q_C}{\omega V^2} \text{ Farads}\]

Power Factor Correction - Benefits

Economic Benefits

  • Reduced line losses: \(P_{loss} = I^2 R\) (lower current)

  • Improved voltage regulation

  • Increased system capacity

  • Reduced penalty charges from utility

Technical Benefits

  • Reduced transformer and cable loading

  • Better voltage stability

  • Improved motor starting capability

  • Reduced reactive power flow

Example

For 100 kW load at 0.8 lagging PF, to improve to 0.95 PF: \(Q_C = 100(\tan 36.87° - \tan 18.19°) = 100(0.75 - 0.33) = 42\) kVAR

Static VAR Compensators (SVC)

  • Components:

    • Thyristor Switched Capacitor (TSC)

    • Thyristor Controlled Reactor (TCR)

    • Fixed Capacitor (FC)

  • Operating Principle:

    \[Q_{SVC} = Q_C - Q_L(\alpha)\]
    where \(\alpha\) is firing angle
  • Advantages:

    • Continuous reactive power control

    • Fast response (2-3 cycles)

    • Voltage regulation capability

    • Harmonic filtering

  • V-I Characteristic: Drooping characteristic for stable operation

Comparison of Compensation Methods

Compensation Methods Comparison
Feature Series Shunt SVC
Primary purpose Power transfer Voltage control Dynamic control
Response time Instantaneous Instantaneous 2-3 cycles
Control Fixed Fixed/Switched Continuous
Cost High Moderate High
Stability impact Transient Voltage Both
Installation Line Substation Substation

Key Formulas Summary for GATE

  1. Economic Dispatch without losses:

    \[\dfrac{dC_i}{dP_{Gi}} = \lambda\]
  2. Economic Dispatch with losses:

    \[\dfrac{dC_i}{dP_{Gi}} = \lambda (1 - \dfrac{\partial P_L}{\partial P_{Gi}})\]
  3. Series Compensation:

    \[\% \text{Compensation} = \dfrac{X_C}{X_L} \times 100,\quad P_{max} = \dfrac{V_S V_R}{X_L - X_C}\]
  4. Power Factor Correction:

    \[Q_C = P (\tan \cos^{-1} \text{PF}_1 - \tan \cos^{-1} \text{PF}_2)\]
  5. Shunt Compensation:

    \[Q_C = V^2 \omega C,\quad \Delta V = \dfrac{Q_C X}{V}\]

GATE Problem-Solving Tips

Economic Load Dispatch

  • Always check if generator limits are violated

  • For losses, use iterative method systematically

  • Remember: \(\lambda\) has units of Rs/MWh

Compensation

  • Series compensation: Focus on power transfer improvement

  • Shunt compensation: Focus on voltage and PF improvement

  • Power factor correction: Use trigonometric identities carefully

Common Mistakes to Avoid

  • Forgetting to include losses in power balance

  • Wrong signs in reactive power calculations

  • Mixing up series and shunt compensation effects