Problem 93 A three -phase source has $$\b... [FREE SOLUTION]

Chapter 5: Problem 93

A three -phase source has $$\begin{aligned} v_{a r}(t) &=100 \cos \left(\omega t-60^{\circ}\right) \mathrm{V} \\ v_{b n}(t) &=100 \cos \left(\omega t+60^{\circ}\right) \mathrm{V} \\ v_{c n}(t) &=-100 \cos (\omega t) \mathrm{V} \end{aligned}$$ Is this a positive-sequence or a negativesequence source? Find time-domain expressions for $v_{a b}(t), v_{b c}(t),$ and $v_{c u}(t)$

Short Answer

Expert verified

The source is a negative-sequence. The voltages are: $v_{ab}(t) = -100 \sqrt{3} \sin(\omega t)$, $v_{bc}(t) = 100 \sqrt{3} \cos(\omega t + 30^\circ)$, $v_{ca}(t) = -100 \sqrt{3} \cos(\omega t - 30^\circ)$.

Step by step solution

Understand Phase Sequence

To determine if the source is positive-sequence or negative-sequence, we analyze the phase angles of the voltages. Positive-sequence has a phase progression of 120掳 in the order of a-b-c, while negative-sequence has the reverse order, c-b-a.

Determine Phase Sequence

The given phase angles are: \[ \begin{aligned}v_{ar} & : -60^\circ, \v_{bn} & : +60^\circ, \v_{cn} & : 180^\circ \end{aligned} \]The order of the phase progression here is a-c-b, indicating a negative-phase sequence.

Find Line-to-Line Voltage $v_{ab}(t)$

The voltage $v_{ab}(t)$ is the difference between $v_{ar}(t)$ and $v_{bn}(t)$:\[ \begin{aligned}v_{ab}(t) & = v_{ar}(t) - v_{bn}(t) \& = 100 \cos(\omega t - 60^\circ) - 100 \cos(\omega t + 60^\circ) \& = 100\left[\cos(\omega t - 60^\circ) - \cos(\omega t + 60^\circ)\right] \& = 100(-2 \sin(\omega t) \sin(60^\circ)) \& = -100 \sqrt{3} \sin(\omega t) \end{aligned} \]

Find Line-to-Line Voltage $v_{bc}(t)$

The voltage $v_{bc}(t)$ is the difference between $v_{bn}(t)$ and $v_{cn}(t)$:\[ \begin{aligned}v_{bc}(t) & = v_{bn}(t) - v_{cn}(t) \& = 100 \cos(\omega t + 60^\circ) + 100 \cos(\omega t) \& = 100 \left[\cos(\omega t + 60^\circ) + \cos(\omega t) \right] \& = 100 \left[2 \cos\left(\omega t + 30^\circ\right) \cos\left(30^\circ\right)\right] \& = 100 \sqrt{3} \cos(\omega t + 30^\circ) \end{aligned} \]

Find Line-to-Line Voltage $v_{ca}(t)$

The voltage $v_{ca}(t)$ is the difference between $v_{cn}(t)$ and $v_{ar}(t)$:\[ \begin{aligned}v_{ca}(t) & = v_{cn}(t) - v_{ar}(t) \& = -100 \cos(\omega t) - 100 \cos(\omega t - 60^\circ) \& = -100 \left[\cos(\omega t) + \cos(\omega t - 60^\circ) \right] \& = -100 \left[2 \cos\left(\omega t - 30^\circ\right) \cos\left(30^\circ\right)\right] \& = -100 \sqrt{3} \cos(\omega t - 30^\circ) \end{aligned} \]

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Key Concepts

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

Phase Sequence

In a three-phase electrical system, the phase sequence refers to the order in which the voltage waves reach their peak values. This order is crucial for determining the operational characteristics of the system. There are two main types of phase sequences: positive-sequence and negative-sequence.

Positive-Sequence: In this sequence, the phase voltages progress in the order of a-b-c with a 120掳 phase shift between each. This sequence is the typical setup used in power systems because it enables efficient energy transfer and produces balanced loads.
Negative-Sequence: This sequence follows the reverse order c-b-a. It can cause imbalances and is generally avoided, but it is essential for certain types of fault analysis.

Analyzing the given problem, the phase angles are:

$v_{ar}$ : -60掳
$v_{bn}$ : +60掳
$v_{cn}$ : 180掳

The progression order of a-c-b indicates a negative-phase sequence for this three-phase source.

Line-to-Line Voltage

Line-to-line voltage is the voltage difference between two phases in a three-phase system. It is essential for understanding how voltage behaves in interconnected systems. Knowing how to calculate line-to-line voltage helps in designing and analyzing circuits correctly. To find the line-to-line voltage, you take the difference between the phase voltages:

v_ab(t): This is the difference between $v_{ar}(t)$ and $v_{bn}(t)$. Calculated as $-100 \sqrt{3} \sin(\omega t)$, it shows the effective potential difference between phases 'a' and 'b'.
v_bc(t): This is obtained by subtracting $v_{cn}(t)$ from $v_{bn}(t)$, resulting in $100 \sqrt{3} \cos(\omega t + 30^\circ)$. This measures the voltage difference from 'b' to 'c'.
v_ca(t): Found by subtracting $v_{ar}(t)$ from $v_{cn}(t)$, giving $-100 \sqrt{3} \cos(\omega t - 30^\circ)$. It describes the voltage across 'c' to 'a'.

Accurate line-to-line voltage calculation is essential for ensuring the system operates correctly under both normal and fault conditions.

Positive-Sequence

Within three-phase electrical systems, positive-sequence refers to a set of balanced, evenly spaced voltage or current that circulates in a cycle with a phase difference of 120掳 and follows the specific order a-b-c. Positive-sequence components are significant in power systems for several reasons:

System Balance: Ensures that power is distributed evenly to all phases, minimizing losses and ensuring the load is balanced as expected.
Optimal Performance: Generally indicates the system is operating under normal conditions with pairs of equal magnitude and are spaced by 120掳.
Fault Analysis: Useful in determining the behavior of the system under fault conditions, especially when balanced systems are temporarily unbalanced.

However, the provided problem exhibits a negative-phase sequence rather than a positive one, which normally indicates fault conditions or asymmetric loading scenarios that warrant careful engineering attention.

Negative-Sequence

The presence of negative-sequence components can drastically affect a three-phase system. These are voltage or current components that circulate in a reverse sequence of c-b-a and suggest that the system is experiencing an imbalance. Negative-sequence has several implications:

Asymmetry Detection: It helps in identifying unbalances in the system which might be due to asymmetrical loads or faulty equipment.
Motor Damage Risk: Negative-sequence currents can overheat motors and decrease their performance due to the production of additional reverse torque.
Protection Systems: Electrical protection schemes utilize negative-sequence components to detect and isolate faults quickly, minimizing potential damages.

In the discussed problem, the sequence noted is a-c-b, implicating negative-sequence properties. It's a noteworthy condition because continuous presence of these components necessitates corrective measures to protect system integrity.

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Short Answer

Step by step solution

Understand Phase Sequence

Determine Phase Sequence

Find Line-to-Line Voltage \(v_{ab}(t)\)

Find Line-to-Line Voltage \(v_{bc}(t)\)

Find Line-to-Line Voltage \(v_{ca}(t)\)

Key Concepts

Phase Sequence

Line-to-Line Voltage

Positive-Sequence

Negative-Sequence

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