Question 16.15: In Figure 16–21(a), the load impedance and line impedances a......
In Figure 16–21(a), the load impedance and line impedances are ZY = 10 + j5 Ω per phase and ZW = 0.15 + j0.85 Ω per phase, respectively. The magnitude of the line voltage at the source is VL = 208 V(rms). Using ∠VA = 0° as the phase reference, find the line current phasors and the line voltage phasors at the load for a positive phase sequence. Validate your results using Multisim. Assume 60 Hz.
In this example, the line voltage at the source is VL = 208 V(rms). The phase voltage magnitude at the source is V_{P} = V_{L}/\sqrt{3} = 120 V(rms). This phase voltage appears across the combined impedance ZW + ZY, so the line current magnitude is found to be
I_{L} = \frac{V_{P}}{|Z_{Y} + Z_{W}|} = \frac{120}{|10.15 + j5.85|} = 10.24 A(rms)
The specified phase reference means that VAN = 120∠0° V(rms). For a positive phase sequence, the other two line voltages are VBN = 120∠– 120° V(rms) and VCN = 120 ∠ – 240° V(rms). The phase voltages at the load are balanced and are denoted Vab, Vbc, and Vca. The three line currents are IA, IB, and IC. The first of these currents is found as
\textbf{I}_{A} = \frac{\textbf{V}_{AN}}{Z_{W} + Z_{Y}} = \frac{120 ∠ 0°}{0.15 + j0.85 + 10 + j5} = \frac{120 ∠ 0°}{11.72 ∠ 29.9°} = 10.24 ∠ −29.9° A(rms)
The remaining two line currents are related by 120°. Hence, IB = 10.24 ∠−149.9° A(rms) and IC = 10.24∠−269.9° A(rms).
The phase voltages at the load are balanced and are denoted Van, Vbn, and Vcn.
The first one is found as
Van = IAZY = (10.24 ∠ −29.9°)(10 + j5) = (10.24 ∠ −29.9°)(11.18 ∠ 26.6°)
Van = 114.5 ∠ − 3.36° V(rms)
Line voltages are \sqrt{3} times as large as the phase voltages and lead the phase voltages by 30°. Hence, we have
\textbf{V}_{ab} = (\sqrt{3} ∠ 30°) (114.5 ∠ −3.36°) = 198.4 ∠ 26.6° V(rms)
The other two line voltages are Vbc = 198.4 ∠ −93.4° and Vca = 198.4 ∠ −213.4° V(rms). The line magnitude at the load (198.4 V(rms)) is less than the line voltage at the source (208 V(rms)) due to the voltage loss across the line impedances.
Multisim allows the circuit to be readily analyzed. Draw the circuit as shown in Figure 16–21(b). Note the impedances need to be converted into the time domain. Hence, the line inductance is j0.85 = j2π × 60L or L = 2.25 mH, and the load inductance is j5 = j2π × 60L or L = 13.3 mH. Ask Multisim to perform a Single Frequency AC analysis at 60 Hz. The desired line voltage phasors are the difference between the load voltages Van, Vbn, and Vcn. These need to be added as expressions for analysis. The line currents are simply the currents through each of the elements in the lines. We selected the current through the line resistors. Grapher View returned the table shown in Figure 16–22(c). Note that Multisim likes to avoid angles less than −180° by adding 360° to the angle. Hence, IC = 10.24 ∠ −269.9° A(rms) results in 10.24 ∠ 90.1° A(rms), and Vca = 198.4 ∠ −213.4° V(rms) is reported as Vca = 198.4∠146.6° V(rms).
