SSE JE Electrical Engineering 10 Dec 2020 Papers

$\mathrm{V}$ = Terminal phase voltage applied to the armature (V)
${\mathrm{I}}_{\mathrm{a}}$ = Armature current/phase (A)
${\mathrm{E}}_{\mathrm{b}}$ = Back EMF to the motor (V)
${\mathrm{R}}_{\mathrm{a}}$ = Armature resistance/phase (Ω)
${\mathrm{X}}_{\mathrm{s}}$ = Synchronous reactance/phase (Ω)
${\mathrm{Z}}_{\mathrm{s}}$ = Impedance/phase (Ω)
$\mathrm{\varphi }$ = Phase angle between line current and line voltage.
$\mathrm{\alpha }$ = Load angle (Load angle = Angle between Back EMF and terminal voltage)
${\mathrm{Z}}_{\mathrm{S}}={\mathrm{R}}_{\mathrm{a}}+\mathrm{j}{\mathrm{X}}_{\mathrm{s}}=\mathrm{j}{\mathrm{X}}_{\mathrm{S}}$ (as resistance is very small so neglected)
${\mathrm{I}}_{\mathrm{a}}=\frac{\mathrm{V}-{\mathrm{E}}_{\mathrm{b}}}{{\mathrm{Z}}_{\mathrm{s}}}=\frac{\mathrm{V}-{\mathrm{E}}_{\mathrm{b}}}{\mathrm{j}{\mathrm{X}}_{\mathrm{s}}}$
Power input to the motor $\left(\mathrm{P}\right)$ $=\mathrm{V}\times {\mathrm{I}}_{\mathrm{a}}\times \cos \mathrm{\varphi }$ W per phase From the phasor
$\mathrm{AB}={\mathrm{E}}_{\mathrm{b}}\sin \mathrm{\alpha }={\mathrm{I}}_{\mathrm{a}}{\mathrm{X}}_{\mathrm{s}}\cos \mathrm{\varphi }$
$\mathrm{P}=\mathrm{V}\times {\mathrm{I}}_{\mathrm{a}}\times \cos \mathrm{\varphi }=\frac{{\mathrm{E}}_{\mathrm{b}}\mathrm{V}}{{\mathrm{X}}_{\mathrm{s}}}\sin \mathrm{\alpha }$
Mechanical losses are neglected then Input electrical power = Mechanical power developed
∴ Per phase developed Mechanical power is P $=\frac{{\mathrm{E}}_{\mathrm{b}}\mathrm{V}}{{\mathrm{X}}_{\mathrm{S}}}\sin \mathrm{\alpha }$