ECNG 3015 - PU system and 3Phase Fault calculation

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ECNG 3015 - PU system and 3Phase Fault calculation

  1. 1. ECNG 3015Industrial and Commercial Electrical Systems<br />Lecturer<br />Prof Chandrabhan Sharma<br /># 1<br />PU System<br />
  2. 2. THE PER UNIT SYSTEM<br />Introduction:The per unit system is fundamental in the analysis of power systems since:<br /> - parameters such as voltage and power are usually in the kilo or mega range in a power system - this is due to the large amount of power transmitted - other values such as current and impedance are usually represented as a percent or per unit - determined by using a base or reference value. <br />
  3. 3. The per unit system has several advantages which include:<br /> 1) Calculations are simplified.<br /> 2) Equipment rating.<br />3) Provides a reference to voltages in the power system.<br /> 4) It is an international standard.<br />Usually base megavolt amperes (MVA) and base voltage in kilovolts are selected to specify the base.<br />
  4. 4. (2) Per Unit Conversion<br />Note: Apu is dimensionless<br />(3) P.U. Impedance<br />
  5. 5. (4) To compare two power systems with different MVAs<br />. <br />
  6. 6. Example:<br />Given a transformer with the following specifications: 100MVA, 66/12 kV, 0.1 p.u. (10%).<br />If the base value chosen is 200, then<br />Note: <br />Zp.u. for transformer is the same whether referred to primary or secondary side since the it has the same MVA rating on both sides and turns ratio cancels<br />
  7. 7. (5) The p.u. system is used to extract the current that would flow <br />when a fault occurs. <br />Consider the system below<br />. <br />
  8. 8. The result is a generator whose voltage is 1p.u. Therefore,<br />. <br />(6) Convert the p.u. values to the actual values<br />
  9. 9. Fault Calculations<br /> Why?<br /> - fault current magnitude gives the current settings for protective relays<br /> - provides the required ratings for the C.B. and associated equipment<br />Faults can be either:<br /> a. Symmetrical Fault – 3 faults i.e. n/w still electrically balanced<br />. <br />
  10. 10. b. Unsymmetrical – network not electrically balanced - SLG, DLG, OC – analysis done by symmetrical components<br />The most common fault is the Single Line to Ground Fault (SLG) which occurs over 90% of the time.<br />. <br />Note: If the system is unbalanced, the relationship V = IR does not hold.<br />The operating time of C.B. for clearing a fault is very important in determining fault levels<br />If C.B. operating time  2 cycles after fault initiation then motor contribution to fault currents cannot be ignored.<br />During the first few cycles of a fault, the motors initially act as generators for which the source of excitation is the stored magnetic energy.<br />
  11. 11. In the 1st half cycle of s/c a motor can feed current of magnitudes reaching 10 times the F.L.C.<br />After 5 cycles this current would have decayed to practically ‘0’.<br />Depending on the analysis time frame (re: motors….)<br /> Xd’’, Xd’ and Xd would be used <br />. <br />
  12. 12. Circuit Breakers:<br /> Circuit breakers are needed to:<br /> - isolate equipment for maintenance and<br /> - interrupt load and fault current<br />. <br />Transient Impedance Graph<br />
  13. 13. . <br />
  14. 14. Fault Calculations:<br />. <br />Fault system<br />
  15. 15. Given: <br /> G1 and G2 = 25 MVA, 0.1pu<br /> G3 = 20MVA, 0.1pu<br /> T/F1 and T/F2 = 50 MVA, 10%<br /> Lines AB, AC, BD, CD = j10Ω and <br /> Line B’C’ = j0.75Ω<br />Note: The impedance between circuit breakers are negligible, therefore the rating is the same for all breakers on the same bus.<br />
  16. 16. Neglecting resistances provides the worst case value of fault current, which is more desirable for the purpose of rating C.B. <br />Therefore, using ‘j’ term gives worse case design.<br />Question: <br />For the system shown above, what is the fault level at D/E?<br />Soln:<br />(1) Let base value = 100 MVA<br />Then convert all values to a common base p.u. value.<br />Voltage in the ring = 33kV <br />
  17. 17. or<br />
  18. 18. (2) Using 100 MVA as the base, the 50 MVA transformers must be converted to its equivalent value.<br />Therefore, the total impedance of line BC = j0.2 + j 0.52 + j0.2 = j0.92<br />(3) Now the generators must be converted to its equivalent p.u. values.<br />
  19. 19. (4) Now that all equipment and lines are in p.u. on a common base, we can draw the equivalent circuit. The neutral of the generators is used as the common point for the diagram.<br />
  20. 20. Equivalent circuit<br />
  21. 21. The previous equivalent ckt can be reduced to this ckt<br />
  22. 22. Applying /Y transformation <br />
  23. 23. This reduces to:<br />
  24. 24.
  25. 25.
  26. 26. SHORT CIRCUIT CALCULATIONS<br />R.T.F : Equivalent circuit referred to kV1<br />Solution:<br />
  27. 27.
  28. 28. Example<br />Question: <br />For the system shown above, what is the fault level at A, B, C and D?<br />
  29. 29. Soln:<br /> Let base value = 100 MVA<br />
  30. 30.  Equivalent circuit becomes:<br />
  31. 31.
  32. 32. With the line impedance included (shown in green) the circuit becomes:<br />The short circuit levels would then be:<br />MVAA = 3508; MVAB = 491; MVAC = 132.7; MVAD = 17.4<br />
  33. 33. Limiting S/C levels<br />There are basically two methods <br /> a. Sectionalizers<br /> b. Reactors<br />a. Sectionalizers<br />N.B.: care must be taken in paralleling cables between buses. <br />
  34. 34. Practical case is where bus interconnect via a cable. <br />If the existing cable is overloaded then a second cable may be installed to assist in carrying the load as shown below<br />This will have the effect of increasing the fault level at B.<br />
  35. 35. b. Reactors<br />- means of increasing the impedance artificially by the introduction of reactors<br />Example:<br />Given an old system with switchgear rated at 100 MVA where we need to expand by adding a new feeder and associated switchgear. New rating of added switchgear is 250 MVA. In order to connect both systems, a series reactor of 15 MVA rating is to be introduced. Find the value of the reactance required.<br />
  36. 36. Soln:<br /> Let base value = 10 MVA<br /> Reactance of G1 & G2 to new base = 0.24 p.u.<br />
  37. 37. Equivalent circuit<br />
  38. 38.
  39. 39. Question:<br />Determine the fault level at the new station busbar (B). <br />(Ans. 126.9 MVA)<br />Disadvantages of adding reactors:<br /> - increased regulation<br /> - lowering of power factor<br /> - increase in system losses<br />

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