10.3 Synchronous Machines (Operating Principles, Power-Angle Curves & V-Curves)

Key Takeaways

  • Synchronous machines operate at fixed synchronous speed (ns=120f/Pn_s = 120f/P), where mechanical power is governed by the power angle δ\delta and reactive power is independently controlled by DC field excitation current (IfI_f).

  • For a cylindrical-rotor machine with negligible stator resistance, active power is P=3EfVtXssin⁡δP = \frac{3 E_f V_t}{X_s} \sin \delta with theoretical steady-state pull-out limit at δ=90∘\delta = 90^\circ, while reactive power is Q=3EfVtXscos⁡δ−3Vt2XsQ = \frac{3 E_f V_t}{X_s} \cos \delta - \frac{3 V_t^2}{X_s}.

  • Overexcited synchronous machines (Efcos⁡δ>VtE_f \cos \delta > V_t) supply reactive power to the system (lagging current for generators, leading current for motors), whereas underexcited machines absorb reactive power.

  • A Synchronous Condenser is an unloaded synchronous motor (δ≈0∘\delta \approx 0^\circ) operated in the overexcited regime to provide dynamically adjustable leading VARs for power factor correction and grid voltage stabilization.

  • V-Curves illustrate armature current IaI_a versus field current IfI_f for constant active power contours, reaching minimum IaI_a at unity power factor (PF=1.0\text{PF} = 1.0).

Last updated: August 2026

10.3 Synchronous Machines (Operating Principles, Power-Angle Curves & V-Curves)

Synchronous machines are the primary source of bulk electrical power generation worldwide and serve as specialized high-horsepower industrial drives and dynamic reactive power compensators. Unlike induction machines, synchronous machines operate at strictly synchronous speed (ns=120f/Pn_s = 120f/P) under steady-state conditions, with real power exchange mediated by the rotor power angle (δ\delta) and reactive power exchange controlled by DC field excitation (IfI_f).


1. Machine Construction & Rotor Topologies

A synchronous machine consists of a stationary 3-phase armature winding on the stator and a DC-excited field winding on the rotor.

Rotor Topologies

  1. Cylindrical (Round) Rotor:

    • Uniform air-gap geometry.
    • High-speed 2-pole or 4-pole machines (1,800 rpm1,800\text{ rpm} or 3,600 rpm3,600\text{ rpm} at 60 Hz60\text{ Hz}).
    • Driven by steam or gas turbines (turbogenerators).
    • Characterized by a single per-phase synchronous reactance: Xs=Xal+XarX_s = X_{al} + X_{ar} (leakage reactance plus armature reaction reactance).
  2. Salient-Pole Rotor:

    • Non-uniform air gap with projecting (salient) pole shoes.
    • Low-speed multipole machines (72 rpm72\text{ rpm} to 900 rpm900\text{ rpm}, e.g., 12 to 72 poles).
    • Driven by hydro turbines or large diesel/gas reciprocating engines.
    • Analyzed using Blondel's Two-Axis Theory: Direct-axis synchronous reactance (XdX_d) along the magnetic pole axis, and Quadrature-axis synchronous reactance (XqX_q) along the interpolar axis (Xd>XqX_d > X_q).

DC Field Excitation Systems

The rotor DC field is supplied either through stationary carbon brushes riding on rotating slip rings or via modern brushless exciters (an inverted alternator with shaft-mounted rotating diode rectifiers). Adjusting the field current IfI_f adjusts the internal generated excitation voltage magnitude (EfE_f) along the open-circuit saturation curve.


2. Per-Phase Equivalent Circuit and Phasor Conventions

For a cylindrical-rotor machine on a per-phase (wye) basis:

  • VtV_t: Terminal phase voltage (Vt=VLL/3∠0∘V_t = V_{LL}/\sqrt{3} \angle 0^\circ).
  • EfE_f: Internal generated excitation voltage (Ef∠δE_f \angle \delta).
  • IaI_a: Armature phase current (Ia∠θI_a \angle \theta).
  • RaR_a: Stator armature winding resistance per phase (often neglected in power calculations, Ra≈0R_a \approx 0).
  • XsX_s: Synchronous reactance per phase (Xs=Xl+XaX_s = X_l + X_a).

Phasor Equations

  • Synchronous Generator Mode (current flowing out of machine to grid):

    Ef=Vt+Ia(Ra+jXs)≈Vt+jIaXs\mathbf{E}_f = \mathbf{V}_t + \mathbf{I}_a (R_a + jX_s) \approx \mathbf{V}_t + j\mathbf{I}_a X_s
    • Power angle δ>0∘\delta > 0^\circ (Ef\mathbf{E}_f leads Vt\mathbf{V}_t in the direction of rotation).
  • Synchronous Motor Mode (current flowing into machine from grid):

    Vt=Ef+Ia(Ra+jXs)≈Ef+jIaXs  ⟹  Ef=Vt−jIaXs\mathbf{V}_t = \mathbf{E}_f + \mathbf{I}_a (R_a + jX_s) \approx \mathbf{E}_f + j\mathbf{I}_a X_s \implies \mathbf{E}_f = \mathbf{V}_t - j\mathbf{I}_a X_s
    • Power angle δ<0∘\delta < 0^\circ (Ef\mathbf{E}_f lags Vt\mathbf{V}_t).

3. Power-Angle Equations and Stability Limits

Neglecting armature resistance (Ra=0R_a = 0), the per-phase complex power delivered by a synchronous generator to an infinite bus is:

Ia=Ef∠δ−Vt∠0∘jXs=Efsin⁡δXs−j(Efcos⁡δ−VtXs)\mathbf{I}_a = \frac{E_f \angle \delta - V_t \angle 0^\circ}{jX_s} = \frac{E_f \sin \delta}{X_s} - j \left( \frac{E_f \cos \delta - V_t}{X_s} \right) S3ϕ=3VtIa∗=3Vt[Efsin⁡δXs+j(Efcos⁡δ−VtXs)]\mathbf{S}_{3\phi} = 3 \mathbf{V}_t \mathbf{I}_a^* = 3 V_t \left[ \frac{E_f \sin \delta}{X_s} + j \left( \frac{E_f \cos \delta - V_t}{X_s} \right) \right]

Active and Reactive Power Formulas

P3ϕ=3EfVtXssin⁡δ=Ef,LLVt,LLXssin⁡δ[W]P_{3\phi} = \frac{3 E_f V_t}{X_s} \sin \delta = \frac{E_{f,LL} V_{t,LL}}{X_s} \sin \delta \quad [\text{W}] Q3ϕ=3EfVtXscos⁡δ−3Vt2Xs[VAR]Q_{3\phi} = \frac{3 E_f V_t}{X_s} \cos \delta - \frac{3 V_t^2}{X_s} \quad [\text{VAR}]

where EfE_f and $V_t in the three-phase formulas are per-phase line-to-neutral magnitudes (or line-to-line magnitudes if the factor of 3 is omitted).

Salient-Pole Power Equation (Reluctance Torque)

In salient-pole machines, the saliency (Xd≠XqX_d \neq X_q) creates a reluctance power component that operates independently of field excitation:

P3ϕ=3EfVtXdsin⁡δ⏟Field Excitation Power+3Vt22(1Xq−1Xd)sin⁡2δ⏟Reluctance PowerP_{3\phi} = \underbrace{\frac{3 E_f V_t}{X_d} \sin \delta}_{\text{Field Excitation Power}} + \underbrace{\frac{3 V_t^2}{2} \left( \frac{1}{X_q} - \frac{1}{X_d} \right) \sin 2\delta}_{\text{Reluctance Power}}

Steady-State Stability Limit (Pull-Out Power)

  • Cylindrical Machine: Maximum theoretical power occurs at δ=90∘\delta = 90^\circ: Pmax,3ϕ=3EfVtXsP_{max,3\phi} = \frac{3 E_f V_t}{X_s} If mechanical turbine power exceeds PmaxP_{max} (or if electrical load increases such that δ>90∘\delta > 90^\circ), the synchronizing power coefficient dPdδ\frac{dP}{d\delta} becomes negative, causing the generator to lose synchronism and "slip poles."
  • Salient-Pole Machine: Peak power occurs at δ<90∘\delta < 90^\circ (typically δ≈65∘−75∘\delta \approx 65^\circ - 75^\circ) due to the second-harmonic reluctance term (sin⁡2δ\sin 2\delta).

4. Excitation Regimes, Reactive Power Control & Synchronous Condensers

By adjusting field current IfI_f (and hence EfE_f), the synchronous machine can supply or absorb reactive power independently of active power delivery:

Operating RegimeExcitation ConditionGenerator BehaviorMotor BehaviorReactive Power (QQ)
OverexcitedEfcos⁡δ>VtE_f \cos \delta > V_tDelivers lagging current (supplies +Q+Q to grid)Draws leading current (supplies +Q+Q to bus)+Q+Q exported to grid (boosts bus voltage)
Normal ExcitationEfcos⁡δ=VtE_f \cos \delta = V_tUnity power factor (PF=1.0\text{PF} = 1.0)Unity power factor (PF=1.0\text{PF} = 1.0)Q=0Q = 0
UnderexcitedEfcos⁡δ<VtE_f \cos \delta < V_tDelivers leading current (absorbs −Q-Q from grid)Draws lagging current (absorbs −Q-Q from bus)−Q-Q imported from grid (lowers bus voltage)

Synchronous Condenser Operation

A Synchronous Condenser is a synchronous machine connected to the electrical grid with no mechanical load on its shaft (Pshaft≈0  ⟹  δ≈0∘P_{shaft} \approx 0 \implies \delta \approx 0^\circ):

Q3ϕ≈3Vt(Ef−Vt)XsQ_{3\phi} \approx \frac{3 V_t (E_f - V_t)}{X_s}
  • When overexcited (Ef>VtE_f > V_t), it supplies variable capacitive VARs to the substation bus, supporting system voltage during heavy transmission loading.
  • When underexcited (Ef<VtE_f < V_t), it acts as an adjustable shunt reactor, absorbing inductive VARs during light load / Ferranti effect conditions.

5. Synchronous Machine V-Curves & Generator Capability Limits

V-Curves (IaI_a vs. IfI_f)

Plotting armature current IaI_a against field current IfI_f for constant mechanical power outputs generates a family of V-shaped curves:

  • The bottom vertex of each V-curve corresponds to unity power factor where armature current IaI_a is minimized for that MW level.
  • Operating to the right of the vertex represents the overexcited regime (lagging PF for generator, leading PF for motor).
  • Operating to the left of the vertex represents the underexcited regime (leading PF for generator, lagging PF for motor).

Generator Capability Curve (P-Q Operating Envelope)

A synchronous generator's safe operating region in the P-Q plane is constrained by three physical limits:

  1. Armature Current Limit (Stator Heating): A circle centered at (P=0,Q=0)(P=0, Q=0) with radius equal to rated MVA: P2+Q2≤Srated2P^2 + Q^2 \le S_{rated}^2.
  2. Field Current Limit (Rotor Heating): A circle centered at (0,−3Vt2/Xs)(0, -3V_t^2/X_s) with radius 3Ef,maxVtXs\frac{3 E_{f,max} V_t}{X_s}. Limits maximum overexcited lagging VAR output.
  3. Stator End-Core Heating & Stability Limit: Limits maximum underexcited leading VAR absorption to prevent localized eddy-current heating in stator core laminations and preserve steady-state transient stability margins.

6. Worked Numeric Example: Synchronous Generator Excitation & Power Angle

Problem Statement

A 3-phase, 25 MVA, 13.8 kV, 60 Hz, Y-connected cylindrical-rotor synchronous generator has a synchronous reactance of Xs=6.4 ΩX_s = 6.4\,\Omega per phase and negligible armature resistance (Ra≈0R_a \approx 0). The generator is connected to an infinite bus operating at rated 13.8 kV and delivers 20.0 MW20.0\text{ MW} at 0.800.80 power factor lagging.

Calculate:

  1. Stator terminal phase voltage VtV_t and armature current phasor Ia\mathbf{I}_a.
  2. Internal excitation voltage phasor Ef\mathbf{E}_f (magnitude in line-to-neutral and line-to-line, and power angle δ\delta).
  3. Total three-phase reactive power Q3ϕQ_{3\phi} supplied by the generator.
  4. Steady-state maximum pull-out power (Pmax,3ϕP_{max,3\phi}) under this field excitation.

Step-by-Step Solution

Step 1: Terminal Voltage and Armature Current Phasors

Terminal line-to-neutral voltage:

Vt=13,8003=7,967.43 V∠0∘V_t = \frac{13,800}{\sqrt{3}} = 7,967.43\text{ V} \angle 0^\circ

Apparent power:

S3ϕ=Pcos⁡θ=20.0 MW0.80=25.0 MVAS_{3\phi} = \frac{P}{\cos \theta} = \frac{20.0\text{ MW}}{0.80} = 25.0\text{ MVA}

Armature current magnitude:

Ia=S3ϕ3VLL=25,000,0003×13,800=1,045.92 AI_a = \frac{S_{3\phi}}{\sqrt{3} V_{LL}} = \frac{25,000,000}{\sqrt{3} \times 13,800} = 1,045.92\text{ A}

Power factor angle for lagging current:

θ=−arccos⁡(0.80)=−36.87∘\theta = -\arccos(0.80) = -36.87^\circ Ia=1,045.92∠−36.87∘ A=(836.74−j627.55) A\mathbf{I}_a = 1,045.92 \angle -36.87^\circ\text{ A} = (836.74 - j627.55)\text{ A}

Step 2: Internal Excitation Voltage Phasor (Ef\mathbf{E}_f)

Ef=Vt+jIaXs\mathbf{E}_f = \mathbf{V}_t + j \mathbf{I}_a X_s jIaXs=j(836.74−j627.55)×6.4=j5,355.14+4,016.32=4,016.32+j5,355.14 Vj \mathbf{I}_a X_s = j (836.74 - j627.55) \times 6.4 = j 5,355.14 + 4,016.32 = 4,016.32 + j 5,355.14\text{ V}

Summing with Vt\mathbf{V}_t:

Ef=7,967.43+4,016.32+j5,355.14=11,983.75+j5,355.14 V\mathbf{E}_f = 7,967.43 + 4,016.32 + j 5,355.14 = 11,983.75 + j 5,355.14\text{ V}

Converting to polar coordinates:

∣Ef∣=(11,983.75)2+(5,355.14)2=143,610,264+28,677,525=172,287,789=13,125.84 V (line-to-neutral)|E_f| = \sqrt{(11,983.75)^2 + (5,355.14)^2} = \sqrt{143,610,264 + 28,677,525} = \sqrt{172,287,789} = 13,125.84\text{ V (line-to-neutral)} δ=arctan⁡(5,355.1411,983.75)=arctan⁡(0.44686)=24.08∘\delta = \arctan\left(\frac{5,355.14}{11,983.75}\right) = \arctan(0.44686) = 24.08^\circ

Line-to-line excitation voltage:

Ef,LL=3×13,125.84 V=22,734.6 V=22.73 kVE_{f,LL} = \sqrt{3} \times 13,125.84\text{ V} = 22,734.6\text{ V} = 22.73\text{ kV} Ef=13.13 kV∠24.08∘\mathbf{E}_f = 13.13\text{ kV} \angle 24.08^\circ

Step 3: Reactive Power Supplied

Q3ϕ=P×tan⁡θ=20.0 MW×tan⁡(36.87∘)=20.0×0.750=15.0 MVARQ_{3\phi} = P \times \tan \theta = 20.0\text{ MW} \times \tan(36.87^\circ) = 20.0 \times 0.750 = 15.0\text{ MVAR}

Verification via reactive power formula:

Q3ϕ=3EfVtXscos⁡δ−3Vt2Xs=3×13,125.84×7,967.436.4cos⁡(24.08∘)−3×(7,967.43)26.4Q_{3\phi} = \frac{3 E_f V_t}{X_s} \cos \delta - \frac{3 V_t^2}{X_s} = \frac{3 \times 13,125.84 \times 7,967.43}{6.4} \cos(24.08^\circ) - \frac{3 \times (7,967.43)^2}{6.4} Q3ϕ=(48,995,780×0.91299)−29,756,238=44,732,860−29,756,238=14,976,622 VAR≈15.0 MVARQ_{3\phi} = (48,995,780 \times 0.91299) - 29,756,238 = 44,732,860 - 29,756,238 = 14,976,622\text{ VAR} \approx 15.0\text{ MVAR}

Step 4: Maximum Pull-Out Power (PmaxP_{max})

Pull-out power occurs at δ=90∘\delta = 90^\circ (sin⁡90∘=1.0\sin 90^\circ = 1.0):

Pmax,3ϕ=3EfVtXs=3×13,125.84×7,967.436.4=48,995,780 W=49.0 MWP_{max,3\phi} = \frac{3 E_f V_t}{X_s} = \frac{3 \times 13,125.84 \times 7,967.43}{6.4} = 48,995,780\text{ W} = 49.0\text{ MW}

7. Common Exam Traps & High-Yield Summary

Warning

Exam Trap 1: Confusing Generator vs. Motor Power Angle Signs In synchronous generators, field excitation leads terminal voltage (δ>0∘\delta > 0^\circ). In synchronous motors, mechanical shaft load retards the rotor so field excitation lags terminal voltage (δ<0∘\delta < 0^\circ).

Warning

Exam Trap 2: Generator vs. Motor Overexcitation Conventions

  • An overexcited generator supplies lagging reactive power (+Q+Q) to the grid.
  • An overexcited motor operates at a leading power factor, which also supplies capacitive reactive power (+Q+Q) to the local distribution bus.

Important

Exam Trap 3: Infinite Bus Voltage & Frequency Invariance An infinite bus has constant voltage and frequency. Increasing turbine governor mechanical input PmechP_{mech} increases active power PP and power angle δ\delta, but does NOT change machine speed. Increasing field current IfI_f increases reactive power QQ and terminal voltage support, but does NOT change active power PP.

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Synchronous Generator Overexcited Phasor Diagram
Test Your Knowledge

A 3-phase cylindrical-rotor synchronous generator is delivering rated active power to an infinite bus. If the rotor field current IfI_f is increased while the prime mover turbine power remains constant, how do the power angle δ\delta and reactive power output QQ change?

A

Power angle δ increases and Q decreases

B

Power angle δ increases and Q increases

C

Power angle δ decreases and Q increases

D

Power angle δ remains unchanged and Q increases

Test Your Knowledge

An industrial distribution bus experiences low voltage and an inductive power factor of 0.72 lagging due to heavy induction motor loading. How should an unloaded synchronous motor (synchronous condenser) connected to this bus be operated to correct the power factor toward unity and raise bus voltage?

A

Operate underexcited to absorb excess reactive VARs from the bus

B

Disconnect DC field excitation completely to operate as an induction motor

C

Maintain field excitation at the exact unity power factor normal level

D

Operate overexcited to supply leading capacitive reactive VARs to the bus

Test Your Knowledge

What is the theoretical steady-state pull-out power limit of a 3-phase, 13.8 kV (line-to-line) cylindrical-rotor synchronous generator having Xs=5.0 Ω/phaseX_s = 5.0\,\Omega/\text{phase} when its internal excitation voltage is Ef=9.2 kVE_f = 9.2\text{ kV} line-to-neutral? (Assume infinite bus at rated voltage and Ra≈0R_a \approx 0).

A

44.0 MW

B

25.4 MW

C

14.7 MW

D

76.2 MW

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