Answer:
a) 4160 V
b) 12 kW and 81 kVAR
c) 54 kW and 477 kVAR
Explanation:
1) The phase voltage is given as:
[tex]V_p=\frac{3810.5}{\sqrt{3} }=2200 V[/tex]
The complex power S is given as:
[tex]S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA[/tex]
[tex]where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA[/tex]
The line current I is given as:
[tex]I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)} =100\angle -36.87^o\ A[/tex]
The phase voltage at the sending end is:
[tex]V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV[/tex]
The magnitude of the line voltage at the source end of the line ([tex]V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V[/tex]
b) The Total real and reactive power loss in the line is:
[tex]S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000[/tex]
The real power loss is 12000 W = 12 kW
The reactive power loss is 81000 kVAR = 81 kVAR
c) The sending power is:
[tex]S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000[/tex]
The Real power delivered by the supply = 54000 W = 54 kW
The Reactive power delivered by the supply = 477000 VAR = 477 kVAR
