# Calculate Technical Losses of Transmission / Distribution Line:

March 1, 2014 18 Comments

## Introduction:

- There are two types of Losses in transmission and distribution Line.
- (1) Technical Losses and
- (2) Commercial Losses.
- It is 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 have following parameter.
- Main length of 11 KV Line is 6.18 Kms.
- Total nos. of Distribution Transformer on Feeder 25 KVA= 3 No, 63 KVA =3 No,100KVA=1No.
- 25KVA 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.
- 100KVA Transformer Iron Losses = 290 W, Copper Losses= 1850 W, LT Line Loss= 1380W.
- Maximum Amp is12 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 Transformer.**

- Total Connected Load= (25×3) + (63×3) + (100×1).
- Total Connected Load=364 KVA.

** Peak Load = 1.732 x Line Voltage x Max Amp **

- Peak Load = 1.732x11x12
- Peak Load =228 KVA.

** Diversity Factor (DF) = Connected Load (In KVA) / Peak Load.**

- Diversity Factor (DF) = 364 /228
- Diversity Factor (DF) =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 / 1.732x11x12x0.8×8760
- Load Factor (LF)=0.3060

** Loss Load Factor (LLF)= (0.8 x LFx LF)+ (0.2 x LF) **

- Loss Load Factor (LLF)= ( 0.8 x 0.3060 x 0.3060 ) + (0.2 x 0.306)
- Loss Load Factor (LLF)= 0.1361

** Calculation of Iron losses:**

**Total Annual Iron loss in Kwh =Iron Loss in Watts X Nos of TC on the feeder****X8760 / 1000**- Total Annual Iron loss (25KVA TC)=100x3x8760 /1000 =2628 Kwh
- Total Annual Iron loss (63KVA TC)=200x3x8760 /1000 =5256 Kwh
- Total Annual Iron loss (100KVA TC)=290x3x8760 /1000 =2540 Kwh
- Total Annual Iron loss =2628+5256+2540 =10424Kwh

** Calculation of Copper losses:**

**Total Annual Copper loss in Kwh =Cu Loss in Watts XNos of TC on the feeder LFX LF****X8760 / 1000**- Total Annual Copper loss (25KVA TC)=720x3x0.3×0.3×8760 /1000 =1771 Kwh
- Total Annual Copper loss (63KVA TC)=1300x3x0.3×0.3×8760 /1000 =3199 Kwh
- Total Annual Copper loss (100KVA TC)=1850x1x0.3×0.3×8760 /1000 =1458 Kwh
- Total Annual Copper loss =1771+3199+1458=6490Kwh

** 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 x1.15 x 2
- HT Line Losses = 831 Kwh

** Peak Power Losses= (3 x Total LT Line Losses) / (PPLxDFxDFx 1000)**

- Peak Power Losses= 3 x (3×63+3×260+1×1380) /1.15 x 1.15 x 1000
- Peak Power Losses= 3.0

** LT Line Losses (Kwh)= (PPL.) x (LLF) x 8760 **

- LT Line Losses =
- LT Line Losses =

** Total Technical Losses= (HT Line Losses + LT Line Losses + Annual Cu Losses + Annual Iron Losses) **

- Total Technical Losses = ( 831+ 3315 + 10424 + 6490)
**Total Technical Losses = 21061 Kwh**

**% Technical Loss= (Total Losses) / (Unit Sent Out Annually) x 100 **

- % Technical Loss= (21061/490335) x100=
**4.30%**

excelente…gracias…

Total Connected Load=364 KVA.

Peak Load = 264 / 1.732x11x12

Peak Load =228

I think there is a mistake in the above if Total connected load =364KVA

But peak load is calculated with 264KVA?

please explain.

Thanks

Dear Joe,

It is typing mistake in peak load calculation

connected load is 25KVA 3nos TC+ 63KVA 3NO TC + 100KVA 1No TC =25×3=63×3=100=364KVA

and Peak Load is 1.732 x KV X Max.Amp=1.732x11x12 =228KVA

Thanks for highlighting…

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I would be glad to receive your affirmative response reply at your earliest possible convenience…

may i know the technical loss on a 16 kms 11 kv double circuit feeder having dog conductor with a load of 10 mw transmitted direct on a bus bar of the grid station.. no distribution transformer installed in between sending point to receiving end i-e bus bar..hoping a positive response from your side.

Calculation method is already explain in detail…

May I know the formula of finding HT losses (0.105 or 1.05).

Now I m studying line loss reduction in distribution line.

The above formulae are very useful for me.

Thanks for ur sharing !

Hello sir,

I want tranmission loss of 1 km of overhaed line for 11KV, 33KV and 132KV with detailed explaination

Yhank you.

Hello sir,

Please tell iron loss & copper loss in 200 kva,250kva & 500 kva transformer

thank you,

Sir,Calculate Technical Losses of Transmission / Distribution Line-HT & LT line losses are not matched with Formula’s,In my amply request you ,please explain step by step procedure calculation of T& D losses

Kindly share the formula of LDF, as per REC it is Load Distribution Factor and you have mentioned it as Load Diversity Factor. As per REC formula LDF is (PxL)/(kVA-km) based on sample calculations of two or three representative feeders. Kindly clarify what is kVA in denominator.

Dear Sir,

can u tell me how the formula is derived for HT and LT line losses?

For working out line losses, we can work out peak line losses in Kw/kM and then using the LLF ,line losses for the length of line (6.18 km) can be calculated. In this

, the peak loss work out to 3x 12 x12x 0.54 =0.23328 kW/KM. Then using LLF, the line losses work out as under:-

Annual energy loss/km 0.23328xLLFX8760

=0.23328X 0.1361X8760

=278.12 kWH

Annual energy loss for 6.18 km of line = 278.12×6.18

=1781.81kWHs

The HT Line losses worked out by you are 831. Please reconcile and comment.Formula used by you is not clear. On what basis factor 0.105 has been used . Please explain.

There is no justification for using 0.105

Kindly tell how to find average LT line losses of 200,400,500 and 630kVA transformer?

Hai sir my Query is why dont we use power factor for peak load calculation.& for resistance/Km For 2km we calculate HT-3ph(3-wire) =2*(3*R) or 2*(1*R).please clarify

Thank You