Motor Circuit Theory Wye-Delta Currents 1

[noel0589]

Hi all,

for 3 phase ac inductions motors

Lets say we run a motor to full speed in wye, and that line current is 120A at motor full speed RPM.
Now lets say we switch to delta. Is line current for delta still 120A, and delta phase current 120/sqrt3 = 69A?
Basically, at full speed rotor rpm, do line currents match up whether wye or delta connected?

 

[noel0589]

the thing that kind of throws me off is that although it looks like line currents should be the same in wye or in delta at full speed rpm, it is not true during starting conditions.
Locked rotor current in wye is 33.3% of the value of LRC in delta. Can someone straighten me out? Thanks!!

 

[subtech]

Let me start by saying I'm definately NOT a motor expert, however, I look at motors somewhat the same as large transformers(where I'm a little more "at home").

1. In Wye connected devices, line current is equal to winding current.

2. In delta connected devices, line current is equal to winding current multiplied by the square root of 3 or 1.732.

So, if current in one of the delta connected motor windings is 100 amps, then line current will be 173.2 amps.

Hope this helps.

 

[noel0589]

Subtech, I know that already...
please re-read the question:
if ac induction motor is at full speed, is line current for wye equal to line current for delta provided that applied voltage across lines is the same?
It is confusing because line currents of wye and delta during motor starting definitely are not equal even if the same voltage is applied across the lines. Thanks

 

[rbulsara]

The line currents will not be identical as a wye connection applies 58% per phase compared to a delta connection. This results in much less toruqe and hence less power produced (Power =torque* speed). Since the net power produced is less in wye for the same line voltage in delta, the line current in wye is proportionately lesser.

 

[bigbillnky]

I will try and see if I can help you understand some of the things you ask.
First, motor speed is determined by the equation: speed=120xfrequency/# of motor poles. Voltage actually does not determine motor speed.
Second, starting a motor wye connected is done to limit inrush current and reduce starting torque. If the motor is ran while connected wye, the actual line current is roughly 1/3 the current the motor will draw when it is ran delta connected. This is due to the fact in a delta connected motor, each phase voltage is dropped thru 2 windings in parallel. We know from theory that the total effect of resistances(impedances in a motor) in parallel are always LESS than the value of 1 resistance. When the motor is wye connected, each phase voltage is dropped thru one winding in series and two windings in parallel. Hence, the lower line current when the motor is connected wye.
Last, WINDING current is 1.73 times higher when the motor is wye connected than when it is delta connected. This makes it imperative that the overloads protecting the motor when it is delta connected are properly sized to provide adequate protection. I hope this helps.

Bigbillnky,C.E.F.....(Chief Electrical Flunky)

 

[aolalde]

If the very same motor is connected wye or delta, the voltage required to produce the same magnetic flux has to change proportional to the squared root of 3, if the frequency is kept constant. For instance a winding designed to operate at 460 volts “delta connected” needs 796.7 volts to produce the same flux when “wye connected”.

Then, if the motor in the above example is left “wye connected” and supplied with 460 volts, it is working with an extreme low voltage. The magnetic field is so weak that the motor will probably stall if the rated full load is applied to the shaft. Now the line current will be the locked rotor current at reduced voltage, almost the same that the motor takes when it starts wye connected. ( 2 times the full load current at 460 volts). The windings will burn in minutes or may be seconds.

Flux/pole= E*10^8/(4.44*N*f*Kp*Kd) (Lines)
Were:
E = RMS phase voltage
N= turns in series per phase
F= frequency in Hertz
Kp & Kd are winding constants.

For a given winding,N will change with the connection, eg series or parallel, wye or delta, etc.

 

[Marke]

Hello noel0589

At full speed, the current is dependant on the shaft load and the magnetising current.

The voltage across a winding in delta is sqr rt 3 times that in star so the magnetising current for that winding would increase by rt3 (assuming that the inductance remains constant) this will result in twice the magnetising current on the incoming phases.

The load current is related to the shaft load in a manner that KWin equals KW out (ignoring the motor losses) The line voltage is constant so we should see the same line current for equal shaft load, but in the case of the star connection all that current is flowing in one winding as opposed to the delta situation where the current is flowing in two windings. This will increase the stator copper loss in star for equal shaft load. The slip will also be higher, resulting in increase rotor losses.

At light load, you will see a noticable higher current in delta. At high load, the star connected motor would stall and draw the higher current.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[aolalde]

noel0589:

If the load is small enough, say 1/3 of the nominal load so that the shaft is allowed to spin close to the synchronous speed for both conditions and the same motor is supplied with the same voltage (Vll) but first “delta connected” and then “wye connected”. The currents should be close but different as function of the losses generated and the power factor (PF).

I = Pe(kVA) / (1.732*Vll)

Pin = Pout + losses (real power, kW)
Pe = Pin/PF (Total electric power in, kVA)
Pout = the same mechanical power taken from the shaft (kW).

For the “Wye Connection”, the PF will be higher due to the low magnetizing current ( weaker field). But the losses will increase due to the increased resistance of the stator winding and the higher rotor current due to the increased rotor slip to produce the load torque with lower air-gap flux density (B). So it depends on the motor design parameters but my bet is that the "wye connection" takes the higher current.