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Solution of Assignment 1

by: Shammya Saha

Solution of Assignment 1 EEE 598

Shammya Saha
GPA 3.83

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This is the solution for Assignment 1
Power System Control and Monitoring
Dr. Undrill
Class Notes
25 ?




Popular in Power System Control and Monitoring

Popular in Electrical Engineering

This 4 page Class Notes was uploaded by Shammya Saha on Saturday March 5, 2016. The Class Notes belongs to EEE 598 at Arizona State University taught by Dr. Undrill in Fall 2015. Since its upload, it has received 30 views. For similar materials see Power System Control and Monitoring in Electrical Engineering at Arizona State University.


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Date Created: 03/05/16
Assignment -1 EEE 598 Power Plant Control and Monitoring Spring 2016 Shammya Saha ID: 1209508778 According to the question statement the mechanical power is proportional to the 2.5 power of the speed. So the differential equation [1] defining the motor with an inertia constant H is ???????? 2.5 2???? × ???????? ???????????????? × ???????? = ???? ???????? ???? ???????? ) Initially, the electrical power is equal to the mechanical power and the initial acceleration is zero. Putting this condition, the value of ???? can be determined considering???????? = 51???????? , ???? =???? 900×2???? 60 ????????????/???? Now, the question statement contains the value of ???? with respect to some specific time values after the machine has been tripped which results a zero electrical power. So the above equations turns out to be ???????? 2.5 ???????? = (−???????? ???? ???? )/(2???? × ???????? ????????????????) As the equation is not normalized and the power is in ???????? unit and ???????? is in MVA, this will ???????????????? result a ???? in the unit of second. The above equation can be used to integrate by numerical integration to calculate the value of ???? by estimating the value of????. A MATLABcode iswritten to solvethe equation to findout thevaluesof ???? for different ???? values. ThentheRMSerrorhasbeencalculatedforeachestimatedvalueof????.ThentheRMSerrorswere plotted using a stem plot to show the error value plot and the H value with the minimum error has been accepted. Then the H value with the minimum error is used to calculate the value of J, moment of inertia. The MATLAB code for calculating the H value is provided in the following page. From the plots it has been found out that ???? = ????.???????????? generates the least RMS error for the given data and estimated data of speed. This value of H is used to calculate the value of J by the following equation, ???? = 2???????????? ????????????????= 4.4135 × 10 4 ???????? The sensitivity analysis is done by solving for the value of ???? by changing the exponent from 2.5 to some other value. Then the change required in the value of ???? to adjust the graph to the original given points is calculated. Exponent ???? 2.5 4.4135 × 10 4 4 2.6 3.938 × 10 2.7 3.5046 × 10 4 2.4 4.9287 × 10 4 4 2.3 5.4909 × 10 The above plot is the plot of the value of J with respect to the exponent ranging from 2.3 to 2.7. As the exponent value is increasing, the speed change is going to have a higher negative value which makes the deaccelerating process faster. As moment of inertia is defined by the concept of required torque to get a desired angular acceleration, it can be concluded that the moment of inertia will get lower as we increase the component. To discuss the sensitivity of the value of the values presented in the table are plotted in MATLAB 4 which gives a graph of almost a straight line with the slope of−4.963 × 10 . So for a change of 0.1 of the exponent value the change of J is pretty large which means that the J value is quite sensitive to the exponent of the speed that relates mechanical power and speed. The values presented in the table were also used to make a polynomial curve fit in MATLAB for more accuracy. It has been found that theyfit a second order polynomial with the coefficients as 21240,−155810,300930. This also shows that the J value is quite sensitive to the value of the exponent chosen. References: [1] P. Kundur “Power System Stability”


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