1.1 We have two loads in delta and another in star in parallel, the impedances of the branches of the delta are equal and we have that RΔ=1p.u. and XΔ=0.75 p.u., while in the star we have equal values ​​of the branches, but formed by admittances where we have the conductance in parallel GY=0.64 p.u. and inductive susceptance of BY=-j0.48 p.u. If the loads are being supplied by a positive sequence balanced three-phase source with line-to-line voltage of phases a and b of Vab=1∠ 30º, if the base values ​​are line-to-line base voltage of 1kV and three-phase base power of 100kW. Determine A) Line and phase voltages, line and phase currents of the load in delta, line and phase voltages of the star, line and phase currents of the star.B) Obtain separately the phasor diagram of the delta and star load with the values ​​obtained in A), indicate on the scale that these are 1 p.u.=5cm for example.

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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1.1 We have two loads in delta and another in star in parallel, the impedances of the branches of the delta are equal and we have that RΔ=1p.u. and XΔ=0.75 p.u., while in the star we have equal values ​​of the branches, but formed by admittances where we have the conductance in parallel GY=0.64 p.u. and inductive susceptance of BY=-j0.48 p.u. If the loads are being supplied by a positive sequence balanced three-phase source with line-to-line voltage of phases a and b of Vab=1∠ 30º, if the base values ​​are line-to-line base voltage of 1kV and three-phase base power of 100kW. Determine
A) Line and phase voltages, line and phase currents of the load in delta, line and phase voltages of the star, line and phase currents of the star.B) Obtain separately the phasor diagram of the delta and star load with the values ​​obtained in A), indicate on the scale that these are 1 p.u.=5cm for example.
C) obtain the power triangle and the power factor separately for each load.

D) Obtain the power triangle and the power factor for the source

XA
M
je.75 P.U.
RA
3x4=0.45 P.U
m
RA=1PN XA=jo.15 Pu
BY=-jo.48 P.U.
GY = 0.64 P.U.
GY=0.64P0
w
m
BY
-50.489.0
RA= 1 P₂U.
XA =j 0.75 P.U.
GY-0.64 PU
fuimt
BY=0.48 P.U.
GY=0.64 P₁0
mm.
BY=jo.48Pu
Transcribed Image Text:XA M je.75 P.U. RA 3x4=0.45 P.U m RA=1PN XA=jo.15 Pu BY=-jo.48 P.U. GY = 0.64 P.U. GY=0.64P0 w m BY -50.489.0 RA= 1 P₂U. XA =j 0.75 P.U. GY-0.64 PU fuimt BY=0.48 P.U. GY=0.64 P₁0 mm. BY=jo.48Pu
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