Calculate Cable Voltage Drop for Street Light Pole
February 8, 2016 26 Comments
Example: Calculate Voltage drop of Cable for Street Light Pole. System Voltage is 230V (P-N), Power Factor=0.75. Allowable Voltage Drop = 4% .The Detail of Pole & cable are
Pole Detail:
- Section feeder Pillar is 50 meter away from Pole-1
- Distance between each Pole is 50 Meter Distance
- Luminar of Each Pole Fitting = 2 No’s
- Luminar Watt =250 Watt
Cable Detail:
- Size of Cable= 4CX10 Sq.mm.
- First Pole is connected in R Phase Next Pole is connected in Y Phase Than Next Pole is connected in B Phase. Next Pole is connected again R Phase.
- Resistance of Cable=3.7 Ω/Km
- Reactant of Cable=0.1 Ω/Km
Calculation:
Load of Each Pole
- Load of Each Pole = (Watt of Each Luminar X No of Luminar ) / Volt X P.F
- Load of Each Pole = (250X2) /(230X0.75)
- Load of Each Pole = 2.9 Amp
For Pole Pole-1:
- Pole Connected on “R” Phase
- Total Distance of Pole for “R” Phase =50 Meter ,
- % Voltage drop of Cable= (Current X (Rcosᴓ + JSinnᴓ) X Distance ) / (Volt X No of Cable X 1000)
- % Voltage drop of Cable= (2.9x(3.7×0.75+0.1×0.66)X50 / (230x1x1000)
- % Voltage drop of Cable= 0.18% ———————————(1)
For Pole Pole-2:
- Pole Connected on “Y” Phase
- Total Distance of Pole for “Y” Phase =50+50=100 Meter ,
- % Voltage drop of Cable= (Current X (Rcosᴓ + JSinnᴓ) X Distance ) / (Volt X No of Cable X 1000)
- % Voltage drop of Cable= (2.9x(3.7×0.75+0.1×0.66)X100 / (230x1x1000)
- % Voltage drop of Cable= 0.36% ———————————(2)
For Pole Pole-3:
- Pole Connected on “B” Phase
- Total Distance of Pole for “B” Phase =50+50+50=150 Meter ,
- % Voltage drop of Cable= (Current X (Rcosᴓ + JSinnᴓ) X Distance ) / (Volt X No of Cable X 1000)
- % Voltage drop of Cable= (2.9x(3.7×0.75+0.1×0.66)X150 / (230x1x1000)
- % Voltage drop of Cable= 0.54% ———————————(3)
For Pole Pole-4:
- Pole Connected on “R” Phase
- Total Distance of Pole for “R” Phase =150+50=200 Meter ,
- % Voltage drop of Cable= (Current X (Rcosᴓ + JSinnᴓ) X Distance ) / (Volt X No of Cable X 1000)
- % Voltage drop of Cable= (2.9x(3.7×0.75+0.1×0.66)X200 / (230x1x1000)
- % Voltage drop of Cable= 0.72% ———————————(4)
For Pole Pole-5:
- Pole Connected on “Y” Phase
- Total Distance of Pole for “Y” Phase =200+50=250 Meter ,
- % Voltage drop of Cable= (Current X (Rcosᴓ + JSinnᴓ) X Distance ) / (Volt X No of Cable X 1000)
- % Voltage drop of Cable= (2.9x(3.7×0.75+0.1×0.66)X250 / (230x1x1000)
- % Voltage drop of Cable= 0.9% ———————————(5)
For Pole Pole-6:
- Pole Connected on “B” Phase
- Total Distance of Pole for “B” Phase =250+50=300 Meter ,
- % Voltage drop of Cable= (Current X (Rcosᴓ + JSinnᴓ) X Distance ) / (Volt X No of Cable X 1000)
- % Voltage drop of Cable= (2.9x(3.7×0.75+0.1×0.66)X300 / (230x1x1000)
- % Voltage drop of Cable= 1.07% ———————————(6)
For Pole Pole-7:
- Pole Connected on “R” Phase
- Total Distance of Pole for “R” Phase =300+50=350 Meter ,
- % Voltage drop of Cable= (Current X (Rcosᴓ + JSinnᴓ) X Distance ) / (Volt X No of Cable X 1000)
- % Voltage drop of Cable= (2.9x(3.7×0.75+0.1×0.66)X350 / (230x1x1000)
- % Voltage drop of Cable= 1.25% ———————————(7)
For Pole Pole-8:
- Pole Connected on “Y” Phase
- Total Distance of Pole for “Y” Phase =350+50=400 Meter ,
- % Voltage drop of Cable= (Current X (Rcosᴓ + JSinnᴓ) X Distance ) / (Volt X No of Cable X 1000)
- % Voltage drop of Cable= (2.9x(3.7×0.75+0.1×0.66)X400 / (230x1x1000)
- % Voltage drop of Cable= 1.43% ———————————(8)
For Pole Pole-9:
- Pole Connected on “B” Phase
- Total Distance of Pole for “B” Phase =400+50=450 Meter ,
- % Voltage drop of Cable= (Current X (Rcosᴓ + JSinnᴓ) X Distance ) / (Volt X No of Cable X 1000)
- % Voltage drop of Cable= (2.9x(3.7×0.75+0.1×0.66)X450 / (230x1x1000)
- % Voltage drop of Cable= 1.61% ———————————(9)
Total Voltage Drop:
- Voltage Drop in “R” Phase = 0.18+0.72+1.25 =2.15 %
- Voltage Drop in “Y” Phase =0.36+0.90+1.43 =2.69 %
- Voltage Drop in “B” Phase =0.54+1.07+1.61 =3.22 %
- % Voltage drop in each Phase is Max 3.22% Which is less than 4%
Results:
| Phase | No of Pole | Load (Amp) | Voltage Drop |
| R | 3 | 9 | 2.15 % |
| Y | 3 | 9 | 2.69 % |
| B | 3 | 9 | 3.22 % |
| Total | 9 | 9 | 2.55 % |
JIGNESH PARMAR
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Jignesh.Parmar posted: “Example: Calculate Voltage drop of Cable for Street Light Pole. System Voltage is 230V (P-N), Power Factor=0.75. Allowable Voltage Drop = 4% .The Detail of Pole & cable are
Pole Detail:
Section feeder Pillar is 50 meter away from Pole-1 Distance b”
what size of cable can i use to connect 30 street lights on 6m pole .
the street light are 40watts led and 18m between poles.
the total length is 540m
dear engineer Jignesh.Parmar
I really appreciate that so much knowledge and experience you gave for all the follower
only i have 1 confuse about the equation of ( r cos phi + j xsin phi ) actually i found some website use the direct summation not vectorial i mean they use rcos phi + xsin phi
instead of sqrt ( (rcos phi) ^2 + ( x sin phi ^2) even the values you used in the example is like that
is it an approximation or what because I think the right is using the vectorial formula
pls advice
best regards
ayman zaher
As more and more number of poles gets connected to each phase, load on the previous sections of the cable also will increase…. it is not reflected here…only cable length increase is noted.
Dear Sir the value of current that you have used for calculating voltage drop is the current that an individual fixture required but before first pole the value of current should be 2.9*3=8.7 amps. Correct me if I’m wrong pls.
Thank you
Your question is right but think again…this is a calculation of voltage drop at each individual pole due to the fixtures on that particular pole and then we sum up all of them. Your point is valid if we were using a single core cable.
Dear Jignesh sir
what should be the resistance & capacitance between a motor to earth. how it is measured
calculation is not perfect please specify at the end how did you calculate because at end it comes 411.945/230000=0.0017910652 % how i find it 18% for 1st pole pleasereply
To find % Multiply with 100.
You will have to multiply with 100 in order to get the % drop. It will become 0.179. Can be rounded off to 0.18.
Calculation is not perfect please specify at the end how did you calculate because at end it comes 411.945/230000=0.0017910652 % how i find it 18% for 1st pole please reply
Referring to %VD for the street lighting, 2.9A current per pole is a single phase current. In this case, you have to multiply the %VD by 2 to comply with single phase VD formula.
Commented by:
Engr. B. M. Rashad
Electrical Section Head
B.Sc.pk., B.Sc.Engg.phil., M.Sc.Engg.phil.
Sr. Electrical Design Engineer
for Building Services & Infrastructure Networks (23 years of experience with US Army Corps of Engineers Design Consultants and European Design Consultants
Contact email: mr@adeng.com.sa
Hello mr B? M.Rashed
Mo current suposed to pass in neutral because its 3 phase balanced load up to the fuse box of the lighting pole
JUN.19, 2016
Dear Engr. Ayman Zaher
It is understood that for 3-phase balanced circuits, the neutral current turns out to be zero, but the issue is not that at all. Since your %VD is based on single phase, so going back to IEC-60364, you have to use single phase formula for calculating the single phase %VD. So, do revise your single phase %VD calculation accordingly to meet with the code requirements of IEC-60364 for the 1-phase %VD of street lighting poles. You cannot change the 1-phase %VD formula.
Awaiting for your feedback.
Engr. BMR
your are right. if single phase so multiply by 2. if 3 phase so multiply by sqrt(3)
JUNE 05, 2016
Dear Engr. Jignesh,
I am a Sr. Electrical Design Engineer. I am into the design of building services and infrastructure networks having 23 years of electrical design experience with international foreign design consultants. The design is based on CIE, IESNA, NFPA, IEEE, IEC standards and codes.
I have gone through your calculations for %VD and my comment is that you have to multiply the %VD by 2 based on VD formula for single phase.
Awaiting for your feedback.
Regards
Engr. B. M. Rashad
Electrical Section Head
B.Sc.(pk.), B.Sc.Engg.(phil.), M.Sc.Engg.(phil.)
Sr. Electrical Design Engineer
for Building Services and Infrastructure Networks
Contact email: mr@adeng.com.sa
thanks for nice reply
no one can change the formula of voltage drop calculation at single phase
BUT IF IT REALLY THE CASE FOR SINGLE PHASE
why vd for single phase is 2*I *L (RCOS PHI + X SIN PHI ) ?
BECAUSE THE CURRENT GO IN LINE AND RETURN IN NEUTRAL SO THE TOTAL PATH IS DOUBLE LENGTH AND WE USE 2 ,
NOTHE THIS IS NOT THE CASE WE DISCUSS
SUPPOSE WE HAVE BALANCED 3 PHASE (NO AMPERS IN NEUTRAL )
THE PATH WILL ONLY BE LINE AND VOLTAGE DROP WILL BE I*L (RCOS PHI + X SIN PHI )
FOR 1 PHASE AS PHASE VALUE
FOR LINE VALUE VOLTAGE DROP = SQRT3 * I * L ( R COS PHI + X SIN PHI)
AND AS YOU CAN NOTICE THIS IS THE FORMULA FOR THREE PHASE ALSO
SO THERE IS NO CONTRADICTION THE CASE OF SINGLE PHASE IN OUR EXAMPLE IS NOT AS IT APPEARS SINGLE PHASE ITS ACTUALLY HALF SINGLE PHASE
YOU CAN SEE THEY ARE IDENTICAL AND THAT IS THE LOGIC
AMAAZ1972@GMAIL.COM
SENIOR ELECTRICAL ENGINEER
KHATIB ALAMI CONSULTANTS
Dear Engr. Ayman Zaher
Pls. refer to the attached VD calculation sheets for your info. and further discussions, if any.
Regards
Engr. BMR
Sir, we are planning to replace over head LT line with UG cables for a length of 2.5KM perimeter lights. can you guide me to estimate the cable sizes and other details
Fyi,Three-phase system with neutral point completely unbalanced is considered single-phase system…thanks.
Dear Sir,
How to calculate the power Losses in Transmission Line. For Ex. i am running 12MVA at 132KV transmission line for 25KM using ACSR Panther Conductor. What will be losses. Is there any Calculation
Going through a document, I discover that the cable used on a circuit for street light were varied; i.e 4 x 25mmsq, 4 x 16mmsq…. Is that standard? at what point do we have to varied the cable if it is a standard
dear Mr. Jingesh,
How to calculate voltage drop for each pole individually without considering the total current entering the cable for the remaining phase poles? regards
thank you very much for your effort here MR. Jignesh .
Dear Mr. Jignesh,
You are considering the 2.9A Load on pole 1 to calculate the voltage drop from feeder pillar to first pole. If you see that on this cable (Feeder Pillar to first pole), load will be come for all the 9 nos. poles.
Regards,
Gajendra Singh
Mob: 8130282606
Sir please help me