3-phase complex power calculations

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roy167

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geneva
Hey guys,

I'm looking for resource/book/practice problems/youtube in solving three phase WYE/DELTA, DELTA/ WYE any combination of line/phase voltage current problems for my PE exam. I have no problem with magnitude, I know when to apply root 3 etc, the problem is with angles.

For e.g. Problem like this.


Xdd8CyO.jpg
 
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adamscb

Senior Member
Location
USA
Occupation
EE
Keep in mind it's been way too long since I've done this kind of math, but since in a delta system Eline = Ephase, wouldn't you just calculate the voltage drop on the feeder resistance/inductance combination and then add that voltage to the Vab given? Someone correct me if I'm wrong.

Ran the numbers and I get 13079.37, which is most close to answer (c)
 
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drktmplr12

Senior Member
Location
South Florida
Occupation
Electrical Engineer
roy167 said:
Hey guys,

I'm looking for resource/book/practice problems/youtube in solving three phase WYE/DELTA, DELTA/ WYE any combination of line/phase voltage current problems for my PE exam.
Click to expand...

Check this post from an old thread. I listed many resources that were helpful for me.

http://forums.mikeholt.com/showthread.php?t=193551&page=2&p=1942434#post1942434

Another resource I did not discover until after the exam:

https://drive.google.com/open?id=1aCbNmsV3mhXHcTASP0WKkcwEubaS7NZU

keep at studying and best of luck in April!! :thumbsup:
 
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Julius Right

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
I think you are right. The voltage sense it is opposite to the current sense-according the sign convention.
We may transform the sketch from:
Voltage of Delta connection 04.jpg
If the IaA argument is -20o then the argument of the vector difference IaA-IbB is +10o.:
Voltage of Delta connection 02.jpg
We may use the sketch from IEEE 141[Figure 3-11Phasor diagram of voltage relations for voltage-drop calculations] modified for φ>0:
Voltage of Delta connection 03.jpg
Horizont=OB=VAB+AB; AB=AE-CD; AB=I*R*cos(φ)-I*X*sin(φ)
Horizont=VAB+I*R*cos(φ)-I*X*sin(φ)[really]
Vert=GB=GC+CB=I*X*cos(φ)+I*R*sin(φ)[imaginary]
Then Vab=VAB+(I*cos(φ)+jI*sin(φ))*(R+jX)
 
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roy167

Member
Location
geneva
Going into phasors etc may be little more complicated for problem like this. If delta impedance was given you could convert delta into WYE and solve it for a phase.
how do you solve it for a phase? They have given the delta line current so it would be easy to find phase current and phase voltage drop. The problem is there line impedance in B phase and this is where I don't know what to do.

Can someone help solve it algebraically? Mathematically? Without phasors etc?
 
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adamscb

Senior Member
Location
USA
Occupation
EE
roy167 said:
Going into phasors etc may be little more complicated for problem like this. If delta impedance was given you could convert delta into WYE and solve it for a phase.
how do you solve it for a phase? They have given the delta line current so it would be easy to find phase current and phase voltage drop. The problem is there line impedance in B phase and this is where I don't know what to do.

Can someone help solve it algebraically? Mathematically? Without phasors etc?
Click to expand...

Going into phasors is required, it's not an option. Since the line impedance isn't purely resistive, phasors will come into play.
 
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Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Ladies & Laddies... Here's a shorter way:

Whenever a balanced delta-load is to be analyzed, then. most of the time, the solution is easier if you convert to a balanced 3ph, 4-wire, wye-load circuit ! Now, the circuit can be reduced to single-phase: having one source-voltage, Van; one load-voltage, VAn = VAB/Sqrt(3)! and one line impedance, ZaA !

Facts, derived from the data:


  1. Sequence is un-necessary because circuit is now single-phase!
  2. Load is inductive (lagging) because line-current has a negative angle!
  3. Fact 2 is important because if the current is lagging, then source-voltage is ‘Higher’, than load-voltage!
  4. Thus, you need only calculate the cable V-drop, Dlta-V = IaA x ZaA, in Volts!
  5. Then add Dlta-V, to VAn... not Vab, nor VAB!
  6. Finally, Vab equals Sqrt(3) x (VAn+Dlta-V) !
  7. Hope it helps!

Regards,, Phil Corso
 
Julius Right

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
Phil Corso said:
Ladies & Laddies... Here's a shorter way:

Whenever a balanced delta-load is to be analyzed, then. most of the time, the solution is easier if you convert to a balanced 3ph, 4-wire, wye-load circuit ! Now, the circuit can be reduced to single-phase: having one source-voltage, Van; one load-voltage, VAn = VAB/Sqrt(3)! and one line impedance, ZaA !

Facts, derived from the data:


  1. Sequence is un-necessary because circuit is now single-phase!
  2. Load is inductive (lagging) because line-current has a negative angle!
  3. Fact 2 is important because if the current is lagging, then source-voltage is ‘Higher’, than load-voltage!
  4. Thus, you need only calculate the cable V-drop, Dlta-V = IaA x ZaA, in Volts!
  5. Then add Dlta-V, to VAn... not Vab, nor VAB!
  6. Finally, Vab equals Sqrt(3) x (VAn+Dlta-V) !
  7. Hope it helps!

Regards,, Phil Corso
Click to expand...

If ZAB=ZBC=ZCA=Z and IaA=I>0; IbB=I>-120; IcC=I>-240; then it could be correct. In any different case you need complex "assistance":weeping:.
 
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