To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are
[tex]V = 690V[/tex]
[tex]P_{real} = 2.3MW[/tex]
Real power in 3 phase
[tex]P_{real} = 3V_{ph}I_{ph} Cos\theta[/tex]
Now the Phase Voltage is,
[tex]V_{ph} = \frac{V}{\sqrt{3}}[/tex]
[tex]V_{ph} = \frac{690}{\sqrt{3}}[/tex]
[tex]V_{ph} = 398.37V[/tex]
The current phase would be,
[tex]P_{real} = 3V_{ph}I_{ph} Cos\theta[/tex]
Rearranging,
[tex]I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}[/tex]
Replacing,
[tex]I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}[/tex]
[tex]I_{ph}= 2.26kA/phase[/tex]
Therefore the current per phase is 2.26kA
