coil impedance of a balanced 3 phase load

[dnem]

This question doesn't involve answering a question about an actual installation. This is purely theoretical.

If I have a 3 phase motor with just 3 coils and the coil impedance is 10 ohms for each coil, what is the total impedance for the entire motor.? Does the impedance change if the motor is connected in delta or wye.?

 

[GoldDigger]

This question doesn't involve answering a question about an actual installation. This is purely theoretical.

If I have a 3 phase motor with just 3 coils and the coil impedance is 10 ohms for each coil, what is the total impedance for the entire motor.? Does the impedance change if the motor is connected in delta or wye.?

The biggest obstacle in answering your question comes in deciding just how to define "total impedance" of a three or four terminal network when the definition of impedance involves only two points!
One candidate would be to divide the nominal applied voltage (L to N or L to L?) by the "total" current (line current or phase current? Sum of magnitudes since vector sum will be zero?).

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[steve66]

Impedance is measured between 2 terminals. So are you asking what is the impedance between any of the two motor terminals?

If so, for a Y connected motor, it would be two windings in series, or 20 ohms.

For a delta connected motor, it would be one winding in parallel with the other two windings in series. Or 10 in parallel with 20, which is 6.67 ohms.

 

[dnem]

Impedance is measured between 2 terminals. So are you asking what is the impedance between any of the two motor terminals?

If so, for a Y connected motor, it would be two windings in series, or 20 ohms.

For a delta connected motor, it would be one winding in parallel with the other two windings in series. Or 10 in parallel with 20, which is 6.67 ohms.

So for Delta, 3 windings at 10 ohms per winding
equals 6.67
We will pretend 480 volts so I can continue.
480 divided by 6.67 equals 71.964 amps.
71.964 divided by the square root of 3 equals 23.988 running amps.
480 times 71.964 amps equals 34,542.72 watts.
All correct so far.?

How about Wye.? 10 ohms plus 10 ohms equals 20 ohms. Then what.?

 

[dnem]

The biggest obstacle in answering your question comes in deciding just how to define "total impedance" of a three or four terminal network when the definition of impedance involves only two points!

So do you disagree with...?

For a delta connected motor, it would be one winding in parallel with the other two windings in series. Or 10 in parallel with 20, which is 6.67 ohms.

 

[GoldDigger]

No, that is not correct since in real life you will have a connection to all three terminals, not just two!

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[dnem]

Impedance is measured between 2 terminals. So are you asking what is the impedance between any of the two motor terminals?

If so, for a Y connected motor, it would be two windings in series, or 20 ohms.

If I have a motor with a coil impedance across each coil of 10 ohms, that motor also has an applied voltage I also can clamp-on amp measure the actual amps. Given equal ohmic value of each coil, lets assume equal amp reading on each phase conductor. The voltage divided by (the measured amps times the square root of 3) will give you a ohmic value. So that ohmic value can be calculated and known even tho there are 4 terminations in the Wye motor.

Or look at it this way:
480 volt Wye motor with 5.7735 amps measured on each phase conductor.
5.7735 times the square root of 3 equals 10.
480 volts divided by 10 amps equals 48 ohms.
What is the 48 a measurement of.?
48 is not the impedance of the individual coils, is it.?

 

[GoldDigger]

If you use the L to N voltage and the line current(s) there is no need to bring the square root of three into the exercise.

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[Smart $]

If I have a motor with a coil impedance across each coil of 10 ohms, that motor also has an applied voltage I also can clamp-on amp measure the actual amps. Given equal ohmic value of each coil, lets assume equal amp reading on each phase conductor. The voltage divided by (the measured amps times the square root of 3) will give you a ohmic value. So that ohmic value can be calculated and known even tho there are 4 terminations in the Wye motor.

Or look at it this way:
480 volt Wye motor with 5.7735 amps measured on each phase conductor.
5.7735 times the square root of 3 equals 10.
480 volts divided by 10 amps equals 48 ohms.
What is the 48 a measurement of.?
48 is not the impedance of the individual coils, is it.?

You're calculating it wrong...

If you have (3) 10 ohm coils connected in a wye configuration on a 480V 3Ø 3W system, each line will conduct (480V ÷ √3 ÷ 10ohm) amps... so the 10 ohms will appear to be the "total" impedance of the motor.

 

[dnem]

If I have a motor with a coil impedance across each coil of 10 ohms, that motor also has an applied voltage I also can clamp-on amp measure the actual amps. Given equal ohmic value of each coil, lets assume equal amp reading on each phase conductor. The voltage divided by (the measured amps times the square root of 3) will give you a ohmic value. So that ohmic value can be calculated and known even tho there are 4 terminations in the Wye motor.

Or look at it this way:
480 volt Wye motor with 5.7735 amps measured on each phase conductor.
5.7735 times the square root of 3 equals 10.
480 volts divided by 10 amps equals 48 ohms.
What is the 48 a measurement of.?
48 is not the impedance of the individual coils, is it.?

You're calculating it wrong...

If you have (3) 10 ohm coils connected in a wye configuration on a 480V 3Ø 3W system, each line will conduct (480V ÷ √3 ÷ 10ohm) amps... so the 10 ohms will appear to be the "total" impedance of the motor.

So there is nothing that is 48 ohms.?
277volts÷10Ω=27.7amps
480volts÷10Ω=48amps÷1.732=27.71amps

So for Wye connection, everything (in this example) is 10Ω and everything measures at 27.71amps.?

 

[Smart $]

So there is nothing that is 48 ohms.?
277volts÷10Ω=27.7amps
480volts÷10Ω=48amps÷1.732=27.71amps

So for Wye connection, everything (in this example) is 10Ω and everything measures at 27.71amps.?

Yes... but it should be 480volts÷1.732÷10Ω=27.71amps

Note (480volts÷1.732) is the line-to-neutral 277.128 volts and from there it is simply E÷Z=I

 

[steve66]

So for Delta, 3 windings at 10 ohms per winding
equals 6.67
We will pretend 480 volts so I can continue.
480 divided by 6.67 equals 71.964 amps.
71.964 divided by the square root of 3 equals 23.988 running amps.
480 times 71.964 amps equals 34,542.72 watts.
All correct so far.?

How about Wye.? 10 ohms plus 10 ohms equals 20 ohms. Then what.?

Its not quite that simple, since like Golddigger said, once you connect the 3rd lead we have to consider what effect that has.

But I'm trying to figure out where you are going with this. If we assume 480 volts, and 277 volts line to neutral, we can simplify things like Golddigger said, and not worry about sqrt(3).

Each winding is 277V/10 ohms = 27.7 amps.
277 * 27.7 is 7479 watts, and total wattage is 3* 7479 = 22437 watts.

This is a motor, so we have really simplified things. Basically, we have ignored all of the motor except the windings, and I'm just wondering if where you are going next will bring that into play.

 

[steve66]

Delta connected is a little different. Each winding is connected line-line, so we would have 480V/10 ohms = 48 amps per winding.

480V * 48 amps * 3 = 69120 watts.

This is much larger than the 22,470 for the wye connected motor, and if a wye motor could be reconnected as a delta, and have the same 480V applied, it would probably burn up really fast.

 

[Ultrafault]

This conversation is like trying to describe how many apple slices will an orange produce. Slice it anyway you want it is still an orange.

 

[Ingenieur]

For wye load
Zeq = sqrt3 x 10 Ohm = 17.32
i line = line coil/branch = 480/17.32 = 27.713 A

for ease of calculation do a wye-delta conversion on the wye
Z delta = Z wye x sqrt3

Delta load
480/10 = 48 A line
convert to delta coil/branch 48/sqrt3 = 27.713 A

this can be proven by using simultaneous equations
delta source/wye load
Z I = V or invZ V = I

Z
-Za Zb 0
0 -Zb Zc
1 1 1
all known

I
ia
ib
ic
unknowns

V
Va
Vb
0
known where
Va = V / 0 deg
Vb = V / 120 deg

once you have ia for example Zeq = Va / ia

 

[Phil Corso]

Come on fellas...

A motor's "impedance" is quite different when its not energized, to when its energized but not turning, and when its operating at some load!

Perhaps you mean Inductance? (But then, the OP did say he's getting paid for it!)

Phil Corso

 

[Ingenieur]

I think he is asking the current magnitude difference/relationship between a wye and delta connected load of 10 Ohm per each coil (x 3)

not necessarily a motor
could be 3 10 Ohm heater coils

 

[Smart $]

Come on fellas...

A motor's "impedance" is quite different when its not energized, to when its energized but not turning, and when its operating at some load!

Perhaps you mean Inductance? (But then, the OP did say he's getting paid for it!)

Phil Corso

Yes, a motor's impedance varies for many different reasons. Deenergized is one of them. A different mechanical load while energized is another. However, the case is presented here as having a specific coil (aka winding) impedance... so all other variables which change the winding impedance are irrelevant.

 

[dnem]

Yes, a motor's impedance varies for many different reasons. Deenergized is one of them. A different mechanical load while energized is another. However, the case is presented here as having a specific coil (aka winding) impedance... so all other variables which change the winding impedance are irrelevant.

"all other variables which change the winding impedance are irrelevant", yes, yes, that was what I was trying to do. And therefore that was why I stated the problem as having a specific impedance. I was trying to eliminate variables but...... as the replies came in, I can see that there has been a "mission creep".

So let me change the question

I think he is asking the current magnitude difference/relationship between a wye and delta connected load of 10 Ohm per each coil (x 3)

not necessarily a motor
could be 3 10 Ohm heater coils

Yes, lets do 3 heater coils of 10 ohms each and they are connected to a 3 phase power supply. If I connect them in a delta or wye formation, I will get different coil voltages, different current flow amounts, different power ratings

Delta connected is a little different. Each winding is connected line-line, so we would have 480V/10 ohms = 48 amps per winding.

480V * 48 amps * 3 = 69120 watts.

This is much larger than the 22,470 for the wye connected motor, and if a wye motor could be reconnected as a delta, and have the same 480V applied, it would probably burn up really fast.

Yes, all this stuff can be quickly calced out using watts and ohms pies but the ohm pie has the ohm value and I was wondering how it would calc out depending on the wye or delta connection.

......(But then, the OP did say he's getting paid for it!)

Phil Corso

I get paid for some stuff that I do but this isn't one of them. This is purely theoretical. Im just trying to understand the math to a basic problem.

Let me simplify it this way:
3 independent heater coils of 10 ohms each
power supply is 3 phase 240 volt
ignore any reactance and just assume 100% power factor (remember simple theoretical not exact real life scenario)

How do you fill in the watts law and ohms law pies if connected wye.?
How do you fill in the watts law and ohms law pies if connected delta.?