Calculating Apparent Power and Impedance

[BatmanisWatching1987]

[FONT=&quot]
[/FONT]Here is the problem.
I am trying to practice solving for Apparent Power and Impedance.
I get different angles when I try solving for the apparent power.
Maybe I am using the Conjugate of the current incorrectly when solving for this problem.
I'm not sure what I am doing incorrectly.



View attachment 20190221224308.pdf

 

[mivey]

View attachment 22393

Here is the problem.
I am trying to practice solving for Apparent Power and Impedance.
I get different angles when I try solving for the apparent power.
Maybe I am using the Conjugate of the current incorrectly when solving for this problem.
I'm not sure what I am doing incorrectly.



View attachment 22392

You forgot the 30d phase shift.

 

[mivey]

See the discussion from the thread below if you want. In that thread, Phil made the same mistake. The point is:

Nope.

This is a basic concept. The line current into the corner of a delta is the sum of the two phase currents out of the corner. That is why I listed the phase currents as leaving the corner. In other words you should recognize that

I_aA = I_AB + I_AC

70@-20d = 40.41@10d + 40.41@-50d



What is not true, and is one error you have repeated, is thinking delta phase current is equal to line current divided by the square root of three. Doing so neglects the phase shift common in any wye-delta conversion. In other words:

I_AB /= IaA ÷ sqrt(3)



What is true for this problem is that delta phase current is equal to line current divided by the square root of three at an angle of -30d. In other words:

I_AB = IaA ÷ sqrt(3)@-30d



I used A-B-C rotation in my calcs.

PS: note the missing shift in your first "S" solution. i.e., you mixed phase voltage angles with line current angles.

 

[Sahib]

View attachment 22393
[FONT=&quot]
[/FONT]Here is the problem.
I am trying to practice solving for Apparent Power and Impedance.
I get different angles when I try solving for the apparent power.
Maybe I am using the Conjugate of the current incorrectly when solving for this problem.
I'm not sure what I am doing incorrectly.



View attachment 22392

Just as in your thread calculating real power, the apparent power is sum of individual phase apparent powers and only the phase angle to be taken into consideration for balanced three phase loads.

 

[Julius Right]

I think S=3*VAB*conj.IAB where IAB=(IaA-IbB)/3=40.414<10
If we consider VA=VAB/√3<-30 VA=7.217<-30 [neglecting voltage drop on Z=5+j10]
S=3*Va*conj(IaA)
If DVA=IaA*(5+j10) actually VA=7.4777<-24.24 and S=1570<-4.24

 

[BatmanisWatching1987]

For the Phase Current, I showed a 30 Degree Phase Shift?
Since
I_L= I_p√3∠-30, so I_aA/√3∠-30 = 40.41∠-10

Also I thought in a delta configuration the line voltage equals the phase voltages.

 

[Julius Right]

 

[mivey]

For the Phase Current, I showed a 30 Degree Phase Shift?
Since
I_L= I_p√3∠-30, so I_aA/√3∠-30 = 40.41∠-10

Also I thought in a delta configuration the line voltage equals the phase voltages.

In your first solution, you used a phase voltage and need to use √3∠-30 instead of just √3 as the multiplier.

In the complex power equation, you can't mix line currents with phase voltages without a conversion factor.

The phase current is through the load element. The phase voltage is across the load element. The line current is in the line to the left. The line voltage is line to neutral but the neutral point is not shown in the circuit diagram.

In the circuit you have, the phase voltage is across the load. The line current coming in is a composite of the two phase currents leaving the corner of the delta.

In a wye winding, phase current through the winding equals the line current leaving the transformer, if we call the winding a phase.

In a delta winding, phase current through the winding does not equal line current leaving the transformer because the line current is the combination of two phase currents.

We normally consider phase voltage to mean line to line and line voltage to mean line to neutral unless it is defined differently during the givens of the problem.

 

[BatmanisWatching1987]

What would be the correct Apparent Power of the Load? (Including angles please)

Since I think I might be inputting the degree shift incorrectly on my calculator.

 

[BatmanisWatching1987]

So

S = √3∠-30*V_L*I_L^* = √3∠-30 * (12.5∠0kV) * (70∠20) = 1515.54∠-10.

So when using the formula S = √3*V_L*I_L, I can just use the equivalent formula which includes the shift as
S = √3∠-30*V_L*I_L^*

Please let me know if I am mistaken.

 

[Julius Right]

 

[mivey]

So

S = √3∠-30*V_L*I_L^* = √3∠-30 * (12.5∠0kV) * (70∠20) = 1515.54∠-10.

So when using the formula S = √3*V_L*I_L, I can just use the equivalent formula which includes the shift as
S = √3∠-30*V_L*I_L^*

Please let me know if I am mistaken.

Equivalent may not be the right term.

The best formula is V*I^* per phase. The conversion is embedded in that we are supposed to know to use the correct voltage and current.

The VI* can be expanded because we know V_ab = V_an x sqrt(3)@30d for positive rotation. We also know V_an = V_ab ÷ sqrt(3)@30d

So we have (if you want to mix phase voltages with line currents):

S_3phase = 3 x V_LN x I_line*
= 3 x V_LL ÷ sqrt(3)@30d x I_line*
= sqrt(3) x sqrt(3) x V_LL ÷ sqrt(3)@30d x I_line*
= sqrt(3)@-30d x V_LL x I_line*

 

[BatmanisWatching1987]

Equivalent may not be the right term.

The best formula is V*I^* per phase. The conversion is embedded in that we are supposed to know to use the correct voltage and current.

The VI* can be expanded because we know V_ab = V_an x sqrt(3)@30d for positive rotation. We also know V_an = V_ab ÷ sqrt(3)@30d

So we have (if you want to mix phase voltages with line currents):

S_3phase = 3 x V_LN x I_line*
= 3 x V_LL ÷ sqrt(3)@30d x I_line*
= sqrt(3) x sqrt(3) x V_LL ÷ sqrt(3)@30d x I_line*
= sqrt(3)@-30d x V_LL x I_line*

Thanks this helped out a lot. I need to study these formulas in a little bit more detail now.


My next question for this problem would be to try solving for the Apparent Power of the source including the line loss that Julius Right did earlier.


Also how would I solve for the Apparent Power of the Load using the following formula, S_3P=V²/Z

 

[mivey]

Thanks this helped out a lot. I need to study these formulas in a little bit more detail now.


My next question for this problem would be to try solving for the Apparent Power of the source including the line loss that Julius Right did earlier.


Also how would I solve for the Apparent Power of the Load using the following formula, S_3P=V²/Z

From the basics, we know:
I_line = V_LN ÷ Z_LN
and
I_line* = V_LN* ÷ Z_LN*

From the basics we also know:
S_3phase = 3 x V_LN x I_line*

Now we derive:
S_3phase = 3 x V_LN x I_line*
= 3 x V_LN x V_LN* ÷ Z_LN*
= 3 x |V_LN|^2 ÷ Z_LN*

You can make a similar derivation for other configurations then you are off to the races for any combination you want.

 

[BatmanisWatching1987]

From the basics, we know:
I_line = V_LN ÷ Z_LN
and
I_line* = V_LN* ÷ Z_LN*

From the basics we also know:
S_3phase = 3 x V_LN x I_line*

Now we derive:
S_3phase = 3 x V_LN x I_line*
= 3 x V_LN x V_LN* ÷ Z_LN*
= 3 x |V_LN|^2 ÷ Z_LN*

You can make a similar derivation for other configurations then you are off to the races for any combination you want.

With
V_LN = 7.22∠-30
Z_Y = 103.11∠-10

I get, S_3phase = 3 x |V_LN|^2 ÷ Z_LN* = 3*(7.22∠-30)² / 103.11∠+10 = 1516∠-70.

Once again, why does my angle not much the pervious angle of S_3phase = 1515.54∠-10.