Alternators

LessonAbout 1.5 h

Alternators

Alternator (or synchronous generator) is an AC rotating machine that converts mechanical energy into alternating current electrical energy at a specific voltage and frequency.

1. Generated EMF and Design Equations

  • Pitch Factor (\(k_p\)): ratio of coil pitch emf to full pitch emf
  • Distribution Factor (\(k_d\)): ratio of vector sum to arithmetic sum of induced emfs
  • Coil Span: distance between two sides of a coil in slots

Pitch Factor and Slot Angle

\[ k_p = \sin(90^\circ \times \text{pitch}) \qquad\qquad \text{pitch} = \dfrac{\text{coil span in number of slots}}{\text{slots per pole}} \]
\[ \delta = \dfrac{180^\circ}{\text{slots per pole}} \]

Where: \(k_p\) = pitch factor or coil span factor, \(\text{pitch}\) = fractional pitch, \(\delta\) = slot angle (deg)

Distribution Factor

\[ k_d = \dfrac{\sin(\frac{n\delta}{2})}{n \sin(\frac{\delta}{2})} \]
\[ n = \dfrac{\text{slots per pole}}{\text{number of phases}} \]

Where: \(k_d\) = belt factor or distribution factor, \(n\) = number of slots per pole per phase, \(\delta\) = slot angle (deg)

Generated EMF Per Phase (Design Equation)

\[ E_p = 2.22 k_p k_d f \phi Z \]

Where: \(E_p\) = generated emf per phase (V), \(k_p\) = pitch factor, \(k_d\) = distribution factor, \(f\) = frequency (Hz), \(\phi\) = flux per pole (Wb), \(Z\) = number of conductors in series per phase

EE Board April 1997

Example. A 6-pole, 3-phase AC generator has 72 slots, the coil span is 12. What is the pitch factor?

Solution.

\[ k_p = \sin\left(90^\circ \times \dfrac{\text{Coil Span}}{\text{Slots/pole}}\right) = \sin\left(90^\circ \times \dfrac{12}{72/6}\right) = 1.000 \]

Answer: \(1.000\)

2. Frequency of Generated EMF

  • Synchronous Speed: speed of rotor rotation tied directly to supply frequency
  • Frequency Dependency: proportional to number of field poles and rotor speed

Frequency Equation

\[ f = \dfrac{PN}{120} \]

Where: \(f\) = frequency (Hz), \(P\) = number of poles, \(N\) = speed of rotor rotation (rpm)

EE Board October 1980

Example. What is the speed of a \(100\ \text{kW}\), \(230\ \text{V}\), three-phase, four-pole, \(60\ \text{Hz}\) alternator?

Solution.

\[ N = \dfrac{120f}{P} = \dfrac{120(60)}{4} = 1800\ \text{rpm} \]

Answer: \(1800\ \text{rpm}\)

3. Generated EMF Per Phase (Electric Circuit Relationship)

  • Lagging Power Factor: armature current lags phase voltage by angle \(\theta\)
  • Leading Power Factor: armature current leads phase voltage by angle \(\theta\)
  • Unity Power Factor: armature current is in phase with terminal voltage

Generated EMF Per Phase (Scalar Relation)

\[ E_p = \sqrt{(V_p \cos\theta + I_p R_a)^2 + (V_p \sin\theta \pm I_p X_{L\text{a}})^2} \]
\[ \alpha = \tan^{-1}\left(\dfrac{|V_p \sin\theta \pm I_p X_{L\text{a}}|}{V_p \cos\theta + I_p R_a}\right) \]

Where: \(E_p\) = generated emf per phase (V), \(V_p\) = load voltage per phase (V), \(I_p\) = armature current per phase (A), \(R_a\) = armature winding resistance per phase (\(\Omega\)), \(X_{L\text{a}}\) = inductive reactance of armature winding per phase (\(\Omega\)), \(\alpha\) = angle between \(E_p\) and \(I_p\) (deg), \(+ I_p X_{L\text{a}}\) if pf is lagging, \(- I_p X_{L\text{a}}\) if pf is leading

Generated EMF Per Phase (Complex Format)

\[ E_p \angle \pm \delta = V_p \angle 0^\circ + (I_p \angle \pm \theta)(R_a + jX_{L\text{a}}) \]

Where: \(E_p\) = generated emf per phase (V), \(V_p\) = terminal phase voltage (V), \(I_p\) = armature phase current (A), \(\delta\) = power angle between \(E_p\) and \(V_p\) (deg), \(+\theta\) if pf is leading, \(-\theta\) if pf is lagging

EE Board April 1995

Example. A \(1200\ \text{kVA}\), \(6600\ \text{V}\), 3-phase Y-connected alternator has \(R_a = 0.40\ \Omega\) and \(X_s = 6\ \Omega\) per phase. It delivers rated current at \(0.80\) lagging power factor. What will be the terminal voltage for the same excitation and current at \(0.80\) leading power factor?

Solution.

\[ E_p = \sqrt{(3810.51(0.8) + 104.97(0.4))^2 + (3810.51(0.6) + 104.97(6))^2} = 4560\ \text{V} \]
\[ V_p = \sqrt{E_p^2 - (I_p X_s \cos\theta - I_p R_a \sin\theta)^2} - (I_p X_s \sin\theta + I_p R_a \cos\theta) \implies V_L = \sqrt{3}(4560) \approx 7898\ \text{V} \]

Answer: \(7898\ \text{V}\)

4. Short Circuit Test on Alternators

  • Armature Shorted: terminals 1, 2, 3 shorted through an ammeter
  • Field Current: reduced field current applied to obtain rated armature current

Short Circuit Test Current

\[ I_{\text{sc}} = A \qquad\qquad (\text{for Wye}) \]
\[ I_{\text{sc}} = \dfrac{A}{\sqrt{3}} \qquad\qquad (\text{for Delta}) \]

Where: \(I_{\text{sc}}\) = short circuit armature current per phase (A), \(A\) = ammeter reading during the test (A)

5. Open Circuit Test on Alternator

  • Open Terminals: armature terminals left open-circuited with line voltmeter
  • Excitation: field current set to same value as short circuit test

Open Circuit Test Voltage

\[ E_{\text{oc}} = \dfrac{V}{\sqrt{3}} \qquad\qquad (\text{for Wye}) \]
\[ E_{\text{oc}} = V \qquad\qquad (\text{for Delta}) \]

Where: \(E_{\text{oc}}\) = open circuit voltage per phase (V), \(V\) = voltmeter reading during the test (V)

6. Alternator Parameters Derived from Test

  • DC Resistance Test: ohmic resistance measured between any two terminals
  • Effective Resistance: AC resistance obtained by multiplying DC resistance by 1.25 to 1.75
  • Synchronous Impedance: total internal per-phase opposition to current flow

DC Resistance Per Phase

\[ R_a = \dfrac{1}{2} R_t \qquad\qquad (\text{for Wye}) \]
\[ R_a = \dfrac{3}{2} R_t \qquad\qquad (\text{for Delta}) \]

Where: \(R_a\) = DC armature resistance per phase (\(\Omega\)), \(R_t\) = terminal DC resistance ohmmeter reading (\(\Omega\))

Synchronous Impedance and Reactance

\[ Z_s = \dfrac{E_{\text{oc}}}{I_{\text{sc}}} \]
\[ X_s = \sqrt{Z_s^2 - R_a^2} \]

Where: \(Z_s\) = synchronous impedance per phase (\(\Omega\)), \(X_s\) = synchronous reactance per phase (\(\Omega\)), \(R_a\) = effective armature resistance per phase (\(\Omega\)), \(E_{\text{oc}}\) = open circuit voltage per phase (V), \(I_{\text{sc}}\) = short circuit armature current per phase (A)

EE Board October 1980

Example. A \(10\ \text{kVA}\), \(110\ \text{V}\), 3-phase, 4-wire alternator generates \(70\ \text{V}\) per phase on open circuit. If the synchronous impedance per phase is \(0.2\ \Omega\), find the short-circuit current in each phase.

Solution.

\[ I_{\text{sc}} = \dfrac{E_{\text{oc}}}{Z_s} = \dfrac{70}{0.2} = 350\ \text{A} \]

Answer: \(350\ \text{A}\)

7. Per Unit Values of the Alternator Parameters

  • Base Impedance: reference impedance derived from rated voltage and kVA
  • Per Unit System: dimensionless ratios of actual parameters to base values

Per Unit Parameters

\[ I_{\text{base}} = \dfrac{S_{\text{base}}}{V_{\text{base}}} \qquad\qquad Z_{\text{base}} = \dfrac{V_{\text{base}}}{I_{\text{base}}} = \dfrac{V_{\text{base}}^2}{S_{\text{base}}} \]
\[ R_{\text{pu}} = \dfrac{R_a}{Z_{\text{base}}} \qquad\qquad X_{\text{pu}} = \dfrac{X_s}{Z_{\text{base}}} \qquad\qquad Z_{\text{pu}} = R_{\text{pu}} \pm j X_{\text{pu}} \]

Where: \(S_{\text{base}}\) = rated apparent power per phase (VA), \(V_{\text{base}}\) = rated voltage per phase (V), \(I_{\text{base}}\) = rated armature current per phase (A), \(Z_{\text{base}}\) = base impedance per phase (\(\Omega\)), \(R_{\text{pu}}\) = per-unit resistance, \(X_{\text{pu}}\) = per-unit reactance, \(Z_{\text{pu}}\) = per-unit impedance

EE Board October 1994

Example. A \(30\ \text{MVA}\), \(15\ \text{kV}\) three-phase alternator has unit per-unit impedance. What is its per phase nominal base impedance?

Solution.

\[ Z_{\text{base}} = \dfrac{V_L^2}{S_{3\phi}} = \dfrac{(15000)^2}{30 \times 10^6} = 7.5\ \Omega \]

Answer: \(7.5\ \text{ohms}\)

8. Voltage Regulation

  • Definition: percentage rise in terminal voltage when full load is removed
  • Lagging Load: positive voltage regulation (terminal voltage drops under load)
  • Leading Load: negative voltage regulation possible due to capacitive boost

Voltage Regulation

\[ \%\text{VR} = \dfrac{E_p - V_p}{V_p} \times 100\% \]
\[ \%\text{VR} = \sqrt{(\cos\theta + \%\text{IR})^2 + (\sin\theta \pm \%\text{IX})^2} - 1 \]

Where: \(E_p\) = no-load terminal voltage per phase (V), \(V_p\) = full-load terminal voltage per phase (V), \(\theta\) = power factor angle (deg), \(\%\text{IR}\) = percent resistance drop, \(\%\text{IX}\) = percent reactance drop, \(+\%\text{IX}\) if pf is lagging, \(-\%\text{IX}\) if pf is leading

Example. A \(2500\ \text{kVA}\), \(6600\ \text{V}\) synchronous turbo alternator operates at full load and \(0.8\) pf lagging. \(X_s = 10.4\ \Omega/\text{phase}\) and \(R_a = 0\). Calculate the percentage voltage regulation.

Solution.

\[ V_p = \dfrac{6600}{\sqrt{3}} = 3810.51\ \text{V} \implies I_p = \dfrac{2500 \times 10^3}{\sqrt{3}(6600)} = 218.69\ \text{A} \]
\[ E_p = \sqrt{(3810.51(0.8))^2 + (3810.51(0.6) + 218.69(10.4))^2} = 5558.82\ \text{V} \]
\[ \%\text{VR} = \dfrac{5558.82 - 3810.51}{3810.51} \times 100\% = 45.88\% \]

Answer: \(45.88\%\)

9. Power Developed in the Armature Per Phase

  • Power Angle (\(\delta\)): angle between generated emf \(E_p\) and terminal voltage \(V_p\)
  • Maximum Power Limit: occurs when power angle \(\delta = 90^\circ\)

Power Developed Per Phase

\[ P_d = \dfrac{E_p V_p}{X_s} \sin\delta \]
\[ P_{d,\text{max}} = \dfrac{E_p V_p}{X_s} \]

Where: \(P_d\) = power developed per phase (W), \(P_{d,\text{max}}\) = maximum power developed per phase (W when \(\delta = 90^\circ\)), \(E_p\) = generated emf per phase (V), \(V_p\) = terminal voltage per phase (V), \(X_s\) = synchronous reactance per phase (\(\Omega\)), \(\delta\) = power angle or torque angle (deg)

10. Power Losses and Efficiency

  • Stator Copper Loss: \(I^2 R\) loss in stator armature windings
  • Rotational Losses: core loss, friction, windage, and ventilation losses
  • Field Copper Loss: excitation loss in rotor field winding

Power Losses and Efficiency

\[ P_{\text{losses}} = P_{\text{cu}} + P_{\text{fw}} + P_{\text{core}} + P_f + P_v \]
\[ \eta = \dfrac{P_{\text{out}}}{P_{\text{out}} + P_{\text{losses}}} \times 100\% \]

Where: \(\eta\) = efficiency (\%), \(P_{\text{out}}\) = output power (W), \(P_{\text{losses}}\) = total power losses (W), \(P_{\text{cu}}\) = armature copper loss (W), \(P_{\text{fw}}\) = friction and windage loss (W), \(P_{\text{core}}\) = core loss (W), \(P_f\) = field loss (W), \(P_v\) = ventilation loss (W)

Output and Loss Formulas

\[ P_{\text{out}(1\phi)} = V_L I_L \cos\theta \qquad\qquad P_{\text{out}(3\phi)} = \sqrt{3} V_L I_L \cos\theta \]
\[ P_{\text{cu}(1\phi)} = I_p^2 R_a \qquad\qquad P_{\text{cu}(3\phi)} = 3 I_p^2 R_a \]
\[ P_f = I_f^2 R_f = I_f V_f = \dfrac{V_f^2}{R_f} \]

Where: \(V_L\) = line-to-line voltage (V), \(I_L\) = line current (A), \(I_p\) = phase current (A), \(R_a\) = armature resistance per phase (\(\Omega\)), \(\cos\theta\) = load power factor, \(I_f\) = field current (A), \(R_f\) = field resistance (\(\Omega\)), \(V_f\) = field voltage (V)

11. Armature Copper Losses at Any Size of Load

  • Variable Loss: copper loss varies directly as square of load current
  • Fractional Load: ratio of actual load kW/kVA to rated full load

Copper Losses at Any Load

\[ P_{\text{cu}_x} = x^2 P_{\text{cu}_{\text{FL}}} \]

Where: \(x\) = fractional loading or percentage loading on the alternator, \(P_{\text{cu}_x}\) = copper losses at \(x\) load (W), \(P_{\text{cu}_{\text{FL}}}\) = full load copper losses (W)

12. Governor’s Speed Regulation

  • Governor Droop: drop in prime mover speed from no-load to full-load
  • Speed Regulation: percentage speed rise when full load is disconnected

Governor’s Speed Regulation (GSR)

\[ \text{GSR} = \dfrac{N_{\text{NL}} - N_{\text{FL}}}{N_{\text{FL}}} \qquad\qquad \text{GSR} = \dfrac{f_{\text{NL}} - f_{\text{FL}}}{f_{\text{FL}}} \]

Where: \(\text{GSR}\) = governor’s speed regulation, \(N_{\text{NL}}\) = no-load speed (rpm), \(N_{\text{FL}}\) = full-load speed (rpm), \(f_{\text{NL}}\) = no-load frequency (Hz), \(f_{\text{FL}}\) = full-load frequency (Hz)

13. Characteristic Triangle of an Alternator

  • Governor Droop Slope: slope of frequency versus power load curve
  • Linear Characteristic: assumes straight-line droop from no-load to full-load

Droop Slope Relations

\[ \dfrac{\Delta P}{\Delta f} = \dfrac{P_{\text{FL}}}{f_{\text{NL}} - f_{\text{FL}}} \qquad\qquad \dfrac{\Delta f}{\Delta P} = \text{GD} = \dfrac{\text{GSR} \times f_{\text{FL}}}{P_{\text{FL}}} \]

Where: \(\Delta f\) = change in frequency due to change in load (Hz), \(\Delta P\) = change in power load (kW or MW), \(P_{\text{FL}}\) = power rating of alternator (kW or MW), \(\text{GD}\) = governor’s droop, \(\text{GSR}\) = governor’s speed regulation

14. Parallel Operation of Alternators

  • Voltage Condition: bus voltage magnitude must equal alternator terminal voltage
  • Frequency Condition: operating frequency of incoming machine must match bus frequency
  • Phase Sequence: phase sequence of incoming alternator must match running bus

14.1 Case 1

  • External Characteristics Given: kW ratings, speed regulations, and bus load given
  • Frequency Droop Slope: load division governed by prime mover speed regulation

Parallel Load Sharing (Case I)

\[ \dfrac{\Delta f}{\Delta P_a} = \dfrac{\text{GSR}_a \times f_{\text{FL}_a}}{P_{\text{FL}_a}} \qquad\qquad \dfrac{\Delta f}{\Delta P_b} = \dfrac{\text{GSR}_b \times f_{\text{FL}_b}}{P_{\text{FL}_b}} \]
\[ \Delta P = \Delta P_a + \Delta P_b \qquad\qquad P_{\text{new}} = P_t \pm \Delta P \qquad\qquad f_{\text{new}} = f \pm \Delta f \]

Where: \(\Delta P\) = change in total bus load (kW), \(\Delta P_a, \Delta P_b\) = change in load for alternators A and B (kW), \(f\) = frequency prior to load change (Hz), \(f_{\text{new}}\) = frequency after load change (Hz)

EE Board March 1998

Example. Two alternators driven by shunt motors with speed droops of \(3\%\) and \(4\%\) operate in parallel carrying \(50\ \text{kW}\) each. What are the resulting loads if an additional \(50\ \text{kW}\) load is switched on?

Solution.

\[ \Delta P_a = \left(\dfrac{1/0.03}{1/0.03 + 1/0.04}\right)(50) = 28.57\ \text{kW} \implies P_a = 50 + 28.57 = 78.57\ \text{kW} \]
\[ P_b = 50 + (50 - 28.57) = 71.43\ \text{kW} \]

Answer: \(78.57\ \text{kW}\ \text{and}\ 71.43\ \text{kW}\)

14.2 Case 2

  • Load Sharing Given: individual load kW, power factors, or currents specified
  • Vector Summation: total complex current and power equal sum of individual shares

Parallel Load Sharing (Case II)

\[ I_t \angle \pm \theta_t = I_a \angle \pm \theta_a + I_b \angle \pm \theta_b \]
\[ S_t \angle \pm \theta_t = S_a \angle \pm \theta_a + S_b \angle \pm \theta_b \]
\[ P_t = P_a + P_b \]

Where: \(P\) = real power (W), \(S\) = apparent power (VA), \(I\) = current (A), \(+\theta\) for leading pf, \(-\theta\) for lagging pf

Example. Two 3-phase alternators operated in parallel supply a total system load of \(1050\ \text{kW}\) at \(75\%\) lagging power factor. Alternator A carries \(700\ \text{kW}\) at \(80\%\) lagging power factor. Determine the kVAR of alternator B.

Solution.

\[ Q_b = Q_t - Q_a = 1050\tan(\cos^{-1} 0.75) - 700\tan(\cos^{-1} 0.80) = 926.1 - 525 = 401.1\ \text{kVAR} \]

Answer: \(401\ \text{kVAR}\)

15. Formula Summary

Formula Name Formula Variables
Pitch Factor \(k_p = \sin(90^\circ \times \text{pitch})\) \(k_p\) = pitch factor
\(\text{pitch}\) = coil span / slots per pole
Slot Angle \(\delta = \dfrac{180^\circ}{\text{slots/pole}}\) \(\delta\) = slot angle (deg)
Distribution Factor \(k_d = \dfrac{\sin(\dfrac{n\delta}{2})}{n \sin(\dfrac{\delta}{2})}\) \(k_d\) = distribution factor
\(n\) = slots/pole/phase
Generated EMF per Phase (Design) \(E_p = 2.22 k_p k_d f \phi Z\) \(E_p\) = generated emf per phase (V)
\(f\) = frequency (Hz)
\(\phi\) = flux per pole (Wb)
\(Z\) = conductors per phase
Frequency of Generated EMF \(f = \dfrac{PN}{120}\) \(f\) = frequency (Hz)
\(P\) = poles
\(N\) = rotor speed (rpm)
Generated EMF per Phase (Circuit) \(E_p = \sqrt{(V_p \cos\theta + I_p R_a)^2 + (V_p \sin\theta \pm I_p X_{L\text{a}})^2}\) \(E_p\) = generated emf per phase (V)
\(V_p\) = phase voltage (V)
\(I_p\) = phase current (A)
\(R_a\) = armature resistance (\(\Omega\))
\(X_{L\text{a}}\) = armature reactance (\(\Omega\))
Complex Generated EMF \(E_p \angle \pm \delta = V_p \angle 0^\circ + (I_p \angle \pm \theta)(R_a + jX_{L\text{a}})\) \(\delta\) = power angle (deg)
\(\theta\) = power factor angle (deg)
Short Circuit Current (Wye) \(I_{\text{sc}} = A\) \(I_{\text{sc}}\) = short circuit current per phase (A)
\(A\) = ammeter reading (A)
Short Circuit Current (Delta) \(I_{\text{sc}} = \dfrac{A}{\sqrt{3}}\) \(I_{\text{sc}}\) = short circuit current per phase (A)
\(A\) = ammeter reading (A)
Open Circuit Voltage (Wye) \(E_{\text{oc}} = \dfrac{V}{\sqrt{3}}\) \(E_{\text{oc}}\) = open circuit voltage per phase (V)
\(V\) = voltmeter reading (V)
Open Circuit Voltage (Delta) \(E_{\text{oc}} = V\) \(E_{\text{oc}}\) = open circuit voltage per phase (V)
\(V\) = voltmeter reading (V)
Armature Resistance (Wye DC) \(R_a = \dfrac{1}{2} R_t\) \(R_a\) = DC phase resistance (\(\Omega\))
\(R_t\) = terminal DC resistance (\(\Omega\))
Armature Resistance (Delta DC) \(R_a = \dfrac{3}{2} R_t\) \(R_a\) = DC phase resistance (\(\Omega\))
\(R_t\) = terminal DC resistance (\(\Omega\))
Synchronous Impedance \(Z_s = \dfrac{E_{\text{oc}}}{I_{\text{sc}}}\) \(Z_s\) = synchronous impedance (\(\Omega\))
\(E_{\text{oc}}\) = open circuit voltage per phase (V)
\(I_{\text{sc}}\) = short circuit current per phase (A)
Synchronous Reactance \(X_s = \sqrt{Z_s^2 - R_a^2}\) \(X_s\) = synchronous reactance (\(\Omega\))
\(R_a\) = effective armature resistance (\(\Omega\))
Base Impedance \(Z_{\text{base}} = \dfrac{V_{\text{base}}^2}{S_{\text{base}}}\) \(Z_{\text{base}}\) = base impedance (\(\Omega\))
\(V_{\text{base}}\) = phase base voltage (V)
\(S_{\text{base}}\) = phase base VA
Per Unit Reactance \(X_{\text{pu}} = \dfrac{X_s}{Z_{\text{base}}}\) \(X_{\text{pu}}\) = per-unit reactance
\(X_s\) = synchronous reactance (\(\Omega\))
Voltage Regulation \(\%\text{VR} = \dfrac{E_p - V_p}{V_p} \times 100\%\) \(\%\text{VR}\) = percent voltage regulation
\(E_p\) = no-load voltage (V)
\(V_p\) = full-load voltage (V)
Power Developed per Phase \(P_d = \dfrac{E_p V_p}{X_s} \sin\delta\) \(P_d\) = developed power per phase (W)
\(\delta\) = power angle (deg)
Maximum Power Developed \(P_{d,\text{max}} = \dfrac{E_p V_p}{X_s}\) \(P_{d,\text{max}}\) = maximum developed power per phase (W)
Three-Phase Output Power \(P_{\text{out}(3\phi)} = \sqrt{3} V_L I_L \cos\theta\) \(P_{\text{out}(3\phi)}\) = total 3-phase output power (W)
\(V_L\) = line voltage (V)
\(I_L\) = line current (A)
Three-Phase Copper Loss \(P_{\text{cu}(3\phi)} = 3 I_p^2 R_a\) \(P_{\text{cu}(3\phi)}\) = total 3-phase copper loss (W)
\(I_p\) = phase current (A)
\(R_a\) = phase armature resistance (\(\Omega\))
Alternator Efficiency \(\eta = \dfrac{P_{\text{out}}}{P_{\text{out}} + P_{\text{losses}}} \times 100\%\) \(\eta\) = efficiency (\%)
\(P_{\text{out}}\) = output power (W)
\(P_{\text{losses}}\) = total power loss (W)
Copper Loss at Fractional Load \(P_{\text{cu}_x} = x^2 P_{\text{cu}_{\text{FL}}}\) \(P_{\text{cu}_x}\) = copper loss at fractional load \(x\) (W)
\(P_{\text{cu}_{\text{FL}}}\) = full-load copper loss (W)
Governor’s Speed Regulation \(\text{GSR} = \dfrac{N_{\text{NL}} - N_{\text{FL}}}{N_{\text{FL}}}\) \(\text{GSR}\) = governor’s speed regulation
\(N_{\text{NL}}\) = no-load speed (rpm)
\(N_{\text{FL}}\) = full-load speed (rpm)
Governor Droop Slope \(\dfrac{\Delta f}{\Delta P} = \dfrac{\text{GSR} \times f_{\text{FL}}}{P_{\text{FL}}}\) \(\Delta f\) = frequency change (Hz)
\(\Delta P\) = power change (kW)
\(P_{\text{FL}}\) = rated capacity (kW)
Parallel Active Power Sharing \(P_t = P_a + P_b\) \(P_t\) = total active power (W)
\(P_a, P_b\) = power shares of alternators A and B (W)
Parallel Complex Current Sharing \(I_t \angle \pm \theta_t = I_a \angle \pm \theta_a + I_b \angle \pm \theta_b\) \(I_t\) = total line current (A)
\(I_a, I_b\) = alternator currents (A)
\(\theta\) = power factor angles (deg)