# Calculate Cable Voltage Drop for Street Light Pole

February 8, 2016 14 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 % |

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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”

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

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

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

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

Fyi,Three-phase system with neutral point completely unbalanced is considered single-phase system…thanks.