Synchronous Motors - Power & Torque | Electrical Machines

Demonstrative Video

SECTION 01

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:

for one armature phase of a syn motor:

SECTION 02

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, \(P_{in}=P_m\)

\[\begin{aligned} &\boxed{P_m = \dfrac{3E_bV}{X_s}\sin\alpha}~~~-\mbox{three phase}\\ &\boxed{T_g = 9.55P_m/N_s}~~~N_s~\mbox{in rpm}\\ & (P_m)_{max} = \dfrac{E_bV}{X_s} \end{aligned}\]

SECTION 03

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