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For each set of voltages, state whether or not the voltages form a balanced three-phase set. If the set is balanced, state whether the phase sequence is positive or negative. If the set is not balanced, explain why. a) \(v_{\mathrm{a}}=339 \cos 377 t \mathrm{V}\) \(v_{\mathrm{b}}=339 \cos \left(377 t-120^{\circ}\right) \mathrm{V}\) \(v_{\mathrm{c}}=339 \cos \left(377 t+120^{\circ}\right) \mathrm{V}\) b) \(v_{\mathrm{a}}=622 \sin 377 t \mathrm{V}\) \(v_{b}=622 \sin \left(377 t-240^{\circ}\right) \mathrm{V}\) \(v_{\mathrm{c}}=622 \sin \left(377 t+240^{\circ}\right) \mathrm{V}\) c) \(v_{\mathrm{a}}=933 \sin 377 t \mathrm{V}\) \(v_{b}=933 \sin \left(377 t+240^{\circ}\right) \mathrm{V}\) \(v_{\mathrm{c}}=933 \cos \left(377 t+30^{\circ}\right) \mathrm{V}\) d) \(v_{a}=170 \sin \left(\omega t+60^{\circ}\right) \mathrm{V}\) \(v_{b}=170 \sin \left(\omega t+180^{\prime \prime}\right) V\) \(v_{c}=170 \cos \left(\omega t-150^{\circ}\right) \mathrm{V}\) e) \(v_{a}=339 \cos \left(\omega t+30^{\circ}\right) \mathrm{V}\) \(v_{b}=339 \cos \left(\omega t-90^{\circ}\right) V\) \(v_{c}=393 \cos \left(\omega t+240^{\circ}\right) \mathrm{V}\) f) \(v_{\mathrm{a}}=3394 \sin \left(\omega t+70^{\circ}\right) \mathrm{V}\) \(v_{b}=3394 \cos \left(\omega t-140^{\circ}\right) \mathrm{V}\) \(v_{c}=3394 \cos \left(\omega t+180^{\circ}\right) \mathrm{V}\)

Short Answer

Expert verified
a) Balanced, positive. b) Balanced, negative. c-f) Not balanced due to mixed functions and phase inconsistencies.

Step by step solution

01

Analyze Set (a) for Balance

For set (a), all voltages have the same amplitude of 339 V. The phase angles are 0°, -120°, and +120°, where 120° is the standard phase shift for a balanced set. The voltages are described using cosine, so these represent sinusoidal waveforms where a full cycle is divided into three phases at equal intervals. This set appears balanced.
02

Determine Phase Sequence (a)

To find the phase sequence, we observe that gold the sequence starts at t=0, with phases va, vb shifted by -120°, and vc by +120°. Since they are equally spaced, we check the order. va leads vb, which leads vc. This is a positive phase sequence (abc).
03

Analyze Set (b) for Balance

In set (b), the amplitudes are constant at 622 V, and the sinusoidal functions are given with phase shifts of 0°, -240°, and +240°. Convert angles to a common equivalent: 0° becomes 0°, -240° is equivalent to -240° + 360° = 120°, and +240° = -120° (mod 360°). The set is balanced with a negative phase sequence (acb).
04

Analyze Set (c) for Balance

In set (c), observe different functions (sine and cosine) being used, and different phases: sine wave at 0°, another sine at +240°, and cosine at +30°. These differences mean the set is not balanced, since not all voltages share the same function and the standard phase differences.
05

Analyze Set (d) for Balance

Set (d) includes sinusoidal and cosine waveforms with differing amplitudes and phase shifts: +60° for va, +180° for vb, and -150° for vc cosine. Similar inconsistencies—mixed functions and incorrect spacing—mean it is not balanced.
06

Analyze Set (e) for Balance

Set (e) presents voltages with differing amplitudes (339 V for va and vb, 393 V for vc) and phase shifts of +30°, -90°, and +240°. Discrepancies in amplitudes and improper 120° phase shifts result in an unbalanced set.
07

Analyze Set (f) for Balance

In set (f), volatage va is given by sine with phase {+70°} but voltages vb and vc by cosine, which means that not the same waveform is used for all voltages. Phases do not have a consistent pattern of offsets as well, leading to it not being a balanced set.

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Key Concepts

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

Balanced systems
In three-phase electrical systems, a balanced system is one where all three phase voltages have equal magnitude and are separated by a phase difference of 120°. This configuration helps to ensure that the power supply remains consistent and stable, which is particularly important for large-scale power distribution.

In a balanced setup, the voltages are designed to cancel out inherent imbalances, ensuring that the system operates efficiently. Balancing minimizes current in the neutral wire, reduces power losses, and improves overall system stability.

For instance, in the provided exercise, Set (a) is an example of a balanced system because all voltage magnitudes are the same (339 V) and phase differences are exactly 120°. This is critical for system reliability and efficiency.
Phase sequence
Phase sequence refers to the order in which the voltages reach their maximum values. It is crucial to understand the phase sequence, as it influences the direction and timing of power flow in an AC system.

In a three-phase system, the phase sequence can be either positive or negative, which is represented as 'abc' or 'acb'.
  • **Positive phase sequence (abc)** involves phase 'a' leading phase 'b', which in turn leads phase 'c'.
  • **Negative phase sequence (acb)** refers to phase 'a' leading phase 'c', which then leads phase 'b'.

For example, in Set (a) of the exercise, the phase sequence is positive because voltage \(v_a\) leads \(v_b\), and \(v_b\) leads \(v_c\), maintaining the 'abc' order.
Sine and cosine representation
In electrical engineering, using sine and cosine functions to represent voltages is common practice. It helps in analyzing how these waveforms interact over time.

Sine and cosine functions describe the sinusoidal waveforms of AC voltages. Their use is essential because they provide a straightforward way to model the time-dependent behavior of AC voltages.

In the exercise, this representation is evident in Set (a) and (b), where cosine and sine functions are used respectively. The choice between sine and cosine may influence the phase angle, due to the inherent phase shift between these two functions: cosine leads sine by 90°.
Amplitude and phase angle
Amplitude represents the peak voltage value in a three-phase system and is crucial for determining the power capacity of the system. All three phase voltages should ideally have the same amplitude in a balanced system. Phase angle, on the other hand, refers to the initial angle at which the waveform commences. It plays a pivotal role in defining the time offset between voltage peaks among the phases. For instance, in the exercise's Set (a), the amplitude is consistent at 339 V, highlighting a balanced system. Phase angles differ by 120°, ensuring the phases are correctly separated to maintain balance.

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Most popular questions from this chapter

A balanced, three-phase circuit is characterized as follows: Source voltage in the b-phase is \(20 \angle-90^{\circ} \mathrm{V}\) Source phase sequence is acb; Line impedance is \(1+j 3 \Omega / \phi\) Load impedance is \(117-j 99 \Omega / \phi\) a) Draw the single phase equivalent for the a-phase. b) Calculated the a-phase line current. c) Calculated the a-phase line voltage for the three-phase load.

The three pieces of computer equipment described below are installed as part of a computation center. Each piece of equipment is a balanced three-phase load rated at \(208 \mathrm{V}\). Calculate (a) the magnitude of the line current supplying these three devices and (b) the power factor of the combined load. Hard Drive: \(5.742 \mathrm{kW}\) at 0.82 pf lag CD/DVD drive: \(18.566 \mathrm{kVA}\) at 0.93 pf lag CPU: line current \(81.6 \mathrm{A}, 11.623 \mathrm{kVAR}\)

The total power delivered to a balanced throe phase load when operating at a line voliage of \(6600 \sqrt{3} \mathrm{V}\) is \(1188 \mathrm{kW}\) at a lagging power factor of \(0.6 .\) The impedance of the distribution line ing the load is \(0.5+j 4 \Omega / \phi\). Under these supply operating conditions, the drop in the magnitude of the line voltage between the sending end and the load end of the line is excessive. To compensate, a bank of A-connected capacitors is placed in parallel with the load. The capacitor bank is designed to furnish \(1920 \mathrm{kVAR}\) of magnetizing reactive power when operated at a line voltage of \(6600 \sqrt{3} \mathrm{V}\) a) What is the magnitude of the voltage at the sending end of the line when the load is operat. ing at a line voltage of \(6600 \sqrt{3} \mathrm{V}\) and the capac. itor bank is disconnected? b) Repeat (a) with the capacitor bank connected c) What is the average power efficiency of the line in (a)? d) What is the average power efficiency in \((b) ?\) e) If the system is operating at a frequency of \(60 \mathrm{Hz}\) what is the size of each capacitor in microfarads?

What is the phase sequence of each of the following sets of voltages? a)$$\begin{array}{l} v_{\mathrm{a}}=120 \cos \left(\omega t+54^{\circ}\right) \mathrm{V} \\ v_{\mathrm{b}}=120 \cos \left(\omega t-66^{\circ}\right) \mathrm{V} \\ v_{\mathrm{c}}=120 \cos \left(\omega t+174^{\circ}\right) \mathrm{V} \end{array}$$ b)$$\begin{aligned} &v_{\mathrm{a}}=3240 \cos \left(\omega t-26^{\circ}\right) \mathrm{V}\\\ &v_{b}=3240 \cos \left(\omega t+94^{\circ}\right) \mathrm{V}\\\ &v_{\mathrm{c}}=3240 \cos \left(\omega t-146^{\circ}\right) \mathrm{V} \end{aligned}$$

In a balanced three-phase system, the source has an abc sequence, is Y-connected, and \(\mathbf{V}_{\mathrm{an}}=120 / 20^{\circ} \mathrm{V} .\) The source feeds two loads, both of which are \(Y\) -connected. The impedance of load 1 is \(8+j 6 \Omega / \phi\). The complex power for the a \(\cdot\) phase of load 2 is \(600 / 36^{\circ} \mathrm{VA}\). Find the total complex power supplied by the source.

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