GATE EE · Power Systems

Economic Power Generation And Load Dispatch – GATE EE Power SystemsPractice Questions & PYQs

Generate GATE-level questions on Economic Load Dispatch. Focus on: 1. Incremental fuel cost and Optimal scheduling of generation without and with losses. 2. Penalty factors and Loss coefficients. 3. Unit commitment and Hydro-thermal scheduling basics.

15 questions · 15 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2024NAT

The incremental cost curves of two generators (Gen A and Gen B) in a plant supplying a common load are shown in the figure. If the incremental cost of supplying the common load is Rs. 7400 per MWh, then the common load in MW is ____ (rounded off to the nearest integer).

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📖 Explanation

$\lambda=7400 \mathrm{Rs} / \mathrm{MWhr}$ $I_{C A}\left(P_{G A}\right)=\frac{2000}{100} P_{G A}+8000$

\begin{aligned} I_{C B}\left(P_{G B}\right) & =40 P_{G B}+6000 \\ \lambda & =I_{C A}=I_{C B}=7400 \\ 200 P_{G A}+8000 & =7400 \\ P_{G A} & =-600 \\ P_{G A} & \\lt0 \\ P_{G B} & =\frac{7400-6000}{40}=35 \\ P_{G A}+P_{G B} & =35 \mathrm{MW} \end{aligned}

Q2GATE 2022MCQ

The fuel cost functions in rupees/hour for two 600 MW thermal power plants are given by $Plant 1: \; C_1=350+6P_1+0.004P_1^2$ $Plant 2: \; C_2=450+aP_2+0.003P_2^2$ where $P_1$ and $P_2$ are power generated by plant 1 and plant 2, respectively, in MW and $a$ is constant. The incremental cost of power $(\lambda )$ is 8 rupees per MWh. The two thermal power plants together meet a total power demand of 550 MW. The optimal generation of plant 1 and plant 2 in MW, respectively, are

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📖 Explanation

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

Q3GATE 2021MCQ

Two generators have cost functions $F_{1}$ and $F_{2}$ . Their incremental-cost characteristics are $\dfrac{dF_{1}}{dP_{1}}=40+0.2P_{1}$ $\dfrac{dF_{2}}{dP_{2}}=32+0.4P_{2}$ They need to deliver a combined load of $260 \text{~MW}$ . Ignoring the network losses, for economic operation, the generations $P_{1}$ and $P_{2}$ (in $\text{MW}$ ) are

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📖 Explanation

\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), $P_{1}=160 \mathrm{MW} ; P_{2}=100 \mathrm{MW}$

Q4GATE 2017NAT

Two generating units rated 300 MW and 400 MW have governor speed regulation of 6% and 4% respectively from no load to full load. Both the generating units are operating in parallel to share a load of 600 MW. Assuming free governor action, the load shared by the larger unit is _______ MW.

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📖 Explanation

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}

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 ______ .

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📖 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}

Q6GATE 2015NAT

The incremental costs (in Rupees/MWh) of operating two generating units are functions of their respective powers $P_1 \; and \; P_2$ in MW, and are given by $\frac{dC_{1}}{dP_{1}}=0.2 P_{1}+50$ $\frac{dC_{2}}{dP_{2}}=0.24P_{2}+40$ where $20MW\leq P_{1}\leq 150MW$ $20 MW \leq P_{2} \leq 150MW$ . For a certain load demand, $P_1 \; and \; P_2$ have been chosen such that $dC_1/d P_1$ =76Rs/MWh and $dC_2/d P_2$ =68.8 Rs/MWh. If the generations are rescheduled to minimize the total cost, then $P_2$ is ________.

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📖 Explanation

\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}

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

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📖 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

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📖 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

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📖 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.

Explanation Figure 2

Q10GATE 2009MCQ

Three generators are feeding a load of 100 MW. The details of the generators are In the event of increased load power demand, which of the following will happen ?

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📖 Explanation

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

Q11GATE 2008MCQ

A loss less power system has to serve a load of 250 MW. There are tow generation (G1 and G2) in the system with cost curves $C_1 \; and \; C_2$ respectively defined as follows: $C_{1}(P_{G1})=P_{G1}+0.055\times P^{2}_{G1}$ $C_{2}(P_{G1})=3 P_{G2}+0.03 \times P^{2}_{G2}$ where $P_{G1} \; and \; P_{G2}$ are the MW injections from generator G1 and G2 respectively. Thus, the minimum cost dispatch will be

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📖 Explanation

\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 $P_{G2}=150MW$

Q12GATE 2007MCQ

The incremental cost curves in Rs/MWhr for two generators supplying a common load of 700 MW are shown in the figures. The maximum and minimum generation limits are also indicated. The optimum generation schedule is :

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Q13GATE 2005MCQ

A load centre is at an equidistant from the two thermal generating stations $G_{1}$ and $G_{2}$ as shown in the figure. The fuel cost characteristic of the generating stations are given by $F_{1}=a+bP_{1}+cP_{1}^{2} Rs/hour$ $F_{2}=a+bP_{2}+2cP_{2}^{2} Rs/hour$ Where $P_1 \; and \; P_2$ are the generation in MW of $G_{1}$ and $G_{2}$ , respectively. For most economic generation to meet 300 MW of load $P_1 \; and \; P_2$ respectively, are

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📖 Explanation

Fuel-cost curve of generating station 1 $F_1=a+bP_1+cP_1^2 \; \text{Rs/hour}$ Fuel-cost curve of generating station 2 $F_2=a+bP_2+2cP_2^2 \; \text{Rs/hour}$ The slope of the fuel-cost curve i.e. $\frac{dF_i}{dP_i}$ is called the incremental fuel-cost (IC) and is expressed in Rs/MWh.

\begin{aligned} IC_1&=\frac{dF_1}{dP_1}=b+2cP_1 \\ IC_2&=\frac{dF_2}{dP_2}=b42cP_2 \end{aligned}

For economic generation

\begin{aligned} IC_1&=IC_2\\ b+2cP_1&=b+4cP_2\\ P_1&=2P_2\;\;...(i) \\ &\text{To meet 300 MW of load}\\ P_1+P_2&=300\;\;...(ii)\\ &\text{Solving eq. (i) and (ii)}\\ P_1&=200MW\\ P_2&=100MW \end{aligned}

Q14GATE 2003MCQ

Incremental fuel costs (in some appropriate unit) for a power plant consisting of three generating units are $IC_{1} = 20 + 0.3P_{1}$ $IC_{2} = 30 + 0.4P_{2}$ $IC_{3} = 30$ Where $P_{1}$ is the power in MW generated by unit i for i=1,2 and 3. Assume that all the three units are operating all the time. Minimum and maximum loads on each unit are 50 MW and 300 MW respectively. If the plant is operating on economic load dispatch to supply the total power demand of 700 MW, the power generated by each unit is

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📖 Explanation

When $P_1$ is minimum i.e. $P_1=50MW$ $IC_1=20+0.3 \times 50=35$ When $P_2$ is minimum i.e. $P_2=50MW$ $IC_2=30+0.4 \times 50=50$ For minimum value of $P_1$ and $P_2$ $IC_1$ and $IC_2 \gt IC_3$ Therefore, $P_3=300MW$ [maximum load is assigned to unit 3] Remaining power is to shared by unit 1 and 2 So, $P_1+P_2 = 700-300=400MW \;\;\;...(i)$ For optimal operation,

\begin{aligned} IC_1&=IC_2 \\ 20+0.3P_1&=30+0.4P_2 \\ 0.3P_1-0.4P_2&=10 \;\;...(ii)\end{aligned}

Solving equation (i) and (ii), we get

\begin{aligned} P_1&=242.86\; MW\\ P_2&=157.14 \; MW \end{aligned}

Q15GATE 2002MCQ

Consider a power system with three identical generators. The transmission losses are negligible. One generator $\left ( G_{1} \right )$ has a speed governor which maintains its speed constant at the rated value, while the other generators ( $G_{2}$ and $G_{3}$ ) have governors with a droop of 5%. If the load of the system is increased, then in steady state.

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