Power Factor Correction
Laboratory 1: Power Factor Correction
Figure 1 Power Factor Correction
Note: The ammeters & Voltmeters read PEAK values, but the Wattmeter reads true rms power.
This circuit is available on StudyNet as pfcor.sch, for use with TinaPlus in D411/D421.
This lab consist of two sessions
Session 1: Preparation (2 Hours). This is an open lab session where you will need to fill up the preparation answer sheet
Session 2: Lab experiment (2 Hours).
This is a supervised session where you will need to perform a simulation and fill up the experimental answer sheet.
Preparation answer sheet (Theoretical calculations)
Within the space provided Calculate: 1. The values of R & L in the load, for both series & parallel connections, to achieve the load conditions (S=10kVA and pf=0.9).
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Calculation Area (5%):
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p.f=cos⏀=0.9
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⏀= cos-1(0.9)=25.84°
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S=10 kVA
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P=S ×p.f=10 ×0.9=9 kW
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Q= S2-P2=10k2-9k2=19=4.3589 kVAr
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( /5)
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Figure 1 Power Factor Correction
Note: The ammeters & Voltmeters read PEAK values, but the Wattmeter reads true rms power.
This circuit is available on StudyNet as pfcor.sch, for use with TinaPlus in D411/D421.
This lab consist of two sessions
Session 1: Preparation (2 Hours). This is an open lab session where you will need to fill up the preparation answer sheet
Session 2: Lab experiment (2 Hours).
This is a supervised session where you will need to perform a simulation and fill up the experimental answer sheet.
Preparation answer sheet (Theoretical calculations)
Within the space provided Calculate: 1. The values of R & L in the load, for both series & parallel connections, to achieve the load conditions (S=10kVA and pf=0.9).
-------------------------------------------------
Calculation Area (5%):
-------------------------------------------------
-------------------------------------------------
-------------------------------------------------
-------------------------------------------------
-------------------------------------------------
p.f=cos⏀=0.9
-------------------------------------------------
⏀= cos-1(0.9)=25.84°
-------------------------------------------------
-------------------------------------------------
S=10 kVA
-------------------------------------------------
P=S ×p.f=10 ×0.9=9 kW
-------------------------------------------------
Q= S2-P2=10k2-9k2=19=4.3589 kVAr
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-------------------------------------------------
-------------------------------------------------
-------------------------------------------------
-------------------------------------------------
-------------------------------------------------
-------------------------------------------------
( /5)
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