Economic Power Generation and Load Dispatch

\frac{dC_1}{dP_1}=IC_1=6+0.008P_1
\frac{dC_2}{dP_2}=IC_2=a+0.006P_2
For optimum generation,
IC_1=IC_2=\lambda
Therefore, 6+0.008P_1=8
P_1=250MW
Given, P_1+P_2=550MW
P_550-250=300MW

\begin{aligned} I C_{1} &=I C_{2} \\ 40+0.2 P_{1} &=32+0.4 P_{2} \\ 0.4 P_{2}-0.2 P_{1} &=8 \\ P_{2}+P_{1} &=260 \end{aligned}
Solving equation (i) and (ii),
P_{1}=160 \mathrm{MW} ; P_{2}=100 \mathrm{MW}

Given: speed regulation of generating unit is from no load to full load.
Hence, the rated frequency (50 Hz) will be common for both units at no load nad will drop to lower value as loaded to full load.
Let, P_1= Load shared by 400 MW machine
P_2= Load shred by 300 MW machine
For 400 MW machine drop in frequency from no load to full load =\frac{50 \times 4}{100}=2 Hz
New frequecy of operation of 400 MW machine, f_1=50-\frac{2}{400}P_1
For 300 MW machine drop in frequency from no load to full load =\frac{50 \times 6}{100}=3 Hz
New frequecy of operation of 300 MW machine, f_2=50-\frac{3}{300}P_2
we also know,
\begin{aligned} P_1+P_2&=600 \\\ (\text{Load shared }& \text{between shared units})\\ P_2&=600-P_1 \\ f_1&=f_2=f \\ (\text{Units operatio }& \text{ is in parallel}) \\ 50-\frac{2}{400}P_1 &=50-\frac{3}{300}(600-P_1) \\ \frac{P_1}{200}&=\frac{600-P_1}{100} \\ 100P_1 &=200(600-P_1) \\ P_1&=1200-2P_1\\ 3P_1&=1200\\ P_1&=400 MW \end{aligned}

\begin{aligned} 0.2P_1+50&=76\\ P_1&=130mW\\ 0.24P-2+40&=68.8\\ P_2&=120mW\\ P_1+P_2&=250\;\;...(i)\\ \text{For minimized}& \text{ schedule,}\\ 0.2P_1+50&=0.24P_2+40\\ 0.2P_1-0.24P_2&=-10\;\;\;..(ii)\\ &\text{From eq. (i) and (ii).}\\ P_1&=113.63MW\\ P_2&=136.36MW \end{aligned}

\begin{aligned} C_1&=0.01P_1^2+30P_1+10\\ \frac{dC_1}{dP_1}&=2 \times 0.01P_1+30\\ &=0.02P_1+30\\ C_2&=0.05P_2^2+10P_2+10\\ \frac{dC-2}{dP_2}&=0.1P_2+10\\ \text{For optimum } &\text{incremental cost,}\\ \frac{dC_1}{dP_1}&=\frac{dC_2}{dP_2}\\ 0.02P_1+30&=0.1P_2+10\\ 0.02P_1-0.1P_2&=-20 \;\;...(i)\\ \text{Given, }P_1+P_2&=200 \;\;...(ii)\\ \text{Solving } & \text{equation (i) and (ii),}\\ P_2&=200, P_1=0\\ \text{But the }& \text{range of }P_1 \text{ and }P_2 \text{ are specified as}\\ 100MW \leq & P_1 \leq 150MW\\ 100MW \leq & P_2 \leq 180MW\\ \text{hence, }& P_2=200 MW \text{ is not valid}\\ \text{so, both }& P_1 \text{ and } P_2 \text{ are operating at}\\ P_1=100,P_2=100\;& \Rightarrow \; P_1+P_2=200\\ IC_1=\frac{dC_1}{dP_1}&=0.02 \times P_1+30=32 Rs/MW\\ IC_2=\frac{dC_2}{dP_2}&=0.1 \times P_2+10=20 Rs/MW \end{aligned}

Given, P_{g1}+P_{g2}=1000 \;\;...(i)
Also, incremental cost of production of
g_1=\frac{dC_1}{dP_{g1}}=0.1P_{g1}+A \;\;\;...(ii)
and incremental cost of production of
g_2=\frac{dC_2}{dP_{g2}}=0.2P_{g2}+3A\;\;\;...(iii)
For optimum load sharing
\begin{aligned} \frac{dC_1}{dP_{g1}}&=\frac{dC_2}{dP_{g2}} \\ 0.1P_{g1}+A&=0.2P_{g2}+3A \\ 0.1P_{g1}-0.2P_{g2}&=2A \\ P_{g1}-2P_{g2}&=20A \;\;\;...(iv) \\ \text{Given, } \frac{dC_1}{dP_{g1}}&=\frac{dC_2}{dP_{g2}}=100\; Rs/MWh \end{aligned}
Putting this value in equation (ii), we have
\begin{aligned} 100&=0.1P_{g1}+A\\ A&=(100-0.1P_{g1})\;\;...(v) \end{aligned}
Using equation (iv) and (v), we have:
\begin{aligned} P_{g1}-2P_{g2}&=20(100-0.1P_{g1})\\ 3P_{g1}-2P_{g2}&=20004\;\;...(vi) \end{aligned} \begin{aligned} P_{g1}&=800MW \text{ and } P_{g2}=200MW\\ \therefore \;P_{g1}:P_{g2}&=800:200=4:1 \end{aligned}

From cordination equation
\begin{aligned} \frac{dF_n}{dP_{Gn}}\cdot \frac{1}{1-\frac{\partial P_L}{\partial P_{Gn}}}&=\lambda \\ \text{Given: }P_L&=0.5P_{G1}^2\\ \frac{\partial P_L}{\partial P_{G1}}=P_{G1} \text{ and } \frac{\partial P_L}{\partial P_{G2}}&= 0\\ \Rightarrow \; \frac{dF_{1}}{dP_{G1}}\frac{1}{1-P_{G1}}&=\frac{dF_2}{dP_{G2}}\frac{1}{1-0}\\ \Rightarrow \; 10000 \times \frac{1}{1-P_{G1}}&=12500\\ 1-P_{G1}=\frac{100}{125}&=\frac{4}{5}\\ P_{G1}&=\frac{1}{5}p.u.\\ \end{aligned}
as the base value is 100 MVA
\begin{aligned} P_{G1}&=\frac{1}{5} \times 100=20MW\\ \Rightarrow \; P_L&=0.5P_{G1}^2\\ &=0.5\left ( \frac{1}{5} \right )^2=0.02p.u.\\ \Rightarrow \; P_L&=0.02 \times 100=2MW\\ P_D&=P_{G1}+P_{G2}-P_L\\ 40&=20+P_{G2}-2\\ \Rightarrow \;\;P_{G2}&=22 MV \end{aligned}

Let load center is connected at point x
If r is resistance/km of line
Resistance of section between 1 nad x
R_1=r \times 25=R
Resistance of section between x nad 2
R_2=r \times 75=3R
Current (I) fed by generator \propto power produced by the generator.
I_1 \propto P_1, I_2 \propto P_2 \text{ and }I_3\propto P_3
Transmission losses
\begin{aligned} P_L&=I_1^2 R+I_2^2 3R+ I_3^23R \\ P_L&\propto (P_1^2+3P_2^2+3P_3^2) \\ P_L&=K[P_1^2+3P_2^2+3P_3^2] \end{aligned}
\begin{aligned} &\text{Option (A)}\\ P_1&=80MW\\ P_2&=20MW\\ P_3&=20MW\\ P_L&=K[80^2+3 \times 20^2+3 \times 20^2]\\ &=8800KMW \end{aligned}
\begin{aligned} &\text{Option (B)}\\ P_1&=60MW\\ P_2&=30MW\\ P_3&=30MW\\ P_L&=K[60^2+3 \times 30^2+3 \times 30^2]\\ &=9000KMW \end{aligned}
\begin{aligned} &\text{Option (C)}\\ P_1&=40MW\\ P_2&=40MW\\ P_3&=40MW\\ P_L&=K[40^2+3 \times 40^2+3 \times 40^2]\\ &=11200KMW \end{aligned}
\begin{aligned} &\text{Option (D)}\\ P_1&=30MW\\ P_2&=45MW\\ P_3&=45MW\\ P_L&=K[30^2+3 \times 45^2+3 \times 45^2]\\ &=13050 KMW \end{aligned}
So, option (A) gives minimum losses.

Let x_1,x_2 \text{ and }x_3 are reactance of generator-1, generator-2 and generator-3 respectively.
Neglecting armature resistance of all the three generators
\begin{aligned} VR_1&=0.02p.u. \\ VR_2&=0.04p.u \\ VR_3&=0.03p.u. \\ VR_1 & \lt VR_3 \lt VR_2 \end{aligned}
Since, voltage regulation (VR) \propto reactance of generator .
X_1 \lt X_3 \lt X_2
Power shared by a generator \propto \frac{1}{\text{reactance of generator}}
Power shared by Generator 1 \gt Generator 2 \gt Generator 3

\begin{aligned} C_1(P_{G1})&=P_{G1}+0.055P_{G1}^2\\ \frac{dC_1}{P_{G1}}&=1+0.11P_{G1}\\ C_2(P_{G2})&=P_{G2}+0.03P_{G2}^2\\ \frac{dC_2}{P_{G2}}&=3+0.06P_{G2} \end{aligned}
For minimum cost analysis
\begin{aligned} \frac{dC_1}{P_{G1}}&=\frac{dC_2}{P_{G2}}\\ 1+0.11P_{G1}&=3+0.06P_{G2}\;\;...(i)\\ P_{G1}+P_{G2}&=250MW\;\;...(ii) \end{aligned}
Solving equation (i) and (ii),
we get, P_{G1}=100MW and P_{G2}=150MW