Introduction
- The goal is to minimize production cost while maintaining voltage magnitudes at each bus.
- Load conditions vary with time (day/night, season), so constant power generation is unrealistic.
- Power generation should follow load patterns, which change with seasons.
- To optimize dispatch, we focus on two aspects:
- Economic operation must adapt to load conditions.
- Turbine-governor control must ensure optimal generation.
Economic Operation of Power Systems
- Power was initially supplied by the most efficient plant at light load.
- As load increased, the most efficient plant supplied until its max efficiency was reached.
- The next most efficient plant then supplied power until its max efficiency.
- This continued from the most to the least efficient plants to meet peak demand.
- However, this method failed to minimize total cost of generation.
- An alternative method is needed to account for total generation costs of all units.
Economic Distribution of Loads between the Units of a Plant
- Variable operating costs of each unit are expressed in terms of power output.
- The fuel cost is the dominant cost in thermal or nuclear units.
- Fuel cost must be expressed as a function of power output.
- Other costs, like operation and maintenance, can also be expressed in terms of power output.
- Fixed costs, like capital and depreciation, are not included in fuel cost.
- Fuel requirement is given in Rupees per hour.
- For unit-\(i\), input cost \(f_i\) (Rs./h) is expressed as:
\[f_i = \left[\dfrac{a_i}{2}P_i^2 + b_iP_i + c_i\right]~Rs/h\]
- Operating cost is approximated by a quadratic function of power (MW) versus cost (Rs).
- Incremental operating cost is calculated as:
\[ \lambda_{i} = \dfrac{df_{i}}{dP_{i}} = \left(a_{i}P_{i} + b_{i}\right)~\text{Rs/MWh} \]
- For two units with different incremental costs, load transfer from higher to lower incremental cost reduces the total cost.
- This continues until incremental costs are equal for both units, which is the optimal point of operation.
- This principle extends to \(N\) units in a plant.
- Total fuel cost \(f_T\) is the sum of individual costs:
\[f_{T}=f_{1}+f_{2}+\cdots+f_{N}={\displaystyle \sum_{k=1}^{N}f_{k}}\]
- Total power \(P_T\) is the sum of powers supplied by each unit:
\[P_{T}= P_{1}+P_{2}+\cdots+P_{N}={\displaystyle \sum_{k=1}^{N}P_{k}}\]
- The objective is to minimize \(f_T\) for a given \(P_T\), achieved when:
\[df_{T}=\dfrac{\partial f_{T}}{\partial P_{1}}dP_{1}+\dfrac{\partial f_{T}}{\partial P_{2}}dP_{2}+\cdots+\dfrac{\partial f_{T}}{\partial P_{N}}dP_{N}=0\]
- Since total power is constant:
\[dP_{T}= dP_{1}+dP_{2}+\cdots+dP_{N}=0\]
- Multiplying by \(\lambda\) and subtracting:
\[\left(\dfrac{\partial f_{T}}{\partial P_{1}}-\lambda\right)dP_{1}+\left(\dfrac{\partial f_{T}}{\partial P_{2}}-\lambda\right)dP_{2}+\cdots+\left(\dfrac{\partial f_{T}}{\partial P_{N}}-\lambda\right)dP_{N}=0\]
- The equality holds when:
\[\dfrac{\partial f_{T}}{\partial P_{i}}-\lambda=0,~i=1,\cdots,N\]
- This leads to:
\[\lambda=\dfrac{df_{1}}{dP_{1}}=\dfrac{df_{2}}{dP_{2}}=\cdots=\dfrac{df_{N}}{dP_{N}} \Leftarrow~\text{coordination equation}\]
Figure 1: Economic Load Optimization Curve
Generating Limits
- Not all units are always available to share the load due to scheduled maintenance.
- It is not necessary to switch off less efficient units during off-peak hours.
- Shutting down and starting up units incurs costs and may take over 8 hours to restore and synchronize.
- To handle sudden power demand changes, extra units may be required, creating a spinning reserve.
- The optimal load dispatch must include startup and shutdown costs while ensuring system security.
- Power generation limits are constrained by:
\[P_{min,i} \leq P_i \leq P_{max,i}, ~ i=1,\dots,N\]
- \(P_{max}\) is the maximum power capacity of each unit.
- \(P_{min}\) ensures thermal stability of the boiler in thermal or nuclear stations.
- A unit must produce a minimum power to maintain the boiler's design operating temperature.
Economic Sharing of Loads between Different Plants
- We previously considered the economic operation of a single plant, focusing on how load is shared among its units.
- This analysis ignored transmission line losses, assuming they were part of the load.
- When considering multiple plants connected by transmission lines, line losses must be included in the economic dispatch.
- Including transmission losses in the economic dispatch:
\[P_T = P_1 + P_2 + \cdots + P_N - P_{Loss}\]Since \(P_T\) is constant, we have:\[0 = dP_1 + dP_2 + \cdots + dP_N - dP_{Loss}\]where:\[dP_{Loss} = \dfrac{\partial P_{Loss}}{\partial P_1}dP_1 + \dfrac{\partial P_{Loss}}{\partial P_2}dP_2 + \cdots + \dfrac{\partial P_{Loss}}{\partial P_N}dP_N\]
- Minimum generation cost implies \(df_T = 0\):
\[0 = \left(\lambda \dfrac{\partial P_{Loss}}{\partial P_1} - \lambda \right) dP_1 + \cdots + \left(\lambda \dfrac{\partial P_{Loss}}{\partial P_N} - \lambda \right) dP_N\]Summing up:\[0 = \sum_{i=1}^{N} \left(\dfrac{\partial f_T}{\partial P_i} + \lambda \dfrac{\partial P_{Loss}}{\partial P_i} - \lambda \right) dP_i\]
- The equation is satisfied when:
\[\dfrac{\partial f_T}{\partial P_i} + \lambda \dfrac{\partial P_{Loss}}{\partial P_i} - \lambda = 0, \quad i = 1, \cdots, N\]
- Thus:
\[\lambda = \dfrac{df_1}{dP_1} L_1 = \dfrac{df_2}{dP_2} L_2 = \cdots = \dfrac{df_N}{dP_N} L_N\]where \(L_i\) is the penalty factor for load-\(i\), given by:\[L_i = \dfrac{1}{1 - \dfrac{\partial P_{Loss}}{\partial P_i}}, \quad i = 1, \cdots, N\]
- For an area with \(N\) units, the power generated is represented by the vector:
\[P = \left[\begin{array}{cccc} P_1 & P_2 & \cdots & P_N \end{array}\right]^T\]
- Transmission losses are expressed as:
\[P_{Loss} = P^T B P\]where \(B\) is a symmetric \(N \times N\) matrix, and \(B_{ij}\) are the loss coefficients.\[B=\left[\begin{array}{cccc} B_{11} & B_{12} & \cdots & B_{1 N} \\ B_{12} & B_{22} & \cdots & B_{2 N} \\ \vdots & \vdots & \ddots & \vdots \\ B_{1 N} & B_{2 N} & \cdots & B_{N W} \end{array}\right]\]
- These coefficients vary with plant loading, but for simplicity, they are often assumed constant when calculating the penalty factor \(L_i\).
Economic Dispatch Problem-1
Consider two units of a power plant with fuel costs given by the following equations:
\[\begin{aligned}
F_1 & = 0.2 P_1^2 + 40 P_1 + 120 ~ \text{Rs}/\text{h}\\
F_2 & = 0.25 P_2^2 + 30 P_2 + 150 ~ \text{Rs}/\text{h}
\end{aligned}\]
- Determine the economic operating schedule and the corresponding cost of generation for a demand of 180 MW.
- If the load is equally shared by both units, determine the savings obtained by loading the units optimally.
Solution
Part 1 - Economic Dispatch
- For economical dispatch, the fuel cost derivative relative to each unit's output power must be equal.
\[\frac{dF_1}{dP_1} = \frac{dF_2}{dP_2}\]
- Taking the derivatives of the fuel cost equations:
\[0.4 P_1 + 40 = 0.5 P_2 + 30\]
- The total demand must be met:
\[P_1 + P_2 = 180\]
- Solving this system of equations:
\[P_1 = 88.89 \, \text{MW}, \quad P_2 = 91.11 \, \text{MW}\]
- The total cost of generation is the sum of the individual costs:
\[F_T = F_1 + F_2 = 10,214.43 \, \text{Rs}/\text{h}\]
Part 2: Equal Load Sharing
- If the load is equally shared by both units:
\[P_1 = 90 \, \text{MW}, \quad P_2 = 90 \, \text{MW}\]
- The total cost of generation in this case is:
\[F_T = 10,215 \, \text{Rs}/\text{h}\]
- Thus, the savings obtained by optimally loading the units is:
\[\text{Savings} = 10,215 - 10,214.43 = 0.57 \, \text{Rs}/\text{h}\]
Economic Dispatch Problem-2
The fuel cost functions for three thermal plants are:
\[\begin{array}{r}
F_1 = 0.4 P_1^2 + 10 P_1 + 25 \, \text{Rs/h} \\
F_2 = 0.35 P_2^2 + 5 P_2 + 20 \, \text{Rs/h} \\
F_3 = 0.475 P_3^2 + 15 P_3 + 35 \, \text{Rs/h}
\end{array}\]
The generation limits of the units are:
\[\begin{aligned}
30 \, \text{MW} &\leq P_1 \leq 500 \, \text{MW} \\
30 \, \text{MW} &\leq P_2 \leq 500 \, \text{MW} \\
30 \, \text{MW} &\leq P_3 \leq 250 \, \text{MW}
\end{aligned}\]
Find the optimum schedule for a load of 1000 MW.
Solution
- For optimum dispatch:
\[\frac{d F_1}{d P_1} = \frac{d F_2}{d P_2} = \frac{d F_3}{d P_3}\]
- This leads to the following equations:
\[\begin{aligned} 0.8 P_1 + 10 & = 0.7 P_2 + 5 \\ 0.7 P_2 + 5 & = 0.9 P_3 + 15 \end{aligned}\]
- The total power balance constraint:
\[P_1 + P_2 + P_3 = 1000\]
- By solving the above system of equations, we get:
\[P_1 = 334.3829 \, \text{MW}, \quad P_2 = 389.2947 \, \text{MW}, \quad P_3 = 276.3224 \, \text{MW}\]
- However, unit 3 violates its maximum limit, so we set:
\[P_3 = 250 \, \text{MW}\]
- The remaining load (750 MW) is distributed optimally between units 1 and 2.
- The equations to solve now are:
\[\begin{array}{r} 0.8 P_1 + 10 = 0.7 P_2 + 5 \\ P_1 + P_2 = 750 \end{array}\]
- Solving these, we get the final load distribution as:
\[P_1 = 346.6667 \, \text{MW}, \quad P_2 = 403.3333 \, \text{MW}, \quad P_3 = 250 \, \text{MW}\]
Economic Dispatch Problem-3
Consider a two bus system.
The incremental fuel cost characteristics of plant 1 and plant 2 are:
\[\begin{aligned}
& \frac{d F_1}{d P_1}=0.025 P_1+14 \mathrm{Rs} / \mathrm{MWHr} \\
& \frac{d F_2}{d P_2}=0.05 P_2+16 \mathrm{Rs} / \mathrm{MWHr}
\end{aligned}\]
If 200 MW of power is transmitted from plant 1 to the load, a transmission loss of 20 MW occurs. Determine the optimal generation schedule and the cost of the received power for a load demand of 204.41 MW.
Solution
\[P_L=\left[\begin{array}{ll}
P_1 & P_2
\end{array}\right]\left[\begin{array}{ll}
B_{11} & B_{12} \\
B_{21} & B_{22}
\end{array}\right]\left[\begin{array}{l}
P_1 \\
P_2
\end{array}\right]\]
- Since the load is at bus \(2, P_2\) will not have any effect on \(P_L\).
\[\begin{aligned} B_{12} & =B_{21}=0 ; \quad B_{22}=0 \\ \Rightarrow~ P_L & =B_{11} P_1^2 \end{aligned}\]
- For 200 MW of \(P_1, P_L=20 \mathrm{MW}\).
\[\begin{gathered} 20=B_{11} 200^2 \\ B_{11}=0.0005 \mathrm{MW}^{-1} \\ P_L=0.0005 P_1^2 \end{gathered}\]
- For optimum dispatch,
\[L_1 \frac{d F_1}{d P_1}=L_2 \frac{d F_2}{d P_2}=\lambda\]
- Since \(P_L\) is a function of \(P_1\) alone,
\[\begin{gathered} L_1=\frac{1}{1-\frac{\partial P_L}{\partial P_1}}=\frac{1}{1-0.001 P_1} \\ L_2=1 \\ \left(\frac{1}{1-0.001 P_1}\right) 0.025 P_1+14=0.05 P_2+16 \end{gathered}\]
- On simplification,
\[\begin{aligned} & 0.041 P_1-0.05 P_2+0.00005 P_1 P_2=2 \\ & P_1+P_2-0.0005 P_1^2=204.41 \end{aligned}\]
- So,
\[\begin{array}{r} f_1\left(P_1, P_2\right)=2 \\ f_2\left(P_1, P_2\right)=204.41 \end{array}\]
- Let us solve them by \(\mathrm{N}-\mathrm{R}\) method.
\[\begin{aligned} \Delta \mathrm{f} & =\mathrm{J} \Delta \mathrm{P} \\ \Delta f &=\left[\begin{array}{c} 2-f_1\left(P_1, P_2\right) \\ 204.41-f_2\left(P_1, P_2\right) \end{array}\right] \\ \mathrm{J} &=\left[\begin{array}{ll} \frac{\partial f_1}{\partial P_1} & \frac{\partial f_1}{\partial P_2} \\ \frac{\partial f_2}{\partial P_1} & \frac{\partial f_2}{\partial P_2} \end{array}\right] \end{aligned}\]
- To find the initial estimate: Let us solve the problem without loss.
\[\begin{gathered} 0.025 P_1+14=0.05 P_2+16 \\ P_1+P_2=204.41 \\ P_1^0=162.94 ; P_2^0=41.47 \end{gathered}\]
- First Iteration:
\[\begin{aligned} \Delta f^0 & =\left[\begin{array}{c} 2-f_1\left(P_1^0, P_2^0\right) \\ 204.41-f_2\left(P_1^0, P_2^0\right) \end{array}\right]=\left[\begin{array}{c} -2.9449 \\ 13.2747 \end{array}\right] \\ J^0 & =\left[\begin{array}{cc} 0.0431 & -0.0419 \\ 0.8371 & 0.9585 \end{array}\right] \\ \left[\begin{array}{l} \Delta P_1^0 \\ \Delta P_2^0 \end{array}\right] & =\left[\begin{array}{c} -29.7060 \\ 39.7906 \end{array}\right] \\ \left[\begin{array}{l} P_1^1 \\ P_2^1 \end{array}\right] & =\left[\begin{array}{l} P_1^0 \\ P_2^0 \end{array}\right]+\left[\begin{array}{c} \Delta P_1^0 \\ \Delta P_2^0 \end{array}\right]=\left[\begin{array}{c} 133.2340 \\ 81.2606 \end{array}\right] \end{aligned}\]
- It took 6 iterations to converge.
\[P_1=133.3153 \mathrm{MW} \quad P_2=79.9812 \mathrm{MW}\]
- The cost of received power is
\[\lambda=L_2 \frac{d F_2}{d P_2}=1 \times(0.05 \times 79.9812+16)=19.9991 \mathrm{Rs} . / \mathrm{MWh}\]