An example of calculating the technical losses of T&D lines
An example of calculating the technical losses of T&D lines (photo credit: paulhasenmeier.wordpress.com)

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%

About Author //

author-pic

Jignesh Parmar

jiguparmar - Jignesh Parmar has completed M.Tech (Power System Control) ,B.E(Electrical). He is member of Institution of Engineers (MIE),India. Membership No:M-1473586.He has more than 13 years experience in Transmission -Distribution-Electrical Energy theft detection-Electrical Maintenance-Electrical Projects (Planning-Designing-Technical Review-coordination -Execution). He is Presently associate with one of the leading business group as a Deputy Manager at Ahmedabad,India. He has published numbers of Technical Articles in "Electrical Mirror", "Electrical India", "Lighting India", "Industrial Electrix"(Australian Power Publications) Magazines. He is Freelancer Programmer of Advance Excel and design useful Excel base Electrical Programs as per IS, NEC, IEC,IEEE codes. He is Technical Blogger and Familiar with English, Hindi, Gujarati, French languages. He wants to Share his experience & Knowledge and help technical enthusiasts to find suitable solutions and updating themselves on various Engineering Topics.

18 Comments


  1. 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 k.vivek78@live.com


  2. Aliyu Mustapha Rufau
    Aug 14, 2015

    Thanks for the calculations


  3. 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


  4. ashoka r
    May 13, 2015

    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.


  5. ashoka r
    May 13, 2015

    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


  6. Aye San Dar
    Mar 08, 2015

    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.


  7. Ashit Kumar Ray
    Mar 07, 2015

    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.


  8. ulaganathan
    Feb 24, 2015

    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.


  9. Sumit
    Feb 04, 2015

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


  10. Dharmin joshi
    Jan 07, 2015

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


  11. Rajasekar
    Oct 13, 2014

    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


  12. zaffar
    Oct 11, 2014

    264÷1.732x11x12=228 how


  13. gaurav kumar goswami
    Aug 06, 2014

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


  14. yeanling
    Jun 11, 2014

    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?


  15. ankit
    Apr 30, 2014

    if cable size is more than actual cable (depend on rated current) so what is loss?
    please give answer in technical and commercial loss.


  16. Ali BY
    Apr 28, 2014

    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)


  17. Ayoyinka Muritala
    Apr 25, 2014

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


  18. Moh'd alqasem
    Apr 25, 2014

    Thank you

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