Answer:
1.23MVA, 226.74kW, 5.16kV
Step-by-step explanation:
Parameters Given
line impedance, Zl = (4 + j6) ohms per phase
load voltage, Vl = 4200V
received complex power, S = 1 × 10⁶VA
power factor, cosФ = 0.75
Ф = 41.41°
sinФ = 0.66
Solution:
S = √3 * Vl * I (that is √3 × line voltage × line current)
1 × 10⁶ = √3 × 4200 × I
I = 137.46A
Vl = 4200∠0
I = 137.46∠- 41.41° lagging
source voltage, Vs = load voltage, Vl + voltage drop along the line, Vd
Vd = Zl * I where ( Zl = 4 + j6, = √(4² + 6²)∠tan⁻¹(6/4), = 7.21∠56.31° )
Vd = 7.21∠56.31° × 137.46∠- 41.41°
= 991.22∠14.9°
Vs = Vl + Vd
= 4200∠0° + 991.22∠14.9°
= 4200(cos 0° + j sin 0°) + 991.22(cos 14.9° + jsin 14.9°)
= 4200 + 957.68 + j254.88
= 5157.68 + j254.88
or
= 5163.97∠2.83° V (line voltage at the sending end of the transmission line)
Sending end current, I = 137.46∠-41.41 A
(a) Complex power = √3 × Vs × I
= √3 × 5163.97∠2.83° ×137.46∠-41.41
= 1229477.76∠-38.58°VA
= 1.23∠-38.58MVA
complex power = 1.23MVA
(b) power loss in the three phase line, Pl = 3 × square of line current, I × line impedance, Rl
Pl =3 × I² × Rl where Zl = R + j X = 4 + j6 hence R = 4
= 3 × 137.46² × 4
= 226743.02W
= 226.74kW
(c) from the above, the line voltage at the sending end of the transmission line is = 5163.97V
= 5.16kW