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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)