Economic Operation of Power Sy

Cheatsheet Content

### Introduction to Economic Dispatch - **Definition:** Economic load dispatch determines the optimal scheduling of generators to minimize operating costs for a given load demand, while satisfying equality and inequality constraints. - **Two separate steps:** - **Unit Commitment (Pre-dispatch):** Selects optimally from available units to meet system load over a period, providing a specified reserve margin. - **Online Load Dispatch:** Distributes load among parallel generating units. ### System Constraints - **Equality Constraints:** Generation must meet demand plus losses. $$\sum_{i=1}^{n} P_{Gi} - P_D - P_{loss} = 0$$ - **Inequality Constraints:** - Generator Constraints, Voltage Constraints, Transformer Tap Settings etc. - **Hard type:** Definite and specific (e.g., tapping range of on-load tap changing transformer). - **Soft type:** Some flexibility (e.g., nodal voltages). ### Cost of Generation - **Cost Function of Unit `i`:** $$C_i = a_i P_{Gi}^2 + b_i P_{Gi} + d_i \quad (1)$$ - **Incremental Fuel Cost (IFC):** Slope of the cost curve. $$\frac{dC_i}{dP_{Gi}} = IFC = 2a_i P_{Gi} + b_i \quad (2)$$ ### Optimal Scheduling Without Losses - **Generator Operating Limits:** $$P_{Gi,min} \le P_{Gi} \le P_{Gi,max} \quad \text{for } i = 1, 2, ..., n$$ - **Objective:** Minimize total operating cost. $$C = \sum_{i=1}^{n} C_i(P_{Gi}) \quad (4)$$ - **Equality Constraint (Load Demand):** $$\sum_{i=1}^{n} P_{Gi} - P_D = 0 \quad (5)$$ - **Lagrange Function:** Solves this constrained optimization problem. $$L = \sum_{i=1}^{n} C_i(P_{Gi}) + \lambda \left( P_D - \sum_{i=1}^{n} P_{Gi} \right)$$ - **Minimization Condition:** $$\frac{\partial L}{\partial P_{Gi}} = 0 \Rightarrow \frac{dC_i}{dP_{Gi}} = \lambda \quad \text{for } i = 1, 2, ..., n \quad (8)$$ This implies that the incremental cost of all generators must be equal to $\lambda$ (Lagrange multiplier). $$\frac{dC_1}{dP_{G1}} = \frac{dC_2}{dP_{G2}} = ... = \frac{dC_n}{dP_{Gn}} = \lambda$$ - **Example Calculation (Load = 50 MW):** Given fuel input functions $F_1 = (8P_1 + 0.024 P_1^2 + 80) \times 2 \text{ Rs/hr}$ and $F_2 = (6P_2 + 0.04 P_2^2 + 120) \times 2 \text{ Rs/hr}$. Incremental costs: $\frac{dC_1}{dP_1} = 0.096 P_1 + 16$ and $\frac{dC_2}{dP_2} = 0.16 P_2 + 12$. For economic loading, $\frac{dC_1}{dP_1} = \frac{dC_2}{dP_2}$ and $P_1 + P_2 = 50 \text{ MW}$. Solving gives $P_1 = 15.625 \text{ MW}$ and $P_2 = 34.375 \text{ MW}$. Costs: $C_1 = 421.736 \text{ Rs/hr}$ and $C_2 = 747 \text{ Rs/hr}$. - **Example Calculation (Load = 150 MW):** $P_1 = 78.126 \text{ MW}$ and $P_2 = 71.874 \text{ MW}$. Costs: $C_1 = 1702.992 \text{ Rs/hr}$ and $C_2 = 1515.74 \text{ Rs/hr}$. - **Total Cost (Economic Operation):** Sum of costs for 50 MW and 150 MW loads, multiplied by 12 hours each. $C_{opt} = (421.736 + 747 + 1702.992 + 1515.74) \times 12 = 52649.61 \text{ Rs/day}$. - **Equal Sharing (Comparison):** - For 50 MW load: $P_1=P_2=25 \text{ MW}$. $C_{1eq}=590 \text{ Rs/hr}$, $C_{2eq}=590 \text{ Rs/hr}$. - For 150 MW load: $P_1=P_2=75 \text{ MW}$. $C_{1eq}=1630 \text{ Rs/hr}$, $C_{2eq}=1590 \text{ Rs/hr}$. - Total cost (Equal Sharing): $C_{eq} = (590+590+1630+1590) \times 12 = 52800 \text{ Rs/day}$. - **Savings:** $C_{eq} - C_{opt} = 52800 - 52649.61 = 150.39 \text{ Rs/day}$. ### Optimal Scheduling With Losses - **Objective:** Minimize total operating cost considering losses. $$C = \sum_{i=1}^{n} C_i(P_{Gi}) \quad (11)$$ - **Equality Constraint (Load Demand and Losses):** $$\sum_{i=1}^{n} P_{Gi} - P_D - P_{loss} = 0 \quad (12)$$ - **Loss Formula:** $$P_{loss} = f(P_{G1}, P_{G2}, ..., P_{Gn})$$ - **Lagrange Function:** $$L = \sum_{i=1}^{n} C_i(P_{Gi}) + \lambda \left( P_D + P_{loss} - \sum_{i=1}^{n} P_{Gi} \right)$$ - **Minimization Condition:** $$\frac{\partial L}{\partial P_{Gi}} = 0 \Rightarrow \frac{dC_i}{dP_{Gi}} - \lambda \left( 1 - \frac{\partial P_{loss}}{\partial P_{Gi}} \right) = 0$$ $$\frac{dC_i}{dP_{Gi}} = \lambda \left( 1 - \frac{\partial P_{loss}}{\partial P_{Gi}} \right) \quad \text{for } i = 1, 2, ..., n \quad (15)$$ - **Incremental Transmission Loss (ITL):** $$ITL_i = \frac{\partial P_{loss}}{\partial P_{Gi}}$$ - **Coordination Equation:** $$\frac{dC_i}{dP_{Gi}} = \lambda (1 - ITL_i) \quad \text{or } (IC)_i = \lambda [1 - (ITL)_i] \quad (17)$$ - **Penalty Factor ($L_i$):** $$L_i = \frac{1}{1 - \frac{\partial P_{loss}}{\partial P_{Gi}}}$$ Thus, $\frac{dC_i}{dP_{Gi}} L_i = \lambda \quad (18)$ - **General Form of Loss Formula (B-coefficients):** $$P_{loss} = \sum_{i=1}^{n} \sum_{j=1}^{n} P_{Gi} B_{ij} P_{Gj} \quad (19)$$ Where $B_{ij}$ are loss coefficients, assumed constant under certain operating conditions. - **Example Calculation:** Given: dF1/dP1 = 0.025 P1 + 15, dF2/dP2 = 0.05 P2 + 20. A load of 125 MW transmitted from plant 1 incurs 15.625 MW loss. $P_L = B_{11}P_1^2 \Rightarrow 15.625 = B_{11} \times 125^2 \Rightarrow B_{11} = 0.001$. Coordination equation for plant 1: $\frac{dF_1}{dP_1} + \lambda \frac{\partial P_L}{\partial P_1} = \lambda$. $\frac{\partial P_L}{\partial P_1} = 0.002 P_1$. $0.025 P_1 + 15 + \lambda(0.002 P_1) = \lambda$. $0.025 P_1 + 15 = \lambda(1 - 0.002 P_1)$. For a given $\lambda$ (cost of received power) = 24 Rs/MWhr: $0.025 P_1 + 15 = 24(1 - 0.002 P_1) \Rightarrow 0.025 P_1 + 15 = 24 - 0.048 P_1$. $0.073 P_1 = 9 \Rightarrow P_1 = 123.28 \text{ MW}$. For plant 2: $0.05 P_2 + 20 = 24 \Rightarrow 0.05 P_2 = 4 \Rightarrow P_2 = 80 \text{ MW}$. Transmission loss: $P_L = 0.001 \times (123.28)^2 = 15.19 \text{ MW}$. Load demand: $P_D = P_1 + P_2 - P_L = 123.28 + 80 - 15.19 = 188.1 \text{ MW}$. ### Transmission Loss Formula - **Current in line b (source 1 only):** $M_{b1} = \frac{I_{b1}}{I_L}$ - **Current in line b (source 2 only):** $M_{b2} = \frac{I_{b2}}{I_L}$ - **Total current in branch b (superposition):** $I_b = M_{b1} I_1 + M_{b2} I_2$ - **Total losses ($P_L$):** $$P_L = \sum_{b=1}^{k} |I_b|^2 R_b$$ If $I_1$ and $I_2$ are currents from plant 1 and 2 respectively, and $M_{b1}, M_{b2}$ are current distribution factors. Assuming X/R ratio is same for all lines, and phase angle of all load currents are same, then $M_{b1}, M_{b2}$ are real. $$P_L = \sum_{b=1}^{k} R_b |M_{b1} I_1 + M_{b2} I_2|^2$$ The general loss formula in terms of power can be derived as: $$P_L = B_{11}P_1^2 + B_{22}P_2^2 + 2B_{12}P_1P_2$$ Where $B_{11}, B_{22}, B_{12}$ are B-coefficients relating to the network configuration and impedance. ### Unit Commitment - **Definition:** The process of determining the optimal schedule (ON/OFF status) of generating units over a specific period (e.g., 24 hours to a week) to meet the forecasted load at minimum total operating cost. - **Purpose:** Unlike Economic Dispatch which determines how much power a committed unit should produce, Unit Commitment determines which units should be available in the first place. - **Key aspects:** - Unit Commitment selects optimally out of the available units to meet the system load and provide a specified margin of reserve. - Energy requirements and demand are constantly changing. - Requires sufficient generating units to meet load economically with enough reserve capacity for abnormal operating conditions (e.g., unit failure). - Ensures reliable and economical supply. ### Constraints on Unit Commitment - **Spinning Reserve:** Total generation available from synchronized units minus load demand and losses. Essential for interconnected systems to handle sudden load changes or unit losses. Needs to be spread among fast and slow responding units. - **Thermal Constraints:** - **Minimum Up-Time:** Units in operation cannot be switched off instantly; they require a minimum up-time. - **Minimum Down-Time:** Units disconnected from the grid require a minimum down-time before reconnection. - **Start-up Costs:** Cost incurred to bring a thermal unit from cold to an operational state. - **Crew Constraints:** Optimum number of operating personnel required. - **Ramp Rates:** Limits on how fast a unit's electrical output can change to prevent turbine damage. - **Hydro Constraints:** - **Water Availability:** Scheduling constrained by reservoir volume and discharge rates. - **Cascading Plants:** Coordination required in "hydro valleys" where water flows from upstream to downstream plants. - **Pumped Storage:** Units that pump water uphill during low-demand periods and generate during peak periods. - **Must-Run Units:** Certain units are kept online regardless of cost for voltage support, grid stability, or industrial supply. - **Fuel Constraints:** Limited fuel supply or "take-or-pay" contracts requiring a specific amount of fuel to be burned. ### Methods for Solving Unit Commitment Problem 1. **Priority List:** Simplest method; units are ranked based on full load average production cost (or other criteria) to create a priority list. 2. **Dynamic Programming** 3. **Lagrange Multiplier** ### Unit Commitment Example - **Question:** A plant has three units with given specifications. Determine which units should be committed to feed a load of 500 MW economically. - Unit 1: Min 160 MW, Max 600 MW, $F_1 = 600 + 7.1 P_1 + 0.00141 P_1^2$. Fuel cost = 1.1 Rs/MBtu. - Unit 2: Min 100 MW, Max 450 MW, $F_2 = 350 + 7.80 P_2 + 0.00195 P_2^2$. Fuel cost = 1.0 Rs/MBtu. - Unit 3: Min 50 MW, Max 250 MW, $F_3 = 80 + 8.0 P_3 + 0.0049 P_3^2$. Fuel cost = 1.2 Rs/MBtu. - **Solution - Average Production Cost Calculation (Full Load):** - **Unit 1 (600 MW):** $F_1 = 600 + 7.1(600) + 0.00141(600)^2 = 5367.6 \text{ MBTu/hr}$. Average cost = $5367.6 \times 1.1 / 600 = 9.84 \text{ Rs/MWhr}$. - **Unit 2 (450 MW):** $F_2 = 350 + 7.80(450) + 0.00195(450)^2 = 4254.87 \text{ MBTu/hr}$. Average cost = $4254.87 \times 1.0 / 450 = 9.455 \text{ Rs/MWhr}$. - **Unit 3 (250 MW):** $F_3 = 80 + 8.0(250) + 0.0049(250)^2 = 2386.25 \text{ MBTu/hr}$. Average cost = $2386.25 \times 1.2 / 250 = 11.45 \text{ Rs/MWhr}$. - **Priority List (based on average cost, ascending):** | Unit | Rs/MWhr | Min MW | Max MW | |------|---------|--------|--------| | 2 | 9.455 | 100 | 450 | | 1 | 9.8455 | 160 | 600 | | 3 | 11.45 | 50 | 250 | - **Unit Commitment Scheme (combinations to meet 500 MW load):** | Combination | Min. MW from combination | Max. MW from combination | |-------------|--------------------------|--------------------------| | 2+1+3 | 310 | 1300 | | 2+1 | 260 | 1050 | | 2 | 100 | 450 | ### What is SCADA - **Definition:** SCADA (Supervisory Control And Data Acquisition) is an industrial control system used to monitor, control, and automate processes in various industries (power generation, water treatment, oil & gas, manufacturing, transportation). - **Functionality:** Integrates hardware and software to collect real-time data. ### Main Functions of SCADA - **Data Acquisition:** Collects data from sensors, meters, and field devices. - **Monitoring:** Displays real-time operational data (pressure, voltage, temperature, flow rate, etc.). - **Control:** Allows operators to remotely open/close valves, start/stop motors, or adjust setpoints. - **Data Storage & Analysis:** Archives historical data for trend analysis and decision-making. - **Alarm Handling:** Alerts operators to abnormal or unsafe conditions. ### Components of SCADA System - **Field Devices (Sensing & Actuation Layer):** - **Sensors/Transducers:** Measure parameters (temperature, pressure, flow, voltage). - **Actuators:** Devices (valves, motors, circuit breakers) that act based on control signals. - **RTUs and PLCs (Data Acquisition & Local Control Layer):** - **RTU (Remote Terminal Unit):** Collects sensor data, sends to control center, executes commands. Designed for real-time operation and high reliability. - **PLC (Programmable Logic Controller):** Performs local automation (logic-based control), often used instead of RTUs. - **Communication Network (Data Transmission Layer):** - Connects field devices/RTUs/PLCs with the control center. - Can be wired (Ethernet, fiber optics) or wireless (radio, satellite, cellular). - Ensures secure and reliable data transfer. - **Control Center (Supervisory & Management Layer):** - **SCADA Master Station / Servers:** Central computer system that receives, processes, and stores data. - **HMI (Human Machine Interface):** Graphical interface for operators to monitor, analyze, and control processes. - **Database / Historian:** Archives real-time and historical data for analysis, reporting, and decision-making. ### Subsynchronous Resonance (SSR) - **Definition:** A phenomenon in power systems where energy is exchanged between the electrical network and turbine generators at frequencies below the synchronous frequency, potentially leading to damaging oscillations. - **Mechanism:** Oscillatory attributes of electrical and mechanical variables associated with turbine generators, when coupled to a series compensated transmission system, lead to lightly damped, undamped, or negatively damped and growing oscillatory energy interchange. - **Series Compensation:** Deployed in transmission lines to enhance power transmission capability. The level of series compensation is the ratio of impedance of the series capacitor. - **SSR Condition:** When multi-mass turbine generators are connected to series compensated lines, sustained or growing oscillations occur due to energy exchange at one or more natural frequencies of electrical and mechanical components of the combined system. - Typically, series compensation ranges from 20–80%. ### References - Nagrath and Kothari, *Modern Power System Analysis*, McGrawHill, New York - C. L. Wadhwa, *Electric Power System*, New Age International (P) Limited Publishers