3 PHASE CIRCUIT PROBLEM - Haaalllppp

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SUMMARY

The discussion centers on solving a three-phase circuit problem involving three loads, each with a resistance of 16 ohms and an inductive reactance of 12 ohms, connected in a wye configuration to a 240V supply. The impedance per phase is calculated using the formula Z = √(R² + XL²), resulting in an impedance of 20 ohms. The current per phase is determined to be 8.6A, leading to a total kVA of 3.5kVA. The power factor and total kW remain to be calculated, with suggestions to use complex numbers for more efficient calculations.

PREREQUISITES
  • Understanding of three-phase circuits
  • Familiarity with impedance calculations (Z = √(R² + XL²))
  • Knowledge of power factor (PF = kW/kVA)
  • Ability to perform calculations involving complex numbers
NEXT STEPS
  • Learn how to calculate power factor in three-phase systems
  • Study the use of complex numbers in electrical engineering calculations
  • Explore the implications of wye vs. delta configurations in three-phase circuits
  • Investigate the relationship between line-to-neutral and line-to-line voltages in three-phase systems
USEFUL FOR

Electrical engineering students, professionals working with three-phase systems, and anyone involved in circuit analysis and design.

JGrecs
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Homework Statement



Three load each having a resistance of 16 ohms and an inductive reactance of 12 ohms are wye connected to a 240V three phase supply. Determine:

a) the impedance per phase
b) the current per phase
c) the kVA total
d) the Power Factor
e) the kW total

Homework Equations



impedance? (Z= √R^2 + XL^2)??
wye: IPhase=ILine=ILoad
kVA= 3 X VPhase X IPhase
kVA= 1.732 X VLine X ILine
PF= kW/kVA
p= 3 X VPhase X IPhase X PF
P= 1.732 X VLine X ILine X PF

The Attempt at a Solution



A) ?

B) VLine= 240V

IPhase= VPhase/RPhase

VPhase= 240V/1.732
VPhase= 138.5V

IPhase= 138.5V/16Ω
IPhase= 8.6A

C) kVA= 3 X VPhase X IPhase

kVA= 3 X 138.5V X 8.6A
= 3.5kVA

D) ?

E) ?


Not sure how to take these ones on. Any tips? formulas to use?
 
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JGrecs said:

Homework Statement



Three load each having a resistance of 16 ohms and an inductive reactance of 12 ohms are wye connected to a 240V three phase supply. Determine:

a) the impedance per phase
b) the current per phase
c) the kVA total
d) the Power Factor
e) the kW total

Homework Equations



impedance? (Z= √R^2 + XL^2)??
wye: IPhase=ILine=ILoad
kVA= 3 X VPhase X IPhase
kVA= 1.732 X VLine X ILine
PF= kW/kVA
p= 3 X VPhase X IPhase X PF
P= 1.732 X VLine X ILine X PF

The Attempt at a Solution



A) ?

B) VLine= 240V

IPhase= VPhase/RPhase

VPhase= 240V/1.732
VPhase= 138.5V

IPhase= 138.5V/16Ω
IPhase= 8.6A

C) kVA= 3 X VPhase X IPhase

kVA= 3 X 138.5V X 8.6A
= 3.5kVA

D) ?

E) ?


Not sure how to take these ones on. Any tips? formulas to use?

I did some browsing and came across this site:

http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE302

with the following graphic:

attachment.php?attachmentid=56802&stc=1&d=1363469614.gif


which seems to indicate that the voltage specification for a Wye generating system specifies the line-to-neutral (or phase) voltage of the generator. If that is true for this problem, then the line-to-line voltage will be in the neighborhood of 416V.

A circuit diagram would look something like:

attachment.php?attachmentid=56805&stc=1&d=1363469921.gif


For the impedance you can use the reactance as you've shown. You can determine the power factor for the load from its real and reactive components.

An alternative is to leave the impedance in complex form and do the calculations in complex form too (phasors). This would automatically take care of all the "1.732" and pf factors and give you the kVA, kW and kVAR values all at once. It helps if your calculator handles complex numbers :smile:
 

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