Recently I came across a statement saying that motors connected in star can increase the effiency than connected in delta mode.Since line currents of both modes are same and voltage being same how the input power for the same output is reduced..?Can anybody shed some light on this......

The simple answer is that you should connect it in the manner shown on the nameplate for the voltage you are using.

But there are some general statements that can be made. Take a given motor driving a constant horsepower load near rated nameplate load and connect in star and delta:

• In star each leg has roughly sqrt(3) lower voltage and roughly sqrt(3) higher current (even though the phase quantities seen outside the motor are roughly the same).
  
• In star you will have higher I^2*R, but less core loss.
  
• In delta you will have higher core loss but lower I^2*R. Also delta may cause undesirable harmonics.

BUT as stated at the beginning, motors are designed for connecting a specific winding winding configruation (detla or star...usually star) for specified terminal voltage and operating conditions (start or run). I would not recommend switching the configuration to anything different than what is on the nameplate for the terminal voltage you are using.

This message has been edited. Last edited by: electricpete,

Dear, delta higher phase current & there for higher i2r should b the loss.....but u told otherwise....

Hi GIRI
I believe it is correct as I wrote it. In words I would say the reason is that the delta connection provides higher voltage across each phase winding and therefore draws lower current thru each phase winding and therefore creates lower I^2*R losses. Below is a more formal discussion:

Assumptions:
* horsepower is constant (this is a reasonable assumption since horsepower is determined primarily by the driven equipment)
* efficiency and power factor are approximately constant. There are some variations between delta and star/wye, but this is a small effect which is neglected for simplicity.
* voltage is balanced

Symbols:
VLL = line-to-line voltage
VLN = line to neutral voltage
Vpw = phase winding voltage (see below for definition of phase winding)
IL = line current
Ipw = phase winding current
Ppw = real power output of a single phase winding
P = total motor power output
pf = power factor
eff = efficiency

The power output by the motor can be determined as follows:
P = sqrt3 * VLL * IL * pf * eff [EQUATION 1]

The motor has three "phase windings". The phase winding is whatever is between T1-T4 or T2-T5 or T3-T6. The real power delived by one phase winding is: Ppw = Vpw * Ipw *pf*eff

The total real power delivered by the sum of the three phase windings is three times as large:
P = 3 * Vpw * Ipw *pf*eff[EQUATION 2]

Connect the motor in wye/star:
we know each phase winding sees a voltage VLN and we know that VLN = VLL / sqrt(3). Substituting Vpw = VLL/sqrt(3) into equation 2 we have:
P = 3 * (VLL/sqrt3)* Ipw *pf*eff= sqrt3 * VLL * Ipw *pf*eff
Equating this to equation 1 we have
P = sqrt3 * VLL * IL * pf * eff = sqrt3 * VLL * Ipw *pf*eff
We solve to find Ipw = IL

Connect the motor in delta:
we know each phase winding sees full voltage VLL. Substituting Vpw = VLL into equation 2 we have:
P = 3 * VLL* Ipw *pf*eff
Equating this to equation 1 we have
P = sqrt3 * VLL * IL * pf * eff = 3 * VLL* Ipw *pf*eff
We solve to find Ipw = sqrt(3) * IL/3 = IL/sqrt(3)

Compare the results:
In star/wye connection, each phase winding carries Ipw = IL
In delta connection, each phase winding carries Ipw = IL/sqrt(3)
We see that the current thru the phase winding is lower in delta
Resistance of each phase winding is unchanged.
So the I^2*R losses will be lower in delta

This message has been edited. Last edited by: electricpete,

Exactly.........but tell me which one would be more efficient star or delta...in delta less winding loss n in star less core loss....but in motor core loss remains the same i believe whatever may be the load...leads to the fact that in higher side load or any load delta would be more beneficial... .

quote:

but in motor core loss remains the same i believe whatever may be the load...

I agree core loss does not vary significantly with load. But it does vary with voltage and we are changing the voltage seen by the winding when we switch between delta and star.

quote:

leads to the fact that in higher side load or any load delta would be more beneficial...

.
imo - could go either way. If the motor was designed to operate in wye for the system voltage and was near saturation to begin with - then changing to delta could cause decrease efficiency and could also possibly jeopardize the core interlaminar insulation. Even though we are decreasing the load component of current we will increase the magnetizing component of current and associate harmonics. So core loss increases and I^2*R might not decrease as much as you think looking at load component of current alone.

The safe bet is to follow the nameplate or manufacturer's instructions.

thank u so much sir.........