Q5GATE 2015NAT

Consider the economic dispatch problem for a power plant having two generating units. The fuel costs in Rs/MWh along with the generation limits for the two units are given below: $C_{1}(P_{1})=0.01{P_{1}}^{2+}30P_{1}+10$ ; $100MW\leq P_{1} \leq 150MW$ $C_{2}(P_{2})=0.05{P_{2}}^{2+}10P_{2}+10$ ; $100MW\leq P_{2} \leq 180MW$ The incremental cost (in Rs/MWh) of the power plant when it supplies 200 MW is ______ .

🚀 Solve in Practice Mode

📖 Explanation

\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}

Q7GATE 2014MCQ

The fuel cost functions of two power plants are Plant $P_1:C_1\equiv 0.05Pg^2_1+APg_1+B$ Plant $P_2:C_2\equiv 0.10Pg^2_2+3APg_2+2B$ where, $Pg_1 \; and \; Pg_2$ are the generated powers of two plants, and A and B are the constants. If the two plants optimally share 1000 MW load at incremental fuel cost of 100 Rs/MWh, the ratio of load shared by plants $P_1 \; and \; P_2$ is

🚀 Solve in Practice Mode

📖 Explanation

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}

Q8GATE 2012MCQ

The figure shows a two-generator system applying a load of $P_D$ = 40MW, connected at bus 2. The fuel cost of generators G1 and G2 are : $C_1(P_{G1})$ =10,000 Rs/MWh and $C_2(P_{G2})$ =12,500 Rs/MWh and the loss in the line is $P_{loass(pu)}=0.5P_{G1(pu)}^2$ , where the loss coefficient is specified in pu on a 100 MVA base. The most economic power generation schedule in MW is

🚀 Solve in Practice Mode

📖 Explanation

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}

Q9GATE 2011MCQ

A load center of 120 MW derives power from two power stations connected by 220 kV transmission lines of 25 km and 75 km as shown in the figure below. The three generators G1, G2 and G3 are of 100 MW capacity each and have identical fuel cost characteristics. The minimum loss generation schedule for supplying the 120 MW load is

🚀 Solve in Practice Mode

📖 Explanation

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.

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