![When a base KVA has been considered, the % reactance
can be found by using the following relation
% reactance at base kVA =
Base kVA
Rated kVA
× % reactance at
rated kVA
Percentage Reactance & Base KVA
EEE436
Navila Rahman Nadi
G=2500 KVA, 10% T=500 KVA, 5%
For example,
The Base KVA = 2500KVA
So, the Percentage Reactance of G=10%
the Percentage Reactance of T= 25%
[
5𝑋2500
500
= 25%] Figure 2.5: Symmetrical fault calculation Method](https://image.slidesharecdn.com/lecture02-221215221859-ff20afc9/85/Lecture-02-pptx-13-320.jpg)
![Line Current Calculation from Base KVA
EEE436
Navila Rahman Nadi
Based on this the line current can be calculated as:
=
𝑩𝒂𝒔𝒆 𝑲𝑽𝑨
𝟑𝑽𝑳−𝑳,
[In 3 Phase System,
Real Power, P= 3𝑉𝑙𝐼𝑙𝑐𝑜𝑠∅
Apparent Power, S= 3𝑉𝑙𝐼𝑙
As, 𝑐𝑜𝑠∅=
𝑃
𝑆
So,
𝐼𝑙 =
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
3𝑉𝑙
]](https://image.slidesharecdn.com/lecture02-221215221859-ff20afc9/85/Lecture-02-pptx-14-320.jpg)
![Assignment 2:
The bus-bars of a power station are in two sections P and Q separated by a
reactor. Connected in section P are two 15 MVA generators of 12 % reactance each and to
Q one 8 MVA generator of 10% reactance. The reactor is rated at 10 MVA and 15%
reactance. Feeders are connected to bus-bar P through transformers, each rated at 5 MVA
and 4% reactance. Determine the maximum short-circuit MVA with which circuit breakers
on the outgoing side of the transformers have to deal.
Examples of Symmetric Fault Calculation
EEE436
Navila Rahman Nadi
[Wrong reactance diagram with
calculation has been done in the
book]](https://image.slidesharecdn.com/lecture02-221215221859-ff20afc9/85/Lecture-02-pptx-33-320.jpg)
Lecture 02.pptx
- 1. Lecture 2 Faults in Power System (Symmetrical Faults) EEE436 Switchgear and Protection
- 2. EEE436 Short Circuit • The most common type of fault occurs in power system is short circuit fault. • Whenever a fault occurs on a network such that a large current flows in one or more phases, a short circuit is said to have occurred. Navila Rahman Nadi
- 3. EEE436 Causes of Short Circuit (i) Internal effects : • Breakdown of equipment or transmission lines • From deterioration of insulation in a generator, transformer etc. (ageing of insulation, inadequate design or improper installation) (ii) External effects: • Insulation failure due to lightning surges • Overloading of equipment causing excessive heating • Mechanical damage by public etc. Navila Rahman Nadi
- 4. EEE436 Effects of Short Circuit (i) The heavy current due to short-circuit causes excessive heating which may result in fire or explosion. For example: a spark on a transmission line not cleared quickly will burn the conductor severely causing it to break, resulting in a long-time interruption of the line. (ii) The low voltage created by the fault has a very harmful effect on the service rendered by the power system. For example: If the voltage remains low for even a few seconds, the consumers’ motors may be shut down and generators on the power system may become unstable. Navila Rahman Nadi
- 5. 3-phase system can be classified into two main categories: (i) Symmetrical faults (ii) Unsymmetrical faults (i) Symmetrical faults That fault which gives rise to symmetrical (same amount of) fault currents (i.e. equal fault currents with 120o displacement) is called a symmetrical fault. The most common example of symmetrical fault is when all the three conductors of a 3-phase line are brought together simultaneously into a short-circuit condition. Faults in a power system Navila Rahman Nadi EEE436
- 6. • Referring to the Figure fault currents IR, IY and IB will be equal in magnitude with 120-degree displacement among them. • Because of balanced nature of fault, only one phase(line-to-neutral basis) need be considered in calculations since condition in the other two phases will also be similar. Symmetrical Faults on 3-Phase EEE436 Navila Rahman Nadi Figure 2.1: Schematic diagram of Symmetrical Fault occurrence
- 7. (ii) Unsymmetrical faults Those faults which give rise to unsymmetrical currents (i.e. unequal line currents with unequal displacement) are called unsymmetrical faults. The unsymmetrical faults may take one of the following forms: (a) Single line-to-ground fault (b) Line-to-line fault (c) Double line-to-ground fault Faults in a power system EEE436 Navila Rahman Nadi Figure 2.2: Schematic diagram of UnSymmetrical Fault occurrence
- 8. EEE436 Importance of Short Circuit Current Calculation (symmetrical fault current) (i) A short-circuit on the power system is cleared by a circuit breaker or a fuse. It is necessary, therefore, to know the maximum possible values of short-circuit current so that switchgear of suitable rating may be installed to interrupt them. (ii) The magnitude of short-circuit current determines the setting and sometimes the types and location of protective system. (iii) The calculation of short-circuit currents enables us to make proper selection of the associated apparatus (e.g. bus-bars, current transformers etc.) so that they can withstand the forces that arise due to the occurrence of short circuits. Navila Rahman Nadi
- 9. Calculation of Symmetrical Fault Current The reactance of generators, transformers, reactors etc. is usually expressed in percentage reactance (instead of ohmic reactance) to permit rapid short circuit calculations EEE436 Navila Rahman Nadi G=2500 KVA, 10% T=500 KVA, 5% Figure 2.3: Schematic diagram of Symmetrical calculation
- 10. • It is the percentage of the total phase voltage drop when full load current flows through it. Percentage Reactance ….Equation (1) ……Equation (3) EEE436 Navila Rahman Nadi
- 11. Percentage Reactance ……Equation (3) ….Equation (2) EEE436 Navila Rahman Nadi Here, KV= line voltage not phase voltage
- 12. • Generally, the various equipments used in the power system have different kVA ratings. • Therefore, it is necessary to find the percentage reactance of all the elements on a common kVA rating. This common kVA rating is known as base kVA. The value of this base kVA can be considered as any number as : (i) equal to that of the largest plant (ii) equal to the total plant capacity (iii) any arbitrary value Base KVA Navila Rahman Nadi EEE436 Navila Rahman Nadi
- 13. When a base KVA has been considered, the % reactance can be found by using the following relation % reactance at base kVA = Base kVA Rated kVA × % reactance at rated kVA Percentage Reactance & Base KVA EEE436 Navila Rahman Nadi G=2500 KVA, 10% T=500 KVA, 5% For example, The Base KVA = 2500KVA So, the Percentage Reactance of G=10% the Percentage Reactance of T= 25% [ 5𝑋2500 500 = 25%] Figure 2.5: Symmetrical fault calculation Method
- 14. Line Current Calculation from Base KVA EEE436 Navila Rahman Nadi Based on this the line current can be calculated as: = 𝑩𝒂𝒔𝒆 𝑲𝑽𝑨 𝟑𝑽𝑳−𝑳, [In 3 Phase System, Real Power, P= 3𝑉𝑙𝐼𝑙𝑐𝑜𝑠∅ Apparent Power, S= 3𝑉𝑙𝐼𝑙 As, 𝑐𝑜𝑠∅= 𝑃 𝑆 So, 𝐼𝑙 = 𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 3𝑉𝑙 ]
- 15. If X is the only reactance element in the circuit, then short- circuit current is given by : 𝐼𝑠𝑐 = 𝑉 𝑋 = 𝑉 %𝑋𝑉 100𝐼 = I × 100 %𝑋 { Replacing the value of X = %𝑋 𝑉 100𝐼 in the above equation} Example: If the percentage reactance of an element is 20% and the full- load current of the circuitry is 50 A, then short-circuit current will be 50 × 100/20 = 250 A when only that element is in the circuit. Short Circuit Current Calculation EEE436 Navila Rahman Nadi
- 16. We know, % reactance at base kVA = Base kVA Rated kVA × % reactance at rated kVA We know, full load current I= 𝐵𝑎𝑠𝑒 𝐾𝑉𝐴 3𝑉𝐿−𝐿, Symmetric Fault Calculation We know, Short Circuit current 𝐼𝑠𝑐 = I × 100 %𝑋 EEE436 Navila Rahman Nadi
- 17. From the above illustration, it is clear that whatever may be the value of base kVA, short-circuit current is the same. Does Different Base KVA values differs the value of short circuit current? EEE436 Navila Rahman Nadi
- 18. Reactor Control of Short-Circuit Current(?) EEE436 Navila Rahman Nadi Find Your Answer from the Book HAPPY HOUR!
- 19. Examples of Symmetric Fault Calculation Example 01: The below figure shows the single line diagram of a 3-phase system. The percentage reactance of each alternator is given. Find the short-circuit current that will occur at the point of F. EEE436 REACTANCE DIAGRAM IS A MUST Considering base KVA as 35,000KVA
- 21. • The product of normal system voltage and short-circuit current at the point of fault expressed in kVA is known as short-circuit kVA • Let V = normal phase voltage in volts I = full-load current in amperes at base kVA %X = percentage reactance of the system on base kVA up to the fault point • So, Short-circuit kVA for 3-phase circuit: Short Circuit KVA As, 𝐼𝑆𝐶 = I × 100 %𝑋 EEE436 Navila Rahman Nadi
- 22. Examples of Symmetric Fault Calculation Example 02: An 11 kV generating station has four identical 3-phase alternators A, B, C and D, each of 10 MVA capacity and 12% reactance. There are two sections of bus-bar, P and Q linked by a reactor rated at 10 MVA with 24% reactance. Alternators A and B are connected to bus-bar section P and alternators C and D to bus-bar section Q. From each section, load is taken through a number of 6 MVA, 11 kV/66 kV step- up transformers, each having a reactance of 3%. Find the short circuit current at the transformer end. . EEE436 Navila Rahman Nadi
- 23. Navila Rahman Nadi Let 10 MVA is the base MVA,
- 24. Example 03: 3-phase transmission line operating at 10 kV and having a resistance of 1Ω and reactance of 4 Ω is connected to the generating station bus-bars through 5 MVA step-up transformer having a reactance of 5%. The bus-bars are supplied by a 10 MVA alternator having 10% reactance. Calculate the short-circuit kVA fed to symmetrical fault between phases if it occurs (i) at the load end of transmission line (ii) at the high voltage terminals of the transformer EEE436 Navila Rahman Nadi Examples of Symmetric Fault Calculation
- 25. Let 10,000 KVA is the base KVA The line impedance is given in ohms. So by using equation
- 27. Example 4: The plant capacity of a 3-phase generating station consists of two 10,000 kVA generators of reactance 12% each and one 5000 kVA generator of reactance 18%. The generators are connected to the station bus-bars from which load is taken through three 5000 kVA step-up transformers each having a reactance of 5%. Determine the maximum fault MVA which the circuit breakers on (i) low voltage side and (ii) high voltage side may have to deal with. EEE436 Navila Rahman Nadi Examples of Symmetric Fault Calculation
- 28. Examples of Symmetric Fault Calculation EEE436 Navila Rahman Nadi
- 29. EEE436 Navila Rahman Nadi Examples of Symmetric Fault Calculation Short circuit current= 𝐹𝑎𝑢𝑙𝑡 𝐾𝑉𝐴 3×𝑙𝑖𝑛𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 Short circuit current= 𝐼𝑙 × 100 %𝑋
- 30. Example 5: A 3-phase, 20 MVA, alternator is connected 10 kV line and has internal reactance of 5% and negligible resistance. Find the External reactance per phase to be connected in series with the alternator so that steady current on short-circuit does not exceed 8 times the full load current. Examples of Symmetric Fault Calculation EEE436 Navila Rahman Nadi Single Line Diagram As, Phase voltage× 3 = 𝐿𝑖𝑛𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 , Short Circuit current 𝐼𝑠𝑐 = I × 100 %𝑋
- 31. Example 5: A 3-phase, 20 MVA, alternator is connected 10 kV line and has internal reactance of 5% and negligible resistance. Find the External reactance per phase to be connected in series with the alternator so that steady current on short-circuit does not exceed 8 times the full load current. Examples of Symmetric Fault Calculation EEE436 Navila Rahman Nadi Single Line Diagram
- 32. EEE436 Assignment 1: A small generating station has two alternators of 3000 kVA and 4500 kVA and percentage reactance's of 7% and 8% respectively. The feeders are provided with the circuit breakers which have a rupturing capacity of 150 MVA. There is an extended system added to the bus bar by a supply from the grid via a transformer of 7500 kVA which has 7·5% reactance. Find the reactance (ohmic) of the reactor connected in the bus-bar section after the two alternators to prevent the circuit breakers being overloaded, if a symmetrical short-circuit occurs on an outgoing feeder connected right after the B alternator. Assume the bus voltage = 3300 V Examples of Symmetric Fault Calculation Short Circuit KVA= 150MVA
- 33. Assignment 2: The bus-bars of a power station are in two sections P and Q separated by a reactor. Connected in section P are two 15 MVA generators of 12 % reactance each and to Q one 8 MVA generator of 10% reactance. The reactor is rated at 10 MVA and 15% reactance. Feeders are connected to bus-bar P through transformers, each rated at 5 MVA and 4% reactance. Determine the maximum short-circuit MVA with which circuit breakers on the outgoing side of the transformers have to deal. Examples of Symmetric Fault Calculation EEE436 Navila Rahman Nadi [Wrong reactance diagram with calculation has been done in the book]
- 34. References & Helpful Site EEE436 Navila Rahman Nadi https://www.electrical4u.com/ V. K. Mehta, “Principles of power system”, Chapter 17. Practice the examples (except 17.8, 17.9 and 17.10)