You’re the operator of a $15000V_{rms}$, $60HZ$ electrical substation. When you get to work one day, you see that the station is delivering $6MW$of power with a power factor of$0.90$.

a. What is the rms current leaving the station?

b. How much series capacitance should you add to bring the power factor up to $1.0$?

c. How much power will the station then be delivering?

We are given that you’re the operator of a $15000V_{rms}/60HZ$ electrical substation. When you get to work one day, you see that the station is delivering $6MW$of power with a power factor of $0.90$.

We need to find that What is the rms current leaving the station?

We know that the avg power is

$$\begin{gathered} P\hspace{0.17em}=I_{rms}{E}_{rms}\cos \varnothing \\ {I}_{rms}=\frac{P}{E_{rms}\cos \varnothing } \\ {I}_{rms}=\frac{6x{10}^6}{15x{10}^{3}x0.9} \\ {I}_{rms}=0.44kA \end{gathered}$$

We are given that you’re the operator of a $15000V_{rms}/60HZ$ electrical substation. When you get to work one day, you see that the station is delivering $6MW$of power with a power factor of $0.90$.

We need to find that How much series capacitance should you add to bring the power factor up to $1$?

In the LCR circuit the value of $Z\hspace{0.17em}=\frac{E_{rms}}{I_{rms}}$

The power factor is given by role="math" localid="1650738938667" $\cos \varnothing =0.9$so the phase angle is $25.{84}^0$, to make power factor is equal to 1 so

$$\begin{gathered} X_c=Z\sin \varnothing \\ {X}_c=14.72ohm \end{gathered}$$

And we know that

$$\begin{gathered} X_c=\frac{1}{2\mathrm{\pi fC}} \\ C=\frac{1}{2{\mathrm{\pi fX}}_{\mathrm{c}}} \\ C=\frac{1}{2x3.14x60x14.72} \\ C=1.80x{10}^{-4}F \end{gathered}$$

you’re the operator of a $15000V_{rms}/60HZ$ electrical substation. When you get to work one day, you see that the station is delivering$6MW$ of power with a power factor of $0.90$.

We need to find How much power will the station then be delivering?

So power dissipated by resistor is

$$\begin{gathered} P=\frac{{V_{rms}}^2}{R}=\frac{{V_{rms}}^2}{Z\cos \varnothing } \\ P=\frac{{15000}^2}{33.78x\cos (25.84)} \\ P=7.4MW \end{gathered}$$