Guideline to Design Electrical Network for Building / Small Area.

 Guideline to Design Electrical Network for Building / Small Area.

 (1)  Calculate Electrical Load:

  • Find out built up area in Sqft.of per flat per House/Dwelling unit.
  • Multiply area in Sqft. by Load/Sqft according to following Table
Type of Load Load/Sqft
Industrial 100 Watt/Sqft
Commercial 30 Watt/Sqft
Domestic 15 Watt/Sqft
  • Apply the diversity factor and Compute the load of all dwelling units in the area.
Type of Load Diversity Factor
Industrial 0.5
Commercial 0.8
Domestic 0.4
  • Add the load of common services such as Auditorium, Street Lights, Lifts and Water Pumps etc. For simplicity purpose 0.5kW/dwelling units may be considered as common load.
  • Compute the “Total Load” of the area by adding load observed at above.
  • Apply the power factor of 0.8 to determine the load in kVA.
  • Compute the Load in kVA= “Total Load”/0.8
  • Take transformer loading of 65% considering the network arrangement Ring Main Circuit.

 (2) Decide voltage grade for Electrical Load:

  • If load is equal to or more than 2.50MVA, the area shall be fed through 33kV feeder. For such loads, the land space for 33/11kV Sub-station shall have to be allocated by builder / Society/ Authority.
  • For load between 1 MVA to 2.5MVA, dedicated 11kV feeder shall be preferred.
  • For load below 1 MVA, existing 11kV feed can be tapped through VCB or RMU.

(3) Decide Size of Transformer:

  • Select T.C Size of 25 KVA,63 KVA,100 KVA,200 KVA or 400 KVA  according to your Load.
  • The maximum capacity of distribution transformer acceptable is 400 kVA as a standard capacity.
  • Only two-no of transformer at one location shall be acceptable. If there is more number of transformers HT shall be required to extend using underground cables to locate additional transformer.

(4) RMU / LT Panel:

  • Either VCB or Ring Main Circuit shall be used to control transformers. There cables should have metering arrangement at 11kV. The protection system at incoming supply shall be using numerical relays.
  • On LT side of transformer, LT main feeder pillar shall be provided. The Incoming shall be protected by MCCB/SFU.
  • The distribution pillar-box shall be connected into Ring Main Unit.
  • The incomer of distribution pillar shall have MCCB / SFU. The outgoing shall have HRC fuses.

(5) The LT cables from T.C to LT panel / Main feeder pillar:

  • Decide Size of LT Cable from T.C to LT Panel as per following Table.

Transformer Size

Cable

630kVA transformers 2 no x 1C x 630 Sq mm, Al, XLPE Cable
400kVA transformers 1 no x 1C x 630 Sq mm, Al, XLPE
250kVA transformers 3 ½ C x 400 Sq mm, Al, XLPE
160kVA transformers 3 ½ C x 300 Sq mm, Al, XLPE
100kVA transformers 3 ½ C x 150 Sq mm, Al, XLPE

(6) Considering various Factors & Length of Cable:

  • The factors for cable loading shall be taken as 70%.
  • The factor for multiplicity of cables from same cable trench shall be 80%.
  • The suggested maximum length of LT cable feeder shall be 250 Mtrs.
  • The LT cables shall be connected in ring main circuit.
  • The load on sub-feeder pillar shall be restricted to 150kW.

(7) LT cables from main feeder pillars to distribution pillar boxes:

Load on distribution pillar

 

LT Cable Size
Up to 50kW 3 ½ C x 150 sqmm, AL, XLPE
Up to 100kW 3 ½ C x 300 sqmm, AL, XLPE
Up to 150 kW 3 ½ C x 400sqmm, AL, XLPE

(8) Calculate Voltage Drop and T&D Losses:

  • The entire system has to be designed for a voltage drop of 2.0% from 11kV Side of transformer to metering equipment at end consumer premises.
  • The entire system has to be designed for T&D losses of service maximum 2.0% from 11kV to end consumer meter including of service cable.

Ref:

  1. NPC Limited.
  2. Electrical code.

Demand Factor-Diversity Factor-Utilization Factor-Load Factor

(1) Demand factor:

  • Demand Factor = Maximum demand of a system / Total connected load on the system
  • Demand factor is always less than one. 
  • Example: if a residence having 6000W equipment connected has a maximum demand of 3000W,Than demand factor = 3000W / 6000W = 55%.
  • The lower the demand factor, the less system capacity required to serve the connected load.
  • Feeder-circuit conductors should have an ampere sufficient to carry the load; the ampere of the feeder-circuit need not always be equal to the total of all loads on all branch-circuits connected to it.
    Remember that the demand factor permits a feeder-circuit ampere to be less than 100% of the sum of all branch-circuit loads connected to the feeder.
  • Example: One Machine Shop has
  1. Fluorescent fixtures=1 No, 5kw each, Receptacle outlets =1 No, 1500w each.
  2. Lathe=1No, 10 Hp, Air Compressor=1 No, 20 Hp, Fire Pump=1 No, 15 Hp.
  • After questioning the customer about the various loads, the information is further deciphered as follows:
  1. The shop lights are on only during the hours of 8 a.m. to 5 p.m.
  2. The receptacle outlets are in the office only, and will have computers and other small loads plugged into them.
  3. The lathe is fully loaded for 5 minutes periods.  The rest of the time is setup time.  This procedure repeats every 15 minutes.
  4. The air compressor supplies air to air tools and cycles off and on about half the time.
  5. The fire pump only runs for 30 minutes when tested which is once a month after hours.
  • Calculation:
  • Lighting Demand Factor = Demand Interval Factor x Diversity Factor.
  • = (15 minute run time/ 15 minutes) x 1.0 = 1.0
  • Lighting Demand Load = 5 kW x 1.0 = 5 kW
  • Receptacle Outlet Demand Factor = Demand Interval Factor x Diversity Factor
  • = (15 minute run time / 15 minutes) x 0.1 = 0.1
  • Receptacle Outlet Demand Load = 15 x 1500 watts x 0.1 = 2.25 kW
  • Lathe Demand Factor = Demand Interval Factor x Diversity Factor.
  • = (5 minute run time / 15 minutes) x 1.0 =0 .33
  • Lathe Demand Load = 10 hp x .746 x .33 = 2.46 kW
  • Air Compressor Demand Factor = Demand Interval Factor x Diversity Factor.
  • = (7.5 minute run time / 15 minutes) x 1.0 = 0.5
  • Air Compressor Demand Load = 20 hp x .746 x .5 = 7.46 kW
  • Fire Pump Demand Factor = Demand Interval Factor x Diversity Factor.
  • = (15 minute run time/ 15 minutes) x 0.0 = 0.0
  • Fire Pump Demand Load = 15 hp x .746 x 0.0 = 0.0 kW
  • Summary of Demand Loads :
Equipment kW D.F. Demand KW
Lighting 5 1 5
Receptacle Outlets 22.5 .1 2.25
Lathe 7.5 .33 2.46
Air Compressor 15 0.5 7.46
Fire Pump 11.25 0.0 0.0
TOTAL   61.25 Kw   17.17 Kw

(2) Diversity factor / simultaneity factor (Ks):

  • Diversity Factor = Sum of Individual Max. Demand. /  Max. Demand on Power Station.
  • Diversity Factor = Installed load. /  Running load.
  • Diversity factor is usually more than one. (Since the sum of individual max. demands >Max. Demand)
  • The load is time dependent as well as being dependent upon equipment characteristics. The diversity factor recognizes that the whole load does not equal the sum of its parts due to this time Interdependence (i.e. diverseness).
  • When the maximum demand of a supply is being assessed it is not sufficient to simply add together the ratings of all electrical equipment that could be connected to that supply. If this is done, a figure somewhat higher than the true maximum demand will be produced. This is because it is unlikely that all the electrical equipment on a supply will be used simultaneously.
  • The concept of being able to De-rate a potential maximum load to an actual maximum demand is known as the application of a diversity factor.
  • 70% diversity means that the device in question operates at its nominal or maximum load level 70% of the time that it is connected and turned on.
  • If total installed full load ampere is twice your running load ampere then the diversity factor is two.
  • If total installed full load ampere is four times your load a ampere then the diversity factor is four.
  • If everything (all electrical equipment) was running at full load at the same time the diversity factor is equal to One
  • Greater the diversity factor, lesser is the cost of generation of power.
  • Diversity factor in a distribution network is the ratio of the sum of the peak demands of the individual customers to the peak demand of the network.
  • This will be determined by the type of service, i.e., residential, commercial, industrial and combinations of such.
  • Example-I: A distribution feeder serves 5 houses, each of which has a peak demand of 5 KW. The feeder peak turns out to be 20 kw. The diversity is then 20/25 or 0.8. This results from the timing differences between the individual heating/cooling, appliance usages in the individual customers.
  • As supply availability decreases, the diversity factor will tend to increase toward 1.00. This can be demonstrated when restoring service after outages (called “cold starts”) as the system initial surge can be much greater than the historical peak loads.
  • Example-II: A sub-station has three outgoing feeders:
  1. feeder 1 has maximum demand 10 MW at 10:00 am,
  2. feeder 2 has maximum demand 12 MW at 7:00 pm and
  3. feeder 3 has maximum demand 15 MW at 9:00 pm,
  4. While the maximum demand of all three feeders is 33 MW at 8:00 pm.
  • Here, the sum of the maximum demand of the individual sub-systems (feeders) is 10 + 12 + 15 = 37 MW, while the system maximum demand is 33 MW. The diversity factor is 37/33 = 1.12. The diversity factor is usually greater than 1; its value also can be 1 which indicates the maximum demand of the individual sub-system occurs simultaneously.
  • Diversity is the relationship between the rated full loads of the equipment downstream of a connection point, and the rated load of the connection point. To illustrate:
  1. The building at these co-ordinates is fitted with a 100A main supply fuse.
  2. The distribution board has 2no. 6A breakers, 1no. 20A breaker and 5no. 32A breakers, a total, potentially, of 192A.
  • Not all these rated loads are turned on at once. If they were, then the 100A supply fuse would rupture, as it cannot pass 192A. So the diversity factor of the distribution board can be said to be 192A/100A, or 1.92, or 52%.
  • Many designers prefer to use unity as the diversity factor in calculations for planning conservatism because of plant load growth uncertainties. Local experience can justify using a diversity factor larger than unity, and smaller service entrance conductors and transformer requirements chosen accordingly.
  • The diversity factor for all other installations will be different, and would be based upon a local evaluation of the loads to be applied at different moments in time. Assuming it to be 1.0 may, on some occasions, result in a supply feeder and equipment rating that is rather larger than the local installation warrants, and an over-investment in cable and equipment to handle the rated load current. It is better to evaluate the pattern of usage of the loads and calculate an acceptable diversity factor for each particular case.
  • In the case of the example given above, achieving a diversity of 1.0 or 100% would require well over twice the cross-sectional area of copper cable to be installed in a deep trench underneath a field, the rebuild of a feeder cabinet to larger dimensions, more substantial overhead supply cables for a distance exceeding 2km northwards and a different tariff, where one pays rather more for a kWh than at present. The investment required to achieve 1.0 simply isn’t justifiable in this particular case.
  • Diversity factor is mostly used for distribution feeder size and transformer as well as to determine the maximum peak load and diversity factor is always based on knowing the process. You have to understand what will be on or off at a given time for different buildings and this will size the feeder. Note for typical buildings diversity factor is always one. You have to estimate or have a data records to create 24 hours load graph and you can determine the maximum demand load for node then you can easily determine the feeder and transformer size.
  • The diversity factor of a feeder would be the sum of the maximum demands of the individual consumers divided by the maximum demand of the feeder. In the same manner, it is possible to compute the diversity factor on a substation, a transmission line or a whole utility system.
  • The residential load has the highest diversity factor. Industrial loads have low diversity factors usually of 1.4, street light practically unity and other loads vary between these limits.

Diversity Factor in distribution Network:

Elements of System

Diversity Factors

Residential Commercial General
Power
Large
Industrial
Between individual users 2.00 1.46 1.45
Between transformers 1.30 1.30 1.35 1.05
Between feeders 1.15 1.15 1.15 1.05
Between substations 1.10 1.10 1.10 1.10
From users to transformers 2.00 1.46 1.44
From users to feeder 2.60 1.90 1.95 1.15
From users to substation 3.00 2.18 2.24 1.32
From users to generating station 3.29 2.40 2.46 1.45

Diversity Factor for distribution switchboards:

            Number of circuits Diversity Factor (ks)
Assemblies entirely tested 2 and 3 0.9
4 and 5 0.8
6 to 9 0.7
10 and more 0.6
Assemblies partially tested in every case choose 1

Diversity Factor for according to circuit function (IEC 60439):

                        Circuits Function Diversity Factor (ks)
Lighting 0.9
Heating and air conditioning 0.8
Socket-outlets 0.7
Lifts and catering hoist
For the most powerful motor 1
For the second most powerful motor 0.75
For all motors 0.8

Diversity Factor for an apartment block:

Apartment Diversity Factor (ks)
2 To 4 1
5To 19 0.78
10To 14 0.63
15To 19 0.53
20To 24 0.49
25To 29 0.46
30 To 34 0.44
35 To 39 0.42
40To 40 0.41
50 To Above 0.40
  • Example: 5 storey apartment building with 25 consumers, each having 6 kVA of installed load.
    The total installed load for the building is: 36 + 24 + 30 + 36 + 24 = 150 kVA
    The apparent-power supply required for the building is: 150 x 0.46 = 69 kVA
  • It is a matter of common experience that the simultaneous operation of all installed loads of a given installation never occurs in practice, i.e. there is always some degree of diversity and this fact is taken into account for estimating purposes by the use of a simultaneity factor / Diversity Factor (ks).
  • The Diversity factor ks is applied to each group of loads (e.g. being supplied from a distribution or sub-distribution board). The determination of these factors is the responsibility of the designer, since it requires a detailed knowledge of the installation and the conditions in which the individual circuits are to be exploited. For this reason, it is not possible to give precise values for general application.

 Designing Size of Electrical Switchgear by use of Demand Factor and Diversity Factor:

  • Diversity factors are used by utilities for distribution transformer sizing and load predictions.
  • Demand factors are more conservative and are used by NEC for service and feeder sizing.
  • Demand factors and diversity factors are used in design.
  • For example, the sum of the connected loads supplied by a feeder is multiplied by the demand factor to determine the load for which the feeder must be sized. This load is termed the maximum demand of the feeder. The sum of the maximum demand loads for a number of sub feeders divided by the diversity factor for the sub feeders will give the maximum demand load to be supplied by the feeder from which the sub feeders are derived.
  • Example-1: Suppose We have four individual feeder-circuits with connected loads of 250 kVA, 200 kVA, 150 kVA and 400 kVA and demand factors of 90%, 80%, 75% and 85% respectively.Use a diversity factor of 1.5.
  • Calculating demand for feeder-circuits
    • 250 kVA x 90% = 225 kVA
    • 200 kVA x 80% = 160 kVA
    • 150 kVA x 75% = 112.5 kVA
    • 400 kVA x 85% = 340 kVA
    • 837.5 kVA
    • The sum of the individual demands is equal to 837.5 kVA.
    • If the main feeder-circuit were sized at unity diversity: kVA = 837.5 kVA ÷ 1.00 = 837.5 kVA.
    • The main feeder-circuit would have to be supplied by an 850 kVA transformer.
    • However, using the diversity factor of 1.5, the kVA = 837.5 kVA ÷ 1.5 = 558 kVA for the main feeder.
    • For diversity factor of 1.5, a 600 kVA transformer could be used.
    • Example-2: A conveyor belt made up of six sections, each driven by a 2 kW motor. As material is transported along this belt, it is first carried by section 1, and then each section in succession until the final section is reached. In this simple example only one section of conveyor is carrying material at any point in time. Therefore five motors are only handling no-load mechanical losses (say .1 kW) keeping the belts moving whilst one motor is handling the load (say 1 kW). The demand presented by each motor when it is carrying its load is 1 kW, the sum of the demand loads is 6 kW but the maximum load presented by the system at any time is only 1.5 kW.
    • Diversity factor =Sum of Individual Max. Demand / Max. Demand = 6 Kw / 1.5 Kw =4.
    • Demand Factor = Maximum demand / Total connected load = 1.5 Kw / 12 Kw = 0.125.

(3) Load factor:

  • Load Factor = Average load. /Maximum load during a given period.
  • It can be calculated for a single day, for a month or for a year.
  • Its value is always less than one. Because maximum demand is always more than avg. demand.
  • It is used for determining the overall cost per unit generated. Higher the load factor, lesser will be the cost per unit.
  • Load Factor = Load that a piece of equipment actually draws / Load it could draw (full load).
  • Example: Motor of 20 hp drives a constant 15 hp load whenever it is on.
  •  The motor load factor is then 15/20 = 75%.
  • Load factor is term that does not appear on your utility bill, but does affect electricity costs. Load factor indicates how efficiently the customer is using peak demand.
  • Load Factor = ( energy (kWh per month) ) / (  peak demand (kW) x hours/month )
  • A high load factor means power usage is relatively constant. Low load factor shows that occasionally a high demand is set. To service that peak, capacity is sitting idle for long periods, thereby imposing higher costs on the system. Electrical rates are designed so that customers with high load factor are charged less overall per kWh.
  • For Example
  • Customer A – High Load Factor
  • 82% load factor = (3000 kWh per month x 100%) / 5 kW x 730 hours/month.
  • Customer B – Low Load Factor
  • 41% load factor = (3000 kWh per month x 100%) / 10kW x 730 hours/month.
  • To encourage the efficient use of installed capacity, electricity rates are structured so the price per kWh above a certain load factor is lower. The actual structure of the price blocks varies by rate.

(4) Utilization factor (Ku):

  • In normal operating conditions the power consumption of a load is sometimes less than that indicated as its nominal power rating, a fairly common occurrence that justifies the application of an utilization factor (ku) in the estimation of realistic values.
  • Utilization Factor = The time that a equipment is in use./ The total time that it could be in use.
  • Example: The motor may only be used for eight hours a day, 50 weeks a year. The hours of operation would then be 2000 hours, and the motor  Utilization factor for a base of 8760 hours per year would be 2000/8760 = 22.83%. With a base of 2000 hours per year, the motor Utilization factor would be 100%. The bottom line is that the use factor is applied to get the correct number of hours that the motor is in use.
  • This factor must be applied to each individual load, with particular attention to electric motors, which are very rarely operated at full load. In an industrial installation this factor may be estimated on an average at 0.75 for motors.
  • For incandescent-lighting loads, the factor always equals 1.
  • For socket-outlet circuits, the factors depend entirely on the type of appliances being supplied from the sockets concerned.

Maximum demand:

  • Maximum demand (often referred to as MD) is the largest current normally carried by circuits, switches and protective devices. It does not include the levels of current flowing under overload or short circuit conditions.
  •  Assessment of maximum demand is sometimes straightforward. For example, the maximum demand of a 240 V single-phase 8 kW shower heater can be calculated by dividing the power (8 kW) by the voltage (240 V) to give a current of 33.3 A. This calculation assumes a power factor of unity, which is a reasonable assumption for such a purely resistive load.
  • There are times, however, when assessment of maximum demand is less obvious. For example, if a ring circuit feeds fifteen 13 A sockets, the maximum demand clearly should not be 15 x 13 = 195 A, if only because the circuit protection will not be rated at more than 32 A. Some 13 A sockets may feed table lamps with 60 W lamps fitted, whilst others may feed 3 kW washing machines; others again may not be loaded at all.
  • Lighting circuits pose a special problem when determining MD. Each lamp-holder must be assumed to carry the current required by the connected load, subject to a minimum loading of 100 W per lamp holder (a demand of 0.42 A per lamp holder at 240 V). Discharge lamps are particularly difficult to assess, and current cannot be calculated simply by dividing lamp power by supply voltage. The reasons for this are:
  1. Control gear losses result in additional current,
  2. the power factor is usually less than unity so current is greater, and
  3. Chokes and other control gear usually distort the waveform of the current so that it contains harmonics which are additional to the fundamental supply current.
  • So long as the power factor of a discharge lighting circuit is not less than 0.85, the current demand for the circuit can be calculated from:
  • current (A) = (lamp power (W) x 1.8) / supply voltage (V)
  • For example, the steady state current demand of a 240 V circuit supplying ten 65 W fluorescent lamps would be: I = 10X65X1.8A / 240 = 4.88A
  • Switches for circuits feeding discharge lamps must be rated at twice the current they are required to carry, unless they have been specially constructed to withstand the severe arcing resulting from the switching of such inductive and capacitive loads.

(5) Coincidence factor:

  • The coincidence factor =Max. demand of a system / sum of the individual maximum demands
  • The coincidence factor is the reciprocal of the diversity factor

Demand Factor & Load Factor according to Type of Industries:

Type of Industry Demand Factor Load Factor Utilization Factor (DF x LF)
Arc Furnace 0.55 0.80 0.44
Induction Furnace 0.90 0.80 0.72
Steel Rolling mills 0.80 0.25 0.20
Mechanical/ Electrical
a) Single Shift 0.45 0.25 0.11
b) Double Shift 0.45 0.50 0.22
Cycle Industry 0.40 0.40 0.16
Wire products 0.35 0.40 0.14
Auto Parts 0.40 0.50 0.20
Forgings 0.50 0.35 0.17
Cold Storage
a) Working Season 0.60 0.65 0.39
b) Non-Working Season 0.25 0.15 0.04
Rice Sheller’s
a) Working Season 0.70 0.80 0.56
b)  Non-Working Season 0.05 0.30 0.01
Ice Candy Units
a) Working Season 0.50 0.65 0.32
b) Non-Working Season 0.50 0.10 0.05
Ice Factories
a) Working Season 0.80 0.65 0.52
b) Non-Working Season 0.80 0.10 0.08
Cotton Ginning
a) Working Season 0.70 0.25 0.17
b) Non-Working Season 0.10 0.10 0.01
Spinning Mills 0.60 0.80 0.48
Textile Industry 0.50 0.80 0.40
Dyeing and Printing 0.40 0.50 0.20
Ghee Mills 0.50 0.50 0.25
Oil Mills 0.70 0.50 0.35
Solvent Extraction Mills 0.45 0.50 0.22
Plastic 0.60 0.25 0.11
Soap 0.50 0.25 0.12
Rubber (Foot Wear) 0.45 0.35 0.16
Distilleries 0.35 0.50 0.17
Chemical Industry 0.40 0.50 0.20
Gas Plant Industry 0.70 0.50 0.35
Pain and Colour Factory 0.50 0.40 0.20
Sugar 0.30 0.45 0.13
Paper 0.50 0.80 0.40
Flour Mills(Single Shift) 0.80 0.25 0.20
Atta Chakies 0.50 0.25 0.12
Milk Plants 0.40 0.80 0.32
Printing Presses 0.35 0.30 0.10
Repair Workshops 0.40 0.25 0.10
Bottling Plants 0.40 0.35 0.14
Radio Stations 0.55 .0.45 0.25
Telephone exchange 0.50 0.90 0.45
Public Water Works 0.75 0.40 0.30
Medical Colleges 0.60 0.25 0.15
Hospitals 0.25 0.90 0.22
Nursing Homes 0.50 0.50 0.25
Colleges and Schools 0.50 0.20 0.10
Hotels and Restaurants 0.75 0.40 0.30
Marriage Palaces 1.00 0.25 0.25

Demand Factor & Load Factor according to Type of Buildings:

Individual Facilities Demand Factor Load Factor
Communications – buildings 60-65 70-75
Telephone exchange building 55-70 20-25
Air passenger terminal building 65-80 28-32
Aircraft fire and rescue station 25-35 13-17
Aircraft line operations building 65-80 24-28
Academic instruction building 40-60 22-26
Applied instruction building 35-65 24-28
Chemistry and Toxicology Laboratory 70-80 22-28
Materials Laboratory 30-35 27-32
Physics Laboratory 70-80 22-28
Electrical and electronics systems laboratory 20-30 3-7
Cold storage warehouse 70-75 20-25
General warehouse 75-80 23-28
Controlled humidity warehouse 60-65 33-38
Hazardous/flammable storehouse 75-80 20-25
Disposal, salvage, scrap building 35-40 25-20
Hospital 38-42 45-50
Laboratory 32-37 20-25
Dental Clinic 35-40 18-23
Medical Clinic 45-50 20-23
Administrative Office 50-65 20-35
Single-family residential housing 60-70 10-15
Detached garages 40-50 2-4
Apartments 35-40 38-42
Fire station 25-35 13-17
Police station 48-53 20-25
Bakery 30-35 45-60
Laundry/dry cleaning plant 30-35 20-25
K-6 schools 75-80 10-15
7-12 schools 65-70 12-17
Churches 65-70 5-25
Post Office 75-80 20-25
Retail store 65-70 25-32
Bank 75-80 20-25
Supermarket 55-60 25-30
Restaurant 45-75 15-25
Auto repair shop 40-60 15-20
Hobby shop, art/crafts 30-40 25-30
Bowling alley 70-75 10-15
Gymnasium 70-75 20-45
Skating rink 70-75 10-15
Indoor swimming pool 55-60 25-50
Theater 45-55 8-13
Library 75-80 30-35
Golf clubhouse 75-80 15-20
Museum 75-80 30-35

Calculate Transverse Load on PCC/RCC/Tublar/RSJ Pole.

Calculate Transverse Load on PCC/RCC/RSJ/Tublar/RSJ Pole.

  • Calculate Crippling load on Pole
  • Calculate Wind Load on Pole.
  • Calculate wind Load on all Conductors.
  • Calculate Transverse Load on Pole
  • Decide spacing between Two Conductor as per IS:5613.
  • Calculate required Voltage level of Overhead Line according to Distance.

FREE DOWNLOAD. 

Working Principle of ELCB and RCB

Working Principle of ELCB and RCB:

An Earth Leakage Circuit Breaker (ELCB) is a device used to directly detect currents leaking to earth from an installation and cut the power and mainly used in TT earthing systems.

There are two types of ELCBs,

  1. Voltage Earth Leakage Circuit Breaker (voltage-ELCB)
  2. Current Earth Leakage Current Earth Leakage Circuit Breaker (Current-ELCB).

Voltage-ELCBs were first introduced about sixty years ago and Current-ELCB was first introduced about forty years ago. For many years, the voltage operated ELCB and the differential current operated ELCB were both referred to as ELCBs because it was a simpler name to remember. But the use of a common name for two different devices gave rise to considerable confusion in the electrical industry. If the wrong type was used on an installation, the level of protection given could be substantially less than that intended. To ignore this confusion, IEC decided to apply the term Residual Current Device (RCD) to differential current operated ELCBs. Residual current refers to any current over and above the load current

 Voltage Base ELCB.

  •  Voltage-ELCB is a voltage operated circuit breaker. The device will function when the Current passes through the ELCB. Voltage-ELCB contains relay Coil which it being connected to the metallic load body at one end and it is connected to ground wire at the other end.
  • If the voltage of the Equipment body is rise (by touching Phase to metal Part or Failure of Insulation of Equipment) which could cause the difference between earth and load body voltage, the danger of electric shock will occur. This voltage difference will produce an electric current from the load metallic body passes the relay loop and to earth. When voltage on the equipment metallic body rose to the danger level which exceed to 50Volt, the flowing current through relay loop could move the relay contact by disconnecting the supply current to avoid from any danger electric shock.
  • The ELCB detects fault currents from live to the earth (ground) wire within the installation it protects. If sufficient voltage appears across the ELCB’s sense coil, it will switch off the power, and remain off until manually reset. A voltage-sensing ELCB does not sense fault currents from live to any other earthed body.

                       

  • These ELCBs monitored the voltage on the earth wire, and disconnected the supply if the earth wire voltage was over 50 volts.
  • These devices are no longer used due to its drawbacks like if the fault is between live and a circuit earth, they will disconnect the supply. However, if the fault is between live and some other earth (such as a person or a metal water pipe), they will NOT disconnect, as the voltage on the circuit earth will not change. Even if the fault is between live and a circuit earth, parallel earth paths created via gas or water pipes can result in the ELCB being bypassed. Most of the fault current will flow via the gas or water pipes, since a single earth stake will inevitably have a much higher impedance than hundreds of meters of metal service pipes buried in the ground.

  • The way to identify an ELCB is by looking for green or green and yellow earth wires entering the device. They rely on voltage returning to the trip via the earth wire during a fault and afford only limited protection to the installation and no personal protection at all. You should use plug in 30mA RCD’s for any appliances and extension leads that may be used outside as a minimum.

 Advantages

  • ELCBs have one advantage over RCDs: they are less sensitive to fault conditions, and therefore have fewer nuisance trips.
  • While voltage and current on the earth line is usually fault current from a live wire, this is not always the case, thus there are situations in which an ELCB can nuisance trip.
  • When an installation has two connections to earth, a nearby high current lightning strike will cause a voltage gradient in the soil, presenting the ELCB sense coil with enough voltage to cause it to trip.
  • If the installation’s earth rod is placed close to the earth rod of a neighboring building, a high earth leakage current in the other building can raise the local ground potential and cause a voltage difference across the two earths, again tripping the ELCB.
  • If there is an accumulated or burden of currents caused by items with lowered insulation resistance due to older equipment, or with heating elements, or rain conditions can cause the insulation resistance to lower due to moisture tracking. If there is a some mA which is equal to ELCB rating than ELCB may give nuisance Tripping.
  • If either of the earth wires become disconnected from the ELCB, it will no longer trip or the installation will often no longer be properly earthed.
  • Some ELCBs do not respond to rectified fault current. This issue is common for ELCBs and RCDs, but ELCBs are on average much older than RCB so an old ELCB is more likely to have some uncommon fault current waveform that it will not respond to.
  • Voltage-operated ELCB are the requirement for a second connection, and the possibility that any additional connection to earth on the protected system can disable the detector.
  • Nuisance tripping especially during thunderstorms.

Disadvantages:

  • They do not detect faults that don’t pass current through the CPC to the earth rod.
  • They do not allow a single building system to be easily split into multiple sections with independent fault protection, because earthing systems are usually use common earth Rod.
  • They may be tripped by external voltages from something connected to the earthing system such as metal pipes, a TN-S earth or a TN-C-S combined neutral and earth.
  • As electrically leaky appliances such as some water heaters, washing machines and cookers may cause the ELCB to trip.
  • ELCBs introduce additional resistance and an additional point of failure into the earthing system.

 Can we assume whether Our Electrical System is protected against Earth Protection or not by only Pressing ELCB Test Switch?

  • Checking the health of the ELCB is simple and you can do it easily by pressing TEST Push Button Switch of ELCB. The test push-button will test whether the ELCB unit is working properly or not. Can we assume that If ELCB is Trip after Pressing TEST Switch of ELCB than your system is protected against earth protection? Then you are wrong.
  • The test facility provided on the home ELCB will only confirm the health of the ELCB unit, but that test does not confirm that the ELCB will trip when an electric shock hazard does occur. It is a really sad fact that all the while this misunderstanding has left many homes totally unprotected from the risk of electric shocks.
  • This brings us or alarming us to think over second basic requirement for earth protection. The second requirement for the proper operation of a home shock protection system is electrical grounding.
  • We can assume that the ELCB is the brain for the shock protection, and the grounding as the backbone. Therefore, without a functional grounding (Proper Earthing of Electrical System) there is totally no protection against electrical shocks in your house even if  You have installed ELCB and its TEST switch show proper result. Looking after the ELCB alone is not enough. The electrical Earthing system must also be in good working order for the shock protection system to work. In addition to routine inspections that should be done by the qualified electrician, this grounding should preferably be inspected regularly at shorter intervals by the homeowner and need to pour Water in Earthing Pit at Regular interval of Time to minimize Earth Resistance.

 Current-operated ELCB (RCB):

  • Current-operated ELCBs are generally known as Residual-current devices (RCD). These also protect against earth leakage. Both circuit conductors (supply and return) are run through a sensing coil; any imbalance of the currents means the magnetic field does not perfectly cancel. The device detects the imbalance and trips the contact.
  • When the term ELCB is used it usually means a voltage-operated device. Similar devices that are current operated are called residual-current devices. However, some companies use the term ELCB to distinguish high sensitivity current operated 3 phase devices that trip in the milliamp range from traditional 3 phase ground fault devices that operate at much higher currents.

  • Typical RCB circuit:

  • The supply coil, the neutral coil and the search coil all wound on a common transformer core.
  • On a healthy circuit the same current passes through the phase coil, the load and return back through the neutral coil. Both the phase and the neutral coils are wound in such a way that they will produce an opposing magnetic flux. With the same current passing through both coils, their magnetic effect will cancel out under a healthy circuit condition.
  • In a situation when there is fault or a leakage to earth in the load circuit, or anywhere between the load circuit and the output connection of the RCB circuit, the current returning through the neutral coil has been reduced. Then the magnetic flux inside the transformer core is not balanced anymore. The total sum of the opposing magnetic flux is no longer zero. This net remaining flux is what we call a residual flux.
  • The periodically changing residual flux inside the transformer core crosses path with the winding of the search coil. This action produces an electromotive force (e.m.f.) across the search coil. An electromotive force is actually an alternating voltage. The induced voltage across the search coil produces a current inside the wiring of the trip circuit. It is this current that operates the trip coil of the circuit breaker. Since the trip current is driven by the residual magnetic flux (the resulting flux, the net effect between both fluxes) between the phase and the neutral coils, it is called the residual current devise.
  • With a circuit breaker incorporated as part of the circuit, the assembled system is called residual current circuit breaker (RCCB) or residual current devise (RCD). The incoming current has to pass through the circuit breaker first before going to the phase coil. The return neutral path passes through the second circuit breaker pole. During tripping when a fault is detected, both the phase and neutral connection is isolated.
    • RCD sensitivity is expressed as the rated residual operating current, noted IΔn. Preferred values have been defined by the IEC, thus making it possible to divide RCDs into three groups according to their IΔn value.
    • High sensitivity (HS): 6- 10- 30 mA (for direct-contact / life injury protection)
    • Standard IEC 60755 (General requirements for residual current operated protective devices) defines three types of RCD depending on the characteristics of the fault current.
    • Type AC: RCD for which tripping is ensured for residual sinusoidal alternating currents

Sensitivity of RCB:

  • Medium sensitivity (MS): 100- 300- 500- 1000 mA (for fire protection)
  • Low sensitivity (LS): 3- 10- 30 A (typically for protection of machine)

Type of RCB:

Type A: RCD for which tripping is ensured

  • for residual sinusoidal alternating currents
  • for residual pulsating direct currents
  • For residual pulsating direct currents superimposed by a smooth direct current of 0.006 A, with or without phase-angle control, independent of the polarity.

Type B: RCD for which tripping is ensured

  • as for type A
  • for residual sinusoidal currents up to 1000 Hz
  • for residual sinusoidal currents superposed by a pure direct current
  • for pulsating direct currents superposed by a pure direct current
  • for residual currents which may result from rectifying circuits
    • three pulse star connection or six pulse bridge connection
    • two pulse bridge connection line-to-line with or without phase-angle monitoring, independently of the polarity
    • There are two groups of devices:

Break time of RCB:

  1. G (general use) for instantaneous RCDs (i.e. without a time delay)
  • Minimum break time: immediate
  • Maximum break time: 200 ms for 1x IΔn, 150 ms for 2x IΔn, and 40 ms for 5x IΔn
  1. S (selective) or T (time delayed) for RCDs with a short time delay (typically used in circuits containing surge suppressors)
  • Minimum break time: 130 ms for 1x IΔn, 60 ms for 2x IΔn, and 50 ms for 5x IΔn
  • Maximum break time: 500 ms for 1x IΔn, 200 ms for 2x IΔn, and 150 ms for 5x IΔn
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