The Y-connected source in Example 11.1 feeds a-connected load through a distribution line having an impedance of 0.3+j0.9 Ω/Φ. The load impedance is 118.5+j85.8Ω/Φ. Use the a-phase internal voltage of the generator as the reference.a) Construct a single-phase equivalent circuit of the three-phase s

Question

The Y-connected source in Example 11.1 feeds a-connected load through a distribution line having an impedance of 0.3+j0.9 Ω/Φ. The load impedance is 118.5+j85.8Ω/Φ. Use the a-phase internal voltage of the generator as the reference.
a) Construct a single-phase equivalent circuit of the three-phase system.
b) Calculate the line currents [latex] \pmb{I}_{aA}[/latex],[latex] \pmb{I}_{bB}[/latex], and [latex] \pmb{I}_{cC}[/latex].
c) Calculate the phase voltages at the load terminals.
d) Calculate the phase currents of the load.
e) Calculate the line voltages at the source terminals.

Accepted Answer

a) Figure 11.14 shows the single-phase equivalent circuit. The load impedance of the Y equivalent is

[latex] \frac{118.5 + j85.8}{3}= 39.5 + j 28.6 Ω/Φ.[/latex].

b) The a-phase line current is

[latex] \pmb{I}_{aA}=\frac{120 \angle 0^{\circ}}{(0.2 + 0.3 + 39.5) + j(0.5 + 0.9 + 28.6)}[/latex]

[latex]=\frac{120 \angle 0^{\circ}}{40+j30}[/latex]

[latex]= 2.4 \angle -36.87^{\circ} A[/latex].

Hence

[latex]\pmb{I}_{bB}= 2.4\angle -156.87^{\circ} A,[/latex]

[latex] \pmb{I}_{cC}= 2.4\angle 83.13^{\circ} A.[/latex]

c) Because the load is connected, the phase voltages are the same as the line voltages. To calculate the line voltages, we first calculate [latex] \pmb{V}_{AN}[/latex]:

[latex] \pmb{V}_{AN}= \left(39.5 + j28.6\right)\left(2.4 \angle -36.87^{\circ}\right)[/latex]

[latex]= 117.04 \angle-0.96^{\circ}V [/latex].

Because the phase sequence is positive, the line voltage is [latex] \pmb{V}_{AB}[/latex]:

[latex] \pmb{V}_{AB}= \left(\sqrt{3} \angle 30^{\circ}\right)\pmb{V}_{AN}[/latex]

[latex] = 202.72 \angle29.04^{\circ} V[/latex] ,

Therefore

[latex] \pmb{V}_{BC}= 202.72\angle -90.96^{\circ} V,[/latex]

[latex] \pmb{V}_{CA}= 202.72\angle 149.04^{\circ} V. [/latex]

d) The phase currents of the load may be calculated directly from the line currents

[latex] \pmb{I}_{AB}= \left(\frac{1}{\sqrt{3}} \angle 30^{\circ}\right)\pmb{I}_{aA}[/latex]

[latex] = 1.39\angle -6.87^{\circ} A[/latex] ,

Once we know [latex] \pmb{I}_{AB}[/latex], we also know the other load phase currents:

[latex] \pmb{I}_{BC}= 1.39 \angle -126.87^{\circ} A,[/latex]

[latex] \pmb{I}_{CA}= 1.39 \angle 113.13^{\circ} A. [/latex]

Note that we can check the calculation of [latex] \pmb{I}_{AB}[/latex] by using the previously calculated [latex] \pmb{V}_{AB}[/latex] and the impedance of the Δ-connected load; that is,

[latex] \pmb{I}_{AB}= \frac{\pmb{V}_{AB}}{Z_{\phi}}=\frac{ 202.72\angle 29.04^{\circ} }{118.5+j85.8}[/latex]

[latex] = 1.39\angle -6.87^{\circ} A[/latex] .

e) To calculate the line voltage at the terminals of the source, we first calculate [latex] \pmb{V}_{an}[/latex] Figure 11.14 shows that [latex] \pmb{V}_{an}[/latex] is the voltage drop across the line impedance plus the load impedance, so

[latex] \pmb{V}_{an}=\left(39.8 + j29.5\right)\left(2.4 \angle -36.87^{\circ}\right)[/latex]

[latex]= 118.90 \angle-0.32^{\circ}V [/latex].

The line voltage [latex] \pmb{V}_{ab}[/latex]is

[latex] \pmb{V}_{ab}= \left(\sqrt{3} \angle 30^{\circ}\right)\pmb{V}_{an}[/latex],

or

[latex] \pmb{V}_{ab}= 205.94 \angle 29.68^{\circ}[/latex]V.

Therefore

[latex] \pmb{V}_{bc}= 205.94 \angle -90.32^{\circ}[/latex]V.

[latex] \pmb{V}_{ca}= 205.94 \angle 149.68^{\circ}[/latex]V.

Question 11.2: The Y-connected source in Example 11.1 feeds a-connected loa...

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