160 V 70 Hz gle-phase load has an active power of P = 2 H mensated to cos = 0.85 using a parallel capa . Reactive power and apparent power befo . Current before compensation

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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PRELIMINARY 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:
Some parameters are given related to your student ID numbers. These are:
V1 = 160 V
f1 = 70 Hz
1) A single-phase load has an active power of P = 2 kW at V1 V @f1Hz 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 Xc and C using 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.
Transcribed Image Text:PRELIMINARY 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: Some parameters are given related to your student ID numbers. These are: V1 = 160 V f1 = 70 Hz 1) A single-phase load has an active power of P = 2 kW at V1 V @f1Hz 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 Xc and C using 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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