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# How to calculate voltage regulation of distribution line

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

## 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 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

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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
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%. ### 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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Get access to premium HV/MV/LV technical articles, electrical engineering guides, research studies and much more! It helps you to shape up your technical skills in your everyday life as an electrical engineer. ### Jignesh Parmar

Jignesh Parmar has completed M.Tech (Power System Control), B.E (Electrical). He is member of Institution of Engineers (MIE), India. He has more than 20 years experience in transmission & distribution-energy theft detection and maintenance electrical projects.

1. Bukar
May 25, 2021

I want to design a 33kV single circuit radiating using 100mm AAC from a 60MVA 132/33kV to a distance of 140km.
I have the problem of controlling the Voltage Regulation of this 33kV line.

2. ozi
Dec 02, 2020

how do I determine the capacity of voltage regulator to use on the secondary side [11kV] of a transformer which is 15MVA capacity

3. Ajay gothwal
Sep 15, 2020

Sir what is VR constant of dog conductor

4. Mokhoele Mahao
Jan 07, 2020

Good day,

1. If I am pulling the 5600 MVA @ 0.98 PF on 11kV system and the distance from main sub to my MV sub is around 1.5KM what size of voltage regulators do I need to install to balance my incoming voltage to the plant

5. Ghazanfar Ali Khan
Aug 01, 2017

How 3rd harmonics are mitigated in tertiary winding, delta connected.
The answer is pending since 2015. Why no body has turned back with a comprehensive reply,including Jignesh Parmar.

6. NP Singh
Feb 08, 2017

One 11 KV HT line feeder is required to provide power supply to the sub divided 85 Nos. industrial plots. Though these plots are not yet sold but considering the futuristic requirement of the power supply of the individual plot holder, an electrification scheme is required to be passed from DHBVN as it is technically not feasible to get power supply from the DHBVN. The total minimum load as per the norms of the dhbvn is 3915 kv, hence we need to apply for the sanctioning of the load. Could anyone suggest, how the demand factor / diversification factor is taken and what would be the load required for applying.
As per my calculation the load is : 3910 divided by 0.9 = 4344 kva
Demand factor would = 3910 x .625 = 2443.75 kva or 2.45 mva approx.
Request confirm whether the demand factor 0.625 will be divided or multiplied to get the total load?

7. Engr. Ghazanfar Ali Khan
Oct 01, 2016

How third harmonics are mitigated by tertiary winding?

8. Narendrasinh Desai
Aug 05, 2016

sir
i am working in Electricity Department Silvassa. recently i have made calculation as per above method which is as under . kindly guide me is it true?
% Voltage Regulation of Distribution Line
A
66/11KV Athal s/s end Load 250KVA=225KW at PF.-09

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) ATHAL FEEDER II
• Current Capacity of ACSR Conductor = 300Amp,
• Resistance = 0.2792Ω and Reactance = 0 Ω,
Permissible limit of % Voltage Regulation at Trail end is 5%.
Load at point A=250KVA
Taking P.F-0.9 of system factor
Load =250 X 0.9=225KW
Length of Load Point from Athal S/s=1.5KM=1500Mtr
Method- (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) =225 / 1.732x11000x0.99 = 13.12 Amp.
• Required No of conductor / Phase =13.12 / 300 =0.044 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 (0.272×0.99+0×0.141)x13.12) / 1×1000)x1500) = 9.21 Volt
• Receiving end Voltage at Point A = Sending end Volt-Voltage Drop= (11000-9.21) = 10990.79 Volt.
• % Voltage Regulation at Point A = ((Sending end Volt-Receiving end Volt) / Receiving end Volt) x100
• % Voltage Regulation at Point A = ((11000-10990.79) / 10990.79 )x100 = 0.084%
• % Voltage Regulation at Point A =0.084 %

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