Which Class of Wire need to be used for House Wiring

Different Class of Conductor

• As per IEC 60228, Electrical wires/cables are classified into different classes according to the conductor’s flexibility, conductor hardness & thermal effects.
• There are four classes of flexibility for electrical cables
• Class 1 = Solid conductor= ideal conductors for permanent installations.
• Class 2 =Stranded conductor= conductors designed for fixed installation.
• Class 5 =Flexible conductor= preferred to used where flexibility is required, for movable equipment , where there is vibration in equipment.
• Class 6 =Very Flexible conductor= highly flexible conductors used in robotics, flexible codes.
• Classes 3 and 4 are not described in IEC 60228.
• The most basic type of conductor is a single, solid wire (Class 1). It provides a smaller diameter, the largest Cross-Sectional Area (CSA), and the clearest signal, it is mechanically fragile and susceptible to breakage after repeated bending cycles.
• To improve flexibility, wires are stranded together (Class-2, Class-5, Class-6). Class 2 is a multi-wired conductor, while classes 5 and 6 are fine or ultra-fine wired conductors. The IEC standard specifies values such as the maximum diameter and maximum resistance for the individual wires.
• The more wires that are stranded together to make a given size, the more flexible the conductor will be. This indicates that a higher class corresponds to a greater number of strands within the conductor. Additionally, stranded wires are significantly easier to manipulate and bend during installation compared to a single wire of equivalent cross-section.
• Classes 1 and 2 are intended for use in cables for fixed installations. On the other hand, Classes 5 and 6 are designed for use in flexible cables and cords but may also be used for fixed installations.

(A) Class 1: Solid Conductors

• Construction: Single Conductor, solid copper wire.
• Flexibility: Rigid and non-flexible. the cable should not be bent more than about four times its diameter
• Characteristics: High electrical conductivity and resistance to corrosion, but less suitable for environments requiring flexibility.
• Advantages: Less expensive than cables with multiple wires
• Disadvantages: Less suitable for applications involving movement.
• Heat and Losses: Class 1 wires are more efficient for fixed wiring due to lower resistance and heat generation. 
• Applications: Typically used in permanent, stationary installations, House wiring where the conductor will not be subject to frequent movement or low flexibility is not a problem such as in building wiring and power distribution.
• They are often used when cables with larger cross-sections are required for fixed installations. They are not suitable for very flexible cables, which are used, for example, in continuously moving objects such as robotic arms in industrial production

(B) Class 2: Stranded Conductors

• Construction: Composed of multiple smaller copper wires twisted or braided together to form a single conductor.
• Flexibility: More flexible than Class 1, allowing for some movement without breaking or damaging the wire.
• Characteristics: Offers a balance of flexibility and durability but may not be as conductive as a solid conductor of the same gauge.
• Advantages:Lower electrical resistance and less heat buildup under load.
• Disadvantages:Less suitable for applications involving movement.
• Heat and Losses:Class 2 wires are more efficient for fixed wiring due to lower resistance and heat generation. 
• Applications: Primarily used for fixed installations like permanent building and house wiring and for industrial applications with increased cable flexibility requirements.

(C) Class 5: Flexible Conductors

• Construction: Consists of many fine copper wires (often tinned for corrosion resistance) twisted together, making the conductor highly flexible.
• Flexibility: Extremely flexible, designed for applications where the conductor needs to withstand frequent movement, bending, or vibration without damage.
• Characteristics: High flexibility, durable against wear and tear, but may have slightly lower conductivity compared to solid conductors due to the finer strands.
• Advantages:Superior flexibility.
• Disadvantages:Higher electrical resistance, which can result in greater heat loss and voltage drops.
• Heat and Losses:Class 5 wires are not efficient for fixed wiring due to higher resistance and heat generation compared to Class-2. 
• Applications: Used in situations where more flexibility is required, such as in circuits that may need to be bent, coiled, or moved occasionally. Ideal for portable appliances and equipment that move constantly like portable cords, flexible cables, and power tools that require a durable, yet highly flexible conductor.

(D) Class 6: Extra Flexible Conductors

• Construction: Made up of very fine copper wires, typically tinned, twisted into a very flexible configuration.
• Compared to class 5, the number of strands and wires arranged around each other is even larger, which can further increase the flexibility of the wire
• Flexibility: The highest level of flexibility among copper conductors, suitable for applications requiring frequent movement or twisting.
• Characteristics: Very high flexibility, ideal for dynamic applications, but may have lower conductivity due to the fine strands.
• Applications: Used in highly flexible cable assemblies, robotics, automobile, machine and tool construction and flexible power cables where the conductor will experience constant movement and mechanical stress.

Which Class of Conductor Used for House hold Wiring:

• To reduce power consumption, eliminate heating of wires, The Selection of House wire is most important.
• The selection of Wires broadly depends on conductor Resistance, Current, Quality of conductor material, Cross section area and power consumption.
• Resistance: A conductor with higher resistance will consume more power (P = I²R, where P is power, I is current, and R is resistance).
• Current: If both conductors are used in the same application with the same current, the one with the higher resistance will consume more power.
• Conductor Quality: Many people believe that wire quality simply by measuring its diameter and doing a mathematical calculation to estimate resistance. Conductor resistance is not just about the size of copper but it also depends on copper purity. For example, impure or recycled copper may have a bigger cross-sectional area but still higher resistance, which means more heat, more energy loss and shorter wire life.
• Conductor Size: The material (copper, aluminum, etc.) and the cross-sectional area of the conductor also significantly affect resistance and power consumption.
• Power Consumption: The power consumption of a conductor is primarily determined by its resistance and the current flowing through it, rather than its classification (Class 2 or Class 5). However, the classification itself does provide some context regarding the conductor’s characteristics:
• There are mainly two types of Conductors solid (Class-1) and stranded (Class-2 & Class-5).

(A) Selection between Class-1 or Class-2 (Solid or Standard):

• Solid conductor (Class-1) has less flexibility hence not easily passing in conceal conduits of house wires and making hot spots at conductor bends. Due to less flexibility easily break conductor at its termination location.
• The cables used in Building wiring switched to Class 2 copper conductors as it offered better flexibility over the Class 1 solid copper conductors. It is also technically superior and avoid hot spots at bends without compromising the current carrying capacity on account of its resistance being the same as specified for Class 1 copper conductors.
• Multi-stranded conductor (Class-2) shall be replaced to single solid conductors (Class-1) for all the House wiring.

(B) Selection between Class-2 & Class-5 (Standard or fine Standard):

• Stranded conductor can be divided broadly in two types one is multi-strand conductor (known as class-2 conductor) other is Flexible stranded conductor (known as class-5 conductor).
• The difference between Class-2 & Class-5 Wires are as under
• Resistance: The conductor resistance of class 5 is high compared to class 2 conductor, the heat generated for the same current loading will be different on both class of conductors

Copper Conductor Resistance based on class (IS:694)
Wire Size	Copper Conductor Resistance (Ω/Km)		Insulation Thickness (mm)		Tensile Strength (N/mm2)
	CLASS-2	CLASS-5	CLASS-2	CLASS-5	CLASS-2	CLASS-5
0.75 Sq.mm	24.5	26	0.7	0.6	12.5	10
1 Sq.mm	18.1	19.5	0.7	0.6	12.5	10
1.5 Sq.mm	12.1	13.3	0.7	0.6	12.5	10
2.5 Sq.mm	7.41	7.98	0.8	0.7	12.5	10
4 Sq.mm	4.61	4.95	0.8	0.8	12.5	10

• Insulation: The insulation thickness for the class 5 conductor cable is lesser than specified for class 2, not better for higher load conditions
• Mechanical Strength: The mechanical strength of insulation for class 5 is lesser in comparison to class 2, this can lead to issues during conduit pull.
• Flexibility: The difference lies mainly in flexibility.
• Class 2 wires have fewer strands (Conductor) of more diameter. Example :14/0.31mm(max.) 14 strands each of 0.31 mm (max.). typically, 7 strands are used which makes them less flexible and more suitable for fixed installations.
• Class 5 wires have more strands (Conductor) of less diameter. Example: 32/0.21mm(max.) 32 strands each of 0.21 mm (max.). Typically, 30 to 50 strands are used, which makes them more flexible and easier to bend.
• Application: For fixed wiring application conductors with Class 2 copper shall be used. Worldwide the usage of class 2 conductors is specified for building wires as it offers lower resistance, mechanical strength is higher.
• Class 5 wires are commonly used in applications where flexibility is important, such as in portable appliances and equipment where the lengths are preferably 1.5 to 2-meter, power tools, panel wiring (As bending and routing of such cables in constricted paths do not stress on the cable and handling and installation of such conductors in confined areas is easier).
• Amount of Copper content: Actually, in class 5 conductors, copper content is less than that of class 2 conductor which make them more flexible wires. The reason is copper wires with class 5 conductor are cheaper.
• Lesser copper content in wire leads to increased cable resistance and which may in turn increase power consumption and loss. on the other side these coper wires with class 5 conductor can bring disaster in a building as it increases the disconnection time of protective device due to its increased resistance. The power loss of class 5 conductor is higher and is against the concepts of energy conservation or sustainability.

Comparison of Class 2 and Class 5 copper conductors
Property	Characteristic	CLASS-2	CLASS-5
Installation	Passing through Conduit	Easy to Pass through Conduits due to less flexible compared to Class-5	Conductor is more Flexible hence change of Cable getting Stuck in Conduit
	Termination	No of Strands are less hence easy to crip Lugs	No of Strands are higher hence difficult to hold all strand under Lugs while crimping
	Maintenance	In case of replacement easy to pull out Wire from Conduit	difficult to pullout from conduit after installation.
Mechanical	Tensile Strength	Higher mechanical Strength to withstand Stress	Mechanically weak compared to Class-2
	Loose Connection	Cable stay firm near its termination in case of vibration	Might get loose in case of vibration
	Conductor Roundness	Conductor bunch is circular due to less number of strands	Due to more no of Strands and it’s arrangement. Not circular as compared to class-2
	Conductor Structure	Fewer Strands (Conductor) of large Size	More Strands (Conductor) of Small Size
Electrical	Resistance	Less conductor Resistance	Higher Conductor Resistance
	Current Capacity	Higher Current carrying capacity	Less Current carrying capacity
	Derating Factor	Better conductor roundness makes symmetrical arrangement in conduit, reduce derating Factor	Higher derating Factor
	Amount of Copper	Use of Copper is higher than Class-5 for the Same Size of Conductor	Use of Copper is less than Class-2 for the Same Size of Conductor
	Power Loss	Less Power Loss due to less resistance	6 to 8% Higher Power Loss compared to Class-2
	Insulation	Insulation thickness is higher than Class-5	Insulation thickness is less than Class-2
	Heat Built up	Minimal (Due to less Resistance)	Heat Faster under Load
Cost	Cheaper	5 to 8% Costly (due to more Copper) than Class-5	Cheaper than Class-2
Application	Fixed /Movable	Typically used fixed / Permeant Wires installation in wall and ceiling where wires are permeant and used regularly	Used for Flexible application like Power cord, extension Wires Board, for Movable parts

• Multi-stranded conductor (Class-2) shall be used for house wiring due to its Less Resistance, Less heat loss, Low Power consumption, better insulation, higher mechanical Strength.

Conclusion:

• Actually, Wires with Class-5 copper are used for appliance wiring and panel wiring (Not Fixed Wiring) only and they are also manufactured accordingly. Wiring with Class-5 copper conductors do not conform to the code of practice of wiring and hence it’s illegal to use them in fixed wiring and In IS 694 specifies panel wire and building wire as a similar group of functioning. This created a confusion and an create opportunity to misuse of Class 5 conductors as building wires because Class-5 Wires are 8 % cheaper than wires with Class-2 copper conductor.
• However, we must avoid to use Class-5 Wires for Wiring Application due to its less copper content (due to its more flexibility) in wire will lead to increase cable resistance and which may in turn increase the power consumption, higher watt loss, higher voltage drops, higher fault loop impedance.
• Higher impedance of the circuit may lead to accidents due to higher disconnection time of protective device. Less mechanical strength, less insulation hence heats up in load and not safe in continuous load.
• For fixed wiring application, House Wiring conductors with Class 2 copper shall be used.

Calculate Diesel Generator Protection Setting

Recommended Generator Protection are

Recommended Generator Protection
ANSI Code	Protection Function
27	Under Voltage
32	Reverse Power
37	Under Power
40	Loss of Excitation
46	Negative Phase Sequence /Un Balance Load
49T	Thermal Overload
50	Instantaneous Over Current
51	Time grade Over Current
51G	Earth Fault Time Overcurrent
50/51V	Voltage Restrained Overcurrent
59	Over voltage
60G	Fuse Failure Monitor
64S	Stator Earth Fault Protection
81	Under / Over Frequency
87	Three Phase Current Differential
87N	Neutral Current Differential
87G	Generator Differential Protection
24G	Over excitation (Volt/Hertz) Protection
21G	Impedance Protection
59N or 64G1	Stator EF protection (0-95%)
27TN or 64G2	Stator EF protection (100%)
50BF	Breaker Failure Protection
24G	Over excitation (Volt/Hertz) Protection
78G	Pole slip protection

Protection Setting Calculation:

(1) Under Voltage Relay (27):

• The Under Voltage Relay measure either phase-to-phase (Ph-Ph) or phase-to-neutral (Ph-N) fundamental RMS voltage depending on the input voltage setting. If the value of measured voltages deviates from the setting values, then these relays will give a trip indication.
• Reason:
• An under-voltage condition in a diesel generator can occur due to several reasons, overloading the generator beyond its capacity, faulty Automatic Voltage Regulator (AVR), issues with the stator windings, problems with the voltage sampling line, loose connections, low engine speed, fuel problems, and issues with the excitation system
• Setting:
• The Typical under-voltage setting is usually 80 % of the normal rated voltage. If the voltage falls below this level for the set amount of time, then the tripping command is issued by the relay and hence the system is isolated. The time setting is used to avoid tripping due to any transient disturbances. the exact setting can vary depending on the specific generator and system requirements.
• Usually, motors stall at below 80% of their rated voltage. An under-voltage element can be set to trip motor circuits once fall below 80% so that on the restoration of supply an overload is not caused by the simultaneous starting of all the motors.
• Normally Generators are designed to operate continuously at minimum voltage of 95% of its rated voltage.
• Two levels of tripping are provided depending on the severity of the condition, these under voltage elements are blocked from tripping when the generator breaker is open to allow for startup conditions.
• Calculation:
• For 415V Diesel Generator
• Level 1 (Slow)= 80% of Rated Voltage
• Level 1 (Slow)= 80% x 415V =332 V
• Time Delay = 5 sec.
• Level 2 (Fast): 70% of Rated Voltage
• Level 2 (Fast)= 70% x 415V =290 V
• Time Delay = 0 sec.

(2) Over Voltage Protection [59]:

• The Over Voltage Relay measure either phase-to-phase (Ph-Ph) or phase-to-neutral (Ph-N) fundamental RMS voltage depending on the input voltage setting. If the value of measured voltages deviates from the setting values, then these relays will give a trip indication.
• Reason:
• System over voltages can damage the insulation of components. Over voltages occur due to sudden loss of load, improper working of tap changer, Generator AVR malfunction, Reactive component malfunctions, etc.
• Setting:
• The Overvoltage setting is usually 110 to 130 % of the normal operating voltage depending on the system requirement.
• If the voltage rises above this level for the set amount of time then the tripping command issued by the relay and hence the system is isolated. The time setting is used to avoid tripping due to any transient disturbances.
• Calculation:
• For 415V Diesel Generator
• Level 1 (Slow)= 110% of Rated Voltage
• Level 1 (Slow)= 110% x 415V =456 V
• Time Delay = 5 sec.
• Level 2 (Fast): 130% of Rated Voltage
• Level 2 (Fast)= 130% x 415V =539 V
• Time Delay = 0 sec.

(3) Reverse Power Protection [32R]:

• Reverse power relay is an electronic, microprocessors-based protection device which is used for monitoring and stopping the power supply flowing grid side to the DG side or generator running in parallel with another generator. If accidentally leakage current is received by generator, then it can start to running as motor. This situation may be very dangerous for generator set.
• The function of the reverse power relay is to prevent a reverse power condition in which power flows from the bus bar into the generator.
• This condition can occur when there is a failure in the prime mover such as an engine or a turbine which drives the generator.
• Relay detects the reverse flow of power from the load back to the generator, which can occur during system faults or abnormal operating conditions. By sensing this reverse power flow, the relay triggers a protective action, typically disconnecting the generator to prevent further issues.
• The generator are classified by their Prime Mover which determine the amount of Reverse power they can motor.

Sr. No	Prime Mover	Motorizing Power in % of Unit Rating
1	Gas Turbine (single shaft)	100%
2	Gas Turbine (Double Shaft)	10-15%
3	4 Cycle Diesel	15%
4	2 Cycle Diesel	25%
5	Hydraulic Turbine	2-100
6	Steam Turbine (Conventional)	1-4%
7	Steam Turbine (Cond Cooled)	0.5 to 1.0%

• Reason:
• When Two or more unit running in parallel
• In LT panel if the DG supply is running then grid supply should be switched off and if the grid supply is running then DG supply should be switched off. When one source is on then second source accidentals starts to leakage current resultant a large fault may be occurred and system can be failed. So, for prevention of other source leakage the RPR relay is used.
• Failure of Speed controller or another breakdown. When the prime mover of a generator running in a synchronized condition fails. There is a condition known as motoring, where the generator draws power from the bus bar, runs as a motor and drives the prime mover. This happens as in a synchronized condition all the generators will have the same frequency. Any drop in frequency in one generator will cause the other power sources to pump power into the generator. The flow of power in the reverse direction is known as the reverse power relay.
• If the frequency of the machine to be synchronized is slightly lesser than the bus bar frequency and the breaker is closed, power will flow from the bus bar to the machine. Hence, during synchronization(forward), frequency of the incoming machine is kept slight higher than that of the bus bar i.e. the synchroscope is made to rotate in the “Too fast” direction. This ensures that the machine takes on load as soon as the breaker is closed.
• Loss of excitation:
• Failure of AVR
• Setting:
• A generator reverse power relay setting is typically set between 2% to 8% of the generator’s rated power, depending on the type of prime mover (like a diesel engine or steam turbine), with diesel engines generally requiring a higher setting (around 8%) compared to turbines (around 2 to 5%) to prevent unnecessary tripping during transient conditions; this setting essentially determines the threshold at which the relay will activate to protect the generator from reverse power flow, which can damage the machine if it becomes too significant. 
• Calculation:
• Generator capacity :500KVA ,415V,0.9 Power factor
• Full Load Current =500×1000/(1.732*415)
• Full Load Current =695A
• Setting at 5%
• Reverse Power = -5%*500*0.9 = -22.5KW
• Relay Setting= Reverse Power / Real Power =-22.5 / 500 =-4.50%
• Relay Setting =-4.50%
• Time delay proposed=5 sec

(4) Negative Phase Sequence (Unbalance Phase) Relay (46):

• The Negative Sequence Overcurrent function provides protection against possible rotor overheating and damage due to unbalanced faults or other system conditions which can cause unbalanced three phase currents in the generator.
• Negative Phase Sequence detects imbalances in the network that does not cause energy loss out of the system.
• Reason:
• Generator or Motor are design to operate in balance three phase loading.
• Generator negative phase sequence currents can result from any unbalance condition on the system including un transposed lines, single phase loads, unbalanced type line faults and open conductors. the unbalance condition leads negative sequence currents having opposite rotation that of power system in generator leads. This reversed rotating current produce double frequency current in rotor structure. This resulting over heating of rotor.
• Setting:
• A generator Negative Phase Sequence (NPS) relay setting is typically set between 2 to 10% of the full load current depending on the specific generator design and manufacturer’s recommendations, aiming to detect significant unbalances in the power system while avoiding unnecessary tripping due to normal load variations; this setting should be based on the generator’s maximum allowable negative sequence I² (current squared) value to prevent excessive rotor heating. 
• Generator withstand limit against negative sequence overcurrent (K) = 10 (As per IEC-60034-1)
• Normally Generator continuous withstand limit: 8 %
• Calculation:
• Generator capacity :500KVA ,415V , CT is 800/1
• Full Load Current =500×1000/(1.732*415)
• Full Load Current =695A
• Setting at 10% .
• Desired pickup current = 10% of rated current
• Relay setting = (0.1 x Rated Current) / CT ratio 
• Relay setting =(0.1×695) / 800
• Relay setting =0.0868A

(5) Thermal Overload Relay (49T):

• In general, generators can operate successfully at rated kVA, frequency, and power factor for a voltage variation of 5% above or below rated voltage. Under emergency condition, it is possible to exceed the continuous output capability for a short time.
• The stator overload function provides protection against possible damage during overload conditions.
• Reason:
• A generator becomes overloaded when too many appliances or devices are plugged in and drawing power simultaneously, exceeding the generator’s rated capacity, often happening when attempting to power heavy appliances like air conditioners, heaters, or electric stoves at the same time; essentially, drawing more power than the generator can supply. 
• Peak usage times: Running multiple high-power appliances simultaneously. 
• Damaged components: Faulty electrical components within the generator can contribute to overload issues. 
• Improper load management: Not prioritizing which appliances to run on the generator. 
• Adding new equipment: Plugging in additional appliances without considering the generator’s capacity. 
• Setting:
• A generator thermal overload relay setting is typically based on a percentage of the motor’s full load current.
• Common settings are:
• For motors with Service Factor (SF) ≥ 1.15, Set to 125% of FLA.
• For motors with Service Factor (SF) < 1.15, Set to 115% of FLA
• As per IEEE Generator short time thermal capability for balanced three-phase loading diagram (Short time capability curve) the wining will withstand 117% rated current for 120 second.
• Calculation:
• Generator capacity :500KVA ,415V , CT is 800/1
• Full Load Current =500×1000/(1.732*415)
• Full Load Current =695A
• Setting at 117% .
• Desired pickup current = 117% of rated current
• Relay setting = (1.17 x Rated Current) / CT ratio 
• Relay setting =(1.17×695) / 800
• Relay setting =1.016A

(6) Generator Under Frequency Protection (81 G):

• Prevents the steam turbine and generator from exceeding the permissible operating time at reduced frequencies.
• Ensures that the generating unit is separated from the network at a preset value of frequency.
• Prevent over fluxing (v/f) of the generator (large over fluxing for short times).
• The stator under frequency relay measures the frequency of the stator terminal voltage.
• Setting Recommendations:
• within 0.2 to 0.5Hz below the nominal frequency
• For Alarm: 48.0 Hz, 2.0 Sec. time delay. 
• For Trip: 47.5 Hz, 1.0 Sec. or as recommended by Generator Manufacturers.

(7) Instantaneous Over Current Relay (50):

• Instantaneous overcurrent protection is where a protective relay initiates a breaker trip based on current exceeding a pre-programmed “pickup” value for any length of time. 
• Setting:
• A generator phase instantaneous overcurrent relay setting is typically set between 2 to 1.5 times the full load current (FLA) of the generator, ensuring quick tripping in case of a severe fault while avoiding unnecessary trips due to momentary current surges during starting or load fluctuations; this setting is usually referred to as the “pickup current” of the relay. 
• This is back up protection for Generator. To avoid unnecessary trip of the generator we recommend making OFF this function in generator protection.
• Calculation:
• Generator full Load Current = 130A & CT is 300/5 =60
• Setting =1.5 times of Full Load Current
• Setting= 1.5×130 =195A
• 51 Current Setting = Setting / CT Ratio = 195/60 =3.25A.
• Time setting =5 Second.
• The proposed above setting is coordinated with other O/C protection setting.

(8) Time grade Over Current Relay (51):

• Time overcurrent protection is where a protective relay initiates a breaker trip based on the combination of overcurrent magnitude and overcurrent duration, the relay tripping sooner with greater current magnitude. This is a more sophisticated form of overcurrent protection than instantaneous.
• Setting:
• This is back-up protection of the generator, for better time gradings the overcurrent setting should be co-ordinate with load connected feeder overcurrent setting.
• A generator Phase Overcurrent (51) setting is typically set between 125% and 150% of the generator’s full load current, however, the exact setting depends on the specific application and should be coordinated with other system protections.
• Calculation:
• Generator full Load Current = 130A & CT is 300/5 =60
• Setting =150% of Full Load Current
• Setting= 1.5×130 =195A
• 51 Current Setting = Setting / CT Ratio = 195/60 =3.25A.
• Time setting =5 Second.
• The proposed above setting is coordinated with other O/C protection setting.

(9) Earth Fault Time Overcurrent (51G)

• This is back-up protection in Earth Fault of generator, for better time gradings the overcurrent setting should be co-ordinate with load connected feeder setting.
• Setting:
• Earth Fault Relay setting shall be 10 to 20 % Full Load Current
• Calculation:
• Generator full Load Current = 130A & CT is 300/5 =60
• Setting =20% of Full Load Current
• Setting= 0.2×130 =26A
• 51G Current Setting = Setting / CT Ratio = 26/60 =0.43A.
• Time setting =5 Second.
• The proposed above setting is coordinated with other O/C protection setting.

(9) Ground Differential (87 N)

• The ground differential element (87N) that operates based on the difference between the measured neutral current and the sum of the three-phase current inputs.
• The 87N element provides sensitive ground fault detection on resistance-grounded particularly where multiple generators are connected directly to a load bus.
• The relay provides two definite-time delayed ground current differential elements designed to detect ground faults on resistance grounded generator.
• The relay uses the neutral CT connected to the relay input to measure the generator neutral current. It then calculates the residual current, which is the sum of the three phase current inputs (from CTs located at generator terminals).
• The relay adjusts the residual current by the ratio of the CTR and CTRN settings to scale the residual current in terms of the secondary neutral current. It then calculates the difference. Normally, under balanced load or external ground fault conditions, the difference current should be zero. In the event of an internal ground fault, the difference current is nonzero. If the difference current magnitude is greater than the element pickup setting, the element picks up and begins to operate the definite time-delay.
• Setting:
• Earth Fault Relay setting shall be 10 to 20 % Maximum Ground Fault Current
• Calculation:
• Generator grounded through 39.8 Ohms Resistance.
• Generator rated Voltage=13800V, Current 130A
• Maximum Earth Fault Current =(138000 / 1.732) / 39.8
• Maximum Earth Fault Current =7967.4 / 39.8
• Maximum Earth Fault Current = 200 A
• 87N pickup current setting = 10% x 200 / CT Ratio
• 87N pickup current setting = 20 / 60
• 87N pickup current setting = 0.3
• 87N Time delay =0.2s

Calculate Earthing Strip Size for Electrical Equipment’s in Power Distribution Network

EXAMPLE:

Calculate Earthing Strip / Cable Size for Electrical Equipment’s / Panels in Power Distribution Networks.

• At RMU
• At Transformer
• At D.G Set
• At Main Distribution Panel
• At Sub Panel-1
• At Sub Panel-2

CALCULATION:

(1) Earthing Strip Size at RMU:

• Shot circuit capacity at RMU is 18.37 KA for 1 Second.
• Corrosion in Strip is 1% per year
• Earthing Strip shall be replaced after 25 Years.
• Safety Factor for Strip is 1.5
• Earthing Strip Material is GI

Calculation

• As per IS: 3043, clause 17.2.2.1:
• Cross section area of Earthing Strip (A) =(Isc x√t)/k
• Where Isc= Shot circuit current capacity in Ampere.
• t= Time for Shot circuit current in Second.
• K= Material Constant

Bare Conductor Material with No Risk of Fire or Danger to any Other Touching or Surrounding Material
TABLE 11A (IS:3043)
Material	K value (1 second)	K value (2 second)
Steel	80	46
Aluminum	126	73
Copper	205	118

• Cross section area of Earthing Strip (A) = (Isc x√t)/k
• Cross section area of Earthing Strip (A) = (18.37×1000 x√1)/80
• Cross section area of Earthing Strip (A) = 229.69 Sq.mm
• Allowable corrosion =1% per Year
• No of Year for replacement = 25 Year
• Allowable corrosion in 25 Years = 229.69x1x25% =57.40 Sq.mm
• Allowable Safety Factor = 229.69×1.5%=3.44 Sq.mm
• Required Earthing Strip size = Cross sectional area + Total Corrosion allowance + Safety factor
• Required Earthing Strip size=229.69+57.40+3.44 Sq.mm
• Required Earthing Strip size=290.47 Sq.mm
• Proposed GI Earthing Strip shall be 50×6 mm = 300 Sq.mm.
• Here Proposed Earthing Strip Size > Required Earthing Strip Size
• Proposed Earthing Strip is OK

(2) Earthing Strip Size at Transformer:

• Shot circuit capacity at Transformer is 25.32 KA for 1 Second.
• Corrosion in Strip is 1% per year
• Earthing Strip shall be replaced after 25 Years.
• Safety Factor for Strip is 1.5
• Earthing Strip Material for Transformer Neutral is Copper
• Earthing Strip Material for Transformer Body is GI

Calculation

• For Neutral
• As per IS: 3043, clause 17.2.2.1:
• Cross section area of Earthing Strip (A) =(Isc x√t)/k
• Where Isc= Shot circuit current capacity in Ampere.
• t= Time for Shot circuit current in Second.
• K= Material Constant

Bare Conductor Material with No Risk of Fire or Danger to any Other Touching or Surrounding Material
TABLE 11A (IS:3043)
Material	K value (1 second)	K value (2 second)
Steel	80	46
Aluminum	126	73
Copper	205	118

• Cross section area of Earthing Strip (A) = (Isc x√t)/k
• Cross section area of Earthing Strip (A) = (25.32×1000 x√1)/205
• Cross section area of Earthing Strip (A) = 125.51 Sq.mm
• Allowable corrosion =1% per Year
• No of Year for replacement = 25 Year
• Allowable corrosion in 25 Years = 125.51 x1x25% =30.87 Sq.mm
• Allowable Safety Factor = 125.51 x1.5%=1.85 Sq.mm
• Required Earthing Strip size = Cross sectional area + Total Corrosion allowance + Safety factor
• Required Earthing Strip size=125.51+30.87+1.85 Sq.mm
• Required Earthing Strip size=156.24 Sq.mm
• Proposed Cu Earthing Strip shall be 32×6 mm = 192 Sq.mm.
• Here Proposed Earthing Strip Size > Required Earthing Strip Size
• Proposed Earthing Strip is OK
• For Body
• As per IS: 3043, clause 17.2.2.1:
• Cross section area of Earthing Strip (A) =(Isc x√t)/k
• Where Isc= Shot circuit current capacity in Ampere.
• t= Time for Shot circuit current in Second.
• K= Material Constant

Bare Conductor Material with No Risk of Fire or Danger to any Other Touching or Surrounding Material
TABLE 11A (IS:3043)
Material	K value (1 second)	K value (2 second)
Steel	80	46
Aluminum	126	73
Copper	205	118

• Cross section area of Earthing Strip (A) = (Isc x√t)/k
• Cross section area of Earthing Strip (A) = (25.32 x1000 x√1)/80
• Cross section area of Earthing Strip (A) = 316.5 Sq.mm
• Allowable corrosion =1% per Year
• No of Year for replacement = 25 Year
• Allowable corrosion in 25 Years = 316.5 x1x25% =79.12 Sq.mm
• Allowable Safety Factor = 316.5 x1.5%=4.74 Sq.mm
• Required Earthing Strip size = Cross sectional area + Total Corrosion allowance + Safety factor
• Required Earthing Strip size=316.5+79.12+4.74 Sq.mm
• Required Earthing Strip size=400.37 Sq.mm
• Proposed GI Earthing Strip shall be 75×6 mm = 450 Sq.mm.
• Here Proposed Earthing Strip Size > Required Earthing Strip Size
• Proposed Earthing Strip is OK

(3) Earthing Cable Size at D.G Set:

• Shot circuit capacity at D.G Set is 10KA for 1 Second.
• Corrosion in Strip is 1% per year
• Earthing Strip shall be replaced after 25 Years.
• Safety Factor for Strip is 1.5
• Earthing Wire Material is Copper, XLPE Insulated

Calculation

• As per IS: 3043, clause 17.2.2.1:
• Cross section area of Earthing Strip (A) =(Isc x√t)/k
• Where Isc= Shot circuit current capacity in Ampere.
• t= Time for Shot circuit current in Second.
• K= Material Constant

Insulated Protective Conductors Not Incorporated in Cables or Bare Conductors Touching Other Insulated Cables
TABLE 11B (IS:3043)
Material	K value (1 second)	K value (3 second)
Copper , PVC Insulated	136	79
Copper, Rubber Insulated	150	92
Copper, XLPE Insulated	170	98
Aluminum, PVC Insulated	90	52
Aluminum, Rubber Insulated	106	61
Aluminum, XLPE Insulated	112	65
Steel, PVC Insulated	49	28
Steel, Rubber Insulated	58	33
Steel, XLPE Insulated	62	36

• Cross section area of Earthing Strip (A) = (Isc x√t)/k
• Cross section area of Earthing Strip (A) = (10×1000 x√1)/170
• Cross section area of Earthing Strip (A) = 58.82 Sq.mm
• Allowable corrosion =1% per Year
• No of Year for replacement = 25 Year
• Allowable corrosion in 25 Years = 58.82x1x25% =14.70 Sq.mm
• Allowable Safety Factor = 58.82 x1.5%=0.88 Sq.mm
• Required Earthing Strip size = Cross sectional area + Total Corrosion allowance + Safety factor
• Required Earthing Strip size=58.82+14.7+0.88 Sq.mm
• Required Earthing Strip size=74.41 Sq.mm
• Proposed Earthing Single Core Copper, XLPE Cable=95 Sq.mm
• Here Proposed Earthing Cable Size > Required Earthing Cable Size
• Proposed Earthing Cable is OK

(4) Earthing Strip Size at Main Panel :

• Shot circuit capacity at Main Panel is 21 KA for 1 Second.
• Corrosion in Strip is 1% per year
• Earthing Strip shall be replaced after 25 Years.
• Safety Factor for Strip is 1.5
• Earthing Strip Material is GI

Calculation

• As per IS: 3043, clause 17.2.2.1:
• Cross section area of Earthing Strip (A) =(Isc x√t)/k
• Where Isc= Shot circuit current capacity in Ampere.
• t= Time for Shot circuit current in Second.
• K= Material Constant

Bare Conductor Material with No Risk of Fire or Danger to any Other Touching or Surrounding Material
TABLE 11A (IS:3043)
Material	K value (1 second)	K value (2 second)
Steel	80	46
Aluminum	126	73
Copper	205	118

• Cross section area of Earthing Strip (A) = (Isc x√t)/k
• Cross section area of Earthing Strip (A) = (21 x1000 x√1)/80
• Cross section area of Earthing Strip (A) = 262.5 Sq.mm
• Allowable corrosion =1% per Year
• No of Year for replacement = 25 Year
• Allowable corrosion in 25 Years = 262.5 x1x25% =65.62 Sq.mm
• Allowable Safety Factor = 262.5 x1.5%=3.93 Sq.mm
• Required Earthing Strip size = Cross sectional area + Total Corrosion allowance + Safety factor
• Required Earthing Strip size=262.5+65.62+3.93 Sq.mm
• Required Earthing Strip size=332.06 Sq.mm
• Proposed GI Earthing Strip shall be 75×6 mm = 450 Sq.mm.
• Here Proposed Earthing Strip Size > Required Earthing Strip Size
• Proposed Earthing Strip is OK

(5) Earthing Strip Size at Sub Panel-1 :

• Shot circuit capacity at Sub Panel-1 is 11 KA for 1 Second.
• Corrosion in Strip is 1% per year
• Earthing Strip shall be replaced after 25 Years.
• Safety Factor for Strip is 1.5
• Earthing Strip Material is GI

Calculation

• As per IS: 3043, clause 17.2.2.1:
• Cross section area of Earthing Strip (A) =(Isc x√t)/k
• Where Isc= Shot circuit current capacity in Ampere.
• t= Time for Shot circuit current in Second.
• K= Material Constant

Bare Conductor Material with No Risk of Fire or Danger to any Other Touching or Surrounding Material
TABLE 11A (IS:3043)
Material	K value (1 second)	K value (2 second)
Steel	80	46
Aluminum	126	73
Copper	205	118

• Cross section area of Earthing Strip (A) = (Isc x√t)/k
• Cross section area of Earthing Strip (A) = (11 x1000 x√1)/80
• Cross section area of Earthing Strip (A) = 137.5 Sq.mm
• Allowable corrosion =1% per Year
• No of Year for replacement = 25 Year
• Allowable corrosion in 25 Years = 137.5 x1x25% =34.37 Sq.mm
• Allowable Safety Factor = 137.5 x1.5%=2.06 Sq.mm
• Required Earthing Strip size = Cross sectional area + Total Corrosion allowance + Safety factor
• Required Earthing Strip size=137.5+34.37+2.06 Sq.mm
• Required Earthing Strip size=173.93 Sq.mm
• Proposed GI Earthing Strip shall be 32×6 mm = 192 Sq.mm.
• Here Proposed Earthing Strip Size > Required Earthing Strip Size
• Proposed Earthing Strip is OK

(5) Earthing Strip Size at Sub Panel-2 :

• Shot circuit capacity at Sub Panel-2 is 6 KA for 1 Second.
• Corrosion in Strip is 1% per year
• Earthing Strip shall be replaced after 25 Years.
• Safety Factor for Strip is 1.5
• Earthing Strip Material is GI

Calculation

• As per IS: 3043, clause 17.2.2.1:
• Cross section area of Earthing Strip (A) =(Isc x√t)/k
• Where Isc= Shot circuit current capacity in Ampere.
• t= Time for Shot circuit current in Second.
• K= Material Constant

Bare Conductor Material with No Risk of Fire or Danger to any Other Touching or Surrounding Material
TABLE 11A (IS:3043)
Material	K value (1 second)	K value (2 second)
Steel	80	46
Aluminum	126	73
Copper	205	118

• Cross section area of Earthing Strip (A) = (Isc x√t)/k
• Cross section area of Earthing Strip (A) = (6 x1000 x√1)/80
• Cross section area of Earthing Strip (A) = 75.0 Sq.mm
• Allowable corrosion =1% per Year
• No of Year for replacement = 25 Year
• Allowable corrosion in 25 Years = 75 x1x25% =18.75 Sq.mm
• Allowable Safety Factor = 75 x1.5%=1.12 Sq.mm
• Required Earthing Strip size = Cross sectional area + Total Corrosion allowance + Safety factor
• Required Earthing Strip size=75+18.75+1.12 Sq.mm
• Required Earthing Strip size=94.87 Sq.mm
• Proposed GI Earthing Strip shall be 25×6 mm = 150 Sq.mm.
• Here Proposed Earthing Strip Size > Required Earthing Strip Size
• Proposed Earthing Strip is OK

CONCLUSION:

• At RMU=GI Earthing Strip: 50x6MM
• At Transformer Neutral=CU Earthing Strip: 32x6MM
• At Transformer Body=GI Earthing Strip: 75x6MM
• At D.G Set=CU, XLPE Earthing Cable: 1Cx95 SQ.MM
• At Main Distribution Panel=GI Earthing Strip: 75x6MM
• At Sub Panel=1=GI Earthing Strip: 32x6MM
• At Sub Panel-2=GI Earthing Strip: 25x6MM.

Calculate Short Circuit Current at Sub Panel (End user side)

EXAMPLE:

• Calculate short circuit current at Electrical Equipment Panel / Distribution Board at end user side.
• HT Power is received by 11KV RMU and given to 11/0.415KV,1000KVA Transformer by 1 no of 3Cx185 Sq.mm 11KV HT Cable of 25 Meter Length.
• Resistance and Reactance of HT cable are 0.21 Ώ / KM and 0.1 Ώ / KM
• Transformer impedance is 6.25%.
• LT Side of Transformer is connected to Main LT Panel by 8 no of 3.5Cx300 Sq.mm cable of 70 Meter.
• Resistance and Reactance of 3.5Cx300 Sq.mm, XLPE cable are 0.129 Ώ / KM and 0.071 Ώ / KM.
• Main LT Panel is connected to Sub Panel-1 by 1 no of 3.5Cx300 Sq.mm cable of 80 Meter.
• Resistance and Reactance of 3.5Cx300 Sq.mm, XLPE cable are 0.129 Ώ / KM and 0.071 Ώ / KM.
• Main LT Panel is connected to Sub Panel-2 by 1 no of 3.5Cx95 Sq.mm cable of 80 Meter.
• Resistance and Reactance of 3.5Cx95 Sq.mm, XLPE cable are 0.409 Ώ / KM and 0.072 Ώ / KM.
• Calculate Short circuit Current at Sub Panel-1 & Sub Panel-2.

CALCUALTION

• Short Circuit Current is calculated at various points of Power Distribution Networks, At Power Receiving Point (11KV), At Main Panel Point (0.415V) and At Sub Distribution Point (0.415V)

(1) SHORT CIRCUIT CURRENT AT 11 KV PANEL / RMU:

• Assume that 11KV Side System Fault MVA is 350 MVA.
• Shor Circuit Current at 11KV System = Fault MVA / 1.732 x System Voltage at Fault Point.
• Shor Circuit Current at 11KV System = 350 / 1.732 x 11.
• Shor Circuit Current at 11KV System = 18.37 KA.

 (2) SHORT CIRCUIT CURRENT AT TRANSFORMER:

• To calculate Short Circuit Current at Transformer Side, First We need to sum Impedance of the system up to Transformer (Impedance of HT Source +Impedance of HT Cable + Impedance of Transformer).
• Transformer Capacity is 1000 KVA
• Base MVA = TC Size / 1000
• Base MVA = 1000 / 1000
• Base MVA = 1 MVA or
• Base KVA = 1000 KVA
• % Impedance of Source = Base MVA x100 / System Fault MVA
• % Impedance of Source = 1 x100 / 350
• % Impedance of Source = 0.29 Ώ / KM.
• CALCULATE % RESISTANCE & REACTANCE OF CABLE ( 185 SQ.MM HT CABLE)
• % Resistance of Cable =Base KVA (TC Capacity) x Cable Resistance x No of Run x Length of Cable / (System Voltage in KV)²x10.
• % Resistance of Cable =1000 x 0.21 x 1 x 25 / 11x11x10
• % Resistance of Cable =0.004339 % Ώ
• % Reactance of Cable =Base KVA (TC Capacity) x Cable Reactance x No of Run x Length of Cable / (System Voltage in KV)²x10.
• % Reactance of Cable =1000 x 0.1 x 1 x 25 / 11x11x10
• % Reactance of Cable =0.002066 % Ώ
• % Impedance of Cable =√ R² + X²
• % Impedance of Cable =√ (0.004339) ² + (0.002066) ²
• % Impedance of Cable =0.00481 % Ώ.
• Considering 10% Tolerance in Transformer Impedance
• % Impedance of Transformer = Impedance of Transformer – 10% Variation of Transformer Impedance
• % Impedance of Transformer = 6.25 -(6.25X10%)
• % Impedance of Transformer =5.63 % Ώ.
• Total % Impedance up to Transformer = % Impedance of Source +%Cable Impedance + % Transformer Impedance
• Total % Impedance up to Transformer = 0.29+0.00481+5.63 % Ώ.
• Total % Impedance up to Transformer = 5.92% Ώ.
• Fault KVA on LT Side of Transformer = Base MVA of Transformer x100 / Total % Impedance up to Transformer.
• Fault KVA on LT Side of Transformer = 1 x100 / 5.92.
• Fault KVA on LT Side of Transformer = 16.9 KVA.
• Short Circuit Current at Transformer LT Side= Fault KVA / 1.732 x System Voltage at Fault Point.
• Short Circuit Current at Transformer LT Side= 16.9 / 1.732 x 0.415
• Short Circuit Current at Transformer LT Side= 25.32 KA

 (3) SHORT CIRCUIT CURRENT AT MAIN LT PANEL:

• To calculate Short Circuit Current at Main LT Panel, we need to sum Impedance of the system up to Main LT Panel (Total Impedance up to Transformer LT Side + Impedance of LT Cable).
• Total % Impedance up to Transformer LT Side= 5.92% Ώ.
• CALCULATE % RESISTANCE & REACTANCE OF CABLE (300 SQ.MM)
• % Resistance of Cable =Base KVA (TC Capacity) x Cable Resistance x Length of Cable / (System Voltage in KV)²x10.
• % Resistance of Cable =1000 x 0.129 x 1 x 70 / 0.415 x 0.415 x10
• % Resistance of Cable =0.66 % Ώ
• % Reactance of Cable =Base KVA (TC Capacity) x Cable Reactance x No of Run x Length of Cable / (System Voltage in KV)² x10.
• % Reactance of Cable =1000 x 0.071 x 8 x 70 / 0.415×0.415×10
• % Reactance of Cable =0.36 % Ώ
• % Impedance of Cable =√ R² + X²
• % Impedance of Cable =√ (0.66) ² + (0.36) ²
• % Impedance of Cable =0.747 % Ώ.——————————–(B)
• Total % Impedance up to Main LT Panel = % Impedance up Transformer +%Cable Impedance
• Total % Impedance up to Main LT Panel = = 5.92+0.747 % Ώ.
• Total % Impedance up to Main LT Panel = = 6.662% Ώ. ————————–(B)
• Fault KVA on Main LT Panel= Base MVA x100 / Total % Impedance.
• Fault KVA on LT Side of Transformer = 1 x100 / 6.662.
• Fault KVA on LT Side of Transformer = 15.01 KVA.
• Short Circuit Current at Main LT Panel = Fault KVA / 1.732 x System Voltage at Fault Point.
• Short Circuit Current at Main LT Panel = 15.01 / 1.732 x 0.415
• Short Circuit Current at Main LT Panel = 20.88 KA.
• Short Circuit Current at Main LT Panel = 21 KA

(4) SHORT CIRCUIT CURRENT AT SUB PANEL-1:

• To calculate Short Circuit Current at Sub Panel-1, We need to sum Total Impedance of the system up to Sub Panel-1 (Total Impedance up to Main LT Panel + Impedance of LT Cable-1).
• Total % Impedance up to Main LT Panel= 6.662 % Ώ. (Calculated as per (B))
• CALCULATE % RESISTANCE & REACTANCE OF CABLE-1 (300 SQ.MM)
• % Resistance of Cable =Base KVA (TC Capacity) x Cable Resistance x Length of Cable / (System Voltage in KV)²x10.
• % Resistance of Cable =1000 x 0.129 x 1 x 80 / 0.415 x 0.415 x10
• % Resistance of Cable =5.99 % Ώ
• % Reactance of Cable =Base KVA (TC Capacity) x Cable Reactance x No of Run x Length of Cable / (System Voltage in KV)² x10.
• % Reactance of Cable =1000 x 0.071 x 1 x 80 / 0.415×0.415×10
• % Reactance of Cable =3.27 % Ώ
• % Impedance of Cable =√ R² + X²
• % Impedance of Cable =√ (5.99) ² + (3.27) ²
• % Impedance of Cable =6.82 % Ώ.
• Total % Impedance up to Sub Panel-1 = % Impedance up Main LT Panel  +%Cable Impedance
• Total % Impedance = 5.92+6.82 % Ώ.
• Total % Impedance = 13.49% Ώ.
• Fault KVA on Sub Panel-1= Base MVA x100 / Total % Impedance.
• Fault KVA on Sub Panel-1 = 1 x100 / 13.49.
• Fault KVA on Sub Panel-1 = 7.41 KVA.
• Short Circuit Current at Sub Panel-1 = Fault KVA on Sub Panel-1 / 1.732 x System Voltage at Fault Point.
• Short Circuit Current at Sub Panel-1 = 7.41 / 1.732 x0.415.
• Short Circuit Current at Sub Panel-1 = 10.33 KA.
• Short Circuit Current at Sub Panel-1 = 11 KA

(5) SHORT CIRCUIT CURRENT AT SUB PANEL-2:

• To calculate Short Circuit Current at Sub Panel-2, We need to sum Total Impedance of the system up to Sub Panel-2 (Total Impedance up to Main LT Panel + Impedance of LT Cable-2).
• Total % Impedance up to Main LT Panel= 6.662 % Ώ. (Calculated as per (B))
• CALCULATE % RESISTANCE & REACTANCE OF CABLE-2 (95 SQ.MM)
• % Resistance of Cable =Base KVA (TC Capacity) x Cable Resistance x Length of Cable / (System Voltage in KV)²x10.
• % Resistance of Cable =1000 x 0.409 x 1 x 80 / 0.415 x 0.415×10
• % Resistance of Cable =19 % Ώ
• % Reactance of Cable =Base KVA (TC Capacity) x Cable Reactance x No of Run x Length of Cable / (System Voltage in KV)² x10.
• % Reactance of Cable =1000 x 0.072 x 1 x 80 / 0.415×0.415×10
• % Reactance of Cable =3.36 % Ώ
• % Impedance of Cable =√ R² + X²
• % Impedance of Cable =√ (19) ² + (3.36) ²
• % Impedance of Cable =19.294 % Ώ.
• Total % Impedance up to Sub Panel-2 = % Impedance up Main LT Panel +%Cable Impedance
• Total % Impedance up to Sub Panel-2 = 6.662+19.294 % Ώ.
• Total % Impedance up to Sub Panel-2 = 25.956% Ώ.
• Fault KVA on Sub Panel-2 = Base MVA x100 / Total % Impedance up to Sub Panel-2.
• Fault KVA on Sub Panel-2 = 1 x100 / 25.956.
• Fault KVA on Sub Panel-2 = 3.85 KVA.
• Short Circuit Current at Sub Panel-2 = Fault KVA on Sub Panel-2 / 1.732 x System Voltage at Fault Point.
• Short Circuit Current at Sub Panel = 3.85 / 1.732 x 415.
• Short Circuit Current at Sub Panel = 5.36 KA.
• Short Circuit Current at Sub Panel-2 = 6 KA

 CONCLUSION:

• Shor Circuit Current at 11KV System = 18.37 KA.
• Short Circuit Current at Transformer LT Side= 25.32 KA
• Short Circuit Current at Main LT Panel = 21 KA
• Short Circuit Current at Sub Panel-1 = 11 KA
• Short Circuit Current at Sub Panel-2 = 6 KA