Question

The fuel-cost curves for a two-generator power system are given as follows:

$$\begin{gathered} C_1\left(P_1\right)=600+15.P_1+0.05.{\left(P_1\right)}^2 \\ {C}_2\left(P_2\right)=700+20.P_2+0.04.{\left(P_2\right)}^2 \end{gathered}$$

While the system losses can be approximated as

$P_L=2\times {10}^{-4}{\left(P_1\right)}^2+3\times {10}^{-4}{\left(P_2\right)}^2-4\times {10}^{-4}{P}_1{P}_{2}MW$

If the system is operating with a marginal cost ($\lambda$) of$\$60/hr$, determine the output of each unit, the total transmission losses, the total load demand, and the total operating cost.

Short answer

The output of unit-1 and unit-2 is $0.124P_1-0.024P_2=45$and $-0.24P_1+0.116P_2=40$.

The total transmission losses are 19.15MW.

The total load demand is 865.91MW.

The total operating cost is $34435.34\frac{\$}{hr}$.

Step-by-step solution

Write the given data from the question.

The fuel-cost curves for a two-generator power system are$$\begin{gathered} C_1\left(P_1\right)=600+15.P_1+0.05.{\left(P_1\right)}^2 \\ {C}_2\left(P_2\right)=700+20.P_2+0.04.{\left(P_2\right)}^2 \end{gathered}$$

The system losses can be given as $P_L=2\times {10}^{-4}{\left(P_1\right)}^2+3\times {10}^{-4}{\left(P_2\right)}^2-4\times {10}^{-4}{P}_1{P}_{2}MW$.

The marginal cost of operating system ($\lambda$) of $60/hr.

Determine the required formula.

Write the formula output of each unit of marginal cost

$\lambda =\frac{d{C}_i}{d{P}_i}\left(\begin{array}{c}1\\ 1-\frac{\partial P_L}{\partial P_1}\end{array}\right)$ …… (1)

Here, $\frac{d{C}_i}{d{P}_i}$is an incremental operating cost and $\frac{\partial P_L}{\partial P_1}$is incremental system loss.

Write the formula of total transmission loss is

$P_L=2\times {10}^{-4}{\left(P_1\right)}^2+3\times {10}^{-4}{\left(P_2\right)}^2-4\times {10}^{-4}{P}_1{P}_2$ …… (2)

Here,$P_1$is power one and $P_2$is power two.

Write the formula of total load demand

$P_T=P_1+P_2-P_L$ …… (3)

Here,$P_L$is power loss,$P_1$is power first,$P_2$is power second and $P_L$is power loss.

Write the formula of total operating cost …… (5)

$C_T=C_1+C_2$ …… (6)

Here, $C_1$is cost one and $C_2$is cost two.

Determine the output marginal cost of each unit.

Determine the fuel-cost curves for a two-generator power system is,

$$\begin{gathered} C_1\left(P_1\right)=600+15P_1+0.05P_1^2 \\ {C}_2\left(P_2\right)=700+20P_2+0.04P_2^2 \end{gathered}$$ …… (7)

Here, $P_1$and $P_2$are the power first and power second.

The incremental operating costs of the units are,

$$\begin{gathered} \frac{d{C}_1}{d{P}_1}=0+15+\left(2\times 0.05\right)P_1\$/MWh \\ =15+0.1P_1\$/MWh \end{gathered}$$ …… (8)

Here, $P_1$is first power.

$$\begin{gathered} \frac{d{C}_2}{d{P}_2}=0+20+\left(2\times 0.05\right)P_2\$/MWh \\ =20+0.08P_2@\$/MWh \end{gathered}$$ …… (9)

Here, $P_2$is second power.

Determine the expression for $\frac{d{P}_L}{d{P}_1}$from equation (2).

$$\begin{gathered} \frac{d{P}_L}{d{P}_L}=2\times 2\times {10}^{-4}{P}_1+0-4\times {10}^{-4}{P}_2 \\ =4\times {10}^{-4}{P}_1-4\times {10}^{-4}{P}_2 \end{gathered}$$

Determine the expression for $\frac{d{P}_L}{d{P}_2}$from equation (2).

$$\begin{gathered} \frac{d{P}_L}{d{P}_2}=0+2\times 3\times {10}^{-4}{P}_2-4\times {10}^{-4}{P}_1 \\ =6\times {10}^{-4}{P}_2-4\times {10}^{-4}{P}_1 \end{gathered}$$

Determine the output marginal cost of each unit.

For unit-1

Substitute 60 for$\lambda$, $15+0.1P_1$for $\frac{d{C}_1}{d{P}_1}$and $4\times {10}^{-4}{P}_1-4\times {10}^{-4}{P}_2$for $\frac{d{P}_L}{d{P}_2}$into equation (1).

role="math" localid="1654802003368" $$\begin{gathered} 60=\left(15+0.1\times P_1\right)\left(\frac{1}{1-4\times {10}^{-4}{P}_1+4\times {10}^{-4}{P}_2}\right) \\ 60-240\times {10}^{-4}{P}_1+240\times {10}^{-4}{P}_2=15+0.1\times P_1 \\ 0.124P_1-0.024P_2=45 \\ {P}_1=447.6MW \end{gathered}$$

For unit-2

Substitute 60 for$\lambda$,$20+0.08P_2$for $\frac{d{C}_2}{d{P}_2}$and $6\times {10}^{-4}{P}_2-4\times {10}^{-4}{P}_1$for $\frac{d{P}_L}{d{P}_2}$into equation (1)

$$\begin{gathered} 60=\left(20+0.08\times P_2\right)\left(\frac{1}{1-6\times {10}^{-4}{P}_2+4\times {10}^{-4}{P}_1}\right) \\ 60-360\times {10}^{-4}{P}_2+240\times {10}^{-4}{P}_1=20+0.08\times P_2 \\ -0.024P_1+0.116P_2=40 \\ -0.024\times 447.6+0.116P_2=40 \end{gathered}$$

As further solve,

$P_2=437.4MW$

Here,$P_1$and $P_2$are the optimal dispatch value of generator 1 and 2.

Determine the total transmission losses.

Determine the system losses.

Substitute 447.6 for $P_1$and 437.4 for $P_2$into equation (2).

$$\begin{gathered} P_L=2\times {10}^{-4}{\left(447.6\right)}^2+3\times {10}^{-4}{\left(437.4\right)}^2-4\times {10}^{-4}\left(447.6\right)\left(437.4\right) \\ =19.15MW \end{gathered}$$

Determine the total load demand.

Determine the total load demand.

Substitute 447.6 for$P_1$, 437.46 for $P_2$and 19.15 for $P_L$into equation (3).

$$\begin{gathered} P_T=447.6+437.46-19.15 \\ =865.91MW \end{gathered}$$

Determine the total operating cost.

Determine the total operating cost.

Substitute $\left[600+15P_1+0.05P_1^2\right]$for $C_1$and $\left[700+20P_2+0.04P_2^2\right]$for $C_2$into equation (4).

$$\begin{gathered} C_T=\left[600+15P_1+0.05P_1^2\right]+\left[700+20P_2+0.04P_2^2\right] \\ =\left[600+15\times \left(447.6\right)+0.05\times {\left(447.6\right)}^2\right]+\left[700+20\times \left(437.46\right)+0.04\times {\left(437.46\right)}^2\right] \\ =17331.29+17104.05 \\ =34435.34\$/hr \end{gathered}$$