- Power Flow in Inductive Circuits: Power flow through an inductive impedance involves voltage sources and a load with a resistor and inductor.
- Phasor Diagrams: Phasor diagrams visually represent voltages and currents in a circuit, helping to understand power conditions.
- Power Equation of Synchronous Generator: The power equation of a synchronous generator relates the voltage, current, and phase angles to determine power output.
- Maximum Power Output Conditions: Maximum power output in alternators and synchronous motors occurs when the load angle equals the impedance angle.
- Reactive Power and Power Factor: Reactive power flow affects the power factor, which can be leading, lagging, or unity based on the relationship between excitation and terminal voltages.
To derive power conditions for alternators and synchronous motors, first consider two sinusoidal sources separated by a series impedance. The circuit contains a voltage source E1, a second source E2 and a branch with one resistor in series with an inductor. Let source E1 lead source E2 by load angle δ, with current defined from the first source toward the second. The per-phase voltage equation is:
Here Z = R + jX, |Z| is its magnitude and θz is its impedance angle.
Solving the voltage equation gives the line current:
Phasor Diagram for above Circuit
The diagram uses the following notation:
θz is the angle between E1 and E1/Z, or between E2 and E2/Z
I is the line current, and δ is the angle between E1 and E2.
Given below is the phasor diagram for the above circuit: Draw the E1 phasor and the component E1/Z, with θz between E1 and E1/Z. Then draw E2 and E2/Z so that E1 leads E2 by δ. Complete the diagram with the voltage drops IR in phase with current and IX leading current by 90 degrees. We can now derive the power supplied by E1.
Let the power supplied by source E1 be P1. Real power equals RMS voltage multiplied by the current component in phase with it. Therefore P1 = E1 times the current component in phase with the voltage source E1. The component in phase with source E1 is:
Substitution gives:
The diagram defines θz = 90o – αz. Substituting θz gives:
This is the source-side real power at E1.
Let the real power received at E2 be P2. It equals the receiving-end voltage multiplied by its in-phase current component:
The current component in phase with the voltage source E2 is:
Substitution gives:
Again use θz = 90o – αz. Substituting θz gives:
This is the receiving-end real power at E2.
Apply the result to a cylindrical-rotor alternator. Set source E1 to excitation voltage Ef, source E2 to terminal voltage Vt, and let synchronous impedance Zs satisfy Zs = ra + jXs. With current directed outward from E1, Pig is electromagnetic power developed at the internal emf, not total shaft input:
The receiving-end expression gives terminal electrical output
The difference between developed power and terminal output is the armature copper loss represented by this circuit. It excludes core, friction and windage losses. Subtracting terminal output from developed power gives:
Expanding the expression gives:
The phasor geometry gives:
Substitution reduces the difference to:
If armature resistance is neglected, αz becomes zero and Zs becomes approximately Xs. Developed power then equals terminal output:
For finite armature resistance, differentiate terminal output with respect to δ and set the result to zero:
The stationary point occurs when load angle δ equals impedance angle θz.
Substitution gives the maximum terminal output:
This is the maximum terminal output for fixed excitation, terminal voltage and synchronous impedance.
The corresponding maximum-output phasor diagram appears above.
To find maximum internal electromagnetic power, differentiate the source-side expression with respect to δ and set the result to zero:
The stationary point occurs at δ = 180 degrees – θz. When armature resistance is neglected, both maximum-power angles reduce to 90 degrees. Substitution gives:
This is the maximum internal electromagnetic power under the fixed-voltage and fixed-impedance assumptions.
The associated phasor diagram gives these power factor cases for the outward generator-current convention:
(a) When (Efcosδ-Iaracosθ) is less than terminal voltage, current leads and power factor is leading.
(b) When (Efcosδ-Iaracosθ) equals terminal voltage, current is in phase and power factor is unity.
(c) When (Efcosδ-Iaracosθ) is greater than terminal voltage, current lags and power factor is lagging.
The terminal reactive-power expression for a synchronous generator, using outward current, is:
Rearranging gives:
When armature resistance is neglected in generating mode, this reduces to:
With generator output defined as positive, the power factor and reactive-power cases are:
(a) When Efcosδ is less than terminal voltage, current leads, power factor is leading and reactive output is negative.
(b) When Efcosδ equals terminal voltage, power factor is unity and reactive output is zero.
(c) When Efcosδ is greater than terminal voltage, current lags, power factor is lagging and reactive output is positive.
Now apply the same result to a cylindrical-rotor synchronous motor. Let source E1 be terminal voltage Vt, let the internal voltage source E2 be counter emf Ef, and let synchronous impedance Zs satisfy Zs = ra + jXs. Because E1 supplies inward current, Pig in the next expression is terminal electrical input:
The receiving-end expression gives electromagnetic power converted by the synchronous motor
The difference between terminal input and converted power is armature copper loss in this equivalent circuit, not total motor loss. Subtracting converted power from terminal input gives:
Expanding the expression gives:
The motor phasor geometry gives:
Substitution reduces the difference to:
If armature resistance is neglected, αz becomes zero and Zs becomes approximately Xs. Terminal input then equals converted electromagnetic power:
To find maximum converted electromagnetic power of the synchronous motor, differentiate the output expression with respect to δ and set the result to zero:
The stationary point occurs when load angle δ equals impedance angle θz. Substitution gives maximum converted power:
This is the motor’s maximum converted-power expression under the stated assumptions.
The corresponding maximum-output phasor diagram appears above.
To find maximum terminal electrical input, differentiate the motor input expression with respect to δ and set the result to zero:
The stationary point occurs at δ = 180 degrees – θz.
Substitution gives maximum terminal input:
This is the maximum terminal-input expression for the synchronous motor.
The corresponding maximum-input phasor diagram appears above.
With motor current directed into the terminals, the diagram gives these power factor cases:
(a) When (Efcosδ+Iaracosθ) is less than terminal voltage, current lags and power factor is lagging.
(b) When (Efcosδ+Iaracosθ) equals terminal voltage, current is in phase and power factor is unity.
(c) When (Efcosδ+Iaracosθ) is greater than terminal voltage, current leads and power factor is leading.
The input-terminal reactive-power expression for a synchronous motor is:
Rearranging gives:
When armature resistance is neglected in motoring mode, this reduces to:
With motor input defined as positive, the power-factor and reactive-power cases are:
(a) When Efcosδ is less than terminal voltage, the power factor is lagging and reactive input is positive.
(b) When Efcosδ equals terminal voltage, power factor is unity and reactive input is zero at the synchronous motor terminals.
(c) When Efcosδ is greater than terminal voltage, power factor is leading and reactive input is negative.
