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Chapter 12: Problem 2

What is the phase sequence of a balanced three-phase circuit for which \mathbf{V}_{a n}=160 / 30^{\circ} \mathrm{V} \text { and } \(\mathbf{V}_{c n}=160 /-90^{\circ} \mathrm{V} ?\) Find \(\mathbf{V}_{b n}\)

Short Answer

Expert verified
The phase sequence is 'acb' and \(\mathbf{V}_{bn} = 160 \angle 150^\circ\ V\).

Step by step solution

01

Understanding the Given Phases

We are given the voltages \( \mathbf{V}_{an} = 160 \angle 30^\circ\ V \) and \( \mathbf{V}_{cn} = 160 \angle -90^\circ\ V \). In a balanced three-phase circuit, the phase voltages are separated by 120掳. This means that each voltage phasor is spaced 120掳 apart from the others.
02

Determine the Phase Sequence

The phase sequence is determined by the order of the phase angles. We have the angle of phase 'a' as 30掳 and phase 'c' as -90掳. In a positive phase sequence (abc), each successive phase leads the previous one by 120掳. In a negative sequence (acb), each phase lags by 120掳.
03

Verifying Phase Sequence: Positive or Negative

Check whether going from phase 'a' to phase 'c' involves a decrease in the angle (lag) or increase (lead). From \(30^\circ\) to \(-90^\circ\), (-90掳 - 30掳 = \(-120掳\), indicating phase 'c' lags 'a' by 120掳. Hence, it is an negative phase sequence (acb).
04

Calculate \\(\mathbf{V}_{bn}\) in a Negative Sequence

In an 'acb' sequence, \(\mathbf{V}_{b}\) lags \(\mathbf{V}_{c}\) by 120掳. Therefore, \(\mathbf{V}_{bn} = 160 \angle (-90掳 - 120掳)\ V = 160 \angle -210掳\ V\). Normalize this angle within the 0掳 to 360掳 range: \(-210掳 + 360掳 = 150掳\).
05

Conclusion

The phase sequence is 'acb' which is a negative sequence. The voltage \(\mathbf{V}_{bn}\) is \(160 \angle 150^\circ\ V\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Sequence
The **phase sequence** is like the rhythm of a three-phase circuit. It designates the order in which the voltages reach their peak values. In a three-phase system, there are generally two sequences: **positive (abc)** and **negative (acb)**.
In the positive sequence (abc), after phase 'a' reaches its peak, phase 'b' follows, and then phase 'c'. Each phase leads the previous one by 120掳. In contrast, a negative sequence (acb) sees phase 'c' following phase 'a', and then phase 'b', lagging each other by 120掳.
  • In a **positive sequence**, the progression is like moving forward: 'a' 鈫� 'b' 鈫� 'c'.
  • In a **negative sequence**, it's like reversing: 'a' 鈫� 'c' 鈫� 'b'.
The phase sequence impacts the operation of equipment connected to the circuit, such as motors, as it influences the direction of rotation.
Voltage Phasors
**Voltage phasors** provide a convenient way to represent sinusoidal voltages in a three-phase circuit. They combine magnitude and direction (angle) into a single entity designed to simplify circuit analysis.
A phasor is essentially a vector representation of a sinusoidal function with a specific magnitude and phase angle. For example, in the problem statement, we have two voltage phasors: \( \mathbf{V}_{an} = 160 \angle 30^\circ \mathrm{V} \) and \( \mathbf{V}_{cn} = 160 \angle -90^\circ \mathrm{V} \). This means:
  • \( \mathbf{V}_{an} \) is 160 volts, oriented at an angle of 30掳 relative to a reference.
  • \( \mathbf{V}_{cn} \) is also 160 volts but directed at -90掳, implying it lags the reference phase or is advanced depending on the context.
Phasors allow engineers to plot these voltages in a circular coordinate system, visualizing the angles and how each phasor relates spatially.
Balanced Circuit Analysis
**Balanced circuit analysis** plays a crucial role in understanding and managing three-phase systems. A circuit is balanced when all three phases deliver equal magnitudes and are separated by 120掳.
This balance ensures all three voltages exhibit the same amplitude and a consistent phase angle difference. This uniformity minimizes circulating currents and maximizes power efficiency.
  • In the current exercise, the given voltages show that the circuit tries to maintain balance with each voltage phasor separated by approximately 120掳.
  • The calculation of the third phasor from \( \mathbf{V}_{an} \) and \( \mathbf{V}_{cn} \) demonstrates how balance dictates the phase angle of \( \mathbf{V}_{bn} \).
In practical scenarios, balanced circuits help avoid issues like overheating and ensure machines operate smoothly by sharing electrical load evenly across all phases.

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Most popular questions from this chapter

A balanced delta-connected source has phase voltage \(\mathbf{V}_{a b}=416 / 30^{\circ} \mathrm{V}\) and a positive phase sequence. If this is connected to a balanced delta-connected load, find the line and phase currents. Take the load impedance per phase as \(60 / 30^{\circ} \Omega\) and line impedance per phase as \(1+j 1 \Omega\)

In a wye-delta three-phase circuit, the source is a balanced, positive phase sequence with \(\mathbf{V}_{a n}=120 / 0^{\circ} \mathrm{V} .\) It feeds a balanced load with \(\mathbf{Z}_{\Delta}=9+\overline{j 12} \Omega\) per phase through a balanced line with \(\mathbf{Z}_{\ell}=1+j 0.5 \Omega\) per phase. Calculate the phase voltages and currents in the load.

A four-wire wye-wye circuit has $$\begin{array}{c} \mathbf{V}_{a n}=120 / 120^{\circ}, \quad \mathbf{V}_{b n}=120 / 0^{\circ} \\ \mathbf{V}_{c n}=120 /-120^{\circ} \mathrm{V} \end{array}$$ If the impedances are \\[ \begin{array}{c} \mathbf{Z}_{A N}=20 / 60^{\circ}, \quad \mathbf{Z}_{B N}=30 / 0^{\circ} \\ \mathbf{Z}_{c n}=40 / 30^{\circ} \Omega \end{array} \\] find the current in the neutral line.

A balanced wye-connected load absorbs \(50 \mathrm{kVA}\) at a 0.6 lagging power factor when the line voltage is \(440 \mathrm{V}\). Find the line current and the phase impedance.

A delta-connected generator supplies a balanced wye-connected load with an impedance of \(30 /-60^{\circ} \Omega .\) If the line voltages of the generator have a magnitude of \(400 \mathrm{V}\) and are in the positive phase sequence, find the line current \(I_{L}\) and phase voltage \(V_{p}\) at the load.
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