Question

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

$$\begin{gathered} {\mathbf{C}}_{\mathbf{1}}\left(P_1\right)\mathbf{=}\mathbf{600}\mathbf{+}\mathbf{15}\mathbf{.}{\mathbf{P}}_{\mathbf{1}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{.}{\left(P_1\right)}^{\mathbf{2}} \\ {\mathbf{C}}_{\mathbf{2}}\left(P_2\right)\mathbf{=}\mathbf{700}\mathbf{+}\mathbf{20}\mathbf{.}{\mathbf{P}}_{\mathbf{2}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{.}{\left(P_2\right)}^{\mathbf{2}} \end{gathered}$$

While the system losses can be approximated as

${\mathbf{P}}_{\mathbf{L}}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\left(P_1\right)}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\left(P_2\right)}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{P}}_{\mathbf{1}}{\mathbf{P}}_{\mathbf{2}}\mathbf{MW}$

If the system is operating with a marginal cost $\mathbf{\left(}\mathit{\lambda }\mathbf{\right)}$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 $0.124P_1-0.024P_2=45\mathrm{and}-0.24P_1+0.116P_2=40.$

The total transmission losses are 19.15MW.

The total load demand is 865.91MW.

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

Step-by-step solution

Write the given data from the question.

The fuel-cost curves for a two-generator power system are

$$\begin{gathered} {\mathbf{C}}_{\mathbf{1}}\mathbf{\left(}{\mathbf{P}}_{\mathbf{1}}\mathbf{\right)}\mathbf{=}\mathbf{600}\mathbf{+}\mathbf{15}\mathbf{.}{\mathbf{P}}_{\mathbf{1}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{.}{\mathbf{\left(}{\mathbf{P}}_{\mathbf{1}}\mathbf{\right)}}^{\mathbf{2}} \\ {\mathbf{C}}_{\mathbf{2}}\mathbf{\left(}{\mathbf{P}}_{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{700}\mathbf{+}\mathbf{20}\mathbf{.}{\mathbf{P}}_{\mathbf{2}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{.}{\mathbf{\left(}{\mathbf{P}}_{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}} \end{gathered}$$

The system losses can be given as$P_L=2\times {10}^{-4}{\left(P_1\right)}^2+3\times {10}^{-4}{\left(P_2\right)}^2-4\times {10}^{-4}{\mathrm{P}}_1{\mathrm{P}}_2\mathrm{MW}.$

system The marginal cost of operating$\left(\lambda \right)\mathrm{of}\$60/\mathrm{hr}.$

Determine the required formula.

Write the formula output of each unit of marginal cost

$$\begin{gathered} \mathit{\lambda }\mathbf{=}\frac{\mathbf{d}{\mathbf{C}}_{\mathbf{i}}}{\mathbf{d}{\mathbf{P}}_{\mathbf{I}}}\left(\frac{1}{1-{\displaystyle \frac{\partial P_L}{\partial P_1}}}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}......\left(1\right) \\ \mathrm{Here},\frac{{\mathbf{dC}}_{\mathbf{i}}}{{\mathbf{dP}}_{\mathbf{i}}}\mathrm{is}\mathrm{an}\mathrm{incremental}\mathrm{operating}\mathrm{cost}\mathrm{and}\frac{\mathit{\partial }{\mathit{P}}_{\mathit{L}}}{\mathit{\partial }{\mathit{P}}_{\mathit{1}}}\mathrm{is}\mathrm{incremental}\mathrm{system}\mathrm{loss}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{total}\mathrm{transmission}\mathrm{loss}\mathrm{is} \\ {\mathit{P}}_{\mathbf{L}}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{\left(}{\mathit{P}}_{\mathbf{1}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\left(P_2\right)}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathit{P}}_{\mathbf{1}}{\mathit{P}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\dots \dots \left(2\right) \\ \mathrm{Here}\mathbf{,}{\mathit{P}}_{\mathit{1}}\mathit{}\mathrm{is}\mathrm{power}\mathrm{one}\mathrm{and}{\mathit{P}}_{\mathit{2}}\mathrm{is}\mathrm{power}\mathrm{two}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{total}\mathrm{load}\mathrm{demand} \\ {\mathit{P}}_{\mathit{T}}\mathit{=}{\mathit{P}}_{\mathit{1}}\mathit{+}{\mathit{P}}_{\mathit{2}}\mathit{-}{\mathit{P}}_{\mathit{L}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}......\left(3\right) \\ \mathrm{Here},{\mathbf{P}}_{\mathbf{L}}\mathrm{is}\mathrm{power}\mathrm{loss}\mathbf{,}{\mathbf{P}}_{\mathbf{1}}\mathbf{}\mathrm{is}\mathrm{power}\mathrm{first},{\mathbf{P}}_{\mathbf{2}}\mathrm{is}\mathrm{power}\mathrm{second}\mathrm{and}{\mathbf{P}}_{\mathbf{L}}\mathrm{is}\mathrm{power}\mathrm{loss}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{total}\mathrm{operating}\mathrm{cost}\dots \dots \left(5\right) \\ {\mathit{C}}_{\mathit{T}}\mathit{=}{\mathit{C}}_{\mathit{1}}\mathit{+}{\mathit{C}}_{\mathit{2}}\mathit{}\dots \dots \left(6\right) \\ \mathrm{Here},{\mathbf{C}}_{\mathbf{1}}\mathrm{is}\mathrm{cost}\mathrm{one}\mathrm{and}{\mathbf{C}}_{\mathbf{2}}\mathrm{is}\mathrm{cost}\mathrm{two}. \end{gathered}$$

Determine the output marginal cost of each unit.

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

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

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

The incremental operating costs of the units are,

$$\begin{gathered} \frac{d{C}_1}{d{P}_1}=0+15+\left(2\times 0.05\right)P_2\$/\mathrm{MWh}\dots \dots \left(8\right) \\ \mathrm{Here},P_1\mathrm{is}\mathrm{first}\mathrm{powe}r. \\ \frac{d{C}_2}{d{P}_2}=0+20+\left(2\times 0.05\right)P_2\$/\mathrm{MWh}\dots \dots \left(9\right) \\ \mathrm{Here},P_{\mathit{2}}\mathit{}\mathit{}\mathrm{is}\mathrm{second}\mathrm{power}. \\ \mathrm{Determine}\mathrm{the}\mathrm{expression}\mathrm{for}\frac{{\mathrm{dP}}_{\mathrm{L}}}{{\mathrm{dP}}_1}\mathrm{from}\mathrm{equation}\left(2\right). \\ \frac{{\mathrm{dP}}_{\mathrm{L}}}{{\mathrm{dP}}_2}=2\times 2\times {10}^{-4}{P}_1+0-4\times {10}^{-4}{P}_2 \\ =4\times {10}^{-4}{P}_1-4\times {10}^{-4}{P}_2 \\ \mathrm{Determine}\mathrm{the}\mathrm{expression}\mathrm{for}\frac{d{P}_L}{d{P}_2}\mathrm{from}\mathrm{equation}\left(2\right). \\ \frac{d{P}_L}{d{P}_2}=0+2+3\times {10}^{-4}{P}_2-4\times {10}^{-4}{P}_{\mathit{1}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}6\times {10}^{\mathit{-}\mathit{4}}{P}_{\mathit{2}}\mathit{-}4\times {10}^{-4}{P}_{\mathit{1}} \\ \mathrm{Determine}\mathrm{the}\mathrm{output}\mathrm{marginal}\mathrm{cost}\mathrm{of}\mathrm{each}\mathrm{unit}. \\ \mathrm{For}\mathrm{unit}-1 \end{gathered}$$

localid="1655437522603" $$\begin{gathered} \mathrm{Substitute}60\mathrm{for}\mathrm{\lambda },15+0.1{\mathrm{P}}_1\mathrm{for}\frac{d{C}_{\mathit{1}}}{d{P}_{\mathit{1}}}\mathrm{and}4\times {10}^{-4}{\mathrm{P}}_1-4\times {10}^{-4}{\mathrm{P}}_2\mathrm{for}\frac{d{P}_L}{d{P}_{\mathit{2}}}\mathrm{into} \\ \mathrm{equation}\left(1\right). \\ 60=\left(15+0.1\times {\mathrm{P}}_1\right)\left(\frac{1}{1-4\times {10}^{-4}{P}_{\mathit{1}}+4\times {10}^{-4}{P}_{\mathit{2}}}\right) \\ 60-240\times {10}^{-4}{P}_1+240\times {10}^{-4}{P}_{\mathit{2}}=15+0.1\times P_1 \\ 0.124P_1-0.024P_2=45 \\ {P}_1=447.6\mathrm{MW} \end{gathered}$$

$$\begin{gathered} \mathrm{For}\mathrm{unit}-2 \\ \mathrm{Substitute}60\mathrm{for}\mathrm{\lambda },20+0.8P_1\mathrm{for}\frac{d{C}_2}{d{P}_2}\mathrm{and}6\times {10}^{-4}{\mathrm{P}}_2-4\times {10}^{-4}{\mathrm{P}}_1\mathrm{for}\frac{d{P}_L}{d{P}_{\mathit{2}}}\mathrm{into} \\ \mathrm{equation}\left(1\right) \\ 60=\left(20+0.08\times P_{\mathit{2}}\right)\left(\frac{1}{1-6\times {10}^{-4}{P}_{\mathit{2}}+4\times {10}^{-4}{P}_{\mathit{1}}}\right) \\ 60-360\times {10}^{-4}{\mathrm{P}}_{\mathit{2}}+240\times {10}^{-4}{\mathrm{P}}_1=20+0.08\times P_{\mathit{2}} \\ -0.024P_{\mathit{1}}\mathit{+}0.116P_{\mathit{2}}\mathit{=}40 \\ -0.024\times 447.26+0.11P_{\mathit{2}}=40 \\ \mathrm{As}\mathrm{further}\mathrm{solve}, \\ {P}_{\mathit{2}}=437.4\mathrm{MW} \\ \mathrm{Here},P_{\mathit{1}}\mathit{}\mathrm{and}{P}_{\mathit{2}}\mathit{}\mathrm{are}\mathrm{the}\mathrm{optimal}\mathrm{dispatch}\mathrm{value}\mathrm{of}\mathrm{generator}1\mathrm{and}2. \end{gathered}$$

Determine the total transmission losses.

Determine the system losses.

Substitute 447.26 for $P_1$and 437.4 for $P_2$into equation (2).

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

Determine the total load demand.

Determinethe total load demand.

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

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

Determine the total operating cost.

Determine the total operating cost.

$$\begin{gathered} \mathrm{Substitute}\left[600+15P_{\mathit{1}}0.05P_{\mathit{1}}^{\mathit{2}}\right]\mathrm{for}{C}_{\mathit{1}}\mathrm{and}\left[700+20P_{\mathit{2}}+0.04P_{\mathit{2}}^{\mathit{2}}\right]\mathrm{for}{\mathrm{C}}_2\mathrm{into} \\ \mathrm{equation}\left(4\right). \\ {C}_{\mathit{T}}=\left[600+15P_{\mathit{1}}+0.05P_{\mathit{1}}^{\mathit{2}}\right]+\left[700+20P_{\mathit{2}}+0.04P_{\mathit{2}}^{\mathit{2}}\right] \\ =\left[600+15\times \left(447.6\right)+0.05\times {\left(447.6\right)}^2\right]+\left[700+20\times \left(437.46\right)+0.04\times {\left(437.46\right)}^2\right] \\ =17331.29+17104.05 \\ =34435.34\$/\mathrm{hr} \end{gathered}$$