A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a synchronous reactance of 1 ohm per phase with negligible armature resistance. The shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in the motor is 2 kW. The magnitude of the per phase excitation emf of the motor, in volts, is ______ (round off to nearest integer).

The correct answer is 245 (accepting values in the range of 240 to 248).

Step-by-Step Explanation:

1. Identify the given parameters:
Line voltage, $V_L=400\text{V}$
Since the motor is star-connected, the phase voltage $V$ is:
$V=\frac{V_L}{\sqrt{3}}=\frac{400}{\sqrt{3}}\approx 230.94\text{V}$
Synchronous reactance, $X_s=1\Omega$
Armature resistance, $R_a=0\Omega$ (negligible)
Shaft load (output power), $P_{\text{out}}=10\text{kW}$
Power factor, $\cos \varphi =0.8\text{(leading)}$
Since $\cos \varphi =0.8$, the sine of the power factor angle is:
$\sin \varphi =\sqrt{1-{0.8}^2}=0.6$
Losses in the motor, $P_{\text{loss}}=2\text{kW}$

2. Calculate the total input power:
$P_{\text{in}}=P_{\text{out}}+P_{\text{loss}}=10\text{kW}+2\text{kW}=12\text{kW}=12000\text{W}$

3. Determine the armature line/phase current:
For a 3-phase system, the input power is given by:
$P_{\text{in}}=\sqrt{3}{V}_L{I}_L\cos \varphi$
Solving for $I_L$:
$I_L=\frac{12000}{\sqrt{3}\times 400\times 0.8}\approx 21.65\text{A}$
Since the motor is star-connected, the phase current $I_a$ is equal to the line current:
$I_a=21.65\text{A}$

4. Calculate the per-phase excitation EMF ($E$):
For a synchronous motor operating at a leading power factor, the phasor relation for the excitation EMF per phase is:
$E=\sqrt{{\left(V\cos \varphi -I_a{R}_a\right)}^2+{\left(V\sin \varphi +I_a{X}_s\right)}^2}$
Substituting $R_a=0$:
$E=\sqrt{{\left(V\cos \varphi \right)}^2+{\left(V\sin \varphi +I_a{X}_s\right)}^2}$
Substitute the values into the equation:
$E=\sqrt{{\left(230.94\times 0.8\right)}^2+{\left(230.94\times 0.6+21.65\times 1\right)}^2}$
$E=\sqrt{{\left(184.75\right)}^2+{\left(138.56+21.65\right)}^2}$
$E=\sqrt{{\left(184.75\right)}^2+{\left(160.21\right)}^2}$
$E=\sqrt{34132.56+25667.24}$
$E=\sqrt{59799.8}\approx 244.54\text{V}$

Rounding off to the nearest integer, the per phase excitation EMF is 245 V.