calculation delta and star motor power?

i am a student. i have a question about three phase power.

question: a three phase delta connected motor draws a line current of 18A∠-85 degree from the 400v supply.

1) how much power does the load dissipate?

my anwer is:
P=√3 VIcosφ=√3*400*18*cos85=1087w



2) calculate the motor output when connected in star?

my anwer is:

first calculated impedence in three winding

IN delta connection:
Ip=18/√3= 10.4∠-55A (line current lag phase current 30 degree in delta)

Z=400/(10.4∠-55)=38.46∠-55 ohm
IN star connection:
I=V/Z=230/38.46∠-55 =6∠-55
P=√3 VIcosφ=√3 *400*6*cos55
= 2384w

problem is power in delta shoud be 3 time more
in star . my anwer is wrong. but i donot known where
is wrong?

 

straw
The equation for (3 phase) real power you state is correct. Whether the motor is connected delta or wye, The V magnitude is V line to line which is still 400 vac and the current magnitude is still the same, as Line current is the current used in your equation which is still 18 Amperes. The angle is the angle between the voltage across one element and the current through that element with voltage as the reference. You stated an angle but did not give the reference (V Line to Line or V phase). Regardless, draw out the problem and reduce it to single phase and you will notice that the relationship is identical whether the motor is connected delta or wye. By the way this is a really lousey Pf for a motor. The result for (3 phase) real power is the same whether the motor is connected delta or wye and stated above. The result for (3 phase) reactive power is the same whether the motor is connected delta or wye (Q=SQRT(3)VISinAngle. The real power for either connection is SQRT(3)VICosAngle or 12,456 Cos (-85) = 10,787 Watts. Simply stated, there is no difference in line current whether the motor is internally connected delta or wye. In fact when you specify a motor, you rarely if ever state the winding type, that is left to the manufacturer.
Kelly

 

When used in the formula P=√3 VIcosφ, the angle φ is the angle of each impedance, regardless of whether they are connected in delta or wye.

The impedance of each winding is 38.49∠85.

When connected in wye, the voltage on each leg is 230.94, and the line current will result in 6A, which then confirms the statement "power in delta shoud be 3 times more in star".

 

Let's try this, and in a different way. It is so difficult to do this without a blackboard or graphics, but let's try. We will do this by magnitude only as the motor is balanced, and therefore, we need not use the phasors.

The three phase motor draws 18 line amperes with a line to line voltage of 400 Volts.
This amounts to an apparent power of SQRT(3) * 400 * 18 or 12, 456 VA for three phases.

Let us now convert the problem to single phase (and in a Wye configuration by the way) where three of these single phase circuits will amount to the three phase circuit from whence we started. The single phase apparent power will be 12,456/3 or 4,152 VA per phase. The phase to ground voltage, as you stated, is 400/SQRT(3) or 231.2 Volts. The impedance of this phase load (Phase voltage divided by Phase current) is 231.2/18 Amperes or 12.83 Ohms/phase. If you connect three of these circuits together with a common neutral, you will have a three phase circuit with 400 volts line to line and 18 Amperes in each line which amounts to 12,456 VA (our original problem). If you look at the very same circuit, you will also note it is composed of three single phase circuits of 231.2 Volts per phase directly across a 12.83 Ohm impedance which also results in a line current of 18 Amperes.

There is a transformation equation for Wye to Delta connected impedances which you can verify on-line quite easily. The equation to transform each impedance from Wye per phase (which we just calculated) to Delta per leg is Rda = (Rw1*Rw2+Rw1*Rw3+Rw2*Rw3)/Rw3 or in our case Rda = (12.83*12.83+12.83*12.83+12.83*12.83)/12.83 = 38.49 Ohms (your Delta impedance per leg).

Therefore the motor consumes 12,456 VA of Apparent power for three phase, the product of which is calculated by VL-L * I Phase or line * SQRT(3) and has three 38.49 Ohm impedances configured in Delta on a 400 Volt source which draws 18 Amperes.

Or

Therefore the motor consumes 12,456 VA of Apparent power for three phase, the product of which is calculated by 3 * VL-G * I Phase or line and has three 12.83 Ohm impedances configured in Wye on a 400 Volt source which draws 18 Amperes.

The power to each motor (Delta or Wye) and whether (Apparent, Real or Reactive) Power is independent of the internal connection, but indeed the impedance, used for the Delta configuration is three times the impedance used for the Wye configuration. The power is the same; the impedances have the three to one relationship.

I noted I used the wrong side of my slide rule to calculate Cos(-85) and came up with 0.87 in my original reply rather than the 0.087 it should be. My apologies.

Kelly

 

Dear kelly you are right and all calculation made by you are correct but for wye-delta starter they said that "there is 1/3 ratio of power or current between wye and delta connection duration." What is hint or manuplation behind this, if you explain I will be happy. thanks

 

Hi,

Just to chime in with an observation - I have seen a five horsepower motor that was started up in delta configuration to limit current draw, and then switched to a star configuration for final acceleration to running speed. The phase-to-phase voltage difference is less in a delta cofiguration than the phase-to-ground in a star. That voltage reduction through the coils has to reduce the current.

 

Sorry I think there is a wrong in your explanation. Phase to phase voltage difference in delta connection is higher than phase to ground voltage difference in star connection, not less. And wye-delta starter starts firstly as wye and than after a time duration, continue as delta....

 

Hi,

We do have a disagreement, then. Some years ago, we were using two legs of three-phase power for a 240 VAC outlet. We had to allow for the actual voatage being only 206 VAC, vice 240. That was why I was saying that phase-to-phase voltage was less than phase-to-ground.

If I'm incorrect, then I will have learned something.

 

You are correct on the delta-wye startup, typically used on larger motors (in my experience). I believe it's referred to as "Reduced Voltage" starting, but my terminology may be off there. It's a while since I have been around that technology.

 

From a practical point of view, when you say line current, this is the current read from the source to the motor windings. So how does the delta connected load's windings affect the line current. The reason I ask is that I have a delta connected 460 volt, 3 phase, 33 amp, 22 kw, 60 hz, motor that drives a compressor, and it is powered by a step up delta to wye transformer. Basically a wye source to a delta load, and in the past 24 months, we have shorted the motor windings on six different occasions. The current readings on the secondary side of the transformer are approximately 33 amps full load, and 15 amps no load. I am assuming we can say that the secondary side of the transformer is the line current. Is this current equal to the current in the delta connected motor windings?

 

i think the current in delta windings/phase shud be I(line-y)/sqrt(3)