## 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%**

Ramprabha

Thank you sir for giving this information. Plz clear. In load factor it’s mentioned 8760. How, what constant.

RAJESH JAIN

Two Distribution transformers of 630 KVA are installed in a residential group housing society. In summer they have 60% loading and in winter they have only 30-35% loading.I want to know in winter if all load will run at one Transformer and one transformer take in idle position , how much losses we can save. LT cables will be same ,only one Transformer will be in idle position for winter period.

sathish

hi friends any buddy have IEC and other standard codes and definition

WILIBALDO TORO

Dear Sir, do you have this article corrected? the question is because I have the same doubts with respect the ecuations and his results.

thanks in advance

Jeffery B. Ballah

Dear Sir. Thanks for your opportunity provided me to read EEP materials on line. I’m praying that one day I can get a job that will enable me subscribe as a member so that I may be able to download and read these materials daily to improve my skills as a student reading Electrical Engineering in a war ravish country Liberia where there is no research lab and internets But thanks to EEP I’m fighting hard to be myself in this my chosen career.

Best regards.

Saif ur Rehman

Dear sir

I want to know that, how to include line losses in grid extension feasibility study please reply

Hafidz

Sir, what refrence for Book of this formula ? is there from iec,ieee,or etc?

vicku

1st of all tell me about max’m Ampere i.e u have taken max’m amp as 12amps , I don’t understand that that max’m ampere is of Line Conductor or Transformer..

how max’m Ampere Will be Calculated if NOT GIVEN.

If someone Know the Answer Than Kindly Reply me… at k.vivek78@live.com

Aliyu Mustapha Rufau

Thanks for the calculations

Temitope

Sir, your materials seem helpful. In your calculation of HT line losses where did the constants 0.105 from and it later transformed to 1.05. please am not too clear on this. what is the name of the constant? or is the constant meant feeder. and if its for 33kv feeder, how what will be the constant

ashoka r

sir i have very lack of knowledge about reactve power so kindly i request u to send some basic details regarding i hope u do the same.

ashoka r

dear sir i have very lack of knowledge about reactive power (kvar) so pls teach me about it ,

i am looking forward for ur valuable reply

Aye San Dar

Dear Sir,

I am very glad to have a chance sharing your knowledge.

let me know one thing from your technical losses calculation that what is the meaning of No of TC on the feeder??

I am looking forward for your valuable reply.

Ashit Kumar Ray

Sir,i want to know the loss calculation in replacement of ab cable in distribution transformer sub station.Plz give an example with % of loss.

thanking you,

Ashit Kumar Ray,CESU,Odisha.

ulaganathan

Sir

I m working as a electrical engineer in transmission line field about 33kv and 220 kv.I have a doubt about line loss per km at both 33&220kv line and how to calculate them.I need your valuable prescription.

Sumit

There seems to be many calculation errors or formula errors. Also please provide clear definition of all the constants. I am very confused.

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