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Capacitive Current And Reactance

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Question: The following phase schematic diagram (FIGURE 4) shows an 11 kV, 50 Hz, 3-phase, short line feeding a load. By calculation or constructing the phasor diagram (use a scale of 1 mm = 2 A) for the load current with VR as reference, determine the capacitive current and

1. Calculate the capacitive reactance/ph such that the load power factor is increased to 0.98 lag

2. Calculate the percentage reduction in line current with this value of capacitive reactance in circuit.

Answer: Conside the figure given below

Load.jpg

When voltage and length of the transmission line increases the capacitance has greater importance.

The phasor diagram can be constructed as follows

Voltage.jpg

Receiving End

VR = 11/(√3) KV

Apparent Power = VR IR cos θ

IR = (Apparent Power)/(VR cos θ)

= (3.2 × 106)/[(11 × (1/√3) × 103 × cosθ)

Now Cos θ = KW/(KW + KVAR) = 3.2/(3.2 + 1.9) = 0.6

Now IR = (3.2 × 106)/(11 × (1/√3) × 103 × 0.6) = 629A

As we know Cos θ = IC/IR

So IC = IR × cosθ = 0.6 × 629 = 377.4A

C = (-jIC)/2πf = (-j × 377.4)/(2 × 3.14 × 50) = -j1.20Ω

When load power factor is increased to 0.98 the capacitive reactance can be calculated as follows

IR = (3.2 × 106)/(11 × (1/√3) × 103 ×0.98) = 514A

IC = IR × cosθ = 0.98 × 514 = 503.72A

C = (-jIC)/2πf = (-j × 503.72)/(2 × 3.14 × 50) = -j1.60Ω

Percentage reduction in line current = (624 - 514)/624= 0.17%

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