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Answered by: NagaseshaReddy.M

• Sep 10th, 2007

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kVA is the unit for apparent power. Apparent power consists of active and reactive power. Active power is the share of the apparent power which transmits energy from the source (generator) to the user. Reactive power is the share of the apparent power which represents a useless oscillation of energy from the source to the user and back again.

It occurs when on account of some »inertia« in the system there is a phase shift between voltage and current. This means that the current does not change polarity synchronous with the voltage. But the heat generated in a winding as well as the eddy current losses generated in a transformer core depend on the current only, regardless of whether it aligns with the voltage or not.

Therefore the heat is always proportional to the square of the current amplitude, irrespective of the phase angle (the shift between voltage and current). So a transformer has to be rated (and selected) by apparent power.

Raghu

• Feb 9th, 2018

In transformer there is two losses
1. Iron loss
2. Copper loss
The iron loss is depands on Voltage
And the Copper loss is depands on current.

Umar Naik

• Nov 26th, 2017

There are cu losses and iron losses which are not dependent on power factor thats why we rated it in KVA.

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yuvaraj moorthy

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• Mar 25th, 2010

ISO is international standard organisation and IS is Indian standarad ...ISO has IEC which is International Electrotechnical Commision will do the standarad for all equipments and each national interp...

vishwas_deshmukh

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• May 6th, 2008

IS rules is international standrds rule adopted for any comissioning,erection,or  any work we want to perform.They have some certain standads according to study of researcher by comparing all geographical and availabe technical parameteres in that concerning area.Which is suitable for any work.

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KaranSorout  

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• Member Since Jul-2006 | Jul 13th, 2006

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The power ratio between Star to Delta is 1:3

Explanation:

 

Let?s suppose V_s be the supply voltage per phase.

So the line voltage of the supply will be ?3V_s.

Now assume any type of load; for simplicity I?m assuming it a only resistive load.

And let it be ?R? per phase.

 

For Delta connected load:

 

Calculation for per phase power; P_D= I²R

 

Where I ? load current (per phase)

And,

I = ?3V_s/R       {as line voltage of the supply is directly applied to the phase of the delta load}

So,

            P_d = (?3V_s/R )²R = 3V_s²/R  watts per phase.

For 3 phases:

            P_3D = 3P_d = 3*3V_s²/R = 9V_s²/R watts.

 

Now for Star connected load:

 

P_S = I²R = (V_s/R)²R = V_s²/R watts

For 3 phases: P_3S = 3P_S = 3 V_s²/R watts

 

Conclusion:

P_3S / P_{3D  =}  3V_s²/R / 9V_s²/R = 1/3

It may be better concluded by drowing figures. If any variations, can connect to me through karan.sorout@gmail.com  

RUP KAMAL KUTUM

• Feb 3rd, 2018

Well first you assumed voltage per phase as Vs and Line voltage as sqrt(3) x Vs: Which is for star connection. Then, you assumed the same thing for Delta Connection(where line and phase voltages are t...

Anbumozhi R

• Dec 3rd, 2017

Power is Same. Whether it is star to Delta conversion or Delta to star conversion. Power is 1:1

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SHATAKSHI

• Sep 2nd, 2011

You should be able to fully explain whatever you have written in your CV.

veena_gyanchandani

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• Feb 24th, 2010

First of all You should be confident for yourself.You should know about the company profile in which you are going, and why you want to join that company, what is so interesting about that company?If ...

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