Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
A single-phase load is supplied through a feeder whose impedance is 95 + j360 N and a 35- kV/2400-V transformer whose equivalent impedance Zeg is 0.23 + j 1.27 Q referred to its low-voltage side. Use the approximate equivalent circuit of the practical transformer. The load is 160 kW at 0.89 leading power factor and 2340 V. Compute the phasor voltage Vs (expressed in polar form, in volts) at the sending end of the feeder. Refer all quantities to the high voltage side. Feoder Transformer Load 95 +j 360 0 160 kW Vs E1 E2 0.89 pf leading V = 2340 + jo v

A single-phase load is supplied through a feeder whose impedance is 95 + j360 N and a 35- kV/2400-V transformer whose equivalent impedance Zeg is 0.23 + j 1.27 Q referred to its low-voltage side. Use the approximate equivalent circuit of the practical transformer. The load is 160 kW at 0.89 leading power factor and 2340 V. Compute the phasor voltage Vs (expressed in polar form, in volts) at the sending end of the feeder. Refer all quantities to the high voltage side. Feoder Transformer Load 95 +j 360 0 160 kW Vs E1 E2 0.89 pf leading V = 2340 + jo v

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
See similar textbooks
icon
Related questions
Question
A single-phase load is supplied through a feeder whose impedance is 95 + j360 2 and a 35- kV/2400-V transformer whose equivalent impedance Zeq is 0.23 +j 1.27 N referred to its low-voltage side. Use the approximate equivalent circuit of the practical transformer. The load is 160 kW at 0.89 leading power factor and 2340 V. Compute the phasor voltage Vs (expressed in polar form, in volts) at the sending end of the feeder. Refer all quantities to the high voltage side. Feeder Transformer Load 95 +j 360 0 160 kW Vs E1 E2 0.89 pf leading VL = 2340 + jo Vv
Transcribed Image Text:A single-phase load is supplied through a feeder whose impedance is 95 + j360 2 and a 35- kV/2400-V transformer whose equivalent impedance Zeq is 0.23 +j 1.27 N referred to its low-voltage side. Use the approximate equivalent circuit of the practical transformer. The load is 160 kW at 0.89 leading power factor and 2340 V. Compute the phasor voltage Vs (expressed in polar form, in volts) at the sending end of the feeder. Refer all quantities to the high voltage side. Feeder Transformer Load 95 +j 360 0 160 kW Vs E1 E2 0.89 pf leading VL = 2340 + jo Vv
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 9 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question

How do I compute voltage at the high voltage terminal of the transformer

 

Solution
Bartleby Expert
SEE SOLUTION
Follow-up Question
K = 95.94 -36.88° A Therefore the load volt Vload - Load Zload (95.94 -36.88° A) (5 236.87⁰ (2) 479.7 Z-0.01° V And the line losses are given by; Ploss = (line) ² Ruine = (9.594 A)²(0.18 (2) 16.7 W 7| Page a. b. 4.0 TRANSFORMERS PDF Notice that by stepping up the transmission voltage of the power system, the transmission losses have been reduced by a factor of 90. In addition, the voltage at the load dropped significantly in the system with transformers compared to the system without transformers. a L Exercise: 95 + £360 A single-phase load is supplied through a 35-kV feeder whose impedance is 2 and a 35-kV:2400-V transformer whose equivalent impedance is (0.23 +1.27) 22 referred to its low-voltage side. The load is 160 kW at 0.89 leading power factor and 2340 V. Compute the voltage at the high-voltage terminals of the transformer. Compute the voltage at the sending end of the feeder. Compute the power and reactive power input at the sending end of the feeder. Main components of a power transformer The Figure below illustrates a cut-away view of an oil-filled transformer. The main components of the transformer are as follows: Core. Provides a route for the magnetic flux and supports the low-voltage and high voltage windings. Low-voltage winding. It has fewer turns compared with the high voltage windings. Its conductor has a large diameter because it carries more current. High-voltage winding. It has a larger number of turns, and its conductor has a smaller diameter than the low-voltage winding conductor. The high-voltage winding is usually wound around the low-voltage winding (only cooling ducts and insulation separate the windings). This is done to minimize the voltage stress on the core insulation. Tank. Houses the windings, core, and oil. It must be strong enough to withstand the gas pressures and electromagnetic forces that could develop when a fault occurs. Oil. It is a good-quality mineral oil. It provides insulation between the windings, core, and transformer tank. It also removes the heat generated. The oil is specially refined, and it must be free from impurities such as water, inorganic acid, alkali, sulphur, vegetable and mineral oil. Note: Some transformers use askarel, which is a nonflammable insulating and cooling medium instead of insulating oil. It gives good fire protection, which is a significant advantage when the transformer is located inside a building. However, askarel contains polychlorinated biphenals (PCBS). They have been linked to cancer-causing substances. Therefore, they have been banned. Most industries are now in the process of replacing transformers containing askarel with dry-type transformers or transformers containing mineral oil. Thermometer. Measures the oil temperature and initiates an alarm when the temperature exceeds the alarm set point. Low/high-voltage bushing. Ceramic bushing that carries the low/high-voltage conductor and incula insulates it from the tank. The high-voltage bushing is usually filled with oil to enhance the heat removal capability. 8| Page Low/high-voltage connection. Connects the low/high-voltage conductor to the circuit. Conservator tank. It contains oil and has the capability of absorbing the swell of the oil when it becomes hot. 7
Transcribed Image Text:K = 95.94 -36.88° A Therefore the load volt Vload - Load Zload (95.94 -36.88° A) (5 236.87⁰ (2) 479.7 Z-0.01° V And the line losses are given by; Ploss = (line) ² Ruine = (9.594 A)²(0.18 (2) 16.7 W 7| Page a. b. 4.0 TRANSFORMERS PDF Notice that by stepping up the transmission voltage of the power system, the transmission losses have been reduced by a factor of 90. In addition, the voltage at the load dropped significantly in the system with transformers compared to the system without transformers. a L Exercise: 95 + £360 A single-phase load is supplied through a 35-kV feeder whose impedance is 2 and a 35-kV:2400-V transformer whose equivalent impedance is (0.23 +1.27) 22 referred to its low-voltage side. The load is 160 kW at 0.89 leading power factor and 2340 V. Compute the voltage at the high-voltage terminals of the transformer. Compute the voltage at the sending end of the feeder. Compute the power and reactive power input at the sending end of the feeder. Main components of a power transformer The Figure below illustrates a cut-away view of an oil-filled transformer. The main components of the transformer are as follows: Core. Provides a route for the magnetic flux and supports the low-voltage and high voltage windings. Low-voltage winding. It has fewer turns compared with the high voltage windings. Its conductor has a large diameter because it carries more current. High-voltage winding. It has a larger number of turns, and its conductor has a smaller diameter than the low-voltage winding conductor. The high-voltage winding is usually wound around the low-voltage winding (only cooling ducts and insulation separate the windings). This is done to minimize the voltage stress on the core insulation. Tank. Houses the windings, core, and oil. It must be strong enough to withstand the gas pressures and electromagnetic forces that could develop when a fault occurs. Oil. It is a good-quality mineral oil. It provides insulation between the windings, core, and transformer tank. It also removes the heat generated. The oil is specially refined, and it must be free from impurities such as water, inorganic acid, alkali, sulphur, vegetable and mineral oil. Note: Some transformers use askarel, which is a nonflammable insulating and cooling medium instead of insulating oil. It gives good fire protection, which is a significant advantage when the transformer is located inside a building. However, askarel contains polychlorinated biphenals (PCBS). They have been linked to cancer-causing substances. Therefore, they have been banned. Most industries are now in the process of replacing transformers containing askarel with dry-type transformers or transformers containing mineral oil. Thermometer. Measures the oil temperature and initiates an alarm when the temperature exceeds the alarm set point. Low/high-voltage bushing. Ceramic bushing that carries the low/high-voltage conductor and incula insulates it from the tank. The high-voltage bushing is usually filled with oil to enhance the heat removal capability. 8| Page Low/high-voltage connection. Connects the low/high-voltage conductor to the circuit. Conservator tank. It contains oil and has the capability of absorbing the swell of the oil when it becomes hot. 7
Solution
Bartleby Expert
SEE SOLUTION
Knowledge Booster
Sinusoids and Phasors of Alternating Circuit
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,
SEE MORE TEXTBOOKS