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WE HAVE TWO TYPES OF LOSSES IN TRANSFORMER…..ie….iron loss and copper loss…one which will be constant for every loads…and the one which vary with bthe loading conditions respectively…..iron loss occurs all the time whether the transformer is loaded or unloaded and hence also called constant loss…and the other occurs only when it is loaded…and vary with the loads..and hence also called variable loss….an open circuit test in which rated voltage is applied to the primary winding and secondary winding is open circuited(test will be carried out in the low voltage side)is used for determining ironloss…..here the wattmeter gives the iron loss…bcos copper loss at no load conduction is negligible…..and the copper loss could be datemined by short circuit test in which rated current is made to flow through the primary winding and secondary winding will be short circuited….here the wattmeter reading gives full load copper loss…and to determine copper loss at x(FULL LOAD) the below formula can be used…

transformer losses

2011/05/26

12:21 pm

12:21 pm

Anil Sabaji PE CEng

Guest

Hi,

Look for Transformer Test report. It will have Iron losses which will remain almost constant through out the loading cycle. Then look for efficiency on the name plate. This corresponds to full load. Now calculate the full load losses. Example if 1000KVA transformer efficiency is 98% then the losses are 2% i.e. 20KW. If the Iron Losses is indicated as 2KW, then the copper losses are 18KW. Now the copper losses are proportional to square of current. Therefore at 50% load copperlosses will be 0.5 sqare times the full load copper loss. i.e. 0.25X18=4.5KW. Therefore at 50% load you will have 6.5KW losses. If you dont know what is the Iron losses of the transformer, then see the noload power consumption of the transformer, it will be very close to iron losses.

2011/05/26

4:30 pm

4:30 pm

kailash

Guest

anil i agree with you.

But i think to know the full load losses you have to go for open ckt test (i.e. iron losses) and short ckt test (i.e copper losses).then to know tfransformer loss at any load= iron loss + copper loss at that load.

e.g. copper loss at 50% load = ( 50% 0f full load current)2*R

=(I/2)2*R

=I2R/4

2011/06/01

8:48 am

8:48 am

Spir Georges GHALI

Member

Forum Posts: 204

Member Since:

2011/03/26

2011/03/26

Offline

**amit kulkarni said: **

hi can u help me @ transformer (3 phase) losses any formulae or thumb rules.if the transformer is 50 % loaded there are some losses but how to calculate? how we to calculate % of transformer loading?

Dear Amit ;

there are 2 kinds of losses : Iron Losse & Cupper losse, the iron loose is not affected by the load, but the cupper load si affected by the load.

Normally, on the name plate of any transformer there's a percentage value " i % " that's the percentage current of the nominal current for the iron losse, so, if we have it we can calculate the iron losse. and also on the name plate there is a value of cupper losse on full load in " watt ", by whici we can calculate the " R " of the transformer then accordingly the load's current and the value of R we can calculate the cupper losse, notint taht the formula by which we calculate the " R " of any transformer is : R = Pcu/3 . In2 ; where :

- Pcu : is the full load cupper losse

- In2 : is the square on the transformer nominal current

Regards.

2011/06/04

11:14 am

11:14 am

Mohsan islam

Member

Forum Posts: 4

Member Since:

2011/05/14

2011/05/14

Offline

**i want**

to calcute my home electricity load and also want to know illuminus calclations

**formuly any bdy hlp me plz?**

**Hello Dear mhrnaseer**,

You can measure the load current of your home using clamp

ammeter.

Multiply load current to standard voltages which are usually

220V,which results your home load.

Illumination is defined as “the emitted luminous flux per

unit area is called illumination”.

Mathematically

E=Φ/A (lux).

Illumination at any surface can be calculated by two laws:

1) inverse square law

2) Lambert’s law

1)inverse square law tells that the illumination of a

surface is inversely proportional to the square

of the distance of the surface

from the source.

i.e E «1/d^2

2) Lambert’s law tells that illumination is directly

proportional to cosine of the angle made by the normal to the illuminated

surface with the direction of the incident flux.

i.e E « I cosΘ where I is luminous intensity.

combining the both laws we get

E «(I/d^2)cosΘ

Or

E=(I/d^2)cosΘ

2011/06/11

1:44 am

1:44 am

akhila....

Guest

**amit kulkarni said: **

hi can u help me @ transformer (3 phase) losses any formulae or thumb rules.if the transformer is 50 % loaded there are some losses but how to calculate? how we to calculate % of transformer loading?

WE HAVE TWO TYPES OF LOSSES IN TRANSFORMER…..ie….iron loss and copper loss…one which will be constant for every loads…and the one which vary with bthe loading conditions respectively…..iron loss occurs all the time whether the transformer is loaded or unloaded and hence also called constant loss…and the other occurs only when it is loaded…and vary with the loads..and hence also called variable loss….an open circuit test in which rated voltage is applied to the primary winding and secondary winding is open circuited(test will be carried out in the low voltage side)is used for determining ironloss…..here the wattmeter gives the iron loss…bcos copper loss at no load conduction is negligible…..and the copper loss could be datemined by short circuit test in which rated current is made to flow through the primary winding and secondary winding will be short circuited….here the wattmeter reading gives full load copper loss…and to determine copper loss at x(FULL LOAD) the below formula can be used…

W(cu,xfl)=X*X*W(cu,fl) where W(cu,fl) is the full load copper loss…… and the total transformer loss will be iron loss+copper loss

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