transformer

[Rei]

The answer is (A) 759

Igenerator = (132/13.2)Iline

However, this is a delta-wye transformer, isn't there a square-root of 3 different? I would consider the given line current of the transmission line equal the phase current, convert to the primary side and then multiply 1.73 to get the line current of the generator.

 

[Flyer_PE]

View attachment 3250
The answer is (A) 759

Igenerator = (132/13.2)Iline

However, this is a delta-wye transformer, isn't there a square-root of 3 different? I would consider the given line current of the transmission line equal the phase current, convert to the primary side and then multiply 1.73 to get the line current of the generator.

Typically, the turns ratio is presented as the line voltage on both sides of the transformer. The only time you have to worry about adjusting the voltage is if they give you the turns ratio of the individual single-phase transformers.

 

[Rei]

2 questions for this problem:

1. primary voltage is given and they asked for secondary line voltage; however, they still convert the given primary line voltage to phase voltage, then convert to the secondary and the convert to line voltage. The answer is different if you keep the primary line voltage and convert to secondary voltage. I always use the wrong value at the beginning which ended up with a wrong answer.

2. I don't understand where they get -80 deg.

Ia=1.73*Iab<30 which would give Ia=300<+30-50= 300<-20deg

 

[Bluekayak]

View attachment 3252
2 questions for this problem:

1. primary voltage is given and they asked for secondary line voltage; however, they still convert the given primary line voltage to phase voltage, then convert to the secondary and the convert to line voltage. The answer is different if you keep the primary line voltage and convert to secondary voltage. I always use the wrong value at the beginning which ended up with a wrong answer.

2. I don't understand where they get -80 deg.

Ia=1.73*Iab<30 which would give Ia=300<+30-50= 300<-20deg

Here's what I got assuming N1/N2=10/1 and ANSI standard wye-delta connection (high voltage phasors lead low voltage phasors by 30 degrees)

Primary side:

IA=17.3<(-arccos(0.94))=17.3<(-20)

VAN=1732/sqrt(3)<(-30)=1000<(-30)

Secondary side:

Iab=IA(N1/N2)=17.3(10/1)<(-20-30)=173<(-50); -30 degree phase shift from high voltage to low voltage side of wye-delta with standard ANSI connection

Ia=173xsqrt(3)<(-50-30)=300<(-80); in delta connected systems the line currents=phase currents x sqrt(3) and lag by 30 degrees

Vab=VAN(N2/N1)=1000(1/10)<(-30-30)=100<(-60); -30 degree phase shift from high voltage to low voltage side of wye-delta

Not sure why the solution is showing -30 degrees for Vab (need -30 degrees from high voltage to low voltage side of wye-delta transformer). Please advise if this makes sense.

 

[Rei]

I am really going nut with these three phase calculations. Anyone out there really got this concept down? This same question also asked for the solution if the connection is delta-delta. Since both primary and secondary connections are delta, I figure converting the line voltage and current from primary to line secondary are just the turn ratio different with NO phase shift. However, the solution had current off by -30 deg...huh....someone please show me some light before the April test.

 

[DK PE]

Sorry I haven't had time to look through this in detail but believe answer to question above is since you assumed high side VAB = at 0 degrees, by ANSI convention, it has to lead the low side Vab by 30 degrees. Therefore, Vab must be at -30 degrees (lagging is moving CW on phasor diagram)

 

[Rei]

Sorry I haven't had time to look through this in detail but believe answer to question above is since you assumed high side VAB = at 0 degrees, by ANSI convention, it has to lead the low side Vab by 30 degrees. Therefore, Vab must be at -30 degrees (lagging is moving CW on phasor diagram)

Really? The primary line voltage VAB and the secondary line voltage Vab for a delta-delta connection is off by -30 deg?

 

[Flyer_PE]

Sorry I haven't had time to look through this in detail but believe answer to question above is since you assumed high side VAB = at 0 degrees, by ANSI convention, it has to lead the low side Vab by 30 degrees. Therefore, Vab must be at -30 degrees (lagging is moving CW on phasor diagram)

Really? The primary line voltage VAB and the secondary line voltage Vab for a delta-delta connection is off by -30 deg?

Nope. The 30o phase shift only happens when a delta-wye transformation occurs. Also, keep in mind that the ANSI standard is just that...a standard. Just because that's the standard, don't automatically assume any system you work with will follow it.

 

[DK PE]

Sorry I haven't had time to look through this in detail but believe answer to question above is since you assumed high side VAB = at 0 degrees, by ANSI convention, it has to lead the low side Vab by 30 degrees. Therefore, Vab must be at -30 degrees (lagging is moving CW on phasor diagram)

Really? The primary line voltage VAB and the secondary line voltage Vab for a delta-delta connection is off by -30 deg?

I was back on original problem statement was high side Y - low side delta... therefore low side lags and that is where I got -30. I haven't had a moment to think of delta-delta question yet. Sorry i read your question too quickly.

 

[Rei]

Nope. The 30o phase shift only happens when a delta-wye transformation occurs. Also, keep in mind that the ANSI standard is just that...a standard. Just because that's the standard, don't automatically assume any system you work with will follow it.

Yeah, sorry for jumping off to many questions. So for a delta-delta connection, why going from line current primary to line current secondary has a 30 deg shift?

 

[HornTootinEE]

View attachment 3250
The answer is (A) 759

Igenerator = (132/13.2)Iline

However, this is a delta-wye transformer, isn't there a square-root of 3 different? I would consider the given line current of the transmission line equal the phase current, convert to the primary side and then multiply 1.73 to get the line current of the generator.

Or just remember that |S| is always the same on both sides of the transformer, regardless of voltage. Using the turns ratio is the most direct way, but if you have been given a line current and a LL voltage on one side of the transformer and a LL voltage on the other side, you can find the missing line current using |S|=SQRT(3)xVLLxILL.

That little equation (when balanced load is given or assumed) is a beauty and can simplify things immensly. It fails though if you need a power factor or phase angle or anything like that. But when looking for straight magnitudes, it's great and fast.