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Feeder Protection, Conductor Sizing, Load Flow and Fault Calculation
CHAPTER
Feeder Protection, Conductor Sizing, Load Flow and Fault Calculation
5
FUSES Low-voltage (LV) distribution systems in offshore installations are protected by fuses in the same way as their onshore counterparts. Although the Institution of Electrical Engineers (IEE) Wiring Regulations specifically excludes offshore installations from its scope (Part 1 paragraph II-3(vii)), the document that it refers to (IEE Recommendations for the Electrical and Electronic Equipment of Mobile and Fixed Offshore Installations) has been replaced by BS IEC 61892, Mobile and fixed offshore units. Electrical installations. The methods of calculation given in the IEE Wiring Regulations are normally adopted when designing lighting and small power distribution systems, particularly for accommodation modules. The use and limitations of high rupturing capacity fuses are discussed in PART 2 Chapter 5.
MINIATURE CIRCUIT BREAKERS In most situations, miniature circuit breakers (MCBs) may be used as an alternative to fuses where the fault rating of the MCB is sufficiently high. However, because of the different shape of the MCB tripping characteristic compared with that of a fuse (see Fig. 4.5.1), it is not advisable to mix fuses with MCBs in the same circuit if discrimination is required. The degree of discrimination between one MCB and another, and between MCBs and fuses, varies with the BS EN 60898-1 MCB type, but generally a better discrimination can be obtained using fuses. MCBs with fault current ratings more than 16 kA are now available and are in common use offshore, where they save weight and reduce the quantity and type of spare fuses stocked. MCBs used offshore must provide positive indication of contact clearance and must be padlockable when in the open position.
OVERCURRENT AND EARTH FAULT PROTECTION Most of the conventional forms of short cable feeder protection may be used offshore. The use of inverse definite minimum time (IDMT) overcurrent and residually connected earth fault relays may give discrimination problems if a large Offshore Electrical Engineering Manual. https://doi.org/10.1016/B978-0-12-385499-5.00025-X Copyright © 2018 Elsevier Inc. All rights reserved.
227
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CHAPTER 5 Conductor Sizing, Load Flow and Fault Calculation
FIGURE 4.5.1 Miniature circuit breaker and fuse characteristic comparison. LV, low voltage; MCCB, moulded case circuit breaker. Courtesy GEC Alstom Measurements, now GE Alstom Grid Solutions.
proportion of the platform load is on the switchboard being fed. This is because the steady-state fault current available from the platform generators will be of the order of three to four times the full-load current; therefore, if fault clearance times are to be kept short, generator and feeder relay current settings will need to be close. Voltage-controlled relays may be used on generator circuit breakers to overcome this problem, as discussed earlier for generator protection (see PART 4 Chapter 1).
Sizing of Conductors
An alternative is to use definite time relays. However, on LV systems, it is less feasible to use definite time relays because, if generator and main switchboard clearance times are to be kept reasonably short, there are likely to be discrimination problems with fuses and MCBs lower down in the system. As cables are more exposed to mechanical damage than switchboard busbars, it is advisable to protect cables which interconnect switchboards by some form of simple unit protection, rather than IDMT relays with intertripping. This has the advantage of faster operation and may also relieve any discrimination problems associated with the unrestricted method of protection.
SIZING OF CONDUCTORS LOAD FLOW When the electrical distribution system has been configured for optimum convenience, safety and reliability, the various busbars and cables should be sized for the maximum continuous load in each system operating condition. The first task is to ensure that the system 24-h load profile and the load schedule are as up to date as possible and that diversity factors and operating modes have been agreed by all parties and ‘frozen’. If the system is simple, with few parallel paths, load flows may be manually calculated. In either steady-state or transient conditions, the power system can be represented by a physical model, such as produced in a network analyser, or by a mathematical model using a digital computer. With the proliferation of desktop computers, the use of network analysers, even on small systems, is now rare. On larger installations, with many parallel paths, computer load flow programs should be used in any case. Such programs are now available for use on desktop microcomputers at prices starting from a few hundred pounds. Load flow calculations by nodal analysis have become firmly established. Such methods involve 1. the solution of a set of linear simultaneous equations which describe the system configuration, 2. the application of restraints at each node to enable the required complex power and voltage conditions to be maintained. The advantages in using nodal voltage analysis are that the number of equations is smaller than that with the alternative mesh current analysis method, and the system may be described in terms of its node numbers and the impedances of the interconnecting branches. In nodal analysis, the node voltages V are related to the nodal injected currents I by the system admittance matrix Y. In the matrix form, [I] = [Y] [V]
229
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CHAPTER 5 Conductor Sizing, Load Flow and Fault Calculation
The voltage V refers to a value between node and earth, and the current I is the injected nodal current. The total nodal injected power S is obtained from the product of voltage and current conjugate, as follows:[S] = [V] [I] * By taking the current conjugate, the reactive power is given the same sign as active power for lagging current. There are three basic types of nodal constraints: 1. fixed complex voltage 2. fixed complex power 3. fixed-voltage modulus with real power. Type (1) constraint is given to the reference node, usually known as the ‘slack’ or ‘swing’ (in the United States) bus. The type (2) constraint represents a load bus; the type (3), a generation bus. The formation of the nodal matrix and methods available for digital iterative sequences of solution are given in Bergen (1986).
BUSBAR SIZING Switchboard main busbars must be rated to carry the maximum continuous load which can flow in any healthy power system operating condition. Transient conditions giving rise to higher currents, such as those due to large motors starting or downstream faults, may be tolerated momentarily, provided the protection devices are incorporated which will ensure that the outgoing equipment is removed from the system before the busbars get overheated. The continuous current rating must be for the busbars as enclosed in their offshore environmental protection, with natural cooling only. This also applies to the switching and isolating devices in the switchboard.
CABLE SIZING Cable sizing is generally carried out in accordance with current versions of the following: IEC 61892: Mobile and fixed offshore units – Electrical installations: Part 2: System design Part 4: Cables Part 6: Installation IEC 60287: Electric cables – Calculation of the current rating. IEC 92-201: Electrical installations in ships Part 201: System design – General Part 202: System design – Protection Part 352: Choice and installation of electrical cables
Worked Example: Fault Calculation
For motor cables, the basis of the calculation is (Vd × 1000) L= (1.732 × I × (RcosΦ + XsinΦ)
where L, cable length in metres; Vd, permitted steady-state volt drop in volts; I, motor full-load current; R, cable resistance in ohms per kilometre; X, cable reactance in ohms per kilometre and cosΦ, motor power factor at full load. The permissible maximum steady-state volt drop is normally 2.5%, whilst the permissible volt drop during starting is 10%. A typical computer spreadsheet generated motor cable sizing chart is shown in Fig. 2.8.1. Derating factors for cables which pass through insulation and for bunching, application of protective devices, etc. need to be considered in accordance with the current edition of the IEE Wiring Regulations.
WORKED EXAMPLE: FAULT CALCULATION The following calculations and information are not exhaustive but are intended to give the reader sufficient knowledge to enable switchgear of adequate load and fault current rating to be specified. The subject may be studied in more detail by reading the relevant documents listed in Appendix 1. The nomenclature used is generally as given in the GE Alstom Grid Solutions Network Protection and Automation Guide (NPAG) and the Electricity Council’s Power System Protection (IET). When a short-circuit occurs in a distribution switchboard, the resulting fault current can be large enough to damage both the switchboard and associated cables owing to thermal and electromagnetic effects. The thermal effects will be proportional to the duration of the fault current to a large extent, and this time will depend on the characteristics of the nearest upstream automatic protective device which should operate to clear the fault. Arcing faults due to water or dirt ingress are most unlikely in the switchboards of land-based installations, but from experience, they need to be catered for offshore. For switchboards operating with generators of 10 MW or more, it is usually not difficult to avoid the problem of long clearance times for resistive faults. However, with smaller generators, clearance times of several seconds may be required because of the relatively low prospective fault currents available (see earlier section on busbar protection). With small emergency generators, pilot exciters are not normally provided and the supply for the main exciter is derived from the generator output. This arrangement is not recommended, as it allows the collapse of generator output current within milliseconds of the onset of a fault. With such small generators, even subtransient fault currents are small, and it is unlikely that downstream protection relays set to operate for ‘normal’ generation will have operated before the output collapse. It is usual to provide a fault current
231
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CHAPTER 5 Conductor Sizing, Load Flow and Fault Calculation
FIGURE 4.5.2 Schematic of small-generator fault current maintenance circuit. AVR, automatic voltage regulator.
maintenance unit as shown in Fig. 4.5.2. This device is basically a compounding circuit which feeds a current proportional to output current back to the exciter field. When the output current reaches a threshold value well above the normal load current, a relay operates, switching in the compounding circuit. Thus a high output current is maintained by this feedback arrangement until the operation of definite time overcurrent protection, which is set to prevent the generator thermal rating being exceeded.
MAIN GENERATOR FAULT CURRENTS In this example, one generator is rated at 15.0 MW. The alternator has the following parameters: rating 17.7 MVA at a power factor of 0.85 and voltage of 6.6 kV;
Worked Example: Fault Calculation
FIGURE 4.5.3 Typical cable sizing spreadsheet (see Web Reference 2).
reactances Xd″ = 0.195 per unit (pu), × Xd′ = 0.31 pu, Xd = 2.2 pu and Xq″ = 0.265; full-load current, 1550 A and neutral earthing resistor, 10 Ω. There are four more generators of a different type, each rated at 3.87 MVA and having parameters as follows: reactances Xd″ = 0.15 pu , Xd′ = 0.18 pu, Xd = 2.05 pu and Xq″ = 0.15; full-load current, 340 A and neutral earthing resistor, 40 Ω (Fig. 4.5.3). The generator reactance values (pu) may be tabulated on their own base as follows: MVA Unit
Xd″
Xq″
Xo″
Rn (Ω)
1 × 177
0.195
0.265
0.09
10
1 × 3.87
0.15
0.15
0.045
40
Referring to a common base of 17.7 MVA, 6.6 kV and considering the units in parallel: MVA Unit
Xd″
Xq″
Xo″
Rn
1 × 3.87 4 × 3.87
0.686 0.1715
0.686 0.1715
0.2058 0.0515
16.25 4.06
1 × 17.7
0.195
0.265
0.09
4.06
233
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CHAPTER 5 Conductor Sizing, Load Flow and Fault Calculation
Summing the reactances, Xd″ =
Xq″ =
Xo″ =
1 1/0.195 + 1/0.1715 1
1/0.265 + 1/0.1715 1 1/0.09 + 1/0.0515
Rn =
= 0.0912 pu
1 1/4.06 + 1/4.06
= 0.104 pu
= 0.0327 pu = 2.63 pu
Hence, Xd″ + Xq″ = 0.0912 + 0.104 = 0.1952 pu
Xd″ + Xq″ = 0.912 + 0.104 + 0.0327 = 0.2279 pu
( )2 2 | Z1 | = Xd″ + Xq″ + Xo″ + (3Rn )
2
2
Z1 = (0.2279) + (3 × 2.03) = 6.09 pu
Therefore the system fault currents (neglecting motor contribution) are 17.7 Three ‐ phase fault current = = 16.97 kA 3 × 6.6 × 0.0912 Phase ‐ to ‐ phase fault current = Earth fault current =
17.7 6.6 × 0.1952
3 × 17.7 6.6 × 6.09
= 13.74 kA
= 0.762 kA
This method of calculation provides low values which, although not suitable for providing fault currents for the selection of switchgear, are useful for relay setting, as neglecting motor fault contributions provides a safety margin. The extra fault current provided by motors should significantly reduce the operating times of all overcurrent and earth fault protection devices, provided the current transformer saturation is not a problem. A check should be made, however, that generator cable reactances do not reduce prospective fault currents by more than a negligible amount. This reduction is likely to be significant for the smaller generators operating at LVs.
SWITCHBOARD FAULT CURRENTS The prospective fault currents for medium-voltage switchboards can be obtained in a similar manner by summing generator and transformer reactances as follows. Take the base again as 17.7 MVA, and take as an example the generator switchboard
Worked Example: Fault Calculation
connected to the LV switchboard by a 2 MVA transformer with 440 V secondary winding. Then the sequence reactances (pu) are as follows:
X1
X2
X0
At transformer rating
0.13
0.11
0.11
On the base MVA
1.1505
0.974
0.974
Summing generator and transformer reactances, X1 = 0.0912 + 1.1505 = 1.2417 pu X2 = 0.104 + 0.974 = 1.078 pu X0 = 0.974 = 0.974 pu Hence,
(X1 + X2 ) = 1.2417 + 1.078 = 2.3197 pu (X1 + X2 + X0 ) = 1.2417 + 1.078 + 0.974 = 3.294 pu
Therefore the LV system fault currents are
Three ‐ phase fault current =
17.7 3 × 0.44 × 1.2417
Phase ‐ to ‐ phase fault current = Earth fault current =
17.7 0.44 × 2.3197
3 × 17.7 0.44 × 3.294
= 18.7 kA = 17.34 kA
= 21.15 kA
The workings required for protection settings of the same scenario are continued in PART 4 Chapter 7.
235
Feeder Protection, Conductor Sizing, Load Flow and Fault Calculation
5
FUSES Low-voltage (LV) distribution systems in offshore installations are protected by fuses in the same way as their onshore counterparts. Although the Institution of Electrical Engineers (IEE) Wiring Regulations specifically excludes offshore installations from its scope (Part 1 paragraph II-3(vii)), the document that it refers to (IEE Recommendations for the Electrical and Electronic Equipment of Mobile and Fixed Offshore Installations) has been replaced by BS IEC 61892, Mobile and fixed offshore units. Electrical installations. The methods of calculation given in the IEE Wiring Regulations are normally adopted when designing lighting and small power distribution systems, particularly for accommodation modules. The use and limitations of high rupturing capacity fuses are discussed in PART 2 Chapter 5.
MINIATURE CIRCUIT BREAKERS In most situations, miniature circuit breakers (MCBs) may be used as an alternative to fuses where the fault rating of the MCB is sufficiently high. However, because of the different shape of the MCB tripping characteristic compared with that of a fuse (see Fig. 4.5.1), it is not advisable to mix fuses with MCBs in the same circuit if discrimination is required. The degree of discrimination between one MCB and another, and between MCBs and fuses, varies with the BS EN 60898-1 MCB type, but generally a better discrimination can be obtained using fuses. MCBs with fault current ratings more than 16 kA are now available and are in common use offshore, where they save weight and reduce the quantity and type of spare fuses stocked. MCBs used offshore must provide positive indication of contact clearance and must be padlockable when in the open position.
OVERCURRENT AND EARTH FAULT PROTECTION Most of the conventional forms of short cable feeder protection may be used offshore. The use of inverse definite minimum time (IDMT) overcurrent and residually connected earth fault relays may give discrimination problems if a large Offshore Electrical Engineering Manual. https://doi.org/10.1016/B978-0-12-385499-5.00025-X Copyright © 2018 Elsevier Inc. All rights reserved.
227
228
CHAPTER 5 Conductor Sizing, Load Flow and Fault Calculation
FIGURE 4.5.1 Miniature circuit breaker and fuse characteristic comparison. LV, low voltage; MCCB, moulded case circuit breaker. Courtesy GEC Alstom Measurements, now GE Alstom Grid Solutions.
proportion of the platform load is on the switchboard being fed. This is because the steady-state fault current available from the platform generators will be of the order of three to four times the full-load current; therefore, if fault clearance times are to be kept short, generator and feeder relay current settings will need to be close. Voltage-controlled relays may be used on generator circuit breakers to overcome this problem, as discussed earlier for generator protection (see PART 4 Chapter 1).
Sizing of Conductors
An alternative is to use definite time relays. However, on LV systems, it is less feasible to use definite time relays because, if generator and main switchboard clearance times are to be kept reasonably short, there are likely to be discrimination problems with fuses and MCBs lower down in the system. As cables are more exposed to mechanical damage than switchboard busbars, it is advisable to protect cables which interconnect switchboards by some form of simple unit protection, rather than IDMT relays with intertripping. This has the advantage of faster operation and may also relieve any discrimination problems associated with the unrestricted method of protection.
SIZING OF CONDUCTORS LOAD FLOW When the electrical distribution system has been configured for optimum convenience, safety and reliability, the various busbars and cables should be sized for the maximum continuous load in each system operating condition. The first task is to ensure that the system 24-h load profile and the load schedule are as up to date as possible and that diversity factors and operating modes have been agreed by all parties and ‘frozen’. If the system is simple, with few parallel paths, load flows may be manually calculated. In either steady-state or transient conditions, the power system can be represented by a physical model, such as produced in a network analyser, or by a mathematical model using a digital computer. With the proliferation of desktop computers, the use of network analysers, even on small systems, is now rare. On larger installations, with many parallel paths, computer load flow programs should be used in any case. Such programs are now available for use on desktop microcomputers at prices starting from a few hundred pounds. Load flow calculations by nodal analysis have become firmly established. Such methods involve 1. the solution of a set of linear simultaneous equations which describe the system configuration, 2. the application of restraints at each node to enable the required complex power and voltage conditions to be maintained. The advantages in using nodal voltage analysis are that the number of equations is smaller than that with the alternative mesh current analysis method, and the system may be described in terms of its node numbers and the impedances of the interconnecting branches. In nodal analysis, the node voltages V are related to the nodal injected currents I by the system admittance matrix Y. In the matrix form, [I] = [Y] [V]
229
230
CHAPTER 5 Conductor Sizing, Load Flow and Fault Calculation
The voltage V refers to a value between node and earth, and the current I is the injected nodal current. The total nodal injected power S is obtained from the product of voltage and current conjugate, as follows:[S] = [V] [I] * By taking the current conjugate, the reactive power is given the same sign as active power for lagging current. There are three basic types of nodal constraints: 1. fixed complex voltage 2. fixed complex power 3. fixed-voltage modulus with real power. Type (1) constraint is given to the reference node, usually known as the ‘slack’ or ‘swing’ (in the United States) bus. The type (2) constraint represents a load bus; the type (3), a generation bus. The formation of the nodal matrix and methods available for digital iterative sequences of solution are given in Bergen (1986).
BUSBAR SIZING Switchboard main busbars must be rated to carry the maximum continuous load which can flow in any healthy power system operating condition. Transient conditions giving rise to higher currents, such as those due to large motors starting or downstream faults, may be tolerated momentarily, provided the protection devices are incorporated which will ensure that the outgoing equipment is removed from the system before the busbars get overheated. The continuous current rating must be for the busbars as enclosed in their offshore environmental protection, with natural cooling only. This also applies to the switching and isolating devices in the switchboard.
CABLE SIZING Cable sizing is generally carried out in accordance with current versions of the following: IEC 61892: Mobile and fixed offshore units – Electrical installations: Part 2: System design Part 4: Cables Part 6: Installation IEC 60287: Electric cables – Calculation of the current rating. IEC 92-201: Electrical installations in ships Part 201: System design – General Part 202: System design – Protection Part 352: Choice and installation of electrical cables
Worked Example: Fault Calculation
For motor cables, the basis of the calculation is (Vd × 1000) L= (1.732 × I × (RcosΦ + XsinΦ)
where L, cable length in metres; Vd, permitted steady-state volt drop in volts; I, motor full-load current; R, cable resistance in ohms per kilometre; X, cable reactance in ohms per kilometre and cosΦ, motor power factor at full load. The permissible maximum steady-state volt drop is normally 2.5%, whilst the permissible volt drop during starting is 10%. A typical computer spreadsheet generated motor cable sizing chart is shown in Fig. 2.8.1. Derating factors for cables which pass through insulation and for bunching, application of protective devices, etc. need to be considered in accordance with the current edition of the IEE Wiring Regulations.
WORKED EXAMPLE: FAULT CALCULATION The following calculations and information are not exhaustive but are intended to give the reader sufficient knowledge to enable switchgear of adequate load and fault current rating to be specified. The subject may be studied in more detail by reading the relevant documents listed in Appendix 1. The nomenclature used is generally as given in the GE Alstom Grid Solutions Network Protection and Automation Guide (NPAG) and the Electricity Council’s Power System Protection (IET). When a short-circuit occurs in a distribution switchboard, the resulting fault current can be large enough to damage both the switchboard and associated cables owing to thermal and electromagnetic effects. The thermal effects will be proportional to the duration of the fault current to a large extent, and this time will depend on the characteristics of the nearest upstream automatic protective device which should operate to clear the fault. Arcing faults due to water or dirt ingress are most unlikely in the switchboards of land-based installations, but from experience, they need to be catered for offshore. For switchboards operating with generators of 10 MW or more, it is usually not difficult to avoid the problem of long clearance times for resistive faults. However, with smaller generators, clearance times of several seconds may be required because of the relatively low prospective fault currents available (see earlier section on busbar protection). With small emergency generators, pilot exciters are not normally provided and the supply for the main exciter is derived from the generator output. This arrangement is not recommended, as it allows the collapse of generator output current within milliseconds of the onset of a fault. With such small generators, even subtransient fault currents are small, and it is unlikely that downstream protection relays set to operate for ‘normal’ generation will have operated before the output collapse. It is usual to provide a fault current
231
232
CHAPTER 5 Conductor Sizing, Load Flow and Fault Calculation
FIGURE 4.5.2 Schematic of small-generator fault current maintenance circuit. AVR, automatic voltage regulator.
maintenance unit as shown in Fig. 4.5.2. This device is basically a compounding circuit which feeds a current proportional to output current back to the exciter field. When the output current reaches a threshold value well above the normal load current, a relay operates, switching in the compounding circuit. Thus a high output current is maintained by this feedback arrangement until the operation of definite time overcurrent protection, which is set to prevent the generator thermal rating being exceeded.
MAIN GENERATOR FAULT CURRENTS In this example, one generator is rated at 15.0 MW. The alternator has the following parameters: rating 17.7 MVA at a power factor of 0.85 and voltage of 6.6 kV;
Worked Example: Fault Calculation
FIGURE 4.5.3 Typical cable sizing spreadsheet (see Web Reference 2).
reactances Xd″ = 0.195 per unit (pu), × Xd′ = 0.31 pu, Xd = 2.2 pu and Xq″ = 0.265; full-load current, 1550 A and neutral earthing resistor, 10 Ω. There are four more generators of a different type, each rated at 3.87 MVA and having parameters as follows: reactances Xd″ = 0.15 pu , Xd′ = 0.18 pu, Xd = 2.05 pu and Xq″ = 0.15; full-load current, 340 A and neutral earthing resistor, 40 Ω (Fig. 4.5.3). The generator reactance values (pu) may be tabulated on their own base as follows: MVA Unit
Xd″
Xq″
Xo″
Rn (Ω)
1 × 177
0.195
0.265
0.09
10
1 × 3.87
0.15
0.15
0.045
40
Referring to a common base of 17.7 MVA, 6.6 kV and considering the units in parallel: MVA Unit
Xd″
Xq″
Xo″
Rn
1 × 3.87 4 × 3.87
0.686 0.1715
0.686 0.1715
0.2058 0.0515
16.25 4.06
1 × 17.7
0.195
0.265
0.09
4.06
233
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CHAPTER 5 Conductor Sizing, Load Flow and Fault Calculation
Summing the reactances, Xd″ =
Xq″ =
Xo″ =
1 1/0.195 + 1/0.1715 1
1/0.265 + 1/0.1715 1 1/0.09 + 1/0.0515
Rn =
= 0.0912 pu
1 1/4.06 + 1/4.06
= 0.104 pu
= 0.0327 pu = 2.63 pu
Hence, Xd″ + Xq″ = 0.0912 + 0.104 = 0.1952 pu
Xd″ + Xq″ = 0.912 + 0.104 + 0.0327 = 0.2279 pu
( )2 2 | Z1 | = Xd″ + Xq″ + Xo″ + (3Rn )
2
2
Z1 = (0.2279) + (3 × 2.03) = 6.09 pu
Therefore the system fault currents (neglecting motor contribution) are 17.7 Three ‐ phase fault current = = 16.97 kA 3 × 6.6 × 0.0912 Phase ‐ to ‐ phase fault current = Earth fault current =
17.7 6.6 × 0.1952
3 × 17.7 6.6 × 6.09
= 13.74 kA
= 0.762 kA
This method of calculation provides low values which, although not suitable for providing fault currents for the selection of switchgear, are useful for relay setting, as neglecting motor fault contributions provides a safety margin. The extra fault current provided by motors should significantly reduce the operating times of all overcurrent and earth fault protection devices, provided the current transformer saturation is not a problem. A check should be made, however, that generator cable reactances do not reduce prospective fault currents by more than a negligible amount. This reduction is likely to be significant for the smaller generators operating at LVs.
SWITCHBOARD FAULT CURRENTS The prospective fault currents for medium-voltage switchboards can be obtained in a similar manner by summing generator and transformer reactances as follows. Take the base again as 17.7 MVA, and take as an example the generator switchboard
Worked Example: Fault Calculation
connected to the LV switchboard by a 2 MVA transformer with 440 V secondary winding. Then the sequence reactances (pu) are as follows:
X1
X2
X0
At transformer rating
0.13
0.11
0.11
On the base MVA
1.1505
0.974
0.974
Summing generator and transformer reactances, X1 = 0.0912 + 1.1505 = 1.2417 pu X2 = 0.104 + 0.974 = 1.078 pu X0 = 0.974 = 0.974 pu Hence,
(X1 + X2 ) = 1.2417 + 1.078 = 2.3197 pu (X1 + X2 + X0 ) = 1.2417 + 1.078 + 0.974 = 3.294 pu
Therefore the LV system fault currents are
Three ‐ phase fault current =
17.7 3 × 0.44 × 1.2417
Phase ‐ to ‐ phase fault current = Earth fault current =
17.7 0.44 × 2.3197
3 × 17.7 0.44 × 3.294
= 18.7 kA = 17.34 kA
= 21.15 kA
The workings required for protection settings of the same scenario are continued in PART 4 Chapter 7.
235