## Introduction to Losses

**There are two types of losses in transmission and distribution line.**

- Technical Losses and
- Commercial Losses.

It’s necessary to calculate technical and commercial losses. Normally technical losses and commercial losses are calculated separately.

**electrical tariff**, but commercial losses are not implemented to all consumers.

Technical losses of the distribution line mostly depend upon electrical load, type and size of conductor, length of line etc.

**Let’s try to calculate technical losses of one of following 11 KV distribution line **

### Example – 11 kV Distribution Line

**11 KV distribution line have following parameters:**

- Main length of 11 KV line is
**6.18 km**. **Total number of distribution transformer on feeder:**

25 KVA = 3 No.

63 KVA = 3 No.

100 KVA = 1 No.**25 KVA transformer:**

- Iron losses = 100 W

- Copper losses = 720 W

- Average LT line loss = 63W**63KVA transformer:**

- Iron losses = 200 W

- Copper losses = 1300 W

- Average LT line loss = 260W**100 KVA transformer:**

- Iron losses = 290 W

- Copper losses = 1850 W

- LT line loss = 1380W- Maximum amp is
**12 Amps.** - Unit sent out during to feeder is
**490335 Kwh** - Unit sold out during from feeder is
**353592 Kwh** - Normative load diversity factor for
**urban feeder is 1.5**and for**rural feeder is 2.0**

## Calculation

#### Total connected load = No’s of connected transformers

Total connected load = (25×3) + (63×3) + (100×1) = **364 KVA**

#### Peak load = 1.732 x Line voltage x Max. amp

Peak load = 264 / 1.732 x 11 x 12 = **228**

#### Diversity factor (DF) = Connected load (in KVA) / Peak load

Diversity factor (DF) = 364 /228 = **1.15**

#### Load factor (LF) =

Unit sent out (in Kwh) / 1.732 x Line voltage x Max. amp. x P.F. x 8760

Load factor (LF) = 490335 / 1.732 x 11 x 12 x 0.8 × 8760 = **0.3060**

#### Loss load factor (LLF) = (0.8 x LF x LF)+ (0.2 x LF)

Loss load factor (LLF) = (0.8 x 0.3060 x 0.3060) + (0.2 x 0.306) = **0.1361**

### Calculation of iron losses

**Total annual iron loss in KWh = **Iron losses in Watts x Nos of TC on the feeder x 8760 / 1000

**Total annual iron loss (25 KVA TC) =**

100 x 3 x 8760 / 1000 = **2628 KWh**

**Total annual iron loss (63 KVA TC) =**

200 x 3 x 8760 / 1000 = **5256 K kWh**

**Total annual iron loss (100 KVA TC) =**

290 x 3 x 8760 / 1000 = **2540 KWh**

**Total annual iron loss =**

2628 + 5256 + 2540 = **10424 KWh**

### Calculation of copper losses

**Total annual copper loss in KWh = **Cu Loss in Watts x Nos of TC on the feeder LFX LF X8760 / 1000

**Total annual copper loss (25 KVA TC) =**

720 x 3 x 0.3 × 0.3 × 8760 / 1000 = **1771 KWh**

**Total annual copper loss (63 KVA TC) =**

1300 x 3 x 0.3 × 0.3 × 8760 / 1000 = **3199 KWh**

**Total annual copper loss (100 KVA TC) =**

1850 x 1 x 0.3 × 0.3 × 8760 / 1000 = **1458 KWh**

**Total annual copper loss =**

1771 + 3199 + 1458 = **6490 KWh**

#### HT line losses (Kwh) =

0.105 x (conn. load x 2) x Length x Resistance x LLF / (LDF x DF x DF x 2)

**HT line losses** = 1.05 x (265 × 2) x 6.18 x 0.54 x 0.1361 /1.5 x 1.15 x 1.15 x 2 = **831 KWh**

#### Peak power losses =

(3 x Total LT line losses) / (PPL x DF x DF x 1000)

**Peak power losses** = 3 x (3 × 63 + 3 × 260 + 1 × 1380) /1.15 x 1.15 x 1000 = **3.0**

#### LT Line losses (KWh) = (PPL) x (LLF) x 8760

**LT Line losses** =** **3 x 0.1361 x 8760 =** ****3315 KWh**

#### Total technical losses =

(HT Line losses + LT Line losses + Annual Cu losses + Annual iron losses)

**Total technical losses** = (831+ 3315 + 10424 + 6490) =** 21061 KWh**

#### % Technical loss = (Total losses) / (Unit sent out annually) x 100

**% Technical loss** = (21061 / 490335) x 100 =** 4.30%**

**% Technical Loss = 4.30%**

Dharmin joshi

how to find out line current for 33kv overhead transmission line for 3Km? if my power plant is 2mw.

pl mail me

Rajasekar

Sir,

What is 0.105 constant in line loss calculation and while calculating it shows as 1.05 and also the loss while calculating it is not 831 KWh How?

please explain it

zaffar

264÷1.732x11x12=228 how

gaurav kumar goswami

hi ur providing a free and very helpfull guidline for elect engg …thanx for that sir

yeanling

364 /228 = 1.15 ? not 1.6 ?

and

HT line losses = 1.05 x (265 × 2) x 6.18 x 0.54 x 0.1361 /1.5 x 1.15 x 1.15 x 2 = 831 KWh

265× 2, where it is from?

ankit

if cable size is more than actual cable (depend on rated current) so what is loss?

please give answer in technical and commercial loss.

Ali BY

I’m working on a project to reduce the losses in distribution lines of the city of Casablanca (Morocco). I need the methodology used in this example to perform the calculations. Thanks to send me the required elements. (aliby25@gmail.com)

Ayoyinka Muritala

Please, kindly assist and furnish me with the extensive list of test adopted universally performed on Power Transformers.

Moh'd alqasem

Thank you