Power Systemanalysis and Control (EET 408) Laboratory Module

Power Systemanalysis and Control (EET 408) Laboratory Module

<p>Power SystemAnalysis and Control (EET 408) Laboratory Module</p><p>EXPERIMENT 4</p><p>OPTIMAL POWER FLOW</p><p>1. OBJECTIVE:</p><p>To analyze the optimal power flow by means of MiPower soft ware</p><p>2. EQUIPMENT: </p><p>MiPower software</p><p>3. INTRODUCTION: Focus of economic operation study is to schedule or arrange power go out from each existing central in system to load on the system so that the amount of cost generating as minimum as possible or maximum efficiency. Maximum efficiency is minimizes the cost of kilowatthour to consumer and the cost to the company of delivering that kilowatthour in the face of constantly rising prices for fuel, labor, supplies and maintenance. When we study about economic dispatch in a power system, assumed the system in a “power balance’ state. In the system consists of n thermal-generating units connected to a single busbar serving a received electrical load PR as Fig.3.3 below</p><p>1 T G P F 1 1</p><p>2 T G P F 2 2 P R</p><p> n T G P F n n</p><p>Fig.3.3 n thermal unit commited to serve a load of P R</p><p>KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 1 Power SystemAnalysis and Control (EET 408) Laboratory Module</p><p>So, total system generated equal to the total system load plus losses in the transmission line including tie line flows out of the system</p><p> n PG   Pi  PR  PL i1 1 where:</p><p>PG = total output generated electrical power </p><p>Pi = output-generated electrical power for the ith generating unit [per- unit] on a common power base</p><p>PR = total system load, including tie line flows,[per-unit]</p><p>PL = total system transmission losses [per-unit] n = number of generators in service in the system A basic constraint is that for each generator: </p><p>P  P  P mini Gi max 2 To determine economic dispatch in the power plants, we can divided to two categories: a) System without transmission losses</p><p> n PG   Pi  PR i1 3 The operating cost of a unit includes fuel cost, labor, supplies and maintenance in expressed as C  F P i i ini 4 where: </p><p>Ci = operating cost of unit i [$/H]</p><p>Fi = operating unit i [$/ M Btu]</p><p>Pin i = input unit i power [M Btu/H]</p><p>The curve Ci can be represented by a second-order expression</p><p>KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 2 Power SystemAnalysis and Control (EET 408) Laboratory Module</p><p>2 Ci   i   i Pi   i Pi 5 where: i = generator i, one of n units</p><p>Ci = operating cost of generator i[$H]</p><p>Pi = electrical power output of generator i [per-unit] on a common power base ,  and  are in units of $/H</p><p> dFi   for Pi,min  Pi  Pi,max dPi</p><p> dFi   for Pi  Pi,max dPi 6</p><p> dFi   for Pi  Pi,min dPi</p><p>This is the requisite to minimize cost of the system. A general solution is called the method of Lagrangian multipliers. If without losses and no generator limits, the most economic operation</p><p>k  k    i Pi  k  2 i   Bii  7 is called Coordination Equation and </p><p>KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 3 Power SystemAnalysis and Control (EET 408) Laboratory Module</p><p> ng  i PD   i1 2   i ng  1 i1 2 i 8</p><p> b) System with transmission losses </p><p>4. PROCEDURE: PART A</p><p>Cost equation and loss co-efficient of different units in a plant are given in Table 1.0. Determined economic generation for a total load demand of 240 MW. Unit No Cost of Fuel in RM/hour Pschedule</p><p>1 2 90 C1=0.05 P1 + 20 P1 + 800 0<= P1 <=100</p><p>2 2 90 C2=0.06 P2 + 15 P2 + 1000 0<= P2 <=100</p><p>3 2 90 C3=0.07 P3 + 18 P3 + 900 0<= P3 <=100</p><p>TABLE 1.0</p><p>B11= 0.0005; B12= 0.00005; B13= 0.0002;</p><p>B22=0.0004; B23=0.00018; B33= 0.0005;</p><p>B21=B12; B23 = B32; B13 = B31;</p><p>1. Invoke MiPower Tools in MiPower main screen and select “Economic Dispatch by B- Coefficient”.</p><p>KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 4 Power SystemAnalysis and Control (EET 408) Laboratory Module</p><p>2. Click button to create new file. 3. After given a file name and specify the directory in MiPower directory, save it. The following windows will appear.</p><p>4. Figure above shows how to enter data for generator no 1. To input data for gen 2 and </p><p>3, simply click and input all necessary data based on the data given in Table 1.0.</p><p>KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 5 Power SystemAnalysis and Control (EET 408) Laboratory Module</p><p>5. To input the B-coefficient data of B11, follow the procedure as shown in windows </p><p> below. Then click the button. 6. Similarly enter the values if B12 to B31.</p><p>7. Click the main button to save all the input data. </p><p>8. After, click button to run the program. 9. After saving the output in a notepad, print it(Print only the last part of the display output).</p><p>Click here to save every B-coefficient data </p><p>Select 1 to enter data of B11</p><p>Enter 0.0005 as B11</p><p>Click here to save all data Click execute button after saving all data</p><p>10. What is Initial Total generation cost of the system? 11. What is the final cost of generation of every generator and total generation? 12. What is the final value of lambda?</p><p>KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 6 Power SystemAnalysis and Control (EET 408) Laboratory Module</p><p>13. Repeat the procedure 1-9 with initial lambda of 30. How many iterations required to get the same result? 14. Repeat the procedure 1-9 for Pschedule of 100MW. Answer the procedure 10- 13m for this case.</p><p>QUESTION</p><p>1. The fuel-cost functions for three thermal plants in $/h are given by:</p><p>2 C1  350  7.20P1  0.0040P1</p><p>2 C2  500  7.30P2  0.0025P2</p><p>2 C3  600  6.74P3  0.0030P3</p><p> where P1 , P2, and P3 in MW. The governors are set such that generators share the load equally. Neglecting line losses and generator limits, find the total cost in $/h when the total load is:</p><p>(i) PD = 750 MW</p><p>(ii) PD = 1335 MW.</p><p>2. What is the difference between economic dispatch and unit commitment? 3. Solve problem 1, if the governor not share the load equally for load 450 MW</p><p>KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 7 Power SystemAnalysis and Control (EET 408) Laboratory Module</p><p>Name: ______Matrix No:______Experiment No:______Date: ______</p><p>RESULTS: 10. Initial Total generation ______</p><p>11. Generator 1 ______Generator 2 ______Generator 3 ______</p><p>12. Final Value of Lambda ______</p><p>13. No of iterations ______</p><p>14.</p><p>KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 8 Power SystemAnalysis and Control (EET 408) Laboratory Module</p><p>PART B </p><p>1.</p><p>DISCUSSION</p><p>CONCLUSION</p><p>KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 9 Power SystemAnalysis and Control (EET 408) Laboratory Module</p><p>KOLEJ UNIVERSITI KEJURUTERAAN UTARA MALAYSIA – Exp.1 (Revision 1) 10</p>

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    10 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us