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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.

Short Answer

Expert verified
The total complex power supplied by the source is 6102.12 VA at 36.29⁰.

Step by step solution

01

Analyze Given Values

We are given a three-phase source with a line-to-neutral voltage \( \mathbf{V}_{\mathrm{an}} = 120 / 20^{\circ} \space \text{V} \). The load impedances are for Load 1 \( Z_1 = 8 + j6 \space \Omega/\phi \) and the complex power for phase a of Load 2 is \( S_{a2} = 600 / 36^{\circ} \space \text{VA} \).
02

Calculate Load 1 Phase Current

The phase current for Load 1 can be calculated using Ohm's law: \( I_a1 = \frac{\mathbf{V}_{\mathrm{an}}}{Z_1} = \frac{120 / 20^{\circ}}{8 + j6} \). First, find the magnitude and phase of the impedance:\[|Z_1| = \sqrt{8^2 + 6^2} = 10 \quad \text{and} \quad \phi_1 = \tan^{-1}\left(\frac{6}{8}\right) = 36.87^{\circ} \]Thus, \( Z_1 = 10 / 36.87^{\circ} \space \Omega \), and \[ I_a1 = \frac{120 / 20^{\circ}}{10 / 36.87^{\circ}} = 12 / (-16.87^{\circ}) \space \text{A} \]
03

Calculate Load 1 Power

Calculate the complex power for Load 1 using \( S_{1} = \mathbf{V}_{\mathrm{an}} \cdot I_{a1}^* \), where \(^*\) denotes the conjugate:\[ S_1 = 120 / 20^{\circ} \times 12 / 16.87^{\circ} = 1440 / 36.87^{\circ} \quad \text{VA} \]
04

Calculate Total Complex Power of Load 1

Since the load is Y-connected and balanced, the total power for Load 1 is three times the phase power:\[ S_{\text{total, 1}} = 3 \times 1440 / 36.87^{\circ} \quad \text{VA} \]
05

Calculate Total Complex Power Supplied

Add the complex power of Load 1 and Load 2:For Load 2, as it only provided power per phase, the total complex power is:\[ S_{\text{total, 2}} = 3 \times 600 / 36^{\circ} \quad \text{VA} \]The total complex power from the source is:\[ S_{\text{total}} = S_{\text{total, 1}} + S_{\text{total, 2}} \]
06

Perform Final Complex Addition

Perform the complex power addition to find the total:\[ S_{\text{total, 1}} = 3 \times 1440 / 36.87^{\circ} = 4320 / 36.87^{\circ} \quad \text{VA} \]\[ S_{\text{total, 2}} = 1800 / 36^{\circ} \quad \text{VA} \]Using rectangular form for addition:\( S_{\text{total, 1}} = 3445.61 + j2593.46 \; VA \)\( S_{\text{total, 2}} = 1454.09 + j1059.39 \; VA \)Final total:\[ S_{\text{total}} = (3445.61 + 1454.09) + j(2593.46 + 1059.39) \]\[ S_{\text{total}} = 4900 + j3652.85 \quad \text{VA} \]
07

Convert to Polar Form

Convert final total back to polar form:\( |S_{\text{total}}| = \sqrt{4900^2 + 3652.85^2} \)\( \phi_{\text{total}} = \tan^{-1}{\left(\frac{3652.85}{4900}\right)} \)Therefore, the polar form is \( S_{\text{total}} = 6102.12 / 36.29^{\circ} \space \text{VA} \).

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Key Concepts

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

Complex Power Calculation
Complex power is a vital concept in electrical engineering, especially when dealing with AC (Alternating Current) circuits. It combines both real power, measured in watts (W), and reactive power, measured in volt-amperes reactive (VAR), into a single complex number. This number is called complex power and is expressed in volt-amperes (VA).
To calculate complex power, you use the formula:
  • \( S = P + jQ \)
  • \( S = VI^* \)
where \( P \) is real power, \( Q \) is reactive power, \( V \) is the voltage, \( I \) is the current, and \( I^* \) is the complex conjugate of the current. The magnitude of the complex power (\( |S| \) ) gives the apparent power in the circuit.
In a three-phase system, particularly when loads are balanced and the sources are symmetric, the complex power per phase can be simplified as proportionate to total power. This simplifies the computation and helps in straightforward analysis.
Y-Connected Loads
Y-connected loads, also known as star-connected loads, form a 'Y' shape in three-phase electrical systems. They are popular in industrial applications because of their efficiency in distributing power.
Each load in the Y configuration connects to a single neutral point, meaning:
  • The voltage across each phase of the load is equal to the line-to-neutral voltage.
  • Total power in such configurations is calculated by multiplying the power in one phase by three.
This connectivity reduces the complexity of the phase current calculation. It is evident in balanced systems, such as those explored in the original exercise. Y-connections make complex power calculation simpler as we scale phase calculations by a factor of three.
Ohm's Law
Ohm's Law plays a foundational role in circuit analysis. It states that the current through a conductor between two points is directly proportional to the voltage across the two points. Ohm's law is typically expressed as:
  • \( V = IZ \)
where \( V \) is voltage, \( I \) is current, and \( Z \) is impedance. In AC circuits, impedance is a complex quantity, consisting of real and imaginary components, which accounts for loss and reactive effects respectively.
In the context of our exercise, Ohm's Law was used to find the phase current by dividing the voltage by the impedance of Load 1. This step is crucial to understanding load behavior and necessary for calculating the complex power. By understanding the relationship between voltage, current, and resistance, students can solve for unknown quantities in circuit analysis.
Impedance
Impedance is a broader concept than resistance, as it applies to AC circuits and includes resistance and reactance. Measured in ohms (\( \Omega \)), impedance \( Z \) can be expressed in complex form as:
  • \( Z = R + jX \)
where \( R \) is resistance and \( X \) is reactance. The magnitude of impedance is given by \(|Z| = \sqrt{R^2 + X^2}\), and the phase angle \( \phi \) is calculated using \( \tan^{-1}(\frac{X}{R}) \).
Understanding impedance helps determine how effectively a circuit resists and stores energy. In the exercise, solving impedance values allowed for current phase angles to be calculated, relevant in power calculations. With this understanding, students tackle circuit analysis confidently by calculating the current and subsequently the power values.
Three-Phase Circuits
Three-phase systems are an integral part of power distribution, particularly in industrial settings. They consist of three electrical circuits carrying current of the same frequency and amplitude but with a phase difference of 120 degrees.
The advantages of three-phase circuits include:
  • More efficient power transmission.
  • Constant power delivery, reducing pulsations found in single-phase systems.
  • Reduced conductor material requirements.
In practice, these systems provide steady and reliable power since any cost-effective and high-power applications depend significantly on them.
The exercise reveals key insights into the balanced three-phase systems where calculations for power require symmetry in voltage and phase angles, simplifying analysis. These circuits prove to sustain efficiency and cost-effectiveness over single-phase circuits due to their inherent design and operational benefits.

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Most popular questions from this chapter

A balanced three-phase source is supplying \(90 \mathrm{kVA}\) at 0.8 lagging to two balanced Y-connected parallel loads. The distribution line connecting the source to the load has negligible impedance. Load 1 is purely resistive and absorbs \(60 \mathrm{kW}\). Find the per-phase impedance of Load 2 if the line voltage is \(415.69 \mathrm{V}\) and the impedance components are in series.

A balanced three-phase circuit has the following characteristics: Y-Y connected; The line voltage at the source, \(V_{b b, i}\) \(120 \sqrt{3} / 0^{\circ} \mathrm{V}\) The phase sequence is positive; The line impedance is \(2+j 3 \Omega / \phi\) The load impedance is \(28+j 37 \Omega /\) a) Draw the single phase equivalent circuit for the a-phase b) Calculated the line current in the a-phase c) Calculated the line voltage at the load in the a-phase

A balanced three-phase distribution line has an impedance of \(1+j 5 \Omega / \phi\). This line is used to supply three balanced three-phase loads that are connected in parallel. The three loads are \(\mathrm{L}_{1}=75 \mathrm{kVA}\) at 0.96 pf \(\operatorname{lead}, \mathrm{L}_{2}=150 \mathrm{kVA}\) at 0.80 pf \(\operatorname{lag},\) and \(\mathrm{L}_{3}=168 \mathrm{kW}\) and \(36 \mathrm{kVAR}(\mathrm{mag}-\) netizing). The magnitude of the line voltage at the terminals of the loads is \(2500 \sqrt{3} \mathrm{V}\) a) What is the magnitude of the line voltage at the sending end of the line? b) What is the percent efficiency of the distribution line with respect to average power?

Show that the total instantaneous power in a balanced three-phase circuit is constant and equal to \(1.5 V_{m} I_{m} \cos \theta_{\phi},\) where \(V_{m}\) and \(I_{m}\) represent the maximum amplitudes of the phase voltage and phase current, respectively.

A three-phase \(\Delta\) -connected generator has an internal impedance of \(0.6+j 4.8 \Omega / \phi .\) When the load is removed from the generator, the magnitude of the terminal voltage is \(34,500 \mathrm{V}\). The generator feeds a \(\Delta\) -connected load through a transmission line with an impedance of \(0.8+j 6.4 \Omega / \phi\). The per-phase impedance of the load is \(2877-j 864 \Omega\) a) Construct a single-phase equivalent circuit. b) Calculate the magnitude of the line current. c) Calculate the magnitude of the line voltage at the terminals of the load. d) Calculate the magnitude of the line voltage at the terminals of the source. e) Calculate the magnitude of the phase current in the load. f ) Calculate the magnitude of the phase current in the source.

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