Active power of three-phase system

[gruba]

Homework Statement

Find active power of the given three-phase system.
$U_{RS}=U_{ST}=U_{TR}=220\sqrt 3 V$
$R=\omega L=\frac{1}{\omega C}=10\Omega$

2. The attempt at a solution
After reducing the circuit to one phase (see attachments), equivalent impedance is $\underline{Z_e}=\sqrt{101}e^{-j84,3^{o}}$.

$$U_{RS}=U_{ST}=U_{TR}=U_{L}=\sqrt{3}U_R\Rightarrow U_R=220 V\Rightarrow \underline{U_R}=220e^{-j\pi/3}\Rightarrow \underline{I_R}=\frac{220}{\sqrt{101}}e^{j54,3^{o}}$$

Active power of each resistor is given by

$$P_1=P_2=P_3=\mathfrak{R}(\underline{U_L}\underline{I_R^{*}})=\sqrt 3 U_RI_R\cos\angle(\underline{U_L}\underline{I_R})\Rightarrow P=3P_1$$

$$\cos\angle(\underline{U_L}\underline{I_R})\approx 0.099\Rightarrow P_1\approx 825.88 W\Rightarrow P=2477.64W$$

Is this correct?

 

[gneill]

Can you show your calculations for the equivalent impedance?

 

[gruba]

Can you show your calculations for the equivalent impedance?

From circuit_2, $\underline{Z_e}=X_C+(\frac{R}{3} || X_L)$ where $X_C=-j\frac{1}{\omega C},X_L=j\omega L$.

$$\underline{Z_e}=3-j9$$

I think this is the correct impedance.

Question: Expect calculations, are the equations for powers, impedance, voltage and current in my original post set correctly?

 

[gruba]

ATTEMPT EDITED (errors in calculation):

After reducing the circuit to one phase (see attachment):Equivalent impedance is $\underline{Z_e}=X_C+(R/3 || X_L)=-j10+(3+j)=3-j9$
$$U_{RS}=U_{ST}=U_{TR}=U_L=\sqrt{3}U_R\Rightarrow \underline{U_R}=220e^{-j\pi/6}\Rightarrow \underline{I_R}=\sqrt{90}e^{-j71.56^{o}}$$

Active power of each resistor is given by:
$$P_1=P_2=P_3=\mathfrak(\underline{U_R}\underline{I_R^{*}})=\sqrt{3}U_RI_R\cos\angle(\underline{U_L}\underline{I_R})$$
$$\cos\angle(\underline{U_L}\underline{I_R})\approx 0.75\Rightarrow P_1=165\sqrt{270}W\Rightarrow P=3P_1=495\sqrt{270}W$$

Question: Could someone check if equations are set correctly?

 

[gneill]

Your equivalent impedance looks right now.

How did you determine the potential across the resistor? In your equivalent circuit the capacitor is in series with the parallel combination of the resistor and inductor. Note that, since in a perfectly symmetric Y configuration the branches are essentially independent, you can ignore the phase angle of the source for a given branch if you're just looking for the real power in its load. Thus you have:

I put the magnitude of the voltage across the resistor at about 73 V, so whatever method you use to find it should find something similar.

 

[gruba]

How did you determine the potential across the resistor?

Why do we need voltage across the resistor?

Total active power is given by $$P=3P_1=3\sqrt{3}U_RI_R\cos\angle(\underline{U_L}\underline{I_R})$$

To use this equation, we need to draw phase diagram and find $\angle(\underline{U_L}\underline{I_R})$ which is $\approx 41.4^{o}$.

 

[gneill]

You don't necessarily need the voltage across the resistor. But it's a simple way to get to the power dissipated since P = V²/R. You should end up with a power (for one branch) of about 73²/(10/3), or about 1600 W.

In your previous attempt you wrote: $\underline{I_R}=\sqrt{90}e^{-j71.56^{o}}$, but that looks more like the branch impedance value than the branch current.

 

[gruba]

In your previous attempt you wrote: $\underline{I_R}=\sqrt{90}e^{-j71.56^{o}}$, but that looks more like the branch impedance value than the branch current.

You are right, it should be $\underline{I_R}=\frac{220}{\sqrt{90}}e^{j41.56^{o}}$.

But applying the formula $P=3P_1=3\sqrt{3}U_RI_R\cos\angle(\underline{U_L},\underline{I_R})$ doesn't give the result you get
(I think that your result is correct, just don't know why it is not the same using complex domain).

 

[gneill]

It must have something to do with the definitions of, or values of, your variables. Let's try a slightly different approach and see how it works out (It's a good idea to be able to check your work using different methods).

If the branch source voltage is U = 220 V and the branch impedance is Z = (3 - j9) Ω, then the branch current is:

$I = \frac{U}{Z} = \frac{220}{3 - j9} = \left(\frac{22}{3} + j22 \right)~A$

Then the complex power is p = VI* , so that:

$p = (220~V) \left(\frac{22}{3} - j22 \right) ~A = \left( \frac{4840}{3} - j4840 \right)~ VA$

The active power is the real component of the complex power, so P₁ = 4840/3 W.

What values are you using for U_R and I_R?edit: Fixed typo in power calculation (current value).

 

[gruba]

What values are you using for U_R and I_R?

I used $U_R=220V,I_R=\frac{220}{\sqrt{90}},\cos\angle(\underline{U_L},\underline{I_R})=0.75$.

 

[gneill]

I have an idea of where your problem lies. I think you're applying a power formula for a Y-connected source with a Δ-connected load. That would explain the presence of the √3 before the voltage variable, converting the Y source voltage to a Δ line-to-line source.

In this case you've already combined the loads into a single Y-connected load, and the given line-to-line voltage to a Y-source of 220 V, so the formula won't apply. What you have is:

 

[gruba]

Thanks for detailed explanations.