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The total apparent power supplied in a balanced. three-phase \(Y\) - \(\Delta\) system is 3600 VA. The line volit age is 208 V. If the line impedance is negligible and the power factor angle of the load is \(25^{\circ}\), determine the impedance of the load.

Short Answer

Expert verified
The impedance of the load is 12.009 Ω.

Step by step solution

01

Understand Apparent Power in Three-Phase Systems

In a balanced three-phase system, the apparent power (S) is given as the total power supplied. Here, the apparent power is 3600 VA. The line voltage in the system is given as 208 V.
02

Calculate the Phase Voltage

For a balanced Y-connected system, the phase voltage (V_ph) is the line voltage divided by the square root of 3. Given line voltage (V_line) is 208 V, the phase voltage can be calculated as: \[ V_{ph} = \frac{V_{line}}{\sqrt{3}} = \frac{208}{\sqrt{3}} \approx 120.09 \text{ V} \]
03

Determine the Power Factor

The power factor angle is given as 25°, which allows us to compute the power factor (cos φ) using the cosine of the angle. \[ \text{Power Factor (PF)} = \cos(25^{\circ}) = 0.9063 \]
04

Calculate the Real Power

The real power (P) can be calculated from apparent power (S) and the power factor (PF):\[ P = S \times \text{PF} = 3600 \times 0.9063 = 3262.68 \text{ W} \]
05

Calculate the Line Current

Since the line impedance is negligible, the line current (I) can be calculated using apparent power and line voltage:\[ S = \sqrt{3} \times V_{line} \times I \] Solving for I:\[ I = \frac{S}{\sqrt{3} \times V_{line}} = \frac{3600}{\sqrt{3} \times 208} \approx 10 \text{ A} \]
06

Calculate Load Impedance

The load impedance (Z_load) can be found using the phase voltage and the line current, as\[ Z_{load} = \frac{V_{ph}}{I} = \frac{120.09}{10} = 12.009 \Omega \]

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Key Concepts

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

Apparent Power
In electrical engineering, apparent power is a crucial concept, especially within three-phase systems. Apparent power, denoted as S, combines both real power and reactive power in a power system. This is expressed in Volt-Amperes (VA). In a three-phase system, the apparent power is given as the total power supplied, which in this scenario is 3600 VA.
- Apparent power signifies the total flowing from the power source to the load.
- It is a measure of overall power without distinguishing how much of that power is used effectively. Understanding apparent power is essential because it helps in analyzing electrical circuits and ensures efficient energy distribution across power networks. Given its total nature, engineers and technicians often use this value to design and assess power systems effectively.
Power Factor
The power factor is a key parameter in power systems that represents the efficiency of an electrical circuit. It is the cosine of the power factor angle, often expressed as \(\cos \phi\), where \(\phi\) is the phase angle between the current and voltage.
- A power factor of 1 indicates all the supplied power is being effectively used.
- In this exercise, the power factor is calculated from the angle of \(25^{\circ}\), resulting in a power factor of approximately 0.9063.A noticeable power factor indicates the degree to which a system's voltage and current are out of phase. In general, having a higher power factor is desired because it means more of the appliance's power is being used to do useful work, which supports energy efficiency goals.
Load Impedance
Load impedance, often referred to as \(Z\), in electrical circuits defines how much opposition there's to the current flow through the load. In systems, especially three-phase, it's crucial for determining how devices and appliances use supplied power. Load impedance is measured in Ohms (\(\Omega\)).
- In a perfectly balanced system, the load impedance provides insight into both resistance and reactance present.
- Calculating the load impedance involves dividing phase voltage by the line current, which yields an impedance of \(12.009 \Omega\) for the given system.This value helps to inform actionable insights into the performance and efficiency of the load, making it fundamental for electrical calculations and assessments. Proper measurement and adjustments of load impedance can prevent issues such as power loss and maximize system efficiency.
Phase Voltage
Phase voltage refers to the voltage across a single component within a phase of the system. In a Y-connected system, it is derived from the line voltage and is integral for calculating many other electrical properties.
- For Y-systems, phase voltage can be calculated by dividing line voltage by the square root of 3.
- Using a line voltage of 208 V, the phase voltage is calculated to roughly 120.09 V. Understanding phase voltage is critical, especially when designing and evaluating three-phase systems, as it directly affects the other parameters of the electrical system, like current and impedance. Ensuring that phase voltage is correct helps maintain balanced systems and prevents malfunctions or inefficiencies.

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Most popular questions from this chapter

The magnitude of the line voltage at the terminals of a balanced \(Y\) -connected load is 12,800 V. The load impedance is \(216+j 63 \Omega / \phi\). The load is fed from a line that has an impedance of \(0.25+j 2 \Omega / \phi\) a) What is the magnitude of the line current? b) What is the magnitude of the line voltage at the source?

The time-domain expressions for three line-to-neutral voltages at the terminals of a Y-connected load are $$\begin{array}{l} v_{\mathrm{AN}}=7967 \cos \omega t \mathrm{V} \\ v_{\mathrm{BN}}=7967 \cos \left(\omega t+120^{\circ}\right) \mathrm{V} \\ v_{\mathrm{CN}}=7967 \cos \left(\omega t-120^{\circ}\right) \mathrm{V} \end{array}$$ What are the time-domain expressions for the three line-to-line voltages \(v_{\mathrm{AB}}, v_{\mathrm{BC}},\) and \(v_{\mathrm{CA}} ?\)

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.

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 \(81.6 \mathrm{A}, 11.623 \mathrm{kVAR}\)

An abc sequence balanced three-phase \(Y\) -connected source supplies power to a balanced, three-phase \(\Delta\) -connected load with an impedance of \(12+j 9 \Omega / \phi\). The source voltage in the a-phase is \(120 / 80^{\circ} \mathrm{V} .\) The line impedance is \(1+\mathrm{j} 1 \Omega / \phi\) Draw the single phase equivalent circuit for the a-phase and use it to find the current in the a-phase of the load.

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