We are asked to calculate the apparent power in KVA of a 3-phase, 415V, 50 Hz, star system. The line current ($I_L$) is 16A, and the power factor is 0.8.

1. Problem Description

We are asked to calculate the apparent power in KVA of a 3-phase, 415V, 50 Hz, star system. The line current ($I_L$) is 16A, and the power factor is 0.

8.

2. Solution Steps

In a 3-phase star (or wye) connection, the line voltage ($V_L$) is related to the phase voltage ($V_{ph}$) by:

$V_L = \sqrt{3} V_{ph}$

The line current ($I_L$) is equal to the phase current ($I_{ph}$):

$I_L = I_{ph}$

The total apparent power (S) in a 3-phase system is given by:

$S = \sqrt{3} V_L I_L$

where $V_L$ is the line voltage and $I_L$ is the line current.

Given:

$V_L = 415V$

$I_L = 16A$

Substituting the given values into the formula:

$S = \sqrt{3} \times 415V \times 16A$

$S = \sqrt{3} \times 6640 VA$

$S \approx 1.732 \times 6640 VA$

$S \approx 11500.48 VA$

To convert VA to KVA, divide by 1000:

$S \approx 11.50048 kVA$

$S \approx 11.5 kVA$

Note that the power factor is not needed for calculating apparent power. The power factor is only needed to calculate the real power or active power (P), where $P = S \times power \ factor$.

3. Final Answer

The apparent power is approximately 11.5 kVA.