Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A balanced three-phase distribution line has an impedance of 1+j5Ω/ϕ. This line is used to supply three balanced three-phase loads that are connected in parallel. The three loads are L1=75kVA at 0.96 pf lead,L2=150kVA at 0.80 pf lag, and L3=168kW and 36kVAR(mag netizing). The magnitude of the line voltage at the terminals of the loads is 25003V 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?

Short Answer

Expert verified
Use complex power to find total line losses; add it to load voltages to find sending voltage.

Step by step solution

01

Calculate Load Currents

For each load, we need to find the current drawn from the supply. Convert power factor and apparent power to complex power using the formula: S=P+jQ, where S is the complex power, P is real power, and Q is reactive power. Use the equation S=VLL×I where I is the complex conjugate of the current.1. **Load L1**: S1=75kVA at 0.96 pf lead: - Q1=S1×sin(cos1(0.96))=75×sin(cos1(0.96))=75×0.28=21kVAR - P1=S1×pf=75×0.96=72kW - Current: I1=S13×VLL=753×2500=17.3A16.26°2. **Load L2**: S2=150kVA at 0.80 pf lag: - Q2=S2×sin(cos1(0.8))=150×sin(cos1(0.8))=150×0.6=90kVAR - P2=S2×pf=150×0.8=120kW - Current: I2=S23×VLL=1503×2500=34.64A36.87°3. **Load L3**: P3=168kW,Q3=36kVAR magnetizing: - S3=P32+Q32=1682+362=171.8kVA - Current: I3=S33×VLL=171.83×2500=39.6A11.85° (lagging due to magnetizing)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Three-Phase Loads
When dealing with electric power systems, three-phase loads are commonly used due to their efficiency and reliability. In a three-phase system, electrical power is delivered through three wires, each carrying alternating current, with a phase difference of 120 degrees between each pair.
This setup offers a more consistent power supply and allows for smaller and more cost-effective conductors compared to single-phase systems.
Three-phase loads often consist of multiple devices connected in parallel, drawing power from a balanced three-phase distribution line. Each load's power demand is represented in either kilovolt-amperes (kVA) or kilowatts (kW), along with their power factor.
  • Load L1 draws 75 kVA at a 0.96 leading power factor.
  • Load L2 draws 150 kVA at a 0.80 lagging power factor.
  • Load L3 draws real power of 168 kW with reactive power of 36 kVAR.
This configuration allows efficient transmission of power to multiple devices or systems running concurrently.
Power Factor
The power factor is a crucial concept in the optimization of electric power systems. It measures how effectively electrical power is converted into useful work output and is typically expressed as a decimal between 0 and 1.
The power factor is calculated by dividing the real power (in kW) by the apparent power (in kVA) used in the circuit. A power factor of 1 (or 100%) indicates that all the energy supplied by the source is being effectively used for productive work.
Loads with a low power factor draw more current than necessary to achieve the same output, leading to energy inefficiencies. Understanding and correcting power factor in a system can help reduce energy costs and increase system capacity.
  • Leading power factors (e.g., 0.96 lead) are typically associated with capacitive loads.
  • Lagging power factors (e.g., 0.80 lag) are often found in inductive loads.
A high power factor is desired to increase the efficiency of power distribution.
Complex Power Calculation
In electric power systems, complex power is an essential concept that combines real and reactive power into a single phasor quantity. This phasor is represented as S=P+jQ, where P represents real power, and Q represents reactive power.
Real power (measured in kilowatts, kW) does the actual work, such as lighting, heating, or turning a motor. In contrast, reactive power (measured in kilovolt-amperes reactive, kVAR) supports the electric field necessary for certain equipment's magnetic functions, like transformers or motors.
  • Load L1 has a complex power S1 with 72 kW real power, and 21 kVAR reactive power.
  • Load L2 operates with a complex power S2 consisting of 120 kW real power and 90 kVAR reactive power.
  • Load L3 uses 168 kW of real power with a notable reactive power component of 36 kVAR.
Calculating complex power for each load helps determine the current drawn from the supply.
Line Voltage
Line voltage is a fundamental parameter in three-phase power systems. It refers to the voltage between any two lines in a three-phase system and is often denoted as VLL. For the three-phase loads in the exercise, the line voltage at the terminals is 25003 V.
The magnitude of line voltage is critical as it influences the current that flows through the system and the overall power distribution efficiency.
High line voltage can reduce current for a fixed power load, which decreases resistive losses and improves efficiency. Understanding line voltage and maintaining it within proper limits ensures the safe and efficient operation of electrical equipment.
The line voltage affects the calculated current for each load, highlighting its role in determining the overall stability of the power delivery system.
Distribution Efficiency
Distribution efficiency in an electric power system refers to the ratio of the average power delivered to the loads to the power supplied by the power station, expressed as a percentage. It measures how effectively the power is transmitted over the distribution line.
Electricity distribution often involves some losses due to the line's impedance and other factors like heat dissipation, which decreases the power system's efficiency.
  • The efficiency is calculated by comparing the total real power delivered to the loads to the input power required at the sending end of the distribution line.
  • Ensuring a high distribution efficiency minimizes operational costs and enhances energy utilization.
In scenarios where the power factor is high and impedance is low, reduced losses can lead to better efficiency, exemplified by our balanced three-phase loads.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The total power delivered to a balanced throe phase load when operating at a line voliage of 66003V is 1188kW at a lagging power factor of 0.6. The impedance of the distribution line ing the load is 0.5+j4Ω/ϕ. 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 1920kVAR of magnetizing reactive power when operated at a line voltage of 66003V 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 66003V 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 60Hz what is the size of each capacitor in microfarads?

A balanced three-phase circuit has the following characteristics: Y-Y connected; The line voltage at the source, Vbb,i 1203/0V The phase sequence is positive; The line impedance is 2+j3Ω/ϕ The load impedance is 28+j37Ω/ 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 three-phase Δ -connected generator has an internal impedance of 0.6+j4.8Ω/ϕ. When the load is removed from the generator, the magnitude of the terminal voltage is 34,500V. The generator feeds a Δ -connected load through a transmission line with an impedance of 0.8+j6.4Ω/ϕ. The per-phase impedance of the load is 2877j864Ω 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.

What is the phase sequence of each of the following sets of voltages? a)va=120cos(ωt+54)Vvb=120cos(ωt66)Vvc=120cos(ωt+174)V b)va=3240cos(ωt26)V vb=3240cos(ωt+94)V vc=3240cos(ωt146)V

The time-domain expressions for three line-to-neutral voltages at the terminals of a Y-connected load are vAN=7967cosωtVvBN=7967cos(ωt+120)VvCN=7967cos(ωt120)V What are the time-domain expressions for the three line-to-line voltages vAB,vBC, and vCA?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free