Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Notes  >  Power Systems  >  Previous Year Questions- Economic Power Generation and Load Dispatch

Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE) PDF Download

Q1: 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).      (2024)
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)(a) 26
(b) 35
(c) 42
(d) 38
Ans:
(b)
Sol: Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
Q2: The fuel cost functions in rupees/hour for two 600 MW thermal power plants are given by
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)where Pand P2 are power generated by plant 1 and plant 2, respectively, in MW and aa is constant. The incremental cost of power (λ) 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       (2022)
(a) 200, 350
(b) 250, 300
(c) 325, 225
(d) 350, 200
Ans:
(b)
Sol: Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)For optimum generation,
 IC1 = IC2 = λ
Therefore, 6 + 0.008P= 8
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
Q3: Two generators have cost functions F1 and F2. Their incremental-cost characteristics are
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)They need to deliver a combined load of 260 MW. Ignoring the network losses, for economic operation, the generations P1 and  P2 (in MW) are       (2021)
(a) P= P= 130
(b) P= 160, P2 = 100
(c) P1 = 140, P2 = 120
(d) P1 = 120, P2 = 140
Ans:
(b)
Sol: Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Solving equation (i) and (ii),
P1 = 160MW; P2 = 100MW

Q4: 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.        (SET-2 (2017))
(a) 50
(b) 100
(c) 200
(d) 400
Ans:
(d)
Sol: 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, P1= Load shared by 400 MW machine
 P2= Load shred by 300 MW machine
For 400 MW machine drop in frequency from no load to full load = Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
New frequecy of operation of 400 MW machine, Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
For 300 MW machine drop in frequency from no load to full load Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
New frequecy of operation of 300 MW machine, Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
we also know,
P1 + P2 = 600
(Load shared between shared units)
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)(Units operatio is in parallel)
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
Q5: The incremental costs (in Rupees/MWh) of operating two generating units are functions of their respective powers P1 and P2 in MW, and are given by
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)where
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)For a certain load demand, Pand P2 have been chosen such that dC1/dP= 76Rs/MWh and dC2/dP= 68.8 Rs/MWh. If the generations are rescheduled to minimize the total cost, then P2 is ________.       (SET-2(2015))
(a) 76.78
(b) 120.24
(c) 136.36
(d) 156.84
Ans:
(c)
Sol: Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)For minimized schedule,
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
Q6: 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:
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)The incremental cost (in Rs/MWh) of the power plant when it supplies 200 MW is ______ .         (SET-1(2015))
(a) 10
(b) 20
(c) 30
(d) 40
Ans:
(b)
Sol: Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)For optimum incremental cost,
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
Q7: The fuel cost functions of two power plants are
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)where, Pg1 and Pg2 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 P1 and P2 is        (SET-1 (2014))
(a) 1:4
(b) 2:3
(c) 3:2
(d) 4:1
Ans:
(d)
Sol: Given, Pg1 + Pg2 = 1000...(i)
Also, incremental cost of production of
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)and incremental cost of production of
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)For optimum load sharing
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Putting this value in equation (ii), we have
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Using equation (iv) and (v), we have:
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
Q8: The figure shows a two-generator system applying a load of PD = 40MW, connected at bus 2.
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)The fuel cost of generators G1 and G2 are :
 C1(PG1) = 10,000 Rs/MWh and C2(PG2) = 12,500 Rs/MWh and the loss in the line is  Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)where the loss coefficient is specified in pu on a 100 MVA base. The most economic power generation schedule in MW is         (2012)

(a) PG1 = 20, PG2 = 22
(b) PG1=22,PG2=20PG1 = 22, PG2 = 20
(c) PG1 = 20, PG2 = 20
(d) PG1 = 0, PG2 = 40
Ans: 
(a)
Sol: From cordination equation
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)as the base value is 100 MVA
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
Q9: 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      (2011)
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)(a) P= 80MW+losses, P2 = 20MW, P3 = 20MW
(b) P= 60MW, P= 30MW + losses, P= 30MW
(c) P1 = 40MW, P2 = 40MW, P3 = 40MW + losses
(d) P1 = 30MW+losses, P2 = 45MW, P3 = 45MW
Ans:
(a)
Sol: Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Let load center is connected at point x
If r is resistance/km of line
Resistance of section between 1 nad x
R1 = r × 25 = R
Resistance of section between x nad 2
R2 = r × 75 = 3R
Current (I) fed by generator ∝∝ power produced by the generator.
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)So, option (A) gives minimum losses.

Q10: Three generators are feeding a load of 100 MW. The details of the generators are
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
In the event of increased load power demand, which of the following will happen ?       (2009)
(a) All the generator will share equal power
(b) Generator-3 will share more power compared to Generator-1
(c) Generator-1 will share more power compared to Generator-2
(d) Generator-2 will share more power compared to Generator-3
Ans: 
(c)
Sol: Let x1, x2 and x3 are reactance of generator-1, generator-2 and generator-3 respectively.
Neglecting armature resistance of all the three generators
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Since, voltage regulation (VR) ∝ reactance of generator .
X1 < X3 < X2
Power shared by a generator Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
Power shared by Generator 1 > Generator 2 > Generator 3.

Q11: 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 C1 and C2 respectively defined as follows:
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)where PG1 and PG2 are the MW injections from generator G1 and G2 respectively. Thus, the minimum cost dispatch will be       (2008)
(a) PG1 = 250MW; PG2 = 0MW
(b) PG1 = 150MW; PG2 = 100MW
(c) PG1 = 100MW; PG2 = 150MW
(d) PG1 = 0MW; PG2 = 250MW
Ans:
(c)
Sol: Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)For minimum cost analysis
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Solving equation (i) and (ii),
we get, PG1 = 100MW and PG2 = 150MW

Q12: 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 :     (2007)
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)(a) Generator A : 400 MW, Generator B : 300 MW
(b) Generator A : 350 MW, Generator B : 350 MW
(c) Generator A : 450 MW, Generator B : 250 MW
(d) Generator A : 425 MW, Generator B : 275 MW
Ans:
(c)
Sol: Maximum incremental cost in Rs/Mwhr for generator A =600 (at 450 MW)
Minimum incremental cost in Rs/Mwhr for generator B = 650 (at 150 MW)
As maximum value of incremental cost of A is less than minimum value of B.
Therefore, generator 'A' will operate at its maximum (o/p) 450 MW and B at (700 - 450) = 250 MW.

Q13: A load centre is at an equidistant from the two thermal generating stations G1 and G2 as shown in the figure. The fuel cost characteristic of the generating stations are given by
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Where P1 and P2 are the generation in MW of G1 and G2, respectively. For most economic generation to meet 300 MW of load  P1 and P2 respectively, are      (2005)
(a) 150, 150
(b) 100, 200
(c) 200, 100
(d) 175, 125
Ans: 
(c)
Sol: Fuel-cost curve of generating station 1
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Fuel-cost curve of generating station 2
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)The slope of the fuel-cost curve i.e. Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE) is called the incremental fuel-cost (IC) and is expressed in Rs/MWh.
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)For economic generation
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
Q14: Incremental fuel costs (in some appropriate unit) for a power plant consisting of three generating units are
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Where P1 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        (2003)
(a) P1P1 = 242.86MW; P2 = 157.14MW; and  P3 = 300MW
(b) P1P= 157.14MW; P2 = 242.86MW; and P3 = 300MW
(c) P1P1 = 300MW; P2 = 300MW; and P= 100MW
(d) P1 = 233.3MW; P2 = 233.3MW; and P3 = 233.4MW
Ans:
(a)
Sol: When P1 is minimum i.e. P= 50MW
IC1 = 20 + 0.3 × 50 = 35
When P2 is minimum i.e. P2 = 50MW
IC2 = 30 + 0.4 × 50 = 50
For minimum value of P1 and P2
IC1 and IC> IC3
Therefore,
P= 300MW
[maximum load is assigned to unit 3]
Remaining power is to shared by unit 1 and 2
So, P+ P2 = 700 − 300 = 400MW...(i)
For optimal operation,
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)Solving equation (i) and (ii), we get
Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)
Q15: Consider a power system with three identical generators. The transmission losses are negligible. One generator (G1) has a speed governor which maintains its speed constant at the rated value, while the other generators (G2 and G3) have governors with a droop of 5%. If the load of the system is increased, then in steady state.        (2002)
(a) generation of G2 and G3 is increased equally while generation of G1 is unchanged
(b) generation of Galone is increased while generation of G2 and G3 is unchanged.
(c) generation of G1, Gand G3 is increased equally.
(d) generation of G1, G2 and G3 is increased in the ratio 0.5:0.25:0.25.
Ans: 
(b)

The document Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Power Systems.
All you need of Electrical Engineering (EE) at this link: Electrical Engineering (EE)
21 videos|67 docs|45 tests

Top Courses for Electrical Engineering (EE)

FAQs on Previous Year Questions- Economic Power Generation and Load Dispatch - Power Systems - Electrical Engineering (EE)

1. What is the importance of economic power generation in load dispatch?
Ans.Economic power generation is crucial in load dispatch as it ensures that electricity is produced at the lowest possible cost while meeting demand. This optimization helps in reducing operational costs, enhancing the efficiency of the power system, and ensuring reliable energy supply to consumers.
2. How does load dispatch contribute to grid stability?
Ans.Load dispatch contributes to grid stability by balancing supply and demand in real-time. It ensures that generation matches consumption, thereby preventing overloading of generators and ensuring that the grid operates within safe limits. This is essential for maintaining the reliability of the power supply.
3. What are the common methods used for load forecasting in economic power generation?
Ans.Common methods for load forecasting include time-series analysis, regression analysis, and machine learning techniques. These methods analyze historical data and trends to predict future electricity demand, which is vital for effective load dispatch and economic power generation planning.
4. What role do renewable energy sources play in economic power generation?
Ans.Renewable energy sources play a significant role in economic power generation by providing sustainable and often cost-effective alternatives to fossil fuels. They contribute to reducing greenhouse gas emissions, enhancing energy security, and diversifying the energy mix, which is essential for economic and environmental sustainability.
5. How does the concept of marginal cost relate to load dispatch in electrical engineering?
Ans.The concept of marginal cost relates to load dispatch as it refers to the cost of producing one additional unit of electricity. In load dispatch, generators are typically scheduled based on their marginal costs, with the aim of minimizing the overall cost of generation while meeting the required load, thus optimizing the economic efficiency of the power system.
21 videos|67 docs|45 tests
Download as PDF
Explore Courses for Electrical Engineering (EE) exam

Top Courses for Electrical Engineering (EE)

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Summary

,

MCQs

,

mock tests for examination

,

Previous Year Questions with Solutions

,

study material

,

Extra Questions

,

shortcuts and tricks

,

Free

,

video lectures

,

Objective type Questions

,

Important questions

,

Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)

,

Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)

,

Viva Questions

,

past year papers

,

Semester Notes

,

Previous Year Questions- Economic Power Generation and Load Dispatch | Power Systems - Electrical Engineering (EE)

,

Sample Paper

,

Exam

,

ppt

,

practice quizzes

,

pdf

;