The line voltage of a balanced three-phase transmission line is 4200 V rms. The transmission line has an impedance of 4 6 Z j l   per phase. Assuming that the load receives a total of 1 MVA at 0.75 power factor lagging, determine (a) the complex power, (b) the power loss in the three-phase line, (c) the line voltage at the sending end of the transmission line.

1.23MVA, 226.74kW, 5.16kV

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  

answer:

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answer:

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