Question 5.13: Analysis of a Balanced Delta–Delta System Consider the circu...

First, we find the wye-connected equivalents for the source and the load. (Actually, we only need to work with one third of the circuit because the other two thirds are the same except for phase angles.) We choose to work with the A phase of the wye-equivalent circuit. Solving Equation 5.103 for V_an, we find that

\mathrm{V}_{ab}=\mathrm{V}_{an}\times \sqrt{3}\angle 30^\circ            (5.103)

\mathrm{V}_{an}=\frac{\mathrm{V}_{ab}}{\sqrt{3}\angle 30^\circ}=\frac{1000\angle 30^\circ}{\sqrt{3}\angle 30^\circ}=577.4\angle 0^\circ

Using Equation 5.106, we have

Z_{\Delta}=3Z_Y       (5.106)

Z_{Y}=\frac{Z_{\Delta}}{3}=\frac{30+j6}{3}=10+j2

Now, we can draw the wye-equivalent circuit, which is shown in Figure 5.46(b).
In a balanced wye–wye system, we can consider the neutral points to be connected together as shown by the dashed line in Figure 5.46(b). This reduces the three-phase circuit to three single-phase circuits. For phase A of Figure 5.46(b), we can write

\mathrm{V}_{an}=(Z_{\mathrm{line}}+Z_Y)\mathrm{I}_{aA}

Therefore,

\mathrm{I}_{aA}=\frac{\mathrm{V}_{an}} {Z_{\mathrm{line}}+Z_Y}=\frac{577.4\angle 0^\circ}{0.3+j0.4+10+j2}=\frac{577.4\angle 0^{\circ}}{10.3+j2.4}=\frac{577.4\angle 0^\circ}{10.58\angle 13.12^\circ}=54.60\angle -13.12^\circ

To find the line-to-neutral voltage at the load, we write

\mathrm{V}_{An}=\mathrm{I}_{Aa}Z_Y=54.60\angle -13.12^\circ \times (10+j2)=54.60\angle -13.12^\circ \times 10.20\angle 11.31^\circ =556.9\angle -1.81^\circ

Now, we compute the line-to-line voltage at the load:

\mathrm{V}_{AB}=\mathrm{V}_{An}\times \sqrt{3}\angle 30^\circ =556.9\angle -1.81^\circ \times \sqrt{3}\angle 30^\circ=964.6\angle 28.19^\circ

The current through phase AB of the load is

\mathrm{I}_{AB}=\frac{\mathrm{V}_{AB}}{Z_\Delta}=\frac{964.6\angle 28.19^\circ }{30+j6}=\frac{964.6\angle 28.19^\circ}{30.59\angle 11.31^\circ}=31.53\angle 16.88^\circ

The power delivered to phase AB of the load is the rms current squared times the resistance:

P_{AB}=I^2_{AB\mathrm{rms}}R=\left(\frac{31.53}{\sqrt{2}} \right)^2 (30)=14.91\mathrm{~kW}

The powers delivered to the other two phases of the load are the same, so the total power is

P=3P_{AB}=44.73\mathrm{~kW}

The power lost in line A is

P_{\mathrm{line}A}=I^2_{aArms}R_{\mathrm{line}}=\left(\frac{54.60}{\sqrt{2}} \right)^2(0.3)=0.447\mathrm{~kW}

The power lost in the other two lines is the same, so the total line loss is

P_{\mathrm{line}}=3\times P_{\mathrm{line}A}=1.341\mathrm{~kW}