5.3 Synchronous Generators: Operation, Excitation, Power-Angle Equations & V-Curves

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5 Chapter 5: Rotating Machines: Induction & Synchronous

5.1 Three-Phase Induction Motors: Principles, Equivalent Circuits & Slip-Torque Characteristics 5.2 Induction Motor Starting, Speed Control & Braking Methods 5.3 Synchronous Generators: Operation, Excitation, Power-Angle Equations & V-Curves 5.4 Synchronous Motors & Condensers for Power Factor Correction

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5.3 Synchronous Generators: Operation, Excitation, Power-Angle Equations & V-Curves

Key Takeaways Explain This Topic AI

Synchronous generators operate in strict lockstep with grid frequency via N_s = 120*f / P; turbo-generators use high-speed cylindrical (round) rotors with uniform air gaps (X_d ≈ X_q = X_s), while hydro-generators use low-speed salient-pole rotors with pronounced pole projections (X_d > X_q).

The per-phase terminal voltage relation E_f = V_t + I_a*(R_a + j*X_s) dictates that overexcited generators operate at a lagging power factor exporting reactive power (+Q) to support grid voltage, whereas underexcited generators operate at a leading power factor importing reactive power (-Q).

For a cylindrical-rotor generator, active power output is P = (3 E_f V_t / X_s) * sin(δ), establishing the static steady-state stability limit at power angle δ = 90° where maximum power P_max = 3 E_f V_t / X_s occurs.

Two-reaction theory for salient-pole machines separates armature reaction into direct-axis (X_d) and quadrature-axis (X_q) reactances, yielding an additional reluctance power component P_rel = (V_t^2 / 2) * (1/X_q - 1/X_d) * sin(2δ) that shifts peak power to δ < 90°.

The generator P-Q capability curve is bounded by three distinct physical constraints: the stator armature winding heating limit (armature current circle), the rotor field winding thermal limit (excitation circle), and the stator core end-region iron heating / steady-state stability limit in the underexcited quadrant.

Last updated: August 2026

5.3 Synchronous Generators: Operation, Excitation, Power-Angle Equations & V-Curves

Synchronous generators (also called alternators ) produce virtually all commercial electrical energy worldwide across thermal, nuclear, gas turbine, and hydroelectric power stations. Unlike induction machines, synchronous machines operate at a fixed, unyielding speed directly proportional to electrical grid frequency ( N = N s = 120 f / P N = N_s = 120 f / P N = N s ​ = 120 f / P ) and utilize an independent Direct Current (DC) field excitation system on the rotor.

Understanding synchronous machine modeling, phasor analysis under varying power factors, active and reactive power-angle equations, saliency effects, excitation V-curves, and P-Q capability boundaries is a primary focus of the NCEES PE Power examination.

1. Machine Construction: Cylindrical vs. Salient-Pole Rotors

Synchronous generators are categorized mechanically into two primary rotor topologies based on prime mover operating speed. +-----------------------------------------------------------------------------+

• | SYNCHRONOUS GENERATOR ROTOR TOPOLOGIES |
• | |
• | CYLINDRICAL / ROUND ROTOR SALIENT-POLE ROTOR |
• | - High-speed (1,800 or 3,600 RPM) - Low-speed (100 - 600 RPM) |
• | - 2 or 4 poles - Many poles (12 to 72 poles|
• | - Steam and gas turbines (Turbo-generators) - Hydro-turbines and diesels|
• | - Solid steel forging, long axial length, - Large diameter, short |
• | small rotor diameter axial length |
• | - Uniform air gap (X_d ≈ X_q = X_s) - Non-uniform air gap |
• | (X_d > X_q) |
• +-----------------------------------------------------------------------------+
• Mechanical Comparison Matrix

Engineering Parameter Cylindrical (Round) Rotor Salient-Pole Rotor

Prime Mover High-speed steam and gas turbines Low-speed hydraulic turbines, reciprocating internal combustion engines

Speed Range ( 60 Hz 60\text{ Hz} 60 Hz ) 1 , 800 RPM 1,800\text{ RPM} 1 , 800 RPM ( 4 4 4 -pole) or 3 , 600 RPM 3,600\text{ RPM} 3 , 600 RPM ( 2 2 2 -pole) 100 to 600 RPM 100\text{ to } 600\text{ RPM} 100 to 600 RPM ( 12 to 72 12\text{ to } 72 12 to 72 -pole)

Rotor Geometry Small diameter ( 1.0 − 1.5 m 1.0 - 1.5\text{ m} 1.0 − 1.5 m ), long axial length ( 5 − 10 m 5 - 10\text{ m} 5 − 10 m ) Large diameter ( 5 − 15 m 5 - 15\text{ m} 5 − 15 m ), short axial length ( 1 − 2 m 1 - 2\text{ m} 1 − 2 m )

Air Gap Profile Completely uniform around rotor circumference Pronounced projecting poles with non-uniform air gap (minimum at pole center)

Reactance Property Single synchronous reactance: X d ≈ X q = X s X_d \approx X_q = X_s X d ​ ≈ X q ​ = X s ​ Distinct reactances: Direct-axis X d X_d X d ​ , Quadrature-axis X q X_q X q ​ ( X d > X q X_d > X_q X d ​ > X q ​ )

Damper Windings Solid steel rotor body provides eddy-current damping Distinct copper amortisseur bars embedded in pole faces

2. Equivalent Circuit & Phasor Analysis

The per-phase steady-state equivalent circuit of a cylindrical-rotor synchronous generator consists of an internal generated EMF E f \mathbf{E}_f E f ​ in series with the synchronous impedance Z s = R a + j X s \mathbf{Z}_s = R_a + j X_s Z s ​ = R a ​ + j X s ​ . PER-PHASE SYNCHRONOUS GENERATOR CIRCUIT

• +------------[ R_a ]-----------[ jX_s ]------------+ A
• | Armature Synchronous |
• | Resistance Reactance |
• ( + ) |
• E_f ( ~ ) Internal Generated EMF I_a | Load Terminals
• (Excitation) ---> | (Grid Bus)
• ( - ) |
• | + |
• | V_t | Terminal Voltage
• | - |
• +--------------------------------------------------+ N
• Synchronous Reactance Components

The total synchronous reactance X s X_s X s ​ accounts for two distinct physical phenomena: X s = X a l + X a r X_s = X_{al} + X_{ar} X s ​ = X a l ​ + X a r ​

Armature Leakage Reactance ( X a l X_{al} X a l ​ ): Leakage flux around stator slot conductors that does not cross the air gap.

Armature Reaction Reactance ( X a r X_{ar} X a r ​ ): Magnetic flux produced by stator load current I a \mathbf{I}_a I a ​ crossing the air gap and modifying the rotor field flux.

Voltage Equation E f = V t + I a ( R a + j X s ) ≈ V t + j I a X s ( since R a ≪ X s ) \mathbf{E}_f = \mathbf{V}_t + \mathbf{I}_a (R_a + j X_s) \approx \mathbf{V}_t + j \mathbf{I}_a X_s \quad (\text{since } R_a \ll X_s) E f ​ = V t ​ + I a ​ ( R a ​ + j X s ​ ) ≈ V t ​ + j I a ​ X s ​ ( since R a ​ ≪ X s ​ ) GENERATOR PHASOR DIAGRAMS

• LAGGING PF (Overexcited, Q > 0) LEADING PF (Underexcited, Q < 0)
• E_f V_t
• / ^ / ^
• / | j*I_a*X_s / | j*I_a*X_s
• / | / |
• / δ | E_f/ δ |
• +------+------> V_t +------+------> I_a (Leads V_t)
• \ θ | \ θ
• \ | \
• v I_a (Lags V_t) v V_t
• |E_f| > |V_t| ===> Supplies VARs |E_f| < |V_t| ===> Absorbs VARs

Operational State Field Current ( I f I_f I f ​ ) Internal EMF ( ∣ E f ∣ \vert E_f\vert ∣ E f ​ ∣ ) Phase Current I a \mathbf{I}_a I a ​ Reactive Power ( Q Q Q ) Grid Impact

Overexcited High ( I f > I f , unity I_f > I_{f,\text{unity}} I f ​ > I f , unity ​ ) ∣ E f ∣ > ∣ V t ∣ \vert E_f\vert> \vert V_t\vert ∣ E f ​ ∣ > ∣ V t ​ ∣ Lags terminal voltage V t \mathbf{V}_t V t ​ Positive ( Q > 0 Q > 0 Q > 0 , Exported) Boosts system voltage; acts as a capacitor to the grid

Unity Power Factor Normal ( I f = I f , unity I_f = I_{f,\text{unity}} I f ​ = I f , unity ​ ) ∣ E f ∣ = V t 2 + ( I a X s ) 2 \vert E_f\vert= \sqrt{V_t^2 + (I_a X_s)^2} ∣ E f ​ ∣ = V t 2 ​ + ( I a ​ X s ​ ) 2 ​ In phase with V t \mathbf{V}_t V t ​ Zero ( Q = 0 Q = 0 Q = 0 ) Transfers purely real MW power

Underexcited Low ( I f < I f , unity I_f < I_{f,\text{unity}} I f ​ < I f , unity ​ ) ∣ E f ∣ < ∣ V t ∣ \vert E_f\vert< \vert V_t\vert ∣ E f ​ ∣ < ∣ V t ​ ∣ Leads terminal voltage V t \mathbf{V}_t V t ​ Negative ( Q < 0 Q < 0 Q < 0 , Imported) Lowers system voltage; absorbs excess VARs from grid

3. Power-Angle Equations & Static Stability Limit

Let the terminal voltage phasor be the reference: V t = V t ∠ 0 ∘ \mathbf{V}_t = V_t \angle 0^\circ V t ​ = V t ​ ∠ 0 ∘ . The internal generated EMF is E f = E f ∠ δ \mathbf{E}_f = E_f \angle \delta E f ​ = E f ​ ∠ δ , where δ \delta δ is the power angle (torque angle / rotor angle). +-----------------------------------------------------------------------------+

• | ROUND-ROTOR POWER-ANGLE FORMULATIONS (3-PHASE) |
• | |
• | Real Electrical Power Output (P): |
• | P = (3 * E_f * V_t / X_s) * sin(δ) [Watts] |
• | |
• | Reactive Electrical Power Output (Q): |
• | Q = (3 * E_f * V_t / X_s) * cos(δ) - (3 * V_t^2 / X_s) [VAR] |
• +-----------------------------------------------------------------------------+

(Note: For per-phase formulations, omit the factor of 3 and use line-to-neutral voltages V t V_t V t ​ and E f E_f E f ​ .) POWER-ANGLE CURVE & STABILITY BOUNDARIES

• Power P
• ^
• | P_max = (3*E_f*V_t / X_s)
• | /---\
• | / \
• | / \
• | P_rated / \
• | *-------------* \
• | | / \
• | | / \
• | | / \
• +------+---------+-------------------+-----------------------------> Angle δ
• 0° δ_rated δ = 90° 180°
• (20°-35°) (Static Limit)
• <-- STABLE REGION --><------ UNSTABLE REGION ----->
• Static Steady-State Stability Limit

Maximum Power Transfer: Differentiating P P P with respect to δ \delta δ yields the Synchronizing Power Coefficient P s y n P_{syn} P sy n ​ : P s y n = d P d δ = 3 E f V t X s cos ⁡ δ P_{syn} = \frac{dP}{d\delta} = \frac{3 E_f V_t}{X_s} \cos\delta P sy n ​ = d δ d P ​ = X s ​ 3 E f ​ V t ​ ​ cos δ

Stability Criterion: For the generator to remain synchronized following a small perturbation, P s y n > 0 P_{syn} > 0 P sy n ​ > 0 . This restricts steady-state operation to: δ < 90 ∘ \delta < 90^\circ δ < 9 0 ∘

Maximum Theoretical Power Capability: P m a x = 3 E f V t X s at δ = 90 ∘ P_{max} = \frac{3 E_f V_t}{X_s} \quad \text{at } \delta = 90^\circ P ma x ​ = X s ​ 3 E f ​ V t ​ ​ at δ = 9 0 ∘

Practical Operating Margin: Commercial utility generators operate at full load with δ \delta δ between 20 ∘ 20^\circ 2 0 ∘ and 35 ∘ 35^\circ 3 5 ∘ , preserving a massive transient stability margin to survive external grid faults and line switching.

4. Salient-Pole Machines & Two-Reaction Theory (Blondel)

In salient-pole generators, the non-uniform air gap causes the magnetic reluctance along the direct pole axis ( d d d -axis) to be far lower than the reluctance along the interpolar quadrature axis ( q q q -axis). +-----------------------------------------------------------------------------+

• | TWO-REACTION IMPEDANCE PARAMETERS |
• | |
• | Direct-Axis Synchronous Reactance (X_d): |
• | - Small air gap at pole center ===> High permeance ===> Large reactance |
• | |
• | Quadrature-Axis Synchronous Reactance (X_q): |
• | - Large air gap between poles ===> Low permeance ===> Smaller reactance |
• | |
• | Typical Saliency Ratio: X_q ≈ 0.60 to 0.75 * X_d |
• +-----------------------------------------------------------------------------+

Armature Current Decomposition I a = I d + I q \mathbf{I}_a = \mathbf{I}_d + \mathbf{I}_q I a ​ = I d ​ + I q ​ E f = V t + j I d X d + j I q X q \mathbf{E}_f = \mathbf{V}_t + j \mathbf{I}_d X_d + j \mathbf{I}_q X_q E f ​ = V t ​ + j I d ​ X d ​ + j I q ​ X q ​

Salient-Pole Power Equation

Resolving the real electrical power output yields two distinct terms: P = E f V t X d sin ⁡ δ ⏟ Excitation Power Component + V t 2 2 ( 1 X q − 1 X d ) sin ⁡ ( 2 δ ) ⏟ Reluctance Power Component P = \underbrace{\frac{E_f V_t}{X_d} \sin\delta}_{\text{Excitation Power Component}} + \underbrace{\frac{V_t^2}{2} \left(\frac{1}{X_q} - \frac{1}{X_d}\right) \sin(2\delta)}_{\text{Reluctance Power Component}} P = Excitation Power Component X d ​ E f ​ V t ​ ​ sin δ ​ ​ + Reluctance Power Component 2 V t 2 ​ ​ ( X q ​ 1 ​ − X d ​ 1 ​ ) sin ( 2 δ ) ​ ​ +-----------------------------------------------------------------------------+

• | RELUCTANCE POWER CHARACTERISTICS |
• | |
• | 1. Excitation Independence: Reluctance power exists even if field |
• | excitation is completely lost (E_f = 0). It is caused by rotor iron |
• | aligning with the stator rotating magnetic field to minimize reluctance.|
• | 2. Double Frequency Angle (2δ): Reluctance power peaks at δ = 45°. |
• | 3. Total Peak Shift: Due to the sin(2δ) component, maximum power P_max |
• | in a salient-pole generator occurs at δ ≈ 70° - 75° (less than 90°). |
• +-----------------------------------------------------------------------------+
• 5. Generator V-Curves & Inverted V-Curves

V-Curves depict armature current ( I a I_a I a ​ ) as a function of field excitation current ( I f I_f I f ​ ) at various constant active power output levels ( P = 0 % , 50 % , 100 % MW P = 0\%, 50\%, 100\%\text{ MW} P = 0% , 50% , 100% MW ). SYNCHRONOUS GENERATOR V-CURVES

• Armature Current I_a
• ^
• | \ UNDEREXCITED / \ OVEREXCITED /
• | \ (Leading PF) / \ (Lagging PF) /
• | \ / \ /
• | \ / 100% MW \ /
• | \ / \ /
• | \ / \ / UNITY POWER FACTOR
• | \ *-------------------*------/---- LOCUS (cos θ = 1.0)
• | \ / 50% MW \ /
• | \ / \ /
• | * * 0% MW (No Load)
• +------------+--------------------------+--------------------------> Field Current I_f
• Under-Excited Over-Excited
• (Absorbs -Q) (Supplies +Q)

Key V-Curve Takeaways:

Minimum Armature Current Points: For any constant real power level, minimum stator current occurs at exactly Unity Power Factor ( cos ⁡ θ = 1.0 \cos\theta = 1.0 cos θ = 1.0 ).

Right Side of Unity Line ( I f > I f , unity I_f > I_{f,\text{unity}} I f ​ > I f , unity ​ ): Generator is overexcited , operating at a lagging power factor , supplying reactive power ( + Q +Q + Q ) to support grid voltage.

Left Side of Unity Line ( I f < I f , unity I_f < I_{f,\text{unity}} I f ​ < I f , unity ​ ): Generator is underexcited , operating at a leading power factor , absorbing reactive power ( − Q -Q − Q ) from the grid.

Inverted V-Curves: Plot power factor ( cos ⁡ θ \cos\theta cos θ ) versus field current ( I f I_f I f ​ ), forming inverted curves peaking at 1.0 1.0 1.0 at nominal excitation.

6. Generator P-Q Capability Curve

The P-Q Capability Curve defines the safe steady-state thermal and stability operating envelope of a utility synchronous generator. SYNCHRONOUS GENERATOR P-Q CAPABILITY CURVE

• Real Power P (MW)
• ^
• P_max| [ Turbine Real Power Rating ]
• | /---------------------------------\
• | / \ Rotor Field
• | / Stator Armature \ Winding Heating
• | / Current Heating \ Limit
• | / Limit Circle \
• | / (Radius = S_rated) \
• | / \
• | / \
• | / Stator End-Iron Heating \
• | / & Stability Limit (MEL) \
• Q (MVAR) +--+---------------------------------------------------+----> +Q (MVAR)
• (Leading / | (Lagging /
• Underexcited) Overexcited)

Three Primary Physical Operating Limits:

Armature Current Heating Limit (Stator Winding Limit):

Symmetrical circle centered at origin ( P = 0 , Q = 0 ) (P=0, Q=0) ( P = 0 , Q = 0 ) with radius equal to rated apparent power: S rated = P 2 + Q 2 S_{\text{rated}} = \sqrt{P^2 + Q^2} S rated ​ = P 2 + Q 2 ​ .

Prevents overheating and insulation breakdown of the three-phase stator copper windings.

Field Current Heating Limit (Rotor Winding Limit):

Circular boundary centered at coordinates ( 0 , − 3 V t 2 X s ) (0, -\frac{3 V_t^2}{X_s}) ( 0 , − X s ​ 3 V t 2 ​ ​ ) with radius 3 E f , m a x V t X s \frac{3 E_{f,max} V_t}{X_s} X s ​ 3 E f , ma x ​ V t ​ ​ .

Dominates the overexcited (lagging) region, limiting maximum continuous DC field current to prevent rotor thermal failure.

Stator End-Iron Core Heating & Steady-State Stability Limit:

Dominates the underexcited (leading) region.

In the underexcited regime, high axial leakage flux enters stator core end laminations perpendicular to the sheets, inducing massive localized eddy currents and extreme heating in the clamping fingers.

Protected by the Minimum Excitation Limiter (MEL) and loss-of-field relays ( ANSI Device 40 ).

7. Step-by-Step Worked Mathematical Example

Problem Statement:

A 3-phase, 13.8 kV (line-to-line), 50 MVA, 60 Hz, Y-connected cylindrical-rotor synchronous generator has a synchronous reactance of X s = 3.20 Ω X_s = 3.20\ \Omega X s ​ = 3.20 Ω per phase (armature resistance R a R_a R a ​ is negligible). The generator delivers rated MVA at a lagging power factor of 0.80 0.80 0.80 into an infinite power grid held at 13.8 kV 13.8\text{ kV} 13.8 kV .

Calculate:

Rated line current I a I_a I a ​ and terminal line-to-neutral voltage V t V_t V t ​ .

Internal generated excitation EMF per phase ( E f E_f E f ​ ), line-to-line excitation EMF ( E f , L L E_{f,LL} E f , LL ​ ), and power angle δ \delta δ .

Total active power ( P P P ) and reactive power ( Q Q Q ) delivered to the grid.

Maximum theoretical steady-state power capability ( P m a x P_{max} P ma x ​ ) at this excitation level and the static pull-out torque at 1 , 800 RPM 1,800\text{ RPM} 1 , 800 RPM .

Step-by-Step Solution:

Step 1: Terminal Quantities & Rated Current Per-phase line-to-neutral terminal voltage reference: V t = 13 , 800 V 3 = 7 , 967.43 V ∠ 0 ∘ V_t = \frac{13,800\text{ V}}{\sqrt{3}} = 7,967.43\text{ V} \angle 0^\circ V t ​ = 3 ​ 13 , 800 V ​ = 7 , 967.43 V ∠ 0 ∘ Rated Armature Current: I a = S 3phase 3 × V LL = 50 × 10 6 VA 3 × 13 , 800 V = 50 , 000 , 000 23 , 902.30 = 2 , 091.85 A \text{Rated Armature Current: } I_a = \frac{S_{\text{3phase}}}{\sqrt{3} \times V_{\text{LL}}} = \frac{50 \times 10^6\text{ VA}}{\sqrt{3} \times 13,800\text{ V}} = \frac{50,000,000}{23,902.30} = 2,091.85\text{ A} Rated Armature Current: I a ​ = 3 ​ × V LL ​ S 3phase ​ ​ = 3 ​ × 13 , 800 V 50 × 1 0 6 VA ​ = 23 , 902.30 50 , 000 , 000 ​ = 2 , 091.85 A Power Factor Angle: θ = arccos ⁡ ( 0.80 ) = 36.87 ∘ ( lagging ⟹ I a = 2 , 091.85 ∠ − 36.87 ∘ A ) \text{Power Factor Angle: } \theta = \arccos(0.80) = 36.87^\circ \quad (\text{lagging } \implies \mathbf{I}_a = 2,091.85 \angle -36.87^\circ\text{ A}) Power Factor Angle: θ = arccos ( 0.80 ) = 36.8 7 ∘ ( lagging ⟹ I a ​ = 2 , 091.85∠ − 36.8 7 ∘ A ) I a = 2 , 091.85 ( cos ⁡ ( − 36.87 ∘ ) + j sin ⁡ ( − 36.87 ∘ ) ) = 1 , 673.48 − j 1 , 255.11 A \mathbf{I}_a = 2,091.85 \left(\cos(-36.87^\circ) + j \sin(-36.87^\circ)\right) = 1,673.48 - j 1,255.11\text{ A} I a ​ = 2 , 091.85 ( cos ( − 36.8 7 ∘ ) + j sin ( − 36.8 7 ∘ ) ) = 1 , 673.48 − j 1 , 255.11 A

Step 2: Calculate Internal Generated EMF ( E f E_f E f ​ ) & Power Angle ( δ \delta δ ) E f = V t + j I a X s = 7 , 967.43 + j ( 3.20 Ω ) ( 1 , 673.48 − j 1 , 255.11 ) \mathbf{E}_f = \mathbf{V}_t + j \mathbf{I}_a X_s = 7,967.43 + j (3.20\ \Omega)(1,673.48 - j 1,255.11) E f ​ = V t ​ + j I a ​ X s ​ = 7 , 967.43 + j ( 3.20 Ω ) ( 1 , 673.48 − j 1 , 255.11 ) E f = 7 , 967.43 + j 5 , 355.14 − j 2 ( 4 , 016.35 ) = ( 7 , 967.43 + 4 , 016.35 ) + j 5 , 355.14 = 11 , 983.78 + j 5 , 355.14 V \mathbf{E}_f = 7,967.43 + j 5,355.14 - j^2 (4,016.35) = (7,967.43 + 4,016.35) + j 5,355.14 = 11,983.78 + j 5,355.14\text{ V} E f ​ = 7 , 967.43 + j 5 , 355.14 − j 2 ( 4 , 016.35 ) = ( 7 , 967.43 + 4 , 016.35 ) + j 5 , 355.14 = 11 , 983.78 + j 5 , 355.14 V Magnitude ∣ E f ∣ = 11 , 983.78 2 + 5 , 355.14 2 = 143 , 611 , 000 + 28 , 677 , 525 = 172 , 288 , 525 = 13 , 125.87 V \text{Magnitude } |E_f| = \sqrt{11,983.78^2 + 5,355.14^2} = \sqrt{143,611,000 + 28,677,525} = \sqrt{172,288,525} = 13,125.87\text{ V} Magnitude ∣ E f ​ ∣ = 11 , 983.7 8 2 + 5 , 355.1 4 2 ​ = 143 , 611 , 000 + 28 , 677 , 525 ​ = 172 , 288 , 525 ​ = 13 , 125.87 V Line-to-Line Generated EMF: E f , L L = 3 × 13 , 125.87 V = 22 , 734.7 V = 22.735 kV \text{Line-to-Line Generated EMF: } E_{f,LL} = \sqrt{3} \times 13,125.87\text{ V} = 22,734.7\text{ V} = 22.735\text{ kV} Line-to-Line Generated EMF: E f , LL ​ = 3 ​ × 13 , 125.87 V = 22 , 734.7 V = 22.735 kV Power Angle: δ = arctan ⁡ ( 5 , 355.14 11 , 983.78 ) = arctan ⁡ ( 0.44686 ) = 24.08 ∘ \text{Power Angle: } \delta = \arctan\left(\frac{5,355.14}{11,983.78}\right) = \arctan(0.44686) = 24.08^\circ Power Angle: δ = arctan ( 11 , 983.78 5 , 355.14 ​ ) = arctan ( 0.44686 ) = 24.0 8 ∘ E f = 13 , 125.87 V ∠ 24.08 ∘ \mathbf{E}_f = 13,125.87\text{ V} \angle 24.08^\circ E f ​ = 13 , 125.87 V ∠24.0 8 ∘

Step 3: Verify Active and Reactive Power Flow P = 3 E f V t X s sin ⁡ δ = 3 ( 13 , 125.87 ) ( 7 , 967.43 ) 3.20 sin ⁡ ( 24.08 ∘ ) = 313 , 738 , 400 3.20 × 0.4080 = 98 , 043 , 250 × 0.4080 = 40 , 001 , 600 W = 40.00 MW P = \frac{3 E_f V_t}{X_s} \sin\delta = \frac{3 (13,125.87)(7,967.43)}{3.20} \sin(24.08^\circ) = \frac{313,738,400}{3.20} \times 0.4080 = 98,043,250 \times 0.4080 = 40,001,600\text{ W} = 40.00\text{ MW} P = X s ​ 3 E f ​ V t ​ ​ sin δ = 3.20 3 ( 13 , 125.87 ) ( 7 , 967.43 ) ​ sin ( 24.0 8 ∘ ) = 3.20 313 , 738 , 400 ​ × 0.4080 = 98 , 043 , 250 × 0.4080 = 40 , 001 , 600 W = 40.00 MW Direct Check: P = S cos ⁡ θ = 50 MVA × 0.80 = 40.00 MW ( Matches exact! ) \text{Direct Check: } P = S \cos\theta = 50\text{ MVA} \times 0.80 = 40.00\text{ MW} \quad (\text{Matches exact!}) Direct Check: P = S cos θ = 50 MVA × 0.80 = 40.00 MW ( Matches exact! ) Q = 3 E f V t X s cos ⁡ δ − 3 V t 2 X s = ( 98.043 × cos ⁡ 24.08 ∘ ) − 3 ( 7 , 967.43 ) 2 3.20 = ( 98.043 × 0.9130 ) − 59.513 = 89.513 − 59.513 = 30.00 MVAR Q = \frac{3 E_f V_t}{X_s} \cos\delta - \frac{3 V_t^2}{X_s} = (98.043 \times \cos 24.08^\circ) - \frac{3 (7,967.43)^2}{3.20} = (98.043 \times 0.9130) - 59.513 = 89.513 - 59.513 = 30.00\text{ MVAR} Q = X s ​ 3 E f ​ V t ​ ​ cos δ − X s ​ 3 V t 2 ​ ​ = ( 98.043 × cos 24.0 8 ∘ ) − 3.20 3 ( 7 , 967.43 ) 2 ​ = ( 98.043 × 0.9130 ) − 59.513 = 89.513 − 59.513 = 30.00 MVAR Direct Check: Q = S sin ⁡ θ = 50 MVA × 0.60 = 30.00 MVAR ( Matches exact! ) \text{Direct Check: } Q = S \sin\theta = 50\text{ MVA} \times 0.60 = 30.00\text{ MVAR} \quad (\text{Matches exact!}) Direct Check: Q = S sin θ = 50 MVA × 0.60 = 30.00 MVAR ( Matches exact! )

Step 4: Maximum Power Capability & Pull-Out Torque P m a x = 3 E f V t X s = 3 ( 13 , 125.87 ) ( 7 , 967.43 ) 3.20 = 98.043 MW P_{max} = \frac{3 E_f V_t}{X_s} = \frac{3 (13,125.87)(7,967.43)}{3.20} = 98.043\text{ MW} P ma x ​ = X s ​ 3 E f ​ V t ​ ​ = 3.20 3 ( 13 , 125.87 ) ( 7 , 967.43 ) ​ = 98.043 MW Synchronous Speed: ω s = 2 π × 1 , 800 60 = 188.496 rad/s \text{Synchronous Speed: } \omega_s = \frac{2\pi \times 1,800}{60} = 188.496\text{ rad/s} Synchronous Speed: ω s ​ = 60 2 π × 1 , 800 ​ = 188.496 rad/s T p u l l − o u t = P m a x ω s = 98 , 043 , 250 W 188.496 rad/s = 520 , 134 N ⋅ m = 520.13 kN ⋅ m T_{pull-out} = \frac{P_{max}}{\omega_s} = \frac{98,043,250\text{ W}}{188.496\text{ rad/s}} = 520,134\text{ N}\cdot\text{m} = 520.13\text{ kN}\cdot\text{m} T p u ll − o u t ​ = ω s ​ P ma x ​ ​ = 188.496 rad/s 98 , 043 , 250 W ​ = 520 , 134 N ⋅ m = 520.13 kN ⋅ m

Steady-State Stability Margin: SSM = P m a x − P r a t e d P r a t e d = 98.043 − 40.00 40.00 = 145.1 % \text{Steady-State Stability Margin: } \text{SSM} = \frac{P_{max} - P_{rated}}{P_{rated}} = \frac{98.043 - 40.00}{40.00} = 145.1\% Steady-State Stability Margin: SSM = P r a t e d ​ P ma x ​ − P r a t e d ​ ​ = 40.00 98.043 − 40.00 ​ = 145.1% $