1. Armature current
2. Generator efficiency
3. Prime mover torque

1. Armature current.

   The line current can be found from the apparent power:

   \[ S=\surd 3 V_{LL} I_L \qquad I_L=\frac{400,000}{\surd 3 \, 480} \qquad I_L=481A \]

   The generator is Y-connected, \(I_L=I_A\) and therefore:

   \[ \color{red}{I_A=481A} \]
2. Generator efficiency.

   To find the efficiency it is first necessary to find the input power. Using simpliifed generator power flow:

   \[ \begin{aligned} P_{in} & =P_{out}+P_{rotational}+3I_A^2R_A \\ P_{out} & = 400*0.9 \qquad P_{out} = 360kW \\ P_{in} & =360kW+10kW+3\,481^2\,0.02 \qquad P_{in} =383.9kW \end{aligned} \]

   And the efficiency is

   \[ \eta = \frac{P_{in}}{P_{out}} \qquad \eta= \frac{360}{383.9} \qquad \color{red}{\eta=93.8\%} \]
3. Prime mover torque

   The input power is mechanically given by

   \[ P_{in}=\tau \omega_s \]

   with

   \[ \omega_s=\frac{4\pi f}{p}\qquad \omega_s=60\pi \]
   \[ \color{red}{\tau=2037Nm} \]

A 600V Y-connected synchronous generator is tested under open and short circuit conditions, with the results shown in the table below:

Under normal operation, the saturated synchronous reactance \(X_S=0.4\Omega\) and \(R_A\) is negligible.

1. Calculate the unsaturated synchronous reactance
2. The generator is operating at 500kVA with power factor 0.9 lagging. Using \(X_S=0.4\Omega\), calculate the field current required and sketch the phasor diagram.

1. Calculate the unsaturated synchronous reactance.

   The unsaturated synchronous reactance is found from the intial table data, when the open circuit voltage changes linearly with field current. At the first data point:

   \[ \begin{align} & V_{LL_{OC}}=314V \qquad E=\frac{V_{LL_{OC}}}{\surd 3} = 181.3V \\ & X_S=\frac{E}{I_{A_{SC}}} \qquad X_S=\frac{181.3}{400} \qquad \color{red}{X_S=0.4532\Omega} \end{align} \]

   As a check, the unsaturated reactance is slightly larger than the saturated value, as expected.

2. Calculate the field current required.

   To find \(I_f\) it is first necessary to find \(V_{LL_{OC}}\), which means we have to calculate \(E\). With negligible \(R_A\)

   \[ \vec{E}=\vec{V}+j\vec{I}_AX_S \]

   with

   \[ \begin{align} \vec{V} & =V\angle0^{\circ} \qquad V=\frac{600}{\surd 3}=346.4V \\ \vec{I}_A & =I_A\angle\theta \\ I_A & = \frac{S}{\surd 3 V_{LL}} = 529.2A\\ \theta & = -\cos^(-1)(0.9) = -25.8^{\circ} \vec{I}_A = 476.3 -j 230.7 A \end{align} \]

   Note that the angle \(theta\) is negative because the power factor is lagging, \(\vec{I}\) lags \(\vec{V}\). Substituing gives:

   \[ \begin{aligned} \vec{E}&=346.4 +\left(476.3 -j230.7\right)\left(j0.4\right) \\ \vec{E}&=439+j191V \\ \vec{E}&=E\angle\delta \\ E&=478V \qquad \delta = 23.5^{\circ} \end{aligned} \]

   The table data gives \(V_{LL_{OC}}\), which is equal to \(\surd 3 E\).

   \[ V_{LL_{OC}} = \surd 3 \,478 =828V \]

   Therefore, from the table, \(\color{red}{I_f=60A}\).The illustration below steps through how to draw the phasor diagram; step 1 is already drawn

   Step 1 Step 2 Step 3 Step 4