GATE EE

Economic Load Dispatch and Compensation GATE EE Exam Quick Notes

Lecture Notes

SEC 01

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

1ELD Without Transmission Losses
1Problem 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: \(\sum P_{Gi} = P_D\)

1ELD Without Losses - Numerical Example
1Example

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

1ELD 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:

    \[\dfrac{\partial P_L}{\partial P_{Gi}} = 2 \sum_{j=1}^n B_{ij} P_{Gj}\]
1ELD With Losses - Solution Algorithm
1Iterative 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

1Important GATE Formula

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

1Generator 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

SEC 02

Compensation Techniques

1Series 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:

    \(V_S\)
    \(\downarrow\)
    \(X_L\)
    \(\downarrow\)
    \(-jX_C\)
    \(\downarrow\)
    \(V_R\)
    1Series Compensation - Benefits and Drawbacks
    1Benefits
    1Drawbacks
    1GATE Important

    Maximum compensation limited to 70% due to SSR concerns

    1Shunt 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
    \(\uparrow\)
    \(I_C\)
    \(\uparrow\)
    \(C\)
    1Key Formula

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

    1Shunt Compensation - Detailed Analysis
    1Power Factor Correction
    1Power Factor Correction - Benefits
    1Economic Benefits
    1Technical Benefits
    1Example

    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

    1Static VAR Compensators (SVC)
    1Comparison 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
    1Key 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:

      \[Q_C = V^2 \omega C,\quad \Delta V = \dfrac{Q_C X}{V}\]
    1GATE Problem-Solving Tips
    1Economic Load Dispatch
    1Compensation
    1Common Mistakes to Avoid