Question:
A 3-phase, 11 kV, 10 MVA synchronous generator is connected to an inductive load of power factor \( \frac{\sqrt{3}}{2} \) via a lossless line with a per-phase inductive reactance of \( 5\Omega \). The per-phase synchronous reactance of the generator is \( 30\Omega \). If the generator is producing the rated current at the rated voltage, then the power factor at the terminal of the generator is:
A 3-phase, 11 kV, 10 MVA synchronous generator is connected to an inductive load of power factor \( \frac{\sqrt{3}}{2} \) via a lossless line with a per-phase inductive reactance of \( 5\Omega \). The per-phase synchronous reactance of the generator is \( 30\Omega \). If the generator is producing the rated current at the rated voltage, then the power factor at the terminal of the generator is:
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\section{ Voltage, Power Factor, and Phase Angle Calculations}
Power Factor (pF):} Lagging power factor is given by:
\[
pF = \frac{\sqrt{3}}{2} = \cos(30^\circ).
\]
Voltage (\(E\)):} Total voltage magnitude is calculated as:
\[
E = \sqrt{(V \cos 30^\circ + IR)^2 + (V \sin 30^\circ + IX)^2}.
\]
Phase Angle (\(\delta\)):} The phase angle is derived from:
\[
(IZ)^2 = E^2 + V^2 - 2EV \cos \delta.
\]
Final Power Factor:} Adjusted power factor is:
\[
pF = \cos(\delta + 30^\circ).
\]
Key Result:} For the given values, \( pF = 0.629 \, \text{lagging} \).
Updated On: Jan 23, 2025
- \( 0.63 \) lagging
- \( 0.87 \) lagging
- \( 0.63 \) leading
- \( 0.87 \) leading
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Solution and Explanation
Step 1: Compute the terminal voltage.
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\section*{Explanation of Power Factor, Voltage, and Angle Calculations}
\begin{align*}
\text{Power Factor (pF)} & = \frac{\sqrt{3}}{2} = 30^\circ \text{ lagging.}
\end{align*} This indicates that the power factor corresponds to a lagging angle of \(30^\circ\), meaning the current lags the voltage by \(30^\circ\). \subsection{Voltage Magnitude (\(E\))} The voltage \(E\) is determined using the following formula: \[ E = \sqrt{(V \cos 30^\circ + IR)^2 + (V \sin 30^\circ + IX)^2}. \] Here: \(V \cos 30^\circ\) and \(V \sin 30^\circ\) are the voltage components in the horizontal and vertical directions, respectively. \(IR\) and \(IX\) are the resistive and reactive voltage drops. Substituting the values: \[ \frac{11K}{\sqrt{3}} = \sqrt{\left(V \cdot \frac{\sqrt{3}}{2}\right)^2 + \left(\frac{V}{2} + 524.86(5)\right)^2}. \] Squaring both sides: \[ \left(\frac{11K}{\sqrt{3}}\right)^2 = V^2 \cdot \frac{3}{4} + \left(\frac{V}{2} + 2624.319\right)^2. \] Expanding the terms: \[ \left(\frac{11K}{\sqrt{3}}\right)^2 = \frac{3}{4}V^2 + \frac{V^2}{4} + 2624.319V + 2624.319^2. \] Solving for \(V\), we find: \[ V_t = 4618.105 \, \text{V}. \] \subsection*{Impedance Voltage (\((IZ)^2\))} The square of the impedance voltage is given by: \[ (IZ)^2 = E^2 + V^2 - 2EV \cos \delta. \] Substituting the known values: \[ (2624.319)^2 = \left(\frac{11K}{\sqrt{3}}\right)^2 + (4618.105)^2 - 2\left(\frac{11K}{\sqrt{3}}\right)(4618.105) \cos \delta. \] Solving for \(\delta\), the phase angle difference: \[ \delta = 20.96^\circ. \] \subsection{Power Factor (pF)} Finally, the overall power factor is calculated as: \[ pF = \cos(20.96^\circ + 30^\circ). \] Simplifying: \[ pF = \cos(50.96^\circ) = 0.629 \, \text{lagging}. \]
\end{align*} This indicates that the power factor corresponds to a lagging angle of \(30^\circ\), meaning the current lags the voltage by \(30^\circ\). \subsection{Voltage Magnitude (\(E\))} The voltage \(E\) is determined using the following formula: \[ E = \sqrt{(V \cos 30^\circ + IR)^2 + (V \sin 30^\circ + IX)^2}. \] Here: \(V \cos 30^\circ\) and \(V \sin 30^\circ\) are the voltage components in the horizontal and vertical directions, respectively. \(IR\) and \(IX\) are the resistive and reactive voltage drops. Substituting the values: \[ \frac{11K}{\sqrt{3}} = \sqrt{\left(V \cdot \frac{\sqrt{3}}{2}\right)^2 + \left(\frac{V}{2} + 524.86(5)\right)^2}. \] Squaring both sides: \[ \left(\frac{11K}{\sqrt{3}}\right)^2 = V^2 \cdot \frac{3}{4} + \left(\frac{V}{2} + 2624.319\right)^2. \] Expanding the terms: \[ \left(\frac{11K}{\sqrt{3}}\right)^2 = \frac{3}{4}V^2 + \frac{V^2}{4} + 2624.319V + 2624.319^2. \] Solving for \(V\), we find: \[ V_t = 4618.105 \, \text{V}. \] \subsection*{Impedance Voltage (\((IZ)^2\))} The square of the impedance voltage is given by: \[ (IZ)^2 = E^2 + V^2 - 2EV \cos \delta. \] Substituting the known values: \[ (2624.319)^2 = \left(\frac{11K}{\sqrt{3}}\right)^2 + (4618.105)^2 - 2\left(\frac{11K}{\sqrt{3}}\right)(4618.105) \cos \delta. \] Solving for \(\delta\), the phase angle difference: \[ \delta = 20.96^\circ. \] \subsection{Power Factor (pF)} Finally, the overall power factor is calculated as: \[ pF = \cos(20.96^\circ + 30^\circ). \] Simplifying: \[ pF = \cos(50.96^\circ) = 0.629 \, \text{lagging}. \]
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