Question

Rework Problem 6.62, except assume that the limit outputs are subject to the following inequality constraints:

$$\begin{gathered} 200\le P_1\le 800MW \\ 100\le P_2\le 400MW \end{gathered}$$

Short answer

The optimal dispatch value for generator $P_1$and $P_2$are 600MW and 400MW. The incremental operating cost for generator 1 and 2 are $66\frac{\$}{MWh}$and $66\frac{\$}{MWh}$respectively. The total operating cost is $39300\frac{\$}{hr}$.

Step-by-step solution

Write the given data from the question:

Write the fuel cost curve for two generators.

$$\begin{gathered} C_1\left(P_1\right)=600+18.P_1-0.04P_1^2 \\ {C}_2\left(P_2\right)=700+20.P_2-0.03P_2^2 \end{gathered}$$

Write the total load for $P_1$and $P_2$,

$$\begin{gathered} P_1+P_2=P_T \\ {P}_T=1000MW \end{gathered}$$

Determine the formulas to calculate the optimal dispatch values, incremental cost and operating cost.

The equation to calculate the incremental operating cost is given as follows.

${\left(\frac{dC}{dP}\right)}_p$ …... (1)

The total operation cost is given as follows.

$$\begin{gathered} C_T=C_1\left(P_1\right)+C_2\left(P_2\right) \\ {C}_T=600+18.P_1-0.04P_1^2+700+20.P_2-0.03P_2^2 \end{gathered}$$ …... (2)

Calculate the optimal dispatch values, incremental cost and operating cost.

The incremental operating cost can be calculated by differentiate the fuel cost curve of the generator.

Calculate the incremental operating cost for generator 1,

$$\begin{gathered} \frac{d{C}_1}{d{P}_1}=\frac{d}{d{P}_1}\left(600+18.P_1+0.04P_1^2\right) \\ \frac{d{C}_1}{d{P}_1}=18+0.08P_1\frac{\$}{MWh} \end{gathered}$$ ……. (3)

Calculate the incremental operating cost for generator 2,

$$\begin{gathered} \frac{d{C}_2}{d{P}_2}=\frac{d}{d{P}_2}\left(700+20.P_1+0.03P_2^2\right) \\ \frac{d{C}_2}{d{P}_2}=20+0.06P_2\frac{\$}{MWh} \end{gathered}$$ ……. (4)

The minimum cost occurs when the incremental cost of both generators becomes equal to each other.

$\frac{d{C}_1}{d{P}_1}=\frac{d{C}_2}{d{P}_2}$

Substitute $18+0.08P_1\frac{\$}{MWh}$for $\frac{d{C}_1}{d{P}_1}$and $20+0.06P_2\frac{\$}{MWh}$for $\frac{d{C}_2}{d{P}_2}$into above equation.

$$\begin{gathered} 18+0.08P_1=20+0.06P_2 \\ 0.08P_1-0.06P_2=20-18 \\ 0.08P_1-0.06P_2=2 \end{gathered}$$ …… (5)

The total for $P_1$and $P_2$is the 1000MW.

$$\begin{gathered} P_1+P_2=1000MW \\ {P}_1=1000-P_2 \end{gathered}$$ …… (6)

Substitute the value of$P_1$from the equation (6) into equation (5).

$$\begin{gathered} 0.08\left(1000-P_2\right)-0.06P_2=2 \\ 80-0.08P_2-0.06P_2=2 \\ 0.14P_2=78 \\ {P}_2=557.14MW \end{gathered}$$

Substitute value of $P_2$into equation (6).

$$\begin{gathered} P_1=1000-557.14 \\ {P}_1=442.86MW \end{gathered}$$

From the above discussion, it is clear that generator 2 is operating higher than its maximum limits that is 400MW.To obtain the optimum operation the generator 2 will operate at its maximum limit.

Therefore, optimal dispatch value of generator 1 and 2 are 600MW and 400MW respectively.

Calculate the incremental cost for generator 1.

Substitute 600MW for $P_1$into equation (3).

$$\begin{gathered} \frac{d{C}_1}{d{P}_1}=18+0.08\times 600 \\ \frac{d{C}_1}{d{P}_1}=66\frac{\$}{MWh} \end{gathered}$$

Calculate the incremental cost for generator 2.

Substitute 400MW for $P_2$into equation (4).

$$\begin{gathered} \frac{d{C}_2}{d{P}_2}=20+0.06\times 400 \\ \frac{d{C}_2}{d{P}_2}=44\frac{\$}{MWh} \end{gathered}$$

Calculate the total operating cost.

Substitute 600MW for $P_1$and 400MW for $P_2$into equation (2).

$$\begin{gathered} C_T=600+18\times 600+0.04\times {600}^2+700+20\times 400+0.03\times {4000}^2 \\ {C}_T=24800+13500 \\ {C}_T=39300\frac{\$}{hr} \end{gathered}$$

Hence the optimal dispatch value for generator $P_1$and $P_2$are 600MW and 400MW. The incremental operating cost for generator 1 and 2 are $66\frac{\$}{MWh}$and $44\frac{\$}{MWh}$respectively. The total operating cost is $39300\frac{\$}{hr}$.