## Voltage regulation

- 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.

Dear Sir,

In your above example you have taken following considerations

1. ACSR DOG conductor

2. Resistance=0.2792 Ohm

3. Reactance= 0 Ohm

4. Line current at load point-A=98 Amp

5. Cos Ø=0.8

6. Sin Ø=0.6

I want to calculate voltage drop at point-a with following parameters

1. ACSR DOG conductor

2. Resistance=0.2792 Ohm

3. Reactance= 0 Ohm

4. Line current at load point-A =98 Amp

5. Cos Ø=0

6. Sin Ø=1

In this case

Voltage drop=98 x (0.2792×0 + 0x1) =0 Volt

Or

If Cos Ø=0.01 then voltage drop with 98 Amp load would be very small

I am confused, kindly guide pl.

I want to know the voltage regulation for a short 33 kV line with the following parameters

Power 5 MW

line length 14 km

Transmission Voltage 33 kv

Conductor Panther

I worked out % regulation with formula (Vs-Vr/Vr.

Vr=Vs-losses at full load current.

Voltage Regulation = 1.4 %, kindly look it if correct

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.

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

Sir what is VR constant of dog conductor

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

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.

Hello Ghazanfar Ali Khan, you posted this message back in 2017, are you still waiting for an answer from a question posted in 2015? Have you come to the conclusion that you are not going to get an answer here? Its ok, have a cookie and everything will be fine.

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?

How third harmonics are mitigated by tertiary winding?

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 %

correct he