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ECNG 3015 - PU system and 3Phase Fault calculation

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ECNG 3015

ECNG 3015

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

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