## Content

- Introduction to voltage regulation
- Voltage Regulation for 11KV, 22KV, 33KV Overhead Line
- Permissible Voltage Regulation (As per REC)
- Voltage Regulation Values
- Required Size of Capacitor
- Optimum location of capacitors
- Voltage Rise due to Capacitor installation
- 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*.

### 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

### Permissible Voltage Regulation (As per REC)

| |||

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 |

### 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.

(-) 12.5% to (+) 10%.**Above 33kV**(-) 9.0% to (+) 6.0%.**Up to 33kV**(-) 6.0% to (+) 6.0%**Low voltage**

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.

### 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.

### Optimum location of capacitors

L = [1 – (KVARC / 2 KVARL) x (2n – 1)]

**Where:**

* L* – distance in per unit along the line from sub-station.

*– Size of capacitor bank*

**KVARC***– KVAR loading of line*

**KVARL***– 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.*

**n**### Voltage Rise due to Capacitor installation:

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

**Where:**

* KVAR (Cap)* – Capacitor KVAR

*– Reactance per phase*

**X***– Length of Line (mile)*

**L***– Phase to phase voltage in kilovolts*

**V**## 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.

### 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.

Engr. Ghazanfar Ali Khan

How third harmonics are mitigated by tertiary winding?

Narendrasinh Desai

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 %

ssrao

sir we facing probem of 33kv line fault at mtabu clouds before monsoon flesh over the linearc devlop betwwen conductor gi pin of pin insulator

Mohamed A Farah

What is the longest distance an 11KV line can cover before you upgrade to 33KV line.

We are using 11KV line for the inner city. We are trying to move our generation machinery to a site that is 7 kms away from the city. Some of our staff are saying we should upgrade to a 33KV line for the distance. Is that acceptable. Or should we stick to the 11KV line we use for our inner city distribution.

In other words, does it mean a higher resistance for the 11KV line to justify the 33KV line.

Thanks;

Mohamed A Farah

Sompower Co.

IRSHAD

how can i calculate 22KV cable voltage drop ??

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 %

danial fazilatpoor

Dear sir

I have question . Why we choose 5 percent for voltage regulation , I mean i khow there is a standard but the question is what is the idea behind this standard.

thanks a lot.

Girish

Dear Sir,

Could you please share the method/formulae for the voltage drop calculation for overhead LT line feeders.

SK Farid

How its possible Voltage drop proper calculation?

Sushil Kumar

Dear Sir, i want to calculate the voltage drop of an overhead line using DOG Conductor of length around 3000 meter and load around 400 KW.

When i m calculating the voltage drop per line then it is over permissible (5%).

Total load of Line = 400 KW

Total length of line = 3000 meter

Transformer capacity = 630 KVA.

Pls guide me for make it under permissible.

Lombard

I would like to pull power from a 33kv line to my camp 17km away. I am looking at a installing a 200Kva Transformer, 3 phase, 380V

1/ What size CROSS SECTION mm of AAC Aluminium Overhead cable would I require?

2/ Would I be able to connect other transformers upto 400Kva, in Total?

Pulin Panchal

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

Pankaj Chavan

Sir,

LT CABLE VOL.DROP FORMULA IS AS BELOW.

% V.D = 1.732 × LOAD CURRENT × LENGTH × ( CABLE RESISTANCE × COS PHI + CABLE REACTANCE × SIN PHI ) ×100 ÷ NO OF RUNS × 415 × 1000

If any other queries in above formula contact me on above mail.

mohamed ASIM

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…

Dimensioning of Power Distribution Systems (2) | EEP

[…] 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. […]

MANISH

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

Edvard

At what voltage level?

joboyouk

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.

However, there’s still much for me to cover just keep up the good work.

jobo3889@gmail.com

amendozc

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

gaurav rana

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

GAURAV SINGH RANA

gsrana89025@gmail.com

TSRG

Dear Sir,

It is more useful datas.

Many more thanks & grateful to you.