 Save 50% on all EEP Academy courses with Enterprise Membership Plan and study specialized LV/MV/HV technical articles & guides.

# An example of calculating the technical losses of T&D lines

Home / Technical Articles / An example of calculating the technical losses of T&D lines

## Introduction to Losses

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

1. Technical Losses and
2. Commercial Losses.

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

Transmission (technical) losses are directly effected on 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 KkWh

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%

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. shozab
Jun 11, 2018

Dear Mr. Jignesh,

It is nice meeting you, i have gone thru your blogs and find certain excel calculation sheet, actually i m looking for the excel tool for Insulation Coordination as per IEC, IEEE, IS.

Also Line loss calculation using UG cable at HV (33kV and 11kV)

You are requested to kindly post the blog on it electrical section or reply to me with returning email. It will be helpful. Alsot kindly let me the buying charges of these.

2. May 23, 2018

Thank you for your guide on calculating technical loss, however please I need clarification on 0.2 value used in loss load factor calculation and 0.105 used in calculating HT line losses(Kwh). Thank you

3. Rajesh kumar
Mar 05, 2018

Respected sir please provide the information of 8760

• Vatan Kuncolienkar
Oct 20, 2019

8760= Number of hours in a year viz.24×365

4. batuhan alkan
Dec 18, 2017

Ht line losses, peak powe losses and lt line losses calculation has to much calculation mistake can you explain how are you finding 831- 3.0 – 3315

5. Dhrumit
Jun 13, 2017

In Uganda my company get form Umeme 33kv. After line loss I get 28 -29 kg And my transformer set on step 5 last stape but I am not geting 415 volt. So what should I do. Please give me solution if you have.

6. Ramprabha
May 22, 2017

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

7. RAJESH JAIN
Jan 11, 2017

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.

8. sathish
Dec 03, 2016

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

9. WILIBALDO TORO
Oct 17, 2016

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

10. Jeffery B. Ballah
Jun 06, 2016

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.

11. Saif ur Rehman
Apr 08, 2016

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

12. Hafidz
Apr 03, 2016

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

13. vicku
Aug 18, 2015

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 [email protected]

14. Aliyu Mustapha Rufau
Aug 14, 2015

Thanks for the calculations

15. Temitope
Aug 10, 2015

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