A single-phase load has an active power of P = 2 kW at 180 V @60HZ and the power factor is cos% = 0.75. This motor is compensated to coso 0.85 using a parallel capacitor (the load is modelled as series RL). Determine: a. Reactive power and apparent power before compensation using power factor b. Current before compensation C. R, XL and L values of the load d. Reactive power and apparent power after compensation e. Find the reactive power difference between compensated and uncompensated status (which will give the capacitor power) and calculate XC and Cusing the power difference value f. Simulate the uncompensated and compensated circuits. Plot voltage and current on components and calculate the phase shift of the signals from Vin g. Plot Vin and lin comparison for both circuits and show/calculate the phase angles for circuits
A single-phase load has an active power of P = 2 kW at 180 V @60HZ and the power factor is cos% = 0.75. This motor is compensated to coso 0.85 using a parallel capacitor (the load is modelled as series RL). Determine: a. Reactive power and apparent power before compensation using power factor b. Current before compensation C. R, XL and L values of the load d. Reactive power and apparent power after compensation e. Find the reactive power difference between compensated and uncompensated status (which will give the capacitor power) and calculate XC and Cusing the power difference value f. Simulate the uncompensated and compensated circuits. Plot voltage and current on components and calculate the phase shift of the signals from Vin g. Plot Vin and lin comparison for both circuits and show/calculate the phase angles for circuits
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...
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PREUMINARY INFORMATION:
Electrical compensation is a major scale subject that every business/facility is Included. The power
factor should be compensated to change the reactive power up to some specific values since some
reactive power limits are not good for consumption and generation of the electricity (this is a far
large topic which also forms some subbranches, for more information and specification please
research yourself). For this reason, it is an obligation to use a compensator system for both
manufacturers (wind farms/hydroelectric power plants/etc.) and consumers
(residents/factories/etc.). These issues are regulated by official directives.
EXERCISE STEPS:
A single-phase load has an active power of P = 2 kW at 180 V @60HZ and the power factor is coso =
0.75. This motor is compensated to coso = 0.85 using a parallel capacitor (the load is modelled as
series RL). Determine:
a. Reactive power and apparent power before compensation using power factor
b. Current before compensation
C. R, XL and L values of the load
d. Reactive power and apparent power after compensation
e. Find the reactive power difference between compensated and uncompensated status (which will
give the capacitor power) and calculate XCand Cusing the power difference value
f. Simulate the uncompensated and compensated circuits. Plot voltage and current on components
and calculate the phase shift of the signals from Vin
g. Plot Vin and lin comparison for both circuits and show/calculate the phase angles for circuits](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92c8096b-22a9-429a-9bfc-3fe70ca98ced%2F18b6c45a-c45c-4353-9847-3961cc797ac7%2F9rufijd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2-0-ea:: anc20Mass:c20Transter-conuştüru ci.pdf
Dersler
Z Zimbra Gelen Kutu..
C eemanonline.com.t.
e Cağrı Merkezi
66%
PREUMINARY INFORMATION:
Electrical compensation is a major scale subject that every business/facility is Included. The power
factor should be compensated to change the reactive power up to some specific values since some
reactive power limits are not good for consumption and generation of the electricity (this is a far
large topic which also forms some subbranches, for more information and specification please
research yourself). For this reason, it is an obligation to use a compensator system for both
manufacturers (wind farms/hydroelectric power plants/etc.) and consumers
(residents/factories/etc.). These issues are regulated by official directives.
EXERCISE STEPS:
A single-phase load has an active power of P = 2 kW at 180 V @60HZ and the power factor is coso =
0.75. This motor is compensated to coso = 0.85 using a parallel capacitor (the load is modelled as
series RL). Determine:
a. Reactive power and apparent power before compensation using power factor
b. Current before compensation
C. R, XL and L values of the load
d. Reactive power and apparent power after compensation
e. Find the reactive power difference between compensated and uncompensated status (which will
give the capacitor power) and calculate XCand Cusing the power difference value
f. Simulate the uncompensated and compensated circuits. Plot voltage and current on components
and calculate the phase shift of the signals from Vin
g. Plot Vin and lin comparison for both circuits and show/calculate the phase angles for circuits
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