Answer to Question #184284 in Electrical Engineering for nuri

Question

A balanced Δ (delta) connected source has an internal impedance of 0.018+j0.162 Ω/phase. At no load, the terminal voltage of the source has magnitude of 380 VLL. The source is connected to a Δ (delta) connected load, having an impedance of 7.92+j6.35 Ω/phase through a distribution line having an impedance of 0.074+j0.296 Ω/phase. a) Draw the single phase equivalent circuit of 3-phase system b) Calculate the three line currents c) Calculate the phase currents of the source d) Calculate the magnitude of the line voltage at the terminals of the source

Answer 1

a) Draw the single-phase equivalent circuit of a 3-phase system

b) Calculate the three line currents

$S_R=\frac{E_R}{Z_S/3+Z_d+Z_L/3}$

$S_R=\frac{219.39 \angle{-30}}{0.006+j0.054+0.074+j0.296+2.64-j2.116} =67.56 \angle 2.99 Amp$

Since loads and source are balanced, all phase current are equal in magnitude but displaced with respect to each other by 120⁰

$I_Y=I_R= \angle -120=67.65\angle2.99-120=67.65\angle-1770.01 A$

$I_B=I_R= \angle -120=67.65\angle122.99$

c) Calculate the phase currents of the source

It is lead line current by 30⁰

$I_R=I_{RY}-I_{BR}=\sqrt{3}I_{RY} \angle 30$

$I_{RY}=\frac{I_R}{\sqrt{3}} \angle 30=\frac{67.65}{\sqrt{3}} \angle 2.99+30=39.05+\angle 32.99 Amp$

$I_{BY}=\frac{I_Y}{\sqrt{3}} \angle 30=\frac{67.65}{\sqrt{3}} \angle -117.01+30=39.05+\angle -87.01 Amp$

$I_{BR}=\frac{I_B}{\sqrt{3}} \angle 30=\frac{67.65}{\sqrt{3}} \angle 122.99+30=39.05+\angle 152.99 Amp$

d) Calculate the magnitude of the line voltage at the terminals of the source

$(V_R)_L=E_{RY}-Z_S \times I_{RY}$

$(V_R)_L=380\angle0-(0.018+j0.162) \times (39.05 \angle 32.99)=382.89 \angle-0.85 V$

$V_Y=382.89\angle -0.85-120=382.89 \angle-120.85 V$

$V_B=382.89\angle -0.85+120=382.89 \angle119.15 V$