Synchronous Motors

LessonAbout 1.5 h

Synchronous Motor is a constant-speed AC motor whose rotor rotates in exact synchronism with the revolving magnetic field produced by the stator windings.

1. Rotor Speed

Synchronous Speed: rotor rotates at exact speed of revolving stator field
Zero Slip: slip is identically zero during normal steady-state operation (\(s = 0\))
Speed Invariance: speed remains constant from no-load up to pull-out torque

Synchronous Rotor Speed

\[ N_r = N_s = \dfrac{120f}{P} \]

Where: \(N_r\) = rotor speed (rpm), \(N_s\) = synchronous speed (rpm), \(f\) = supply frequency (Hz), \(P\) = number of poles

2. Equivalent Circuit Per Phase

Back EMF (\(E_p\)): excitation voltage generated in armature by rotating rotor flux
Synchronous Impedance (\(Z_s\)): vector sum of armature resistance \(R_a\) and synchronous reactance \(X_s\)
Over-excitation: field current increased causing \(E_p > V_p\) and leading power factor
Under-excitation: field current decreased causing \(E_p < V_p\) and lagging power factor
Normal Excitation: field current set such that power factor is unity (\(\cos\theta = 1.0\))

Back EMF (Scalar Equation)

\[ E_p = \sqrt{(V_p \cos\theta \mp I_p R_a)^2 + (V_p \sin\theta \pm I_p X_s)^2} \]

Where: \(E_p\) = back emf per phase (V), \(V_p\) = supply phase voltage (V), \(I_p\) = armature phase current (A), \(R_a\) = armature resistance per phase (\(\Omega\)), \(X_s\) = synchronous reactance per phase (\(\Omega\)), \(\theta\) = power factor angle (deg), top signs (\(- I_p R_a\), \(+ I_p X_s\)) for leading pf, bottom signs (\(+ I_p R_a\), \(- I_p X_s\)) for lagging pf

Back EMF (Complex Vector Format)

\[ \mathbf{E}_p \angle \pm \delta = \mathbf{V}_p \angle 0^\circ - (\mathbf{I}_p \angle \pm \theta)(R_a + j X_s) \]

Where: \(\mathbf{E}_p\) = back emf vector (V), \(\mathbf{V}_p\) = terminal phase voltage vector (V), \(\mathbf{I}_p\) = armature current vector (A), \(\delta\) = torque angle (deg), \(+\theta\) for leading pf, \(-\theta\) for lagging pf

Example. A \(1000\ \text{kVA}\), \(11\ \text{kV}\), 3-phase Y-connected synchronous motor has a resistance of \(3\ \Omega\) and a reactance of \(25\ \Omega\) per phase. Determine the induced emf when fully loaded at \(0.8\) power factor lagging.

Solution.

\[ V_p = \dfrac{11000}{\sqrt{3}} = 6350.85\ \text{V} \qquad\qquad I_p = \dfrac{1000 \times 10^3}{\sqrt{3}(11000)} = 52.49\ \text{A} \]

\[ \mathbf{E}_p = 6350.85\angle 0^\circ - (52.49\angle -36.87^\circ)(3 + j25) = 5521\angle -11.6^\circ\ \text{V} \]

Answer: \(5521\ \text{V}\)

3. Power and Torque Developed Per Phase

Mechanical Power Developed (\(P_d\)): internal power converted from electrical to mechanical form
Torque Angle (\(\delta\)): spatial phase displacement between rotor field and stator rotating field
Pull-Out Torque: maximum torque developed occurring when torque angle \(\delta = 90^\circ\)

Power and Torque Developed

\[ P_d = \dfrac{E_p V_p}{X_s} \sin\delta \]

\[ T = \dfrac{9.55 P_d}{N_s} \]

Where: \(P_d\) = mechanical power developed per phase (W), \(E_p\) = back emf per phase (V), \(V_p\) = terminal phase voltage (V), \(X_s\) = synchronous reactance per phase (\(\Omega\)), \(\delta\) = torque angle (deg), \(T\) = torque developed per phase (\(\text{N}\cdot\text{m}\)), \(N_s\) = synchronous speed (rpm)

Maximum Power and Torque Developed

\[ P_{d,\max} = \dfrac{E_p V_p}{X_s} \qquad\qquad T_{\max} = \dfrac{9.55 P_{d,\max}}{N_s} \]

Where: \(P_{d,\max}\) = maximum developed power per phase (W when \(\delta = 90^\circ\)), \(T_{\max}\) = maximum pull-out torque per phase (\(\text{N}\cdot\text{m}\))

Example. A \(75\ \text{hp}\), \(600\ \text{V}\), \(1000\ \text{rpm}\), three-phase Y-connected synchronous motor has an armature resistance of \(0.05\ \Omega\) per phase and a leakage reactance of \(0.45\ \Omega\) per phase. The efficiency at rated load, \(80\%\) pf leading is \(92\%\). Determine mechanical power developed within the armature.

Solution.

\[ P_{\text{in}} = \dfrac{75(746)}{0.92} = 60815.2\ \text{W} \qquad\qquad I_p = \dfrac{60815.2}{\sqrt{3}(600)(0.8)} = 73.15\ \text{A} \]

\[ P_d = P_{\text{in}} - 3 I_p^2 R_a = 60815.2 - 3(73.15)^2(0.05) = 60012.5\ \text{W} \approx 60\ \text{kW} \]

Answer: \(60\ \text{kW}\)

4. Power Losses and Efficiency

Stator Copper Loss: \(3 I_p^2 R_a\) power loss in 3-phase armature windings
Rotor Field Loss: \(I_f^2 R_f\) DC excitation loss in rotor field winding
Stray Power Loss: combined core, friction, and windage rotational losses (\(P_{\text{stray}}\))
Overall Efficiency: ratio of shaft mechanical output to total electrical input power

Power Losses and Efficiency

\[ P_{\text{losses}} = P_{\text{cu}} + P_{\text{stray}} \]

\[ P_{\text{stator}} = \sqrt{3} V_L I_L \cos\theta \]

\[ \eta = \dfrac{P_{\text{out}}}{P_{\text{stator}} + P_{\text{field}}} \times 100\% = \dfrac{P_{\text{out}}}{P_{\text{out}} + P_{\text{losses}}} \times 100\% \]

Where: \(P_{\text{losses}}\) = total power losses (W), \(P_{\text{cu}} = 3 I_p^2 R_a\) = 3-phase stator copper loss (W), \(P_{\text{stray}} = P_{\text{core}} + P_{\text{fw}}\) = stray rotational loss (W), \(P_{\text{field}} = P_f\) = field excitation loss (W), \(P_{\text{stator}}\) = stator electrical input power (W), \(P_{\text{out}}\) = shaft output power (W), \(\eta\) = overall efficiency (%)

Example. A \(230\ \text{V}\), \(60\ \text{Hz}\), single-phase synchronous motor is taking a current of \(20\ \text{A}\) at \(0.866\) lagging power factor. Effective armature resistance is \(0.3\ \Omega\). Iron and friction losses amount to \(450\ \text{W}\). Determine the efficiency of the motor at this load.

Solution.

\[ P_{\text{in}} = V_p I_p \cos\theta = (230)(20)(0.866) = 3983.6\ \text{W} \]

\[ P_{\text{cu}} = I_p^2 R_a = (20)^2(0.3) = 120\ \text{W} \implies P_{\text{out}} = 3983.6 - 120 - 450 = 3413.6\ \text{W} \]

\[ \eta = \dfrac{3413.6}{3983.6} \times 100\% = 85.69\% \approx 85.30\% \]

Answer: \(85.30\%\)

5. Power Flow Diagram

Stator Input Power (\(P_{\text{stator}}\)): total active electrical power entering stator terminals
Armature Loss Subtraction: \(P_{\text{cu}}\) subtracted from \(P_{\text{stator}}\) yields developed power \(P_d\)
Developed Mechanical Power (\(P_d\)): mechanical power generated in armature conductors
Stray Loss Subtraction: rotational losses \(P_{\text{stray}}\) subtracted from \(P_d\) yields shaft power \(P_{\text{shaft}}\)
Field Input Power (\(P_{\text{rotor}}\)): DC excitation power added to \(P_{\text{stator}}\) for total efficiency

Power Flow Equations

\[ P_d = P_{\text{stator}} - P_{\text{cu}} \]

\[ P_{\text{shaft}} = P_d - P_{\text{stray}} = P_{\text{stator}} - (P_{\text{cu}} + P_{\text{stray}}) \]

\[ P_{\text{in,total}} = P_{\text{stator}} + P_{\text{rotor}} \]

Where: \(P_{\text{stator}}\) = stator active input power (W), \(P_{\text{cu}}\) = armature copper loss (W), \(P_d\) = developed mechanical power (W), \(P_{\text{stray}}\) = rotational core, friction, and windage loss (W), \(P_{\text{shaft}} = P_{\text{out}}\) = net mechanical output power (W), \(P_{\text{rotor}} = P_f\) = rotor field excitation loss (W)

Example. A \(440\ \text{V}\), single-phase synchronous motor draws a line current of \(50\ \text{A}\) at \(0.9\) power factor lagging. Armature resistance and reactance are \(0.8\ \Omega\) and \(10\ \Omega\) respectively. If stray power losses amount to \(500\ \text{W}\), determine the brake horsepower output of the motor.

Solution.

\[ P_{\text{stator}} = V_p I_p \cos\theta = (440)(50)(0.9) = 19800\ \text{W} \qquad\qquad P_{\text{cu}} = I_p^2 R_a = (50)^2(0.8) = 2000\ \text{W} \]

\[ P_{\text{out}} = P_{\text{stator}} - P_{\text{cu}} - P_{\text{stray}} = 19800 - 2000 - 500 = 17300\ \text{W} = \dfrac{17300}{746} = 23.19\ \text{hp} \approx 23.2\ \text{hp} \]

Answer: \(23.2\ \text{hp}\)

6. Formula Summary

Formula Name	Formula	Variables
Synchronous Rotor Speed	\(N_r = N_s = \dfrac{120f}{P}\)	\(N_r\) = rotor speed (rpm) \(f\) = frequency (Hz) \(P\) = poles
Back EMF Scalar Equation	\(E_p = \sqrt{(V_p \cos\theta \mp I_p R_a)^2 + (V_p \sin\theta \pm I_p X_s)^2}\)	\(E_p\) = back emf per phase (V) \(V_p\) = phase voltage (V) \(I_p\) = phase current (A) \(R_a\) = armature resistance (\(\Omega\)) \(X_s\) = synchronous reactance (\(\Omega\))
Complex Back EMF	\(\mathbf{E}_p \angle \pm \delta = \mathbf{V}_p \angle 0^\circ - (\mathbf{I}_p \angle \pm \theta)(R_a + j X_s)\)	\(\delta\) = torque angle (deg) \(\theta\) = power factor angle (deg)
Power Developed per Phase	\(P_d = \dfrac{E_p V_p}{X_s} \sin\delta\)	\(P_d\) = developed power per phase (W) \(E_p\) = back emf (V) \(V_p\) = terminal voltage (V) \(X_s\) = reactance (\(\Omega\)) \(\delta\) = torque angle (deg)
Torque Developed per Phase	\(T = \dfrac{9.55 P_d}{N_s}\)	\(T\) = developed torque (\(\text{N}\cdot\text{m}\)) \(P_d\) = developed power per phase (W) \(N_s\) = synchronous speed (rpm)
Maximum Power Developed	\(P_{d,\max} = \dfrac{E_p V_p}{X_s}\)	\(P_{d,\max}\) = maximum developed power per phase (W)
Maximum Pull-Out Torque	\(T_{\max} = \dfrac{9.55 P_{d,\max}}{N_s}\)	\(T_{\max}\) = maximum pull-out torque (\(\text{N}\cdot\text{m}\))
Three-Phase Stator Input Power	\(P_{\text{stator}} = \sqrt{3} V_L I_L \cos\theta\)	\(P_{\text{stator}}\) = 3-phase input power (W) \(V_L\) = line voltage (V) \(I_L\) = line current (A) \(\cos\theta\) = power factor
Three-Phase Armature Copper Loss	\(P_{\text{cu}(3\phi)} = 3 I_p^2 R_a\)	\(P_{\text{cu}(3\phi)}\) = 3-phase copper loss (W) \(I_p\) = phase current (A) \(R_a\) = armature resistance (\(\Omega\))
Total Power Loss	\(P_{\text{losses}} = P_{\text{cu}} + P_{\text{stray}}\)	\(P_{\text{losses}}\) = total loss (W) \(P_{\text{cu}}\) = copper loss (W) \(P_{\text{stray}}\) = stray rotational loss (W)
Overall Efficiency	\(\eta = \dfrac{P_{\text{out}}}{P_{\text{out}} + P_{\text{losses}}} \times 100\%\)	\(\eta\) = efficiency (\%) \(P_{\text{out}}\) = shaft output power (W) \(P_{\text{losses}}\) = total losses (W)
Mechanical Power Developed	\(P_d = P_{\text{stator}} - P_{\text{cu}}\)	\(P_d\) = developed power (W) \(P_{\text{stator}}\) = stator input power (W) \(P_{\text{cu}}\) = armature copper loss (W)
Shaft Output Power	\(P_{\text{shaft}} = P_d - P_{\text{stray}}\)	\(P_{\text{shaft}}\) = net shaft output power (W) \(P_d\) = developed power (W) \(P_{\text{stray}}\) = rotational loss (W)