4.3 Economic Dispatch, Energy Efficiency & Load Profile Analysis

Economic Dispatch, Energy Efficiency & Load Profile Analysis

Economic dispatch and industrial energy optimization bridge power generation operations with facility-level energy management. On the NCEES PE Electrical: Power examination, engineers are tested on allocating electrical generation across multiple units at minimum total operating cost, analyzing system load curves, and applying fluid machine affinity laws to calculate drive energy savings.

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1. System Load Curves & Generation Classification

Power system demand fluctuates dynamically throughout the day, month, and year. Two primary graphical tools characterize these fluctuations:

  CHRONOLOGICAL LOAD CURVE (24-Hour)           LOAD DURATION CURVE (LDC)
   Power (MW)                                  Power (MW)
       ^
   Pmax|      /\                               Pmax|~~~\  <-- Peaking (SCGT / Hydro)
       |     /  \     /\                           |    \ 
       |    /    \   /  \                          |=====\ <-- Intermediate (CCGT)
       |___/      \_/    \_                        |      \ 
   Pmin|====================                   Pmin|=======\__ <-- Base Load (Nuclear)
       +------------------------> Time (Hours)     +------------------------> Hours / %
       0           12        24                    0                      8760 (100%)

1. Chronological Load Curve

Plots system electrical power demand ($P$ in $\text{MW}$) as a function of chronological time (hours). The integral (area under the curve) represents total electrical energy generated in megawatt-hours ($\text{MWh}$):

$E_{total} = \int_{0}^{T} P(t)\, dt$

2. Load Duration Curve (LDC)

Constructed by rearranging the chronological load data in descending order of magnitude. The vertical axis represents power ($\text{MW}$), and the horizontal axis represents the number of hours (or percentage of time) during the year that system load equaled or exceeded that value.

Generation Categories on the LDC

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2. Classical Economic Dispatch (Equal Incremental Cost Criterion)

The objective of Economic Dispatch is to allocate a total power demand ($P_D$) among $N$ operating generating units such that total fuel cost is minimized without violating operational limits.

Generator Fuel Cost Curves

The operating cost of a thermal generating unit is conventionally modeled as a quadratic function of its real electrical power output ($P_i$ in $\text{MW}$):

$C_i(P_i) = a_i P_i^2 + b_i P_i + c_i\quad [\text{USD}/\text{hour}]$

Where $a_i$ ($\text{USD}/\text{MW}^2\text{h}$), $b_i$ ($\text{USD}/\text{MWh}$), and $c_i$ ($\text{USD}/\text{h}$, no-load cost) are empirical cost coefficients.

Incremental Fuel Cost ($\lambda_i$)

The incremental fuel cost is the first derivative of the cost curve with respect to power output, representing the cost to produce one additional $\text{MWh}$ of electrical energy:

$\lambda_i = \frac{dC_i}{dP_i} = 2 a_i P_i + b_i\quad [\text{USD}/\text{MWh}]$

The Equal Incremental Cost Criterion (Neglecting Losses)

To minimize total fuel cost $\sum C_i(P_i)$ subject to the power balance constraint $\sum P_i = P_D$, the Lagrangian multiplier method dictates that all units not at a generation limit must operate at the identical incremental cost ($\lambda$):

$\lambda_1 = \lambda_2 = \cdots = \lambda_N = \lambda_{system}$

$\frac{dC_1}{dP_1} = \frac{dC_2}{dP_2} = \cdots = \frac{dC_N}{dP_N} = \lambda$

   Incremental Cost λ ($/MWh)
       ^
       |                          / Unit 2: λ2 = 2 a2 P2 + b2
   λsys|-------------------------+---------
       |                        /|        /
       |          Unit 1: λ1   / |       /  λ1 = 2 a1 P1 + b1
       |               ======+/  |      /
       |                    /|   |     /
     b2|.................../ |   |    /
     b1|................../  |   |   /
       +---------------------+---+-------------------------> Generation (MW)
       0                    P1  P2  P_total = P1 + P2 = PD

Handling Generator Operational Constraints

Each generator is constrained by mechanical and thermal limits: $P_{i,min} \le P_i \le P_{i,max}$.

• If an unconstrained solution yields $P_i < P_{i,min}$, set $P_i = P_{i,min}$ (unit operates at maximum allowed efficiency floor; $\lambda_i > \lambda$).
• If an unconstrained solution yields $P_i > P_{i,max}$, set $P_i = P_{i,max}$ (unit is pinned at its ceiling; $\lambda_i < \lambda$).
• The remaining unconstrained units are then redispatched to supply the balance of load: $P_{D,remaining} = P_D - \sum P_{pinned}$.

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3. Economic Dispatch with Transmission Losses

In large transmission networks, transmission line $I^2 R$ losses ($P_{loss}$) cannot be neglected. Kron's Loss Formula models transmission losses using $B$-coefficients:

$P_{loss} = \sum_{i=1}^{N} \sum_{j=1}^{N} P_i B_{ij} P_j + \sum_{i=1}^{N} B_{0i} P_i + B_{00}$

The Exact Coordination Equation & Penalty Factors

When transmission losses are included, the economic dispatch coordination equation becomes:

$\lambda_i \times L_i = \lambda_{system}$

Where $L_i$ is the Transmission Loss Penalty Factor for generator $i$:

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

$\left( 2 a_i P_i + b_i \right) \times \left( \frac{1}{1 - \frac{\partial P_{loss}}{\partial P_i}} \right) = \lambda_{system}$

Where $\frac{\partial P_{loss}}{\partial P_i}$ is the incremental transmission loss (ITL) of unit $i$. Generators close to major load centers have small or negative ITLs ($L_i \approx 1.0$), while remote generators have large positive ITLs ($L_i > 1.0$), forcing them to generate less power than in the lossless case.

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4. Industrial Energy Efficiency: Motors & Affinity Laws

In industrial facilities, electric motor-driven systems (pumps, fans, compressors) account for over $65%$ of all consumed electrical energy.

Motor Life Cycle Cost (LCC) & NEMA Premium Efficiency

Over an average 15- to 20-year operating life of an industrial induction motor:

• Purchase Price: $\approx 2%\text{--}4%$ of total life cycle cost
• Maintenance & Repair: $\approx 1%\text{--}3%$ of total life cycle cost
• Electrical Energy Consumed: $\approx 93%\text{--}97%$ of total life cycle cost!

NEMA Premium (NEMA MG 1 Table 12-12) and IEEE 841 severe-duty motors achieve higher efficiency by using thinner, higher-grade silicon steel laminations (reducing core hysteresis/eddy losses), larger copper conductor cross-sections (reducing stator $I^2 R$ losses), and optimized rotor bar geometries.

   Induction Motor Loss Breakdown
   +--------------------------------------------------------------------------+
   | Stator I²R (35-40%) | Rotor I²R (15-20%) | Core (15-20%) | F&W | Stray (10%)|
   +--------------------------------------------------------------------------+

Affinity Laws for Centrifugal Pumps and Fans

For centrifugal machines (pumps, fans, blowers), operating parameters vary with rotational speed ($n$, in RPM or Hz) according to the Affinity Laws:

$\text{Law 1 (Volumetric Flow Rate, } Q\text{):}\quad \frac{Q_1}{Q_2} = \frac{n_1}{n_2}$

$\text{Law 2 (Head / Differential Pressure, } H\text{):}\quad \frac{H_1}{H_2} = \left( \frac{n_1}{n_2} \right)^2$

$\text{Law 3 (Brake Shaft Power, } P\text{):}\quad \frac{P_1}{P_2} = \left( \frac{n_1}{n_2} \right)^3$

   Power (kW / hp)
       ^
   100%|----------------------------------------- Full Speed (Throttling Valve)
       |                                       / 
    80%|                                      /   Wasted Throttling Energy
       |                                     /  
    50%|                     +..............+ <-- VFD Power (Cube Law: P ∝ n³)
       |                    / (51.2% Power at 80% Speed)
       |                   / 
       +------------------+---------------------------------> Flow Rate (Q)
       0                 80%               100%

VFD Variable Speed Control vs. Mechanical Throttling

• Mechanical Throttling (Control Valve / Damper): Motor runs at full speed ($100%$ RPM). Throttling introduces a large pressure drop across the valve. Power drops only marginally ($P \propto Q^{0.8\text{--}1.0}$).
• Variable Frequency Drive (VFD): Modulates motor inverter frequency ($f$), reducing shaft speed ($n$). Shaft power drops with the cube of speed ($P \propto n^3$).
• Energy Savings Calculation: Reducing flow to $80%$ ($Q_2/Q_1 = 0.80$) via VFD requires: $P_2 = P_1 \times (0.80)^3 = 0.512 \times P_1\quad (51.2\%\text{ of rated power})$ Energy savings = $100% - 51.2% = 48.8%$!

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5. Step-by-Step Worked Calculation Example

Problem Statement

Two thermal generating units supply a total system load demand of $P_D = 400.0\text{ MW}$. Transmission losses are neglected.

Generator 1:

• Cost function: $C_1(P_1) = 0.004, P_1^2 + 8.0, P_1 + 200.0\quad [\text{USD}/\text{h}]$
• Operating limits: $50.0\text{ MW} \le P_1 \le 300.0\text{ MW}$

Generator 2:

• Cost function: $C_2(P_2) = 0.006, P_2^2 + 7.0, P_2 + 150.0\quad [\text{USD}/\text{h}]$
• Operating limits: $40.0\text{ MW} \le P_2 \le 250.0\text{ MW}$

Calculate:

1. The incremental cost functions ($\lambda_1$ and $\lambda_2$).
2. The system incremental cost ($\lambda_{sys}$) and the optimal power dispatch ($P_1$ and $P_2$).
3. Verify that operating limits are satisfied.
4. The total hourly operating cost ($C_{total}$) of the system.

Solution Walkthrough

Step 1: Formulate Incremental Cost Functions

$\lambda_1 = \frac{dC_1}{dP_1} = 2(0.004) P_1 + 8.0 = 0.008\, P_1 + 8.0\quad [\text{USD}/\text{MWh}]$

$\lambda_2 = \frac{dC_2}{dP_2} = 2(0.006) P_2 + 7.0 = 0.012\, P_2 + 7.0\quad [\text{USD}/\text{MWh}]$

Step 2: Apply Equal Incremental Cost Criterion

Set $\lambda_1 = \lambda_2 = \lambda$ and solve for each generator's output as a function of $\lambda$:

$P_1 = \frac{\lambda - 8.0}{0.008} = 125.0\, \lambda - 1000.0$

$P_2 = \frac{\lambda - 7.0}{0.012} = 83.3333\, \lambda - 583.3333$

Apply the total power balance constraint ($P_1 + P_2 = P_D = 400.0\text{ MW}$):

$(125.0\, \lambda - 1000.0) + (83.3333\, \lambda - 583.3333) = 400.0$

$208.3333\, \lambda - 1583.3333 = 400.0$

$208.3333\, \lambda = 1983.3333$

$\lambda_{sys} = \frac{1983.3333}{208.3333} = \mathbf{9.520\text{ USD/MWh}}$

Now calculate individual generator power outputs:

$P_1 = 125.0(9.520) - 1000.0 = 1190.0 - 1000.0 = \mathbf{190.0\text{ MW}}$

$P_2 = 83.3333(9.520) - 583.3333 = 793.333 - 583.333 = \mathbf{210.0\text{ MW}}$

Power Balance Check: $P_1 + P_2 = 190.0\text{ MW} + 210.0\text{ MW} = 400.0\text{ MW}$ (Exact match).

Step 3: Verify Unit Operating Limits

• Generator 1: $50.0\text{ MW} \le 190.0\text{ MW} \le 300.0\text{ MW}$ (Within limits).
• Generator 2: $40.0\text{ MW} \le 210.0\text{ MW} \le 250.0\text{ MW}$ (Within limits).

Since neither limit is violated, the unconstrained optimal dispatch is valid.

Step 4: Calculate Total Hourly Operating Cost

$C_1(190.0) = 0.004(190.0)^2 + 8.0(190.0) + 200.0 = 0.004(36{,}100) + 1520.0 + 200.0 = 144.40 + 1520.0 + 200.0 = 1{,}864.40\text{ USD/h}$

$C_2(210.0) = 0.006(210.0)^2 + 7.0(210.0) + 150.0 = 0.006(44{,}100) + 1470.0 + 150.0 = 264.60 + 1470.0 + 150.0 = 1{,}884.60\text{ USD/h}$

$C_{total} = C_1 + C_2 = 1{,}864.40 + 1{,}884.60 = \mathbf{3{,}749.00\text{ USD/hour}}\quad (3{,}749.00\text{ USD/hour})$

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6. Common NCEES Exam Pitfalls

Pitfall 1: Equating Total Costs Rather Than Incremental Costs
Never set total cost functions equal ($C_1 = C_2$). Economic dispatch requires equating the derivatives (marginal costs $\lambda_1 = \lambda_2$). Total cost values have no bearing on marginal optimality.

Pitfall 2: Ignoring Generation Boundary Violations
Always check unit limits! If the equal lambda equation yields $P_1 = 320\text{ MW}$ for a unit rated at $300\text{ MW}$, you must fix $P_1 = 300\text{ MW}$ and redispatch remaining units to supply $P_D - 300\text{ MW}$.

Pitfall 3: Applying Linear Flow Relationships to Fan/Pump Power
Remember the Affinity Laws: Flow varies linearly ($Q \propto n$), Pressure varies quadratically ($H \propto n^2$), and Power varies cubically ($P \propto n^3$). Halving flow ($50% flow) reduces power to $(0.5)^3 = 0.125$ ($12.5% of original), NOT $50%!$