Demonstrative Video
Power Flow within Synchronous Motor
\[\begin{aligned}
I_{a} &
=\dfrac{E_{R}}{Z_{s}}=\dfrac{V-E_{b}}{Z_{s}}=\dfrac{V-E_{b}}{R_{a}+jX_{s}}
\end{aligned}\] 
Cu loss in rotor is not met by motor ac input, but by the dc source used for rotor excitation
for one armature phase of a syn motor:
Power developed by Synchronous Motor
Since, \(X_s >> R_a\)
\[\begin{aligned}
E_{b}\sin\alpha & =I_{a}X_{s}\cos\phi\\
\Rightarrow VE_{b}\sin\alpha & =VI_{a}X_{s}\cos\phi\\
\Rightarrow VI_{a}\cos\phi &
=\dfrac{E_{b}V}{X_{s}}\sin\alpha\\
P_{in} & =\dfrac{E_{b}V}{X_{s}}\sin\alpha
\end{aligned}\]
Since, stator Cu losses is neglected, \(P_{in}=P_m\)
\[\begin{aligned}
&\boxed{P_m = \dfrac{3E_bV}{X_s}\sin\alpha}~~~-\mbox{three
phase}\\
&\boxed{T_g = 9.55P_m/N_s}~~~N_s~\mbox{in rpm}\\
& (P_m)_{max} = \dfrac{E_bV}{X_s}
\end{aligned}\] 
Torques of a Synchronous Motor
\(T_d\) when full voltage is applied to stator winding
also called breakaway torque
10% of F.L torque in centrifugal pumps to 200% loaded reciprocating compressors
\(T_d\) developed in running condition
determined by horse-power and speed of the driven machine
must have a running torque greater than the maximum torque required in order to avoid stalling
Syn motor is started as IM till it runs 2 to 5% below \(N_s\)
Afterwards, excitation is switched on and the rotor pulls into step with synchronously rotating stator field
Torque at which the motor will pull into step is called pull-in torque
\(T_{max}\) can develop without pulling out of step or synchronism