Table of Contents

1.0 Introduction:...................................................................................3

2.0 Voltage drop formula derivation:....................................................3

3.0 Calculation Methods.......................................................................4

3.1 Input Data:....................................................................................4

3.2 Approximation Method:...............................................................4

3.3 Precise Calculation Method:.........................................................5

4.0 Worked out Examples:....................................................................8

4.1 Case 1:..........................................................................................8

4.2 Case 2:........................................................................................10

5.0 CONCLUSION:................................................................................12

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Road Lighting System

1.0 Introduction:

Road lighting or Street lighting system inside any in any industry / refinery or petrochemical complex provide required illumination levels for vehicle movement. Usually a 3 phase TPN supply cable is laid from Sub Station Lighting Distribution Board for selected road length and number of poles as one feeder. In each lighting pole single phase supply is taken from this TPN supply cable laid along the road. First pole will be on R phase and the second pole on Y phase and the third pole will be on B phase and so on balancing the loads on all 3 phases. The permitted feeder length or number of lighting poles in one 3 phase feeder circuit depends on selected cable current carrying capacity and permissible voltage drop.

Generally it is ensured that the voltage drop at the last pole fixture terminals ( Where voltage drop will me maximum ) shall not be greater than 5%( i.e. lamp should get 95% of its rated voltage).As part of exercise an attempt has been made to calculate voltage drop using two typical methods. It is to be acknowledged the load current in each phase conductor cable is varying in different segments. Also it is to be noted in each R, Y, B triplet poles, the neutral current becomes zero at the end of each triplet due to balanced load. To start with a general expression has been derived and using that voltage drops has been calculated in two typical live examples.

2.0 Voltage drop formula derivation:

So, Voltage Drop:

ΔV = 2 * I * L * (R cosφ + X sinφ) /1000

Let K=(R cosφ + X sinφ) /1000

So

ΔV = 2 * I * L * K

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3.0 Calculation Methods

3.1 Input Data:

Supply phase voltage=Vph Lamp Current for pole fixture, I=power consumed /(voltage per phase*efficiency*power factor It is assumed Lamp wattage in each pole fixture is same. Let Distance between poles = L mts No of triplets=T (Each triplet consists of 3 consecutive poles on R.Y.B phase supply ) Distance between first pole and Lighting Distribution Box (LDB) in Substation =LF

The circuit diagram below shows the arrangement how the fixtures are connected.

Fig 1

3.2 Approximation Method:

In usual practice, what is done is for complete lighting load , average lighting load per phase is worked out and for the given feeder length voltage drop is calculated. This is illustrated below.

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And load current is calculated as:

I= √((Total KW)2 + (Total KVAR)2) /√3 VL

And finally voltage drop is calculated at last pole using formula:

Clearly this is an approximate calculation due to the following reasons.

1. Load current in each section is not same 2. Neutral current also becomes zero vectorial addition after each Triplet.

3.3 Precise Calculation Method:

Let us start with two triplets. We have to calculate Voltage drop in the last pole fixture which is connected to Blue phase.

Voltage drop has been divided in two parts as under.

Voltage drop in phase conductor from Sub-station to third pole which is on B phase (First Triplet last pole )

Voltage drop in the neutral conductor.

3.3.1 Calculation for Voltage drop in phase Conductor:

Current flowing in first segment is 2I amps

Current in the second segment that is from third pole to sixth pole current is I amperes.

First segment feeder length = LF+2L meters

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Second segment feeder length = 3L meters

As seen in fig (2) voltage drop in phase conductor can be calculated by summing up voltage drops in the two segments.

So voltage drop in phase conductor will be :

[2I*(LF+2L)*K] +[I*(3L)*K] ( where K=(Rcosφ + Xsinφ)/1000 )

Similarly for 3 triplets it will be

[(3I)*(LF+2L)*K] +[(2I)*(3L)*K]+[I*3L*K]

For 4 triplets

[(4I)*(LF+2L)*K] +[(3I)*(3L)*K]+[(2I)*(3L)*K]+[I*(3L)*K] and so on as illustrated below.

Fig 2

As we can see in figure (2)

Progressing from first triplet to second and then third we can see that current can be generalized as:

(T-n)*I where n range from 0 to T-1.

So we can generalize our formula for voltage drop in phase as:

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ΔVph=T*I*(LF+2L)*K + ∑ (T-n)*I*3*L*K

Or IK [ T (LF +2L) + 3L∑(T-n) ]

ΔVph= IK [ T (LF +2L) + 3L∑(T-n) ]

Where n ranges from 1 to T-1 (not from 0 as first term is out of summation due to unequal length)

3.3.2 Neutral voltage drop calculation:

IBY=I in magnitude

As IB and Iy are equal phasor 120 degrees apart. 6 segments of neutral are shown in this figure. Out of which voltage drop occurs only in four segments.

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Voltage drop in one segment of neutral = I * L * K

As there are 2 segments in one triplet where voltage drop occurs

So voltage drop in one triplet in neutral is = 2 * I * L * K

As there are n triplets so voltage drop in neutral is:

3.3.3 Total Voltage Drop:

So total voltage drop in last fixture is:

ΔVph+ ΔVn= IK [T (LF +2L) + 3L∑ (T-n)] + 2 T I L K

Where T=number of triplets

I=current per fixture (in Amperes)

L=distance between poles (in meters)

LF= Distance of First 3rd pole from sub station(in meters)

n ranges from 1 to T-1

K= (R cosφ + X sinφ) /1000 [R=Resistance of cable in per km, X=Reactance of cable in per km, cosφ=power factor]

4.0 Worked out Examples:

Using the above formulae voltage drop is calculated in two cases as given below.

4.1 Case 1:

6 street lights 800 W each rated voltage 240V having power factor 0.9 and an efficiency 0.9. Distance between poles and from Lighting Distribution box is given in figure. Calculate Voltage drop at the last fixture. Cable used is 4 core,16 sqmm copper conductor ,resistance per km=2.33Ω and reactance per km=0.111Ω. (Cable catalog data )

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ΔVn = 2 T I L K

Sol: Precise method:

.Current per fixture I=power/ (voltage per phase*efficiency*power factor)=800/ (V*0.9*0.9)

=4.1235 A

Cable size 4 core 16 sq mm copper cable

K= (Rcosφ + Xsinφ)/1000Where R=Resistance per km=2.33Ω X=Reactance per km=0.1111Ω Cosφ=power factor=0.9

So, K= (2.33 * 0.9 + 0.1111 * 0.436)/1000=0.002146

T=no of triplets=2

From given figureLF=200 L=50

Voltages drop in phase conductor:

ΔVph= IK [ T (LF +2L) + 3L∑(T-i) ] (Where i ranges from 1 to T-1)

ΔVph=4.1235 * 0.002146 [2(200+2*50) + 3*50 ( 2-1) ] ΔVph=6.6367 V

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Voltage drop in neutral

ΔVn = 2xTxIxLxK =2 * 2 * 4.1235 * 50 * 0.002146 ΔVn=1.769 V

Hence total voltage drop:

ΔV = ΔVph + ΔVn =6.6367+1.769

ΔV =8.406 V

Sol: Approximate Method:

Load centre = (L0 + L1 + L2 + L3 + L4 + L5)/6 = (200+250+300+350+400+450)/6 =325 m

Current = √ ((Total KW) 2 + (Total KVAR) 2) /√3 VL

= sqrt ((6*888.84)2 + (6*430.84)2)

=8.24 A

So voltagde drop:

= 8.24 * (2.33 * 0.9 + 0.1111 * 0.44) * 325/1000

=9.96V

Calculation Result:

Voltage drop (approximate method) is 9.96V

Precise calculation method is 8.406 V.

4.2 Case 2:

To Calculate Voltage drop at last fixture for ten numbers of Triplets. This means there are 30 poles in one 3 phase feeders.

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Input data:

Current per phase per fixture, I=power/(voltage per phase*efficiency*power factor)

=150/ ((415/1.73)*0.9*0.9) =0.77289 A Cable size 4 core 16 sq mm copper cable

K= (Rcosφ + Xsinφ)/1000 Where R=Resistance per km=2.33Ω X=Reactance per km=0.1111Ω

So K=0.002146

LF=250mL=35mT=no of triplets =10

So phase voltage drop:

ΔVph= IK [T (LF +2L) + 3L∑ (T-i)]

=0.77289 * 0.002146 * [10(250+2*35) + 3*35 (9+8+7+6+5+4+3+2+1) ] ΔVph =13.144 V

Voltage Drop in NeutralΔVn=2xTxIxLxK =2 * 10 * 0.77289 * 35 * 0.002146

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ΔVn =1.16 V

Total Voltage drop

ΔV=ΔVph + ΔVn =13.144+1.16

ΔV=14.304 V

5.0 CONCLUSION:

An attempt has been made in this report to calculate the voltage drop occurring bat the last pole, where the voltage drop will be maximum .Two methods have been used:.i.)approximate and ii.precise method and the results are compared. Voltage drop comes out to be less when using precise method. Approximate method is hence more conservative.

Also formula for precise method are not lengthy and can be easily implemented in excel sheet. So use of precise method instead of approximate calculation one can obtain more precise value of voltage drop.

This method will be hence beneficial to address on issues where voltage drop stipulation is critical.

For calculation , excel pre- programmed file is attached here with .The user has to enter the basic data as asked in the program for calculation of voltage drop.

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