Electrical Notes and Articles

Total Losses in Power Distribution and Transmission Lines-Part 2

JULY 2, 2013

2 COMMENTS

3 Votes

(2) Non-Technical (Commercial Losses):

Non-technical losses are at 16.6%, and related to meter reading, defective meter and error in meter reading, billing of

customer energy consumption, lack of administration, financial constraints, and estimating unmetered supply of energy

as well as energy thefts.

Main Reasons for Non-Technical Losses:

(1) Power Theft :

Theft of power is energy delivered to customers that is not measured by the energy meter for the customer. Customer

tempers the meter by mechanical jerks, placement of powerful magnets or disturbing the disc rotation with foreign

matters, stopping the meters by remote control.

(2) Metering Inaccuracies:

Losses due to metering inaccuracies are defined as the difference between the amount of energy actually delivered

through the meters and the amount registered by the meters.

All energy meters have some level of error which requires that standards be established. Measurement Canada,

formerly Industry Canada, is responsible for regulating energy meter accuracy.

Statutory requirements5 are for meters to be within an accuracy range of +2.5% and – 3.5%. Old technology meters

normally started life with negligible errors, but as their mechanisms aged they slowed down resulting

in under-recording. Modern electronic meters do not under-record with age in this way.

Consequently, with the introduction of electronic meters, there should have been a progressive reduction in meter errors.

Increasing the rate of replacement of mechanical meters should accelerate this process

(3) Un metered Losses for very small Load:

Unmetered losses are situations where the energy usage is estimated instead of measured with an energy meter. This

happens when the loads are very small and energy meter installation is economically impractical. Examples of this are

street lights and cable television amplifiers.

(4) Un metered supply:

Unmetered supply to agricultural pumps is one of the major reasons for commercial losses. In most states, the

agricultural tariff is based on the unit horsepower (H.P.) of the motors. Such power loads get sanctioned at the low load

declarations.

Once the connections are released, the consumers increasing their connected loads, without obtaining necessary

sanction, for increased loading, from the utility.

Further estimation of the energy consumed in unmetered supply has a great bearing on the estimation of T&D losses on

account of inherent errors in estimation.

Most of the utilities deliberately overestimate the unmetered agricultural consumption to get higher subsidy from the

State Govt. and also project. reduction in losses. In other words higher the estimates of the unmetered consumption,

lesser the T&D loss figure and vice versa.

pattern, ground water level, seasonal variation, hours of operation etc.

(5) Error in Meter Reading:

Proper Calibrated Meter should be used to measure Electrical Energy. Defective Energy Meter should be replaced

immediately.

The reason for defective meter are Burning of meters, Burn out Terminal Box of Meter due to heavy load, improper

C.T.ratio and reducing the recording, Improper testing and calibration of meters.

(6) Billing Problems:

Faulty and untimely serving Bill should be main part of non-Technical Losses.

Normal Complain regarding Billing are Not Receipt of Bill, Late Receipt of Bill, Receiving wrong Bill , Wrong Meter

Reading, Wrong Tariff, wrong Calculations.

How to reduce Technical Losses:

(1) Converting LV Line to HV Line:

Many Distribution pockets of Low Voltage (430V) in Town are surrounded by higher voltage feeders. At this lower voltage,

more conductor current flows for the same power delivered, resulting in higher I2R losses.

Converting old LV (430V) feeders to higher voltage the Investment Cost is high and often not economically justifiable but

If parts of the LV (430V) Primary feeders are in relatively good condition, installing multiple step-down power

transformers at the periphery of the 430 volt area will reduce copper losses by injecting load current at more points (i.e.,

reducing overall conductor current and the distance traveled by the current to serve the load).

(2) Large Commercial / Industrial Consumer get direct Line from Feeder:

Design the distribution network system in such a way that if it is Possible than large consumer gets direct Power Line

from feeder.

(3) Adopting High Voltage Distribution Service (HVDS) for Agricultural

Customer:

In High Voltage direct service (HVDS) ,11KV line direct given to cluster of 2 to 3 Agricultural Customer for Agricultural

Pump set and employed small distribution Transformer (15KVA) for given these 2 to 3 customer through smallest (

almost negligible) LT distribution Lines.

In HVDS there is less distribution losses due to minimum length of Distribution Line, High quality of Power Supply with

no Voltage drop, Less Burn out of motor due to less voltage fluctuation and Good quality of Power, to avoid overloading of

Transformer.

(4) Adopting Arial Bundle Conductor (ABC):

Where LT Line are not totally avoidable use Arial Bundle Conductor to minimize faults in Lines, to avoid direct theft from

Line (Tampering of Line).

(5) Reduce Number of Transformer:

Reduce the number of transformation steps.

Transformers are responsible for almost half of network losses.

High efficiency distribution transformers can make a large impact on reduction of Distribution Losses

(6) Utilize Feeder on its Average Capacity:

By overloading of Distribution Feeder Distribution Losses will be increase.

The higher the load on a power line, the higher its variable losses. It has been suggested that the optimal average

utilization rate of distribution network cables should be as low as 30% if the cost of losses is taken into account.

(7) Replacements of Old Conductor/Cables:

By using the higher the cross-section area of Conductor / cables the losses will be lower but the same time cost will be

high so by forecasting the future Load an optimum balance between investment cost and network losses should be

maintained.

(8) Feeder Renovation / Improvement Program:

Re conductoring of Transmission and Distribution Line according to Load.

Identification of the weakest areas in the distribution system and strengthening /improving them.

Reducing the length of LT lines by relocation of distribution sub stations or installations of additional new distribution

transformers.

Installation of lower capacity distribution transformers at each consumer premises instead of cluster formation and

substitution of distribution transformers with those having lower no load losses such as amorphous core transformers.

Installation of shunt capacitors for improvement of power factor.

Installation of single-phase transformers to feed domestic and nondomestic load in rural areas.

Providing of small 25kVA distribution transformers with a distribution box attached to its body, having provision for

installation of meters, MCCB and capacitor.

Lying of direct insulated service line to each agriculture consumer from distribution transformers

Due to Feeder Renovation Program T&D loss may be reduced from 60-70 % to 15-20 %.

(9) Industrial / Urban Focus Program:

Separations of Rural Feeders from Industrial Feeders.

Instantly release of New Industrial or HT connections.

Identify and Replacement of slow and sluggish meters by Electronics type meters.

In Industrial and agricultural Consumer adopt One Consumer, one Transformer scheme with meter should be

Introduced.

Change of old Service Line by armored cable.

Due to Feeder Renovation Program T&D loss may be reduced from 60-70 % to 15-20 %.

(10) Strictly Follow Preventive Maintenance Program:

Required to adopt Preventive Maintenance Program of Line to reduce Losses due to Faulty / Leakage Line Parts.

Required to tights of Joints, Wire to reduce leakage current.

How to reduce Non-Technical Losses:

(1) Making mapping / Data of Distribution Line:

Mapping of complete primary and secondary distribution system with all parameters such as conductor size, line lengths

etc.

Compilation of data regarding existing loads, operating conditions, forecast of expected loads etc.

Preparation of long-term plans for phased strengthening and improvement of the distribution systems along with

transmission system.

It should be obligatory for all big industries and utilities to carry out Energy Audits of their system.

Further time bound action for initiating studies for realistic assessment of the total T&D Losses into technical and

non-technical losses has also to be drawn by utilities for identifying high loss areas to initiate remedial measures to reduce

the same.

The realistic assessment of T&D Loss of a utility greatly depends on the chosen sample size which in turn has a bearing

on the level of confidence desired and the tolerance limit of variation in results.

In view of this it is very essential to fix a limit of the sample size for realistic quick estimates of losses.

(3) Mitigating power theft by Power theft checking Drives:

Theft of electric power is a major problem faced by all electric utilities. It is necessary to make strict rule by State

Government regarding Power theft. Indian Electricity Act has been amended to make theft of energy and its abatement as

a cognizable offense with deterrent punishment of up to 3 years imprisonment.

The impact of theft is not limited to loss of revenue, it also affects power quality resulting in low voltage and voltage dips.

Required to install proper seal management at Meter terminal Box, at CT/PT terminal to prevent power theft. Identify

Power theft area and required to expedite power theft checking drives.

Installation of medium voltage distribution (MVD) networks in theft-prone areas, with direct connection of each consumer

to the low voltage terminal of the supply transformer.

All existing un metered services should be immediately stopped.

(4) Replacement of Faulty/Sluggish Energy Meter:

It is necessary to replacement of Faulty or sluggish Meter by Distribution Agency to reduce un metered Electrical energy.

Required to test Meter periodically for testing of accuracy of meter. Replacement of old erroneous electromechanical

meters with accurate Electro static Meter (Micro presser base) for accurate measurement of energy consumption.

Use of Meter boxes and seals them properly to ensure that the meters are properly sealed and cannot be tampered.

(5) Bill Collection facility:

Increase Bill’s Payment Cells, Increasing drop Box facility in all Area for Payment Collection.

E-Payment facility gives more relief to Customer for bill Payment and Supply agency will get Payment regularly and

speedily from Customer.

Effectively disconnect the connection of defaulter Customer who does not pay the Bill rather than give them chance to pay

the bill.

(6) Reduce Debit areas of Sub Division:

Recovery of old debts in selected cases through legal, communication and judicial actions.

Ensuring police action when required to disconnect connection of defaulter Consumer.

(7) Watchdog effect on users.

Users must aware that the distribution Agency can monitor consumption at its convenience. This allows the company

fast detection of any abnormal consumption due to tampering or by-passing of a meter and enables the company to take

corrective action.

The result is consumer discipline. This has been shown to be extremely effective with all categories of large and

medium consumers having a history of stealing electricity. They stop stealing once they become aware that the utility has

the means to detect and record it.

These measures can significantly increase the revenues of utilities with high non-technical losses.

(8) Loss Reduction Programmed:

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The increased hours of supply to Agriculture and Rural domestic consumers have resulted in higher loss levels.

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Total Losses in Power Distribution & Transmission Lines-Part 1

JULY 1, 2013

5 COMMENTS

1 Vote

Introduction:

Power generated in power stations pass through large & complex networks like transformers, overhead lines, cables &

other equipments and reaches at the end users. It is fact that the Unit of electric energy generated by Power Station does

not match with the units distributed to the consumers. Some percentage of the units is lost in the Distribution network.

This difference in the generated & distributed units is known as Transmission and Distribution loss.

Transmission and Distribution loss are the amounts that are not paid for by users.

T&D Losses= (Energy Input to feeder(Kwh)-Billed Energy to Consumer(Kwh)) / Energy Input kwh x100

Distribution Sector considered as the weakest link in the entire power sector. Transmission Losses is approximate 17%

while Distribution Losses is approximate 50%.

There are two types of Transmission and Distribution Losses

1.

Technical Losses

2.

Non Technical Losses (Commercial Losses)

(1) Technical Losses:

The technical losses are due to energy dissipated in the conductors, equipment used for transmission Line,

Transformer, sub- transmission Line and distribution Line and magnetic losses in transformers.

Technical losses are normally 22.5%, and directly depend on the network characteristics and the mode of operation.

The major amount of losses in a power system is in primary and secondary distribution lines. While transmission and

sub-transmission lines account for only about 30% of the total losses. Therefore the primary and secondary distribution

systems must be properly planned to ensure within limits.

The unexpected load increase was reflected in the increase of technical losses above the normal level

Losses are inherent to the distribution of electricity and cannot be eliminated.

There are two Type of Technical Losses.

(a) Permanent / Fixed Technical losses:

Fixed losses do not vary according to current. These losses take the form of heat and noise and occur as long as a

transformer is energized.

Between 1/4 and 1/3 of technical losses on distribution networks are fixed losses. Fixed losses on a network can be

influenced in the ways set out below.

Corona Losses.

Leakage Current Losses.

Dielectric Losses.

Open-circuit Losses.

Losses caused by continuous load of measuring elements

Losses caused by continuous load of control elements.

(b) Variable Technical losses

Variable losses vary with the amount of electricity distributed and are, more precisely, proportional to the square of the

current. Consequently, a 1% increase in current leads to an increase in losses of more than 1%.

Between 2/3 and 3/4 of technical (or physical) losses on distribution networks are variable Losses.

By increasing the cross sectional area of lines and cables for a given load, losses will fall. This leads to a direct trade-off

between cost of losses and cost of capital expenditure. It has been suggested that optimal average utilization rate on a

distribution network that considers the cost of losses in its design could be as low as 30 per cent.

joule losses in lines in each voltage level

impedance losses

Losses caused by contact resistance.

Main Reasons for Technical Losses:

(1) Lengthy Distribution lines:

In practically 11 KV and 415 volts lines, in rural areas are extended over long distances to feed loads scattered over large

areas. Thus the primary and secondary distributions lines in rural areas are largely radial laid usually extend over long

distances. This results in high line resistance and therefore high I2R losses in the line.

Haphazard growths of sub-transmission and distribution system in to new areas.

Large scale rural electrification through long 11kV and LT lines.

(2) Inadequate Size of Conductors of Distribution lines:

The size of the conductors should be selected on the basis of KVA x KM capacity of standard conductor for a required

voltage regulation but rural loads are usually scattered and generally fed by radial feeders. The conductor size of these

feeders should be adequate.

(3) Installation of Distribution transformers away from load centers:

Distribution Transformers are not located at Load center on the Secondary Distribution System.

In most of case Distribution Transformers are not located centrally with respect to consumers. Consequently, the farthest

consumers obtain an extremity low voltage even though a good voltage levels maintained at the transformers secondary.

This again leads to higher line losses. (The reason for the line losses increasing as a result of decreased voltage at the

consumers end Therefore in order to reduce the voltage drop in the line to the farthest consumers, the distribution

transformer should be located at the load center to keep voltage drop within permissible limits.

(4) Low Power Factor of Primary and secondary distribution system:

In most LT distribution circuits normally the Power Factor ranges from 0.65 to 0.75. A low Power Factor contributes

towards high distribution losses.

For a given load, if the Power Factor is low, the current drawn in high And the losses proportional to square of the

current will be more. Thus, line losses owing to the poor PF can be reduced by improving the Power Factor. This can be

done by application of shunt capacitors.

Shunt capacitors can be connected either in secondary side (11 KV side) of the 33/11 KV power transformers or at

various point of Distribution Line.

The optimum rating of capacitor banks for a distribution system is 2/3rd of the average KVAR requirement of that

distribution system.

The vantage point is at 2/3rd the length of the main distributor from the transformer.

A more appropriate manner of improving this PF of the distribution system and thereby reduce the line losses is to

connect capacitors across the terminals of the consumers having inductive loads.

By connecting the capacitors across individual loads, the line loss is reduced from 4 to 9% depending upon the extent of

PF improvement.

(5) Bad Workmanship:

Bad Workmanship contributes significantly role towards increasing distribution losses.

Joints are a source of power loss. Therefore the number of joints should be kept to a minimum. Proper jointing

techniques should be used to ensure firm connections.

Connections to the transformer bushing-stem, drop out fuse, isolator, and LT switch etc. should be periodically

inspected and proper pressure maintained to avoid sparking and heating of contacts.

Replacement of deteriorated wires and services should also be made timely to avoid any cause of leaking and loss of

power.

(6) Feeder Phase Current and Load Balancing:

One of the easiest loss savings of the distribution system is balancing current along three-phase circuits.

Feeder phase balancing also tends to balance voltage drop among phases giving three-phase customers less voltage

unbalance. Amperage magnitude at the substation doesn’t guarantee load is balanced throughout the feeder length.

Feeder phase unbalance may vary during the day and with different seasons. Feeders are usually considered

“balanced” when phase current magnitudes are within 10.Similarly, balancing load among distribution feeders will also

lower losses assuming similar conductor resistance. This may require installing additional switches between feeders to

allow for appropriate load transfer.

Bifurcation of feeders according to Voltage regulation and Load.

(7) Load Factor Effect on Losses:

Power consumption of Customer varies throughout the day and over seasons. Residential customers generally draw

their highest power demand in the evening hours. Same commercial customer load generally peak in the early

afternoon. Because current level (hence, load) is the primary driver in distribution power losses, keeping power

consumption more level throughout the day will lower peak power loss and overall energy losses. Load variation is

Called load factor and It varies from 0 to 1.

Load Factor=Average load in a specified time period / peak load during that time period.

For example, for 30 days month (720 hours) peak Load of the feeder is 10 MW. If the feeder supplied a total energy of

5,000 MWH, the load factor for that month is (5,000 MWh)/ (10MW x 720) =0.69.

Lower power and energy losses are reduced by raising the load factor, which, evens out feeder demand variation

throughout the feeder.

The load factor has been increase by offering customers “time-of-use” rates. Companies use pricing power to influence

consumers to shift electric-intensive activities during off-peak times (such as, electric water and space heating, air

conditioning, irrigating, and pool filter pumping).

With financial incentives, some electric customers are also allowing utilities to interrupt large electric loads remotely

through radio frequency or power line carrier during periods of peak use. Utilities can try to design in higher load factors

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by running the same feeders through residential and commercial areas

(8) Transformer Sizing and Selection:

Distribution transformers use copper conductor windings to induce a magnetic field into a grain-oriented silicon steel

core. Therefore, transformers have both load losses and no-load core losses.

Transformer copper losses vary with load based on the resistive power loss equation (P loss = I2R).

For some utilities, economic transformer loading means loading distribution transformers to capacity-or slightly above

capacity for a short time-in an effort to minimize capital costs and still maintain long transformer life.

However, since peak generation is usually the most expensive, total cost of ownership (TCO) studies should take into

account the cost of peak transformer losses. Increasing distribution transformer capacity during peak by one size will

often result in lower total peak power dissipation-more so if it is over Loaded.

Transformer no-load excitation loss(iron loss) occurs from a changing magnetic field in the transformer core whenever it

is energized. Core loss varies slightly with voltage but is essentially considered constant. Fixed iron loss depends on

transformer core design and steel lamination molecular structure. Improved manufacturing of steel cores and

introducing amorphous metals (such as metallic glass) have reduced core losses.

(9) Balancing 3 phase loads

Balancing 3-phase loads periodically throughout a network can reduce losses significantly. It can be done relatively

easily on overhead networks and consequently offers considerable scope for Cost effective loss reduction, given

suitable incentives.

(10) Switching off transformers

One method of reducing fixed losses is to switch off transformers in periods of low demand. If two transformers of a

certain size are required at a substation during peak periods, only one might be required during times of low demand so

that the other transformer might be switched off in order to reduce fixed losses.

This will produce some offsetting increase in variable losses and might affect security and quality of supply as well as

the operational condition of the transformer itself. However, these trade-offs will not be explored and optimized unless

the cost of losses are taken into account.

(11) Other Reasons for Technical Losses:

Unequal load distribution among three phases in L.T system causing high neutral currents.

leaking and loss of power

Over loading of lines.

Abnormal operating conditions at which power and distribution transformers are operated

Low voltages at consumer terminals causing higher drawl of currents by inductive loads.

Poor quality of equipment used in agricultural pumping in rural areas, cooler air-conditioners and industrial loads in

urban areas.

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Calculate TC Size & Voltage Drop due to starting of Large Motor

JUNE 5, 2013

1 COMMENT

10 Votes

Calculate TC Size & Voltage Drop due to starting of Large Motor

Calculate Voltage drop in Transformer ,1000KVA,11/0.480KV,impedance 5.75%, due to starting of 300KW,460V,0.8

Power Factor, Motor code D(kva/hp).Motor Start 2 times per Hour and The allowable Voltage drop at Transformer

Secondary terminal is 10%.

Motor current / Torque:

Motor Full Load Current= (Kwx1000)/(1.732x Volt (L-L)x P.F)

Motor Full Load Current=300×1000/1.732x460x0.8= 471 Amp.

Motor Locked Rotor Current =Multiplier x Motor Full Load Current

Locked Rotor Current (Kva/Hp)

Motor Code

Min

Max

A

3.15

B

3.16

3.55

C

3.56

4

D

4.1

4.5

E

4.6

5

F

5.1

5.6

G

5.7

6.3

H

6.4

7.1

J

7.2

8

K

8.1

9

L

9.1

10

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M

10.1

11.2

N

11.3

12.5

P

12.6

14

R

14.1

16

S

16.1

18

T

18.1

20

U

20.1

22.4

V

22.5

Min Motor Locked Rotor Current (L1)=4.10×471=1930 Amp

Max Motor Locked Rotor Current(L2) =4.50×471=2118 Amp

Motor inrush Kva at Starting (Irsm)=Volt x locked Rotor Current x Full Load Currentx1.732 / 1000

Motor inrush Kva at Starting (Irsm)=460 x 2118x471x1.732 / 1000=1688 Kva

Transformer:

Transformer Full Load Current= Kva/(1.732xVolt)

Transformer Full Load Current=1000/(1.732×480)=1203 Amp.

Short Circuit Current at TC Secondary (Isc) =Transformer Full Load Current / Impedance.

Short Circuit Current at TC Secondary= 1203/5.75= 20919 Amp

Maximum Kva of TC at rated Short Circuit Current (Q1) = (Volt x Iscx1.732)/1000.

Maximum Kva of TC at rated Short Circuit Current (Q1)=480x20919x1.732/1000= 17391 Kva.

Voltage Drop at Transformer secondary due to Motor Inrush (Vd)= (Irsm) / Q1

Voltage Drop at Transformer secondary due to Motor Inrush (Vd) =1688/17391 =10%

Voltage Drop at Transformer Secondary is 10% which is within permissible Limit.

Motor Full Load Current<=65% of Transformer Full Load Current

471 Amp <=65%x1203 amp = 471 Amp<= 781 Amp

Here Voltage Drop is within Limit and Motor Full Load Current<=TC Full Load Current.

Size of Transformer is Adequate.

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Calculate Size of Contactor, Fuse, C.B, Over Load Relay of DOL Starter

JUNE 2, 2013

3 COMMENTS

4 Votes

Calculate Size of Contactor, Fuse, C.B, O/L of DOL Starter

Calculate Size of each Part of DOL starter for The System Voltage 415V ,5HP Three Phase House hold Application

Induction Motor ,Code A, Motor efficiency 80%,Motor RPM 750 ,Power Factor 0.8 , Overload Relay of Starter is Put before

Motor.

Basic Calculation of Motor Torque & Current:

Motor Rated Torque (Full Load Torque) =5252xHPxRPM

Motor Rated Torque (Full Load Torque) =5252x5x750=35 lb-ft.

Motor Rated Torque (Full Load Torque) =9500xKWxRPM

Motor Rated Torque (Full Load Torque) =9500x(5×0.746)x750 =47 Nm

If Motor Capacity is less than 30 KW than Motor Starting Torque is 3xMotor Full Load Current or 2X Motor Full Load

Current.

Motor Starting Torque=3xMotor Full Load Current.

Motor Starting Torque==3×47=142Nm.

Motor Lock Rotor Current =1000xHPx figure from below Chart/1.732×415

Locked Rotor Current

Code

Min

Max

A

1

3.14

B

3.15

3.54

C

3.55

3.99

D

4

4.49

E

4.5

4.99

F

5

2.59

G

2.6

6.29

H

6.3

7.09

I

7.1

7.99

K

8

8.99

L

9

9.99

M

10

11.19

N

11.2

12.49

P

12.5

13.99

R

14

15.99

S

16

17.99

T

18

19.99

U

20

22.39

V

22.4

As per above chart Minimum Locked Rotor Current =1000x5x1/1.732×415=7 Amp

Maximum Locked Rotor Current =1000x5x3.14/1.732×415=22 Amp.

Motor Full Load Current (Line) =KWx1000/1.732×415

Motor Full Load Current (Line) = (5×0.746)x1000/1.732×415=6 Amp.

Motor Full Load Current (Phase)=Motor Full Load Current (Line)/1.732

Motor Full Load Current (Phase)==6/1.732=4Amp

Motor Starting Current =6 to 7xFull Load Current.

Motor Starting Current (Line)=7×6=45 Amp

(1) Size of Fuse:

Fuse as per NEC 430-52

Type of Motor

Time Delay Fuse

Non-Time Delay Fuse

Single Phase

300%

175%

3 Phase

300%

175%

Wound Rotor

150%

150%

Direct Current

150%

150%

Maximum Size of Time Delay Fuse =300% x Full Load Line Current.

Maximum Size of Time Delay Fuse =300%x6= 19 Amp.

Maximum Size of Non Time Delay Fuse =1.75% x Full Load Line Current.

Maximum Size of Non Time Delay Fuse=1.75%6=11 Amp.

(2) Size of Circuit Breaker:

Circuit Breaker as per NEC 430-52

Type of Motor

Instantaneous Trip

Inverse Time

Single Phase

800%

250%

3 Phase

800%

250%

Synchronous

800%

250%

Wound Rotor

800%

150%

Direct Current

200%

150%

Maximum Size of Instantaneous Trip Circuit Breaker =800% x Full Load Line Current.

Maximum Size of Instantaneous Trip Circuit Breaker =800%x6= 52 Amp.

Maximum Size of Inverse Trip Circuit Breaker =250% x Full Load Line Current.

Maximum Size of Inverse Trip Circuit Breaker =250%x6= 16 Amp.

(3) Thermal over Load Relay:

Thermal over Load Relay (Phase):

Min Thermal Over Load Relay setting =70%xFull Load Current(Phase)

Min Thermal Over Load Relay setting =70%x4= 3 Amp

Max Thermal Over Load Relay setting =120%xFull Load Current(Phase)

Max Thermal Over Load Relay setting =120%x4= 4 Amp

Thermal over Load Relay (Phase):

Thermal over Load Relay setting =100%xFull Load Current (Line).

Thermal over Load Relay setting =100%x6= 6 Amp

Application

Contactor

Making Cap

Non-Inductive or Slightly Inductive ,Resistive

Load

AC1

1.5

Slip Ring Motor

AC2

4

Squirrel Cage Motor

AC3

10

Rapid Start / Stop

AC4

12

Switching of Electrical Discharge Lamp

AC5a

3

Switching of Electrical Incandescent Lamp

AC5b

1.5

Switching of Transformer

AC6a

12

Switching of Capacitor Bank

AC6b

12

Slightly Inductive Load in Household or same

type load

AC7a

1.5

Motor Load in Household Application

AC7b

8

Hermetic refrigerant Compressor Motor with

Manual O/L Reset

AC8a

6

Hermetic refrigerant Compressor Motor with

Auto O/L Reset

AC8b

6

Control of Restive & Solid State Load with opto

coupler Isolation

AC12

6

Control of Restive Load and Solid State with

T/C Isolation

AC13

10

Control of Small Electro Magnetic Load (

<72VA)

AC14

6

Control of Small Electro Magnetic Load (

>72VA)

AC15

10

As per above Chart

Type of Contactor= AC7b

Size of Main Contactor = 100%X Full Load Current (Line).

Size of Main Contactor =100%x6 = 6 Amp.

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Making/Breaking Capacity of Contactor=8×6= 52 Amp.

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Calcualte Number of Lighting Fixtures & Lux Level

JUNE 1, 2013

5 COMMENTS

3 Votes

Calculate required No of Fixtures.

Calculate required No of Lamps

.

Calculate Watts / Sq.foot

.

Calculate Energy Cost / Year

.

Calculate Actual Lux Level.

DownLoad

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Calculate Voltage Regulation of Distribution Line

MAY 12, 2013

5 COMMENTS

5 Votes

Introduction:

Voltage regulation or Load Reguation is to maintain a fixed voltage under different load.Voltage regulation is limiting

factor to decide the size of either conductor or type of insulation.

In circuit current need to be lower than this in order to keep the voltage drop within permissible values. The high voltage

circuit should be carried as far as possible so that the secondary circuit have small voltage drop.

Voltage Regulation for 11KV, 22KV, 33KV Overhead Line (As per REC):

% Voltage Regulation= (1.06xPxLxPF) / (LDFxRCxDF)

Where

P=Total Power in KVA

L= Total Length of Line from Power Sending to Power Receiving in KM.

PF= Power Factor in p.u

RC= Regulation Constant (KVA-KM) per 1% drop.

RC=(KVxKVx10)/( RCosΦ+XSinΦ)

LDF= Load Distribution Factor.

LDF= 2 for uniformly distributed Load on Feeder.

LDF>2 If Load is skewed toward the Power Transformer.

LDF= 1 To 2 If Load is skewed toward the Tail end of Feeder.

DF= Diversity Factor in p.u

Permissible Voltage Regulation (As per REC):

Maximum Voltage Regulation at any Point of Distribution Line

Part of Distribution

System

Urban Area

(%)

Suburban Area

(%)

Rural Area (%)

Up to Transformer

2.5

2.5

2.5

Up to Secondary Main

3

2

0.0

Up to Service Drop

0.5

0.5

0.5

Total

6.0

5.0

3.0

Voltage Regulation Values:

The voltage variations in 33 kV and 11kV feeders should not exceed the following limits at the farthest end under peak

load conditions and normal system operation regime.

Above 33kV (-) 12.5% to (+) 10%.

Up to 33kV (-)9.0% to (+)6.0%.

Low voltage (-)6.0% to (+) 6.0%

In case it is difficult to achieve the desired voltage especially in Rural areas, then 11/0.433kV distribution transformers(in

place of normal 11/0.4kV DT’s) may be used in these areas.

Required Size of Capacitor:

Size of capacitor for improvement of the Power Factor from Cos ø1 to Cos ø2 is

Required size of Capacitor(Kvar) = KVA1 (Sin ø1 – [Cos ø1 / Cos ø2] x Sin

ø2)

Where KVA1 is Original KVA.

Optimum location of capacitors:

L = [1 – (KVARC / 2 KVARL) x (2n-1)]

Where,

L = distance in per unit along the line from sub-station.

KVARC = Size of capacitor bank

KVARL = KVAR loading of line

n = relative position of capacitor bank along the feeder from sub-station if the total capacitance is to be divided into more

than one Bank along the line. If all capacitance is put in one Bank than values of n=1.

Voltage Rise due to Capacitor installation:

% Voltage Rise = (KVAR(Cap)x Lx X) / 10xVx2

Where,

KVAR(Cap)=Capacitor KVAR

X = Reactance per phase

L=Length of Line (mile)

V = Phase to phase voltage in kilovolts

Calculate % Voltage Regulation of Distribution Line :

Calculate Voltage drop and % Voltage Regulation at Trail end of following 11 KV Distribution system , System have ACSR

DOG Conductor (AI 6/4.72, GI7/1.57),Current Capacity of ACSR Conductor =205Amp,Resistance =0.2792Ω and

Method-1 (Distance Base):

Voltage Drop = ( (√3x(RCosΦ+XSinΦ)x I ) / (No of Conductor/Phase

x1000))x Length of Line

Voltage drop at Load A

Load Current at Point A (I) = KW / 1.732xVoltxP.F

Load Current at Point A (I) =1500 / 1.732x11000x0.8 = 98 Amp.

Required No of conductor / Phase =98 / 205 =0.47 Amp =1 No

Voltage Drop at Point A = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line

Voltage Drop at Point A =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x1500) = 57 Volt

Receiving end Voltage at Point A = Sending end Volt-Voltage Drop= (1100-57) = 10943 Volt.

% Voltage Regulation at Point A = ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100

% Voltage Regulation at Point A = ((11000-10943) / 10943 )x100 = 0.52%

% Voltage Regulation at Point A =0.52 %

Voltage drop at Load B

Load Current at Point B (I) = KW / 1.732xVoltxP.F

Load Current at Point B (I) =1800 / 1.732x11000x0.8 = 118 Amp.

Distance from source= 1500+1800=3300 Meter.

Voltage Drop at Point B = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line

Voltage Drop at Point B =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x3300) = 266 Volt

Receiving end Voltage at Point B = Sending end Volt-Voltage Drop= (1100-266) = 10734 Volt.

% Voltage Regulation at Point B= ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100

% Voltage Regulation at Point B= ((11000-10734) / 10734 )x100 = 2.48%

% Voltage Regulation at Point B =2.48 %

Voltage drop at Load C

Load Current at Point C (I) = KW / 1.732xVoltxP.F

Load Current at Point C (I) =2000 / 1.732x11000x0.8 = 131 Amp

Distance from source= 1500+1800+2000=5300 Meter.

Voltage Drop at Point C = ( (√3x(RCosΦ+XSinΦ)xI ) / (No of Conductor/Phase x1000))x Length of Line

Voltage Drop at Point C =((1.732x (0.272×0.8+0×0.6)x98) / 1×1000)x5300) = 269 Volt

Receiving end Voltage at Point C = Sending end Volt-Voltage Drop= (1100-269) = 10731 Volt.

% Voltage Regulation at Point C= ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100

% Voltage Regulation at Point C= ((11000-10731) / 10731 )x100 = 2.51%

% Voltage Regulation at Point C =2.51 %

Here Trail end Point % Voltage Regulation is 2.51% which is in permissible limit.

Method-2 (Load Base):

% Voltage Regulation =(I x (RcosǾ+XsinǾ)x Length ) / No of Cond.per

Phase xV (P-N))x100

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Voltage drop at Load A

Load Current at Point A (I) = KW / 1.732xVoltxP.F

Load Current at Point A (I) =1500 / 1.732x11000x0.8 = 98 Amp.

Distance from source= 1.500 Km.

Required No of conductor / Phase =98 / 205 =0.47 Amp =1 No

Voltage Drop at Point A = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100

Voltage Drop at Point A =((98x(0.272×0.8+0×0.6)x1.5) / 1×6351) = 0.52%

% Voltage Regulation at Point A =0.52 %

Voltage drop at Load B

Load Current at Point B (I) = KW / 1.732xVoltxP.F

Load Current at Point B (I) =1800 / 1.732x11000x0.8 = 118 Amp.

Distance from source= 1500+1800=3.3Km.

Required No of conductor / Phase =118 / 205 =0.57 Amp =1 No

Voltage Drop at Point B = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100

Voltage Drop at Point B =((118x(0.272×0.8+0×0.6)x3.3)/1×6351) = 1.36%

% Voltage Regulation at Point A =1.36 %

Voltage drop at Load C

Load Current at Point C (I) = KW / 1.732xVoltxP.F

Load Current at Point C (I) =2000 / 1.732x11000x0.8 = 131Amp.

Distance from source= 1500+1800+2000=5.3Km.

Required No of conductor / Phase =131/205 =0.64 Amp =1 No

Voltage Drop at Point C = (I x (RcosǾ+XsinǾ)x Length ) / V (Phase-Neutral))x100

Voltage Drop at Point C =((131x(0.272×0.8+0×0.6)x5.3)/1×6351) = 2.44%

% Voltage Regulation at Point A =2.44 %

Here Trail end Point % Voltage Regulation is 2.44% which is in permissible limit.

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Calculate Size of Transformer and Voltage drop due to starting of large

size motor

MAY 7, 2013

10 COMMENTS

3 Votes

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Calculate voltage drop in transformer due starting of Single/Multiple motor

Calculate Full Load Current of Transformer.

Calculate Short Circuit Current at Transformer Secondary.

Calculate Max KVA of Transformer at Short Circuit Current.

Calculate Motor In Rush KVA At Starting.

Calculate Voltage Drop at Transformer Secondary Due to Motor in Rush

Calculate Transformer Secondary Voltage

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Design DOL / Star-Delta Starter

MAY 2, 2013

5 COMMENTS

2 Votes

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Calculate Lock Rotor Current

.

Calculate Motor Starting Current.

Calculate Motor Full Load Current

.

Calculate Non-time Delay/Time Delay fuse Size

.

Calculate Circuit Breaker Size

.

Calculate Overload Relay setting

.

Calculate Type of Contactor

.

Calculate Size of Main Contactor

.

Calculate Size of Star Contactor

.

Calculate Size of Delta Contactor.

Standard Arrangement of Main and Auxiliary Contactor

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Calculate Size of Motor-Pump

APRIL 14, 2013

9 COMMENTS

5 Votes

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Calculate Motor-Pump size:

Calculate Pump Hydraulic Power

Calculate Motor-Pump Shaft Power

Calculate Pump Size.

Calculate Motor Size.

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Calculate Numbers of Plate/Pipe/Strip Earthings (Part-3)

APRIL 1, 2013

6 COMMENTS

4 Votes

Calculate Min. Cross Section area of Earthing Conductor:

Cross Section Area of Earthing Conductor As per IS 3043

Cross Section Area of Earthing Conductor (A) =(If x√t) / K

Where t= Fault current Time (Second).

K= Material Constant.

Example:

Calculate Cross Section Area of GI Earthing Conductor for System has 50KA Fault Current for 1 second. Corrosion will

be 1.0 % Per Year and No of Year for Replacement is 20 Years.

Cross Section Area of Earthing Conductor (A) =(If x√t) / K

Here If=50000 Amp

T= 1Second

K=80 (Material Constant, For GI=80, copper K=205, Aluminium K=126).

Cross Section Area of Earthing Conductor (A) =(50000×1)/80

Cross Section Area of GI Earthing Conductor (A)=625 Sq.mm

Allowance for Corrosion = 1.0 % Per Year & Number of Year before replacement say = 20 Years

Total allowance = 20 x 1.0% = 20%

Safety factor = 1.5

Required Earthing Conductor size = Cross sectional area x Total allowance x Safety factor

Required Earthing Conductor size = 1125 Sq.mm say 1200 Sq.mm

Hence, Considered 1Nox12x100 mm GI Strip or 2Nox6 x 100 mm GI Strips

Thumb Rule for Calculate Number of Earthing Rod:

The approximate earth resistance of the Rod/Pipe electrodes can be calculated by

Earth Resistance of the Rod/Pipe electrodes R= K x ρ/L

Where ρ = Resistivity of earth in Ohm-Meter

L= Length of the electrode in Meter.

d= Diameter of the electrode in Meter.

K=0.75 if 25< L/d < 100.

K=1 if 100 < L/d < 600

K=1.2 o/L if 600 < L/d < 300

Number of Electrode if find out by Equation of R(d) =(1.5/N) x R

Where R(d) = Desired earth resistance

R= Resistance of single electrode

N= No. of electrodes installed in parallel at a distance of 3 to 4 Meter interval.

Example:

Calculate Earthing Pipe Resistance and Number of Electrode for getting Earthing Resistance of 1 Ω ,Soil Resistivity of

ρ=40, Length=2.5 Meter, Diameter of Pipe= 38 mm.

Here L/d = 2.5/0.038=65.78 so K=0.75

The Earth Resistance of the Pipe electrodes R= K x ρ/L =0.75×65.78=12 Ω

One electrode the earth resistance is 12 Ω.

To get Earth resistance of 1 Ω the total Number of electrodes required =(1.5×12)/1 =18 No

Calculating Resistance & Number of Earthing Rod:

Reference: As per EHV Transmission Line Reference Book page: 290 and Electrical Transmission & Distribution

Reference Book Westinghouse Electric Corporation, Section-I Page: 570-590.

Earthing Resistance of Single Rods: R = ρx[ln (2L/a)-1]/(2×3.14xL)

Earthing Resistance of Parallel Rods: R = ρx[ln (2L/A]/ (2×3.14xL)

Where L= length of rod in ground Meter,

a= radius of rod Meter

ρ = ground resistivity, ohm- Meter

A= √(axS)

S= Rod separation Meter

Factor affects on Ground resistance:

The NEC code requires a minimum ground electrode length of 2.5 meters (8.0 feet) to be in contact with the soil. But,

there are some factor that affect the ground resistance of a ground system:

Length / Depth of the ground electrode: double the length, reduce ground resistance by up to 40%.

Diameter of the ground electrode: double the diameter, lower ground resistance by only 10%.

Number of ground electrodes: for increased effectiveness, space additional electrodes at least equal to the depth of the

ground electrodes.

Ground system design: single ground rod to ground plate.

The GI Earthing Conductor sizes for various Equipments:

No

Equipments

Earth Strip Size

1

HT switchgear, structures, cable trays &

fence, rails, gate and steel column

55 X 6 mm (GI)

2

_{Lighting Arrestor}

25 X 3 mm (Copper)

3

_{PLC Panel}

25 X 3 mm (Copper)

4

_{DG & Transformer Neutral}

50X6 mm (Copper)

5

_{Transformer Body}

50X6 mm (GI)

6

_{Control & Relay Panel}

25 X 6 mm (GI)

7

_{Lighting Panel & Local Panel}

25 X 6 mm (GI)

8

_{Distribution Board}

25 X 6 mm (GI)

9

_{Motor up to 5.5 kw}

4 mm2 (GI)

11

_{Motor 22 kw to 55 kw}

40 X 6 mm (GI)

12

_{Motor Above 55 kw}

55 X 6 mm (GI)

Selection of Earthing System:

Installations/

Isc Capacity

IR Value

Required

Soil Type/ Resistivity

Earth System

House hold

earthing/3kA

8 ohm

Normal Soil/ up to 50

ohm-meter

Single

Electrode

Sandy Soil/ between 50 to

2000 ohm- meter

Single

Electrode

Rocky Soil/ More than

2000 ohm- meter

Multiple

Electrodes

Commercial

premises,Office /

5kA

2 ohm

Normal Soil/ up to 50

ohm-meter

Single

Electrode

Sandy Soil/ between 50 to

2000 ohm- meter

Multiple

Electrodes

Rocky Soil/ More than

2000 ohm- meter

Multiple

Electrodes

Transformers,

substationearthing,

LT line equipment/

15kA

less than

1 ohm

Normal Soil/ up to 50

ohm-meter

Single

Electrode

Sandy Soil/ between 50 to

2000 ohm- meter

Multiple

Electrodes

Rocky Soil/ More than

2000 ohm- meter

Multiple

Electrodes

LA, High current

Equipment./ 50kA

less than

1 ohm

Normal Soil/ up to 50

ohm-meter

Single

Electrode

Sandy Soil/ between 50 to

2000 ohm- meter

Multiple

Electrodes

Rocky Soil/ More than

2000 ohm- meter

Multiple

Electrodes

PRS, UTS, RTUs,

Data processing

centre etc./5KA

less than

0.5 ohm

Normal Soil/ up to 50

ohm-meter

Single

Electrode

Sandy Soil/ between 50 to

2000 ohm- meter

Multiple

Electrodes

Rocky Soil/ More than

2000 ohm- meter

Multiple

Electrodes

Size of Earthing Conductor:

Ref IS 3043 &Handbook on BS 7671: The Lee Wiring Regulations by Trevor E. Marks.

Size of Earthing Conductor

Area of Phase

Conductor S (mm2)

Area of Earthing

conductor (mm2) When It

is Same Material as

Phase Conductor

Area of Earthing

conductor (mm2) When It

is Not Same Material as

Phase Conductor

S < 16 mm2

S

SX(k1/k2)

16 mm2<S< 35 mm2

16 mm2

16X(k1/k2)

S > 35 mm2

S/2

SX(k1/2k2)

K1 is value of Phase conductor,k2 is value of earthing conductor

Value of K for GI=80, Alu=126,Cu=205 for 1 Sec

Standard Earthing Strip/Plate/Pipe/wire Weight:

GI Earthing Strip:

Size (mm2)

Weight

20 x 3

500 gm Per meter

25 x 3

600 gm Per meter

25 x 6

1/200 Kg Per meter

32 x 6

1/600 Kg Per meter

40 x 6

2 Kg Per meter

50 x 6

2/400 Kg Per meter

65 x 10

5/200 Kg Per meter

75 x 12

7/200 Kg Per meter

GI Earthing Plate:

Plate

Weight

600 x 600 x 3 mm

10 Kg App.

600 x 600 x 4 mm

12 Kg App.

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600 x 600 x 5 mm

15 Kg App.

600 x 600 x 6 mm

18 Kg App.

600 x 600 x 12 mm

36 Kg App.

1200 x 1200 x 6 mm

70 Kg App.

1200 x 1200 x 12 mm

140 Kg App.

GI Earthing Pipe:

Pipe

Weight

3 meter Long BISE

5 Kg App.

3 meter r Long BISE

9 Kg App.

4.5 meter (15′ Long BISE)

5 Kg App.

4.5 meter (15′ Long BISE)

9 Kg App.

4.5 meter (15′ Long BISE)

14 Kg App

GI Earthing Wire:

Plate

Weight

6 Swg

5 meter in 1 Kg

8 Swg

9 meter in 1 Kg

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Calculate Numbers of Plate/Pipe/Strip Earthings (Part-2)

MARCH 20, 2013

4 COMMENTS

8 Votes

(2)Calculate Number of Plate Earthing:

The Earth Resistance of Single Plate electrode is calculated as per IS 3040:

R=ρ/A√(3.14/A)

Where ρ=Resistivity of Soil (Ω Meter),

A=Area of both side of Plate (m2),

Example: Calculate Number of CI Earthing Plate of 600×600 mm, System has Fault current 65KA for 1 Sec and Soil

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Resistivity is 100 Ω-Meters.

Current Density At The Surface of Earth Electrode (As per IS 3043):

Max. Allowable Current Density I = 7.57×1000/(√ρxt) A/m2

Max. Allowable Current Density = 7.57×1000/(√100X1)=757 A/m2

Surface area of both side of single 600×600 mm Plate= 2 x lxw=2 x 0.06×0.06 = 0.72 m2

Max. current dissipated by one Earthing Plate = Current Density x Surface area of electrode

Max. current dissipated by one Earthing Plate =757×0.72= 545.04 Amps

Resistance of Earthing Plate (Isolated)(R)=ρ/A√(3.14/A)

Resistance of Earthing Plate (Isolated)(R)=100/0.72x√(3.14/.072)=290.14 Ω

Number of Earthing Plate required =Fault Current / Max.current dissipated by one Earthing Pipe.

Number of Earthing Plate required= 65000/545.04 =119 No’s.

Total Number of Earthing Plate required = 119 No’s.

Overall resistance of 119 No of Earthing Plate=290.14/119=2.438 Ω.

(3)Calculating Resistance of Bared Earthing Strip:

1)Calculation for earth resistance of buried Strip (As per IEEE):

The Earth Resistance of Single Strip of Rod buried in ground is

R=ρ/Px3.14xL (loge (2xLxL/Wxh)+Q)

Where ρ=Resistivity of Soil (Ω Meter),

h=Depth of Electrode (Meter),

w=Width of Strip or Diameter of Conductor (Meter)

L=Length of Strip or Conductor (Meter)

P and Q are Coefficients

2)Calculation for earth resistance of buried Strip(As per IS 3043)

:

The Earth Resistance of Single Strip of Rod buried in ground is

R=100xρ/2×3.14xL (loge (2xLxL/Wxt))

Where ρ=Resistivity of Soil (Ω Meter),

L=Length of Strip or Conductor (cm)

w=Width of Strip or Diameter of Conductor (cm)

t= Depth of burial (cm)

Example :

Calculate Earthing Resistance of Earthing strip/wire of 36mm Diameter, 262 meter long buried at 500mm depth in

ground, soil Resistivity is 65 Ω Meter.

Here R = Resistance of earth rod in W.

r = Resistivity of soil(Ω Meter) = 65 Ω Meter

l = length of the rod (cm) = 262m = 26200 cm

d = internal diameter of rod(cm) = 36mm = 3.6cm

h = Depth of the buried strip/rod (cm)= 500mm = 50cm

Resistance of Earthing Strip/Conductor (R)=ρ/2×3.14xL (loge (2xLxL/Wt))

Resistance of Earthing Strip/Conductor (R)=65/2×3.14x26200xln(2x26200x26200/3.6×50)

Resistance of Earthing Strip/Conductor (R)== 1.7 Ω

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Calculate Numbers of Plate/Pipe/Strip Earthings (Part-1)

MARCH 8, 2013

19 COMMENTS

13 Votes

Introduction:

Number of Earthing Electrode and Earthing Resistance depends on the resistivity of soil and time for fault Current to

pass through (1 sec or 3 sec). If we divide the area for earthing required by the area of one earth plate gives the no of

Earth pits required.

There is no general rule to calculate the exact no of earth Pits and Size of Earthing Strip, But discharging of leakage

current is certainly dependent on the cross section area of the material so for any equipment

the earth strip size is

calculated on the current to be carried by that strip.

First the leakage current to be carried is calculated and then size of

the strip is determined.

For most of the Electrical equipments like Transformer, DG set etc., the General concept is to have 4 no earth pits.2 no’s

for body earthing With 2 separate strips with the pits shorted and 2 nos for Neutral with 2 separate strips with the pits

shorted.

The Size of Neutral Earthing Strip should be Capable to carry neutral current of that equipment.

The Size of Body Earthing should be capable to carry half of neutral Current.

For example for 100kVA transformer, the full load Current is around 140A.The strip connected should be Capable to carry

at least 70A (neutral current) which means a Strip of GI 25x3mm should be enough to carry the current And for body a

strip of 25×3 will do the needful.

Normally we consider the strip size that is generally used as Standards. However a strip with lesser size which can carry

a current of 35A can be used for body earthing. The reason for using 2 earth pits for each body and neutral and then

shorting them is to serve as back up. If one strip gets Corroded and cuts the continuity is broken and the other Leakage

current flows through the other run thery by completing the circuit. Similarly for panels the no of pits should be 2 nos. The

size can be decided on the main incomer Breaker.

For example if main incomer to breaker is 400A, then Body earthing for panel can have a strip size of 25×6 mm Which

can easily carry 100A.

Number of earth pits is decided by considering the total Fault current to be dissipated to the ground in case of Fault

and the current that can be dissipated by each earth Pit.

Normally the density of current for GI strip can be roughly 200 amps per square cam. Based on the length and dia of the

Pipe used the Number of Earthing Pits can be finalized.

(1) Calculate Numbers of Pipe Earthing:

(A) Earthing Resistance & No of Rod for Isolated Earth Pit (Without Buried

Earthing Strip):

The Earth Resistance of Single Rod or Pipe electrode is calculated as per BS 7430:

Where ρ=Resistivity of Soil (Ω Meter),

L=Length of Electrode (Meter),

D=Diameter of Electrode (Meter)

Example:

Calculate Isolated Earthing Rod Resistance. The Earthing Rod is 4 Meter Long and having 12.2mm Diameter, Soil

Resistivity 500 Ω Meter.

R=500/ (2×3.14×4) x (Loge (8×4/0.0125)-1) =156.19 Ω.

The Earth Resistance of Single Rod or Pipe electrode is calculated as per IS 3040:

R=100xρ/2×3.14xL (loge(4xL/d))

Where ρ=Resistivity of Soil (Ω Meter),

L=Length of Electrode (cm),

D=Diameter of Electrode (cm)

Example:

Calculate Number of CI Earthing Pipe of 100mm diameter, 3 Meter length. System has Fault current 50KA for 1 Sec and

Soil Resistivity is 72.44 Ω-Meters.

Current Density At The Surface of Earth Electrode (As per IS 3043):

Max. Allowable Current Density I = 7.57×1000/(√ρxt) A/m2

Max. Allowable Current Density = 7.57×1000/(√72.44X1)=889.419 A/m2

Surface area of one 100mm dia. 3 meter Pipe= 2 x 3.14 x r x L=2 x 3.14 x 0.05 x3 = 0.942 m2

Max. current dissipated by one Earthing Pipe = Current Density x Surface area of electrode

Max. current dissipated by one Earthing Pipe = 889.419x 0.942 = 837.83 A say 838 Amps

Number of Earthing Pipe required =Fault Current / Max.current dissipated by one Earthing Pipe.

Number of Earthing Pipe required= 50000/838 =59.66 Say 60 No’s.

Total Number of Earthing Pipe required = 60 No’s.

Resistance of Earthing Pipe (Isolated) R=100xρ/2×3.14xLx(loge (4XL/d))

Resistance of Earthing Pipe (Isolated) R=100×72.44/2×3.14x300x(loge (4X300/10))=7.99 Ω/Pipe

Overall resistance of 60 No of Earthing Pipe=7.99/60=0.133 Ω.

(B) Earthing Resistance & No of Rod for Isolated Earth Pit (With Buried

Earthing Strip):

Resistance of Earth Strip(R) As per IS 3043

R=ρ/2×3.14xLx (loge (2xLxL/wt)).

Example:

Calculate GI Strip having width of 12mm , length of 2200 Meter buried in ground at depth of 200mm,Soil Resistivity is

72.44 Ω-Meter

Resistance of Earth Strip(Re)=72.44/2×3.14x2200x(loge (2x2200x2200/.2x.012))= 0.050 Ω

From above Calculation Overall resistance of 60 No of Earthing Pipe (Rp) = 0.133 Ω. And it connected to bury Earthing

Strip. Here Net Earthing Resistance =(RpxRe)/(Rp+Re)

Net Earthing Resistance= =(0.133×0.05)/(0.133+0.05)= 0.036 Ω

(C) Total Earthing Resistance & No of Electrode for Group of Electrode

(Parallel):

In cases where a single electrode is not sufficient to provide the desired earth resistance, more than one electrode shall

be used. The separation of the electrodes shall be about 4 M.

configuration of electrode the array.

The Total Resistance of Group of Electrode in different configurations as per BS 7430:

Ra=R (1+λa/n) Where a= ρ/2X3.14XRXS

Where S= Distance between adjustment Rod (Meter),

λ =Factor Given in Table,

n= Number of Electrode,

ρ=Resistivity of Soil (Ω Meter),

R=Resistance of Single Rod in Isolation (Ω)

Factors for parallel electrodes in line (BS 7430)

Number of electrodes (n)

Factor (λ)

2

1.0

3

1.66

4

2.15

5

2.54

6

2.87

7

3.15

8

3.39

9

3.61

10

3.8

For electrodes equally spaced around a hollow square, e.g. around the perimeter of a building, the equations given

above are used with a value of λ taken from following Table.

For three rods placed in an equilateral triangle, or in an L formation, a value of λ = 1.66 may be assumed.

Factors for electrodes in a hollow square (BS 7430)

Number of electrodes (n)

Factor (λ)

2

2.71

4

5.48

5

6.13

6

6.63

7

7.03

8

7.36

9

7.65

10

7.9

12

8.3

14

8.6

16

8.9

18

9.2

20

9.4

For Hollow Square Total Number of Electrode (N) = (4n-1).

The rule of thumb is that rods in parallel should be spaced at least twice their length to utilize the full benefit of the

additional rods.

If the separation of the electrodes is much larger than their lengths and only a few electrodes are in parallel, then the

resultant earth resistance can be calculated using the ordinary equation for resistances in parallel.

In practice, the effective earth resistance will usually be higher than Calculation. Typically, a 4 spike array may provide an

improvement 2.5 to 3 times. An 8 spike array will typically give an improvement of maybe 5 to 6 times.

The Resistance of Original Earthing Rod will be lowered by Total of 40% for Second Rod, 60% for third Rod,66% for forth

Rod

Example:

Calculate Total Earthing Rod Resistance of 200 Number arranges in Parallel having 4 Meter Space of each and if it

connects in Hollow Square arrangement. The Earthing Rod is 4 Meter Long and having 12.2mm Diameter, Soil

Resistivity 500 Ω.

First Calculate Single Earthing Rod Resistance

R=500/ (2×3.14×4) x (Loge (8×4/0.0125)-1) =136.23 Ω.

Now Calculate Total Resistance of Earthing Rod of 200 Number in Parallel condition.

a=500/(2×3.14x136x4)=0.146

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If Earthing Rod is connected in Hollow Square than Rod in Each side of Square is 200=(4n-1) so n=49 No.

Ra (In Hollow Square) =136.23x (1+9.4×0.146/200) =1.61 Ω.

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Size and Location of Capacitor in Electrical System-(Part 2)

MARCH 3, 2013

8 COMMENTS

9 Votes

Size of Circuit Breaker, Fuse and Conductor of Capacitor Bank:

(A) Thermal and Magnetic setting of a Circuit breaker:

(1) Size of Circuit Breaker:

1.3 to 1.5x Capacitor Current (In) for Standard Duty/Heavy Duty/Energy Capacitors

1.31×In for Heavy Duty/Energy Capacitors with 5.6% Detuned Reactor(Tuning Factor 4.3)

1.19×In for Heavy Duty/Energy Capacitors with 7% Detuned Reactor(Tuning Factor 3.8)

1.12×In for Heavy Duty/Energy Capacitors with 14% Detuned Reactor(Tuning Factor 2.7)

Note: Restrictions in Thermal settings of system with Detuned reactors are due to limitation of IMP (Maximum

Permissible current) of the Detuned reactor.

(2) Thermal Setting of Circuit Breaker:

1.5x Capacitor Current (In) for Standard Duty/Heavy Duty/Energy Capacitors

(3) Magnetic Setting of Circuit Breaker:

5 to10 x Capacitor Current (In) for Standard Duty/Heavy Duty/Energy Capacitors

Example :150kvar,400v, 50Hz Capacitor

Us = 400V, Qs = 150kvar,Un = 400V, Qn = 150kvar

In = 150000/400√3 = 216A

Circuit Breaker Rating = 216 x 1.5 = 324A

Select a 400A Circuit Breaker.

Circuit Breaker thermal setting = 216 x 1.5 = 324 Amp

Conclusion:- Select a Circuit Breaker of 400A with Thermal Setting at 324A and

Magnetic Setting ( Short Circuit ) at 3240A

(B) Fuse Selection

The rating must be chosen to allow the thermal protection to be set to:

1.5 to 2.0 x Capacitor Current (In) for Standard Duty/Heavy Duty/Energy Capacitors.

1.35×In for Heavy Duty/Energy Capacitors with 5.7% Detuned Reactor (Tuning Factor 4.3)

1.2×In for Heavy Duty/Energy Capacitors with 7% Detuned Reactor (Tuning Factor 3.8)