How to calculate voltage regulation of distribution line

How to calculate voltage regulation of distribution line (on photo: Distribution Lines - Oaxaca, Mexico, 2013 via FlickR)

Content

  1. Introduction to voltage regulation
  2. Voltage Regulation for 11KV, 22KV, 33KV Overhead Line
  3. Permissible Voltage Regulation (As per REC)
  4. Voltage Regulation Values
  5. Required Size of Capacitor
  6. Optimum location of capacitors
  7. Voltage Rise due to Capacitor installation
  8. Calculate % Voltage Regulation of Distribution Line

Introduction to voltage regulation

Voltage (load) regulation 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.

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Voltage Regulation for 11KV, 22KV, 33KV Overhead Line

% Voltage Regulation = (1.06 x P x L x PF) / (LDF x RC x DF)

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 = (KV x KV x 10) / ( 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

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Permissible Voltage Regulation (As per REC)

Maximum  Voltage Regulation at any Point of Distribution Line

Part of Distribution SystemUrban Area (%)Suburban Area (%)Rural Area (%)
Up to Transformer2.52.52.5
Up to Secondary  Main320.0
Up to Service Drop0.50.50.5
Total6.05.03.0

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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 11/0.433 kV distribution transformers(in place of normal 11/0.4 kV DT’s) may be used in these areas.

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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.

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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.

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Voltage Rise due to Capacitor installation:

% Voltage Rise = (KVAR(Cap) x Lx X) / 10 x Vx2

Where:

KVAR (Cap) – Capacitor KVAR
X – Reactance per phase
L – Length of Line (mile)
V – Phase to phase voltage in kilovolts

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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 Reactance = 0 Ω,

Permissible limit of % Voltage Regulation at Trail end is 5%.

Calculate percentage of Voltage Regulation of Distribution Line


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.

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Method-2 (Load Base)

% Voltage Regulation =(I x (RcosǾ+XsinǾ)x Length ) / No of Cond.per Phase xV (P-N))x100

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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author-pic

jiguparmar - Jignesh Parmar has completed his B.E(Electrical) from Gujarat University. He is member of Institution of Engineers (MIE),India. Membership No:M-1473586.He has more than 12 years experience in Transmission -Distribution-Electrical Energy theft detection-Electrical Maintenance-Electrical Projects (Planning-Designing-Technical Review-coordination -Execution). He is Presently associate with one of the leading business group as a Assistant Manager at Ahmedabad,India. He has published numbers of Technical Articles in "Electrical Mirror", "Electrical India", "Lighting India", "Industrial Electrix"(Australian Power Publications) Magazines. He is Freelancer Programmer of Advance Excel and design useful Excel base Electrical Programs as per IS, NEC, IEC,IEEE codes. He is Technical Blogger and Familiar with English, Hindi, Gujarati, French languages. He wants to Share his experience & Knowledge and help technical enthusiasts to find suitable solutions and updating themselves on various Engineering Topics.



9 Comments


  1. Pulin Panchal
    May 01, 2014

    Can you give calculation of Voltage Drop of LV system having 2 or 3 run Parallel Power Cable connection ?


  2. mohamed ASIM
    Feb 08, 2014

    Dear sir,

    In calculating voltage drop at the point B, the load current is 118 Amps,(in your example).But while calculating the voltage drop at the point B,you have taken the current value of point A( 98Amps). is that correct..can you explain…

  3. […] The maximum permissible load current Iz of the selected transmission medium (cable or busbar) must be above the conventional tripping current I2-/1.45 of the selected device. […]


  4. MANISH
    Nov 02, 2013

    what is maximum loading in 100 KVA and 63 KVA DTR in HP?


  5. joboyouk
    Oct 16, 2013

    I just presently left my banking job to return to field of engineering. Your work on this site has been a tremendous refresher for me and even much more. I just want to say thank you for being part of those helping my confidence of finding a place in the field.
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  6. amendozc
    Oct 07, 2013

    Buenos días Dr Jignesh Parmar

    Es usted muy gentil con su publicación. Sin embargo, agradezco por favor resolverme las siguientes inquietudes:

    1. Indicarme la razón por la cual, en los ejemplos de regulación, no se acumuló en la carga A, la carga B y C, teniendo en cuenta que se debería acumular el momento eléctrico al momento de cálculo.

    2. Indicarme por favor bibliografía adicional (normatividad) para este tipo de cálculo.

    Gracias Dr


  7. gaurav rana
    Aug 27, 2013

    dear sir!
    with due to respect i want to say that you done a fantastic job on this portal and its very helpful to all engineers,technician and all people who related to electrical engineering field.
    at last my one request to you that give me more info about the interconnecting transformer ICT using auto transformer, please!
    rest is fine and thanking you

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  8. TSRG
    Jun 11, 2013

    Dear Sir,
    It is more useful datas.
    Many more thanks & grateful to you.

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