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Operator Generic Fundamentals Components - AC Motors and Generators © Copyright 2016 Operator Generic Fundamentals 2 Terminal Learning Objectives At the completion of this training session, the trainee will demonstrate mastery of this topic by passing a written exam with a grade of ≥ 80 percent on the following Terminal Learning Objectives (TLOs): 1. Describe the construction, operating characteristics and limitations for an AC generator. 2. Explain the theory of operation of selected types of AC motors. 3. Explain operating characteristics and limitations on AC motors. 4. Analyze the operating characteristics and interactions between two AC generators operating in parallel. © Copyright 2016 Intro Operator Generic Fundamentals 3 AC Generator TLO 1 – Describe the construction, operating characteristics, and limitations for an AC generator. 1.1 Describe the theory of operation of an AC generator. 1.2 State the purpose of the following components of an AC generator: Field, Armature, Prime mover, Rotor, Stator, and Slip rings. 1.3 Define the following electrical terms: volts, amps, watts, inductance, capacitance, VARs, and hertz. 1.4 Describe a stationary field, rotating armature AC generator and a rotating field, stationary armature AC generator. © Copyright 2016 TLO 1 Operator Generic Fundamentals 4 AC Generator Theory ELO 1.1 – Describe the theory of operation of an AC generator. • A simple AC generator consists of: – A conductor or loop of wire in a magnetic field – The loop ends connect to slip rings that are in contact with brushes – When the loop rotates, it cuts magnetic lines of force • As conductor passes through magnetic field – A voltage is induced in the conductor and transferred through slip rings as voltage output Figure: Simple AC Generator © Copyright 2016 ELO 1.1 Operator Generic Fundamentals 5 AC Generator Theory Magnitude of Generated Voltage • Dependent on field strength and speed of rotor • Most generators at constant speed; generated voltage depends on field excitation, or field strength Developing an AC Sine Wave Voltage • Coil rotates, each side cuts the magnetic lines of force in opposite directions • Direction (polarity) of induced voltages depends on direction of movement of coil Figure: Simple AC Generator © Copyright 2016 ELO 1.1 Operator Generic Fundamentals 6 AC Generator Theory • Magnetic field • Produces: Current moving through the conductor • Conductor • Relative motion Resultant Induced (+)Voltage Direction of Magnetic Field Direction of Motion ( ) © Copyright 2016 ELO 1.1 Operator Generic Fundamentals 7 AC Generator Theory Developing an AC Sine Wave Voltage • In vertical position, 0°, coils are moving parallel to magnetic field • As coil rotates, voltages are additive, slip ring X positive (+) and Y is negative • Potential across resistor R causes current flow from Y to X through resistor • Current increases until reaches maximum value (90°) – perpendicular to magnetic field Figure: Simple AC Generator © Copyright 2016 ELO 1.1 Operator Generic Fundamentals 8 AC Generator Theory Developing an AC Sine Wave Voltage • As coil continues to turn, induced voltage and current decrease until both reach zero, where coil is again in the vertical position (180°) • Next half rev produces equal voltage, except with reversed polarity (270° and 360°) • Current flow through R is now from X to Y Figure: Developing an AC Sine Wave Voltage © Copyright 2016 ELO 1.1 Operator Generic Fundamentals 9 AC Generator Theory Period and Frequency • As coil rotates 360°, output voltage goes through one cycle • Period is time required for generator to complete one cycle • Frequency (measured in hertz) is the number of cycles per second 𝑁𝑃 𝑓= 120 • Where: f = frequency (Hz) P = total number of poles N = rotor speed (rpm) 120 = conversion from minutes to seconds and from poles to pole pairs © Copyright 2016 ELO 1.1 Operator Generic Fundamentals 10 AC Generator Theory Peak Voltage and Current • Emax occurs at 90° • Value is termed peak voltage • Can quantify AC voltage or current as peak voltage (Ep) or peak current (Ip) Peak to Peak Voltage and Current • Another common term associated with AC is peak-topeak value (Ep-p or Ip-p) • Peak to peak is magnitude of voltage, or current © Copyright 2016 Figure: AC Sine Wave Voltage ELO 1.1 Operator Generic Fundamentals 11 Effective Value of AC • Effective value is most commonly used for quantifying AC • Amount of AC that produces same heating effect as equal amount of DC • Effective value, because it is the root of the mean (average) square of the currents, or the root-mean-square, or RMS value • RMS = average value X 1.73 Figure: AC Voltage Sine Wave © Copyright 2016 ELO 1.1 Operator Generic Fundamentals 12 AC Generator Theory • Dashed line is average of I2 values, square root of that value is RMS, or effective value (square root of mean square deviation of a waveform) • Average value is ½ Imax2 • RMS value is equal to 0.707 (square root of ½) • Effective value of voltage or current for an AC sine wave can be found by using: Effective Value (RMS) = Peak Value x 0.707 © Copyright 2016 ELO 1.1 Figure: Effective Value of AC Current Operator Generic Fundamentals 13 Phase Angle Guidelines • Phase angle is fraction of a cycle, in degrees, that has gone by since a voltage or current has passed through a given value • Phase difference is common term for phase angle – Describes two different voltages that have same frequency, which pass through zero values in same direction at different times • Angles along axis indicate phases of voltages e1 and e2 • At 120°, e1 passes through zero value 60° ahead of e2 (e2 equals zero at 180°) • Describes as voltage e1 leads e2 by 60 electrical degrees, or voltage e2 lags e1 by 60 electrical degrees © Copyright 2016 ELO 1.1 Figure: Phase Relationship Operator Generic Fundamentals 14 Generator Major Components ELO 1.2 – State the purpose of the following components of an AC generator: field, armature, prime mover, rotor, stator, and slip rings. Field • Coils of conductors receive voltage from a source (excitation) and produce a magnetic flux Armature • Voltage is produced, consists of coils of wire large enough to carry full-load current of generator output Armature Figure: Basic AC Generator © Copyright 2016 ELO 1.2 Operator Generic Fundamentals 15 Generator Components Prime Mover • Component that drives the AC generator; rotating machine, such as diesel engine, steam turbine, or a motor Rotor • Driven by prime mover, can be armature or field (armature when voltage output is generated there, field when field excitation is applied) Stator • Stationary part of the machine, can be either armature or field Slip Rings • Transfer power to or from the rotor, consist of a circular conducting material mounted on rotor shaft • Connected electrically to rotor windings, insulated from shaft © Copyright 2016 ELO 1.2 Operator Generic Fundamentals 16 Generator Components Magnetic Field (Rotating Field Machine) • Current flow through field coil of rotor produces strong magnetic field • Slip rings and brushes conduct excitation to field coil in rotor Continuous Connection • Brushes are spring-held in contact with slip rings to provide continuous connection between field coil and excitation circuit Conductors • Each time rotor makes one complete revolution, one complete cycle of AC is developed • A generator has many turns of wire wound into slots of stator © Copyright 2016 ELO 1.2 Operator Generic Fundamentals 17 Common Electrical Terms ELO 1.3 – Define the following electrical terms: volts, amps, watts, inductance, capacitance, VARs, and hertz. Unit of Measurement Symbol Unit is Used to Measure: Ampere I or A Electrical current Volt E or V Electrical potential difference Hertz Hz Frequency (f) Ohm R or Ω Resistance to current flow Watt W Power (P) Henry H Electrical inductance (L) Farad F Electrical capacitance (C) VAR VAR Reactive power (Q) © Copyright 2016 ELO 1.3 Operator Generic Fundamentals 18 Common Electrical Terms Voltage (Volt) • Potential difference that causes electrons to move in a conductor • E in electrical formulas, V on lab equipment or schematic diagrams Current (Ampere or Amp) • Movement of free electrons through a conductor • I in electrical formulas, A in lab or on schematic diagrams Resistance (Ohm) • One ohm is amount of resistance that limits current in conductor to one ampere when potential difference (voltage) applied to conductor is one volt • R in electrical formulas, capital omega (Ω) in other instances • Relationship between these parameters is Ohm’s law E = I × R © Copyright 2016 ELO 1.3 Operator Generic Fundamentals 19 Common Electrical Terms Power (Watt) • Electricity performs work • Power (P in equations) is rate of performing work, or rate of heat generation • Common unit to specify electric power is watt (W in equations) • Power also described as current (I) in a circuit times voltage (E) across circuit P = I ×E or P = I E • Using Ohm’s Law for voltage, E = I × R and using substitution laws: P = I × (I ×R) • Therefore, power also equals current in a circuit squared multiplied by resistance of circuit: P = 𝐼2 × R © Copyright 2016 ELO 1.3 Operator Generic Fundamentals 20 Common Electrical Terms Inductance (Henry) • Ability of a coil to store energy, induce a voltage in itself, and oppose changes in current flowing through it • Symbol used for inductance is L, Henries (H) are the units • One henry is amount of inductance (L) that permits one volt to be induced when current through coil changes at a rate of one ampere per second • Rate of change in current through a coil per unit time: ∆𝐼 ∆𝑡 • Voltage VL induced in a coil with inductance L: 𝑉𝐿 = −𝐿 ∆𝐼 ∆𝑡 • Negative sign indicates voltage induced opposes change in current through coil per unit time © Copyright 2016 ELO 1.3 Operator Generic Fundamentals 21 Common Electrical Terms Capacitance (Farad) • Ability to store an electric charge (C) • Measured in farads, equal to amount of charge (Q) that can be stored in a device or capacitor divided by voltage (E) applied across the device or capacitor plates when charge was stored 𝑄 𝐶= 𝐸 Frequency (Hertz) • Measured in hertz, number of alternating voltage or current cycles completed per second © Copyright 2016 ELO 1.3 Operator Generic Fundamentals 22 Common Electrical Terms Volt Ampere Reactive (VAR) • Circulating current in a circuit that does no useful work • In AC circuits not purely resistive, voltage and current are out of phase • Power is exchanged in these circuits as inductive fields form and collapse, and capacitors charge and discharge • Relationship between reactive power, apparent power, and true power expressed in power triangle, which equates AC to DC power by showing relations between: – Generator output (Apparent Power - S) in volt-amperes (VA) – Usable power (True Power - P) in watts – Wasted/stored power (Reactive Power - Q) in volt-amperesreactive (VAR) © Copyright 2016 ELO 1.3 Operator Generic Fundamentals 23 Common Electrical Terms • Phase angle (θ) represents inefficiency of AC circuit • Represents comparable values used to find efficiency level of generated power to usable power, expressed as power factor • Power factor is the ratio of True Power/Apparent Power Figure: Power Triangle – “How much you get out for what you have to put in” © Copyright 2016 ELO 1.3 Operator Generic Fundamentals 24 Types of Generators ELO 1.4 – Describe the following types of AC generators: stationary field/rotating armature; rotating field/stationary armature. Stationary Field / Rotating Armature • Small AC generators usually have stationary field and rotating armature • Disadvantage is that slip-ring and brush assembly is in series with load circuits • If components are worn or dirty, flow of current to loads may be reduced or interrupted • For large generators, slip rings and brushes are limiting factor on machine capacity © Copyright 2016 ELO 1.4 Figure: Stationary Field with Rotating Armature Generator Operator Generic Fundamentals 25 Types of Generators Rotating Field / Stationary Armature • When DC field excitation connected to rotor, stationary coils have AC induced into them • Used where large power generation is needed – DC source is supplied to rotating field coils (via slip rings and brushes), which produces magnetic field around rotating element • As prime mover turns rotor, field rotates across conductors of stationary armature and an EMF is induced into armature windings © Copyright 2016 ELO 1.4 Figure: Rotating Field with Stationary Armature Generator Operator Generic Fundamentals 26 Types of Generators Rotating Field / Stationary Armature Advantages over stationary field, rotating armature AC generator: 1. A load can be connected to armature without having moving contacts (slip rings and brushes) in circuit 2. Much easier to insulate stator conductors than rotating conductors 3. Much higher output voltages and currents can be generated © Copyright 2016 Figure: Rotating Field with Stationary Armature Generator ELO 1.4 Operator Generic Fundamentals 27 Types of AC Motors TLO 2 – Explain the theory of operation of selected types of AC motors. 2.1 2.2 2.3 2.4 Describe how a rotating magnetic field is produced in an AC motor. Define slip and explain its effect on AC induction motor operation. Describe how torque is produced in an AC motor. Explain differences between starting and running current for an AC induction motor and the operating limits used to mitigate the effects of starting current. 2.5 Describe how an AC synchronous motor is started and its operating characteristics. © Copyright 2016 TLO 2 Operator Generic Fundamentals 28 Motor Rotating Field ELO 2.1 – Describe how a rotating magnetic field is produced in an AC motor. • Basic principle of operation for AC motors is interaction of revolving magnetic field created in stator by AC current with an opposing magnetic field in rotor • Magnetic field in rotor comes from one of two sources: – Induced from the rotating stator – Provided by a separate DC current source © Copyright 2016 ELO 2.1 Operator Generic Fundamentals 29 Motor Rotating Field Rotating Magnetic Field • Figure illustrates three-phase stator with three-phase AC current applied • Windings connect in a wye configuration • Two windings in each phase wind in same direction • At any instant in time, magnetic field generated by one phase will depend on current flow through that phase • If current flow through that phase is zero, resulting magnetic field is zero • If current flow is at maximum value, resulting field is at maximum value © Copyright 2016 ELO 2.1 Figure: Three-Phase Motor Stator Operator Generic Fundamentals 30 Motor Rotating Field • As magnetic field rotates in stator, rotor also rotates in an attempt to maintain its alignment with stator’s magnetic field • Figure on next slide shows stator’s magnetic field "stopped" at six selected positions • Instances are marked off at 60° intervals representing current flowing in phases A, B, and C © Copyright 2016 ELO 2.1 Operator Generic Fundamentals 31 Motor Rotating Magnetic Field N N C B’ A S N A’ N S S S B C’ C B’ A A’ B C’ N S N S T1 A C B’ A N N N A’ S N S N C B’ A A A’ B C’ A’ B C’ S N T4 N S N S S B’ C A’ A N B C’ S N S S T5 T6 T7 T3 C B’ N T3 Phase B T2 S N B C’ N B C’ T2 N A’ N S Time 1 C B’ S S S S Phase A T4 T5 T6 T7 Phase C © Copyright 2016 ELO 2.1 Operator Generic Fundamentals 32 C C’ Phase A Phase B Time 1 T2 T3 T4 T5 T6 T7 Phase C © Copyright 2016 ELO 2.1 Operator Generic Fundamentals 33 Slip Effects on AC Motor ELO 2.2 – Define slip and explain its effect on AC induction motor operation. • Slip Effects on AC Induction Motor Operation – Not possible for rotor of an AC induction motor to turn at same speed as rotating magnetic field – If speed of rotor were same as that of stator, no relative motion would exist and there would be no induced EMF in rotor – Rotor must rotate at some speed less than that of magnetic field in stator for relative motion to exist © Copyright 2016 ELO 2.2 Operator Generic Fundamentals 34 Slip Effects on AC Motor • Slip is percentage difference between speed of rotor and rotating magnetic field in stator • Amount of torque on rotor will change as slip ratio changes • As slip increases from zero to about 20%, torque increases linearly • As load and slip increase beyond full-load torque, torque will reach a maximum value at about 25% slip • Typical values of between 1% and 8% – From no-load running current to full-load running current © Copyright 2016 Figure: Torque versus Slip for an AC Induction Motor ELO 2.2 Operator Generic Fundamentals 35 Developing Torque in AC Induction Motor ELO 2.3 – Describe how torque is produced in an AC motor. Torque Production • When alternating current flows through stator windings of an AC induction motor, a rotating magnetic field results • Rotating magnetic field cuts conductors of rotor and induces a current in them due to generator action © Copyright 2016 ELO 2.3 Operator Generic Fundamentals 36 Developing Torque in AC Induction Motor Torque Production • Induced current produces a magnetic field, opposite in polarity of stator field, around conductors of rotor • Magnetic field induced in rotor will try to line up with magnetic field produced by stator • Rotor unable to line up, or lock onto, stator field; therefore, must follow behind it Figure: Induced Rotor Magnetic Field • Figure shows force directions of induced rotor magnetic field © Copyright 2016 ELO 2.3 Operator Generic Fundamentals 37 Developing Torque in AC Induction Motor Breakdown Torque • Max torque motor can produce without stalling is breakdown torque • If load increases beyond this point, motor will stall, come to a rapid stop • Typical breakdown torque varies from 200% to 300% of full-load torque Starting Torque • Value of torque at 100% slip and normally 150% to 200% of full-load torque • As rotor accelerates, torque will increase and then decrease to value required to carry load on motor at constant speed, between 1-8 % slip © Copyright 2016 ELO 2.3 Figure: Torque versus Slip for an AC Induction Motor Operator Generic Fundamentals 38 Developing Torque in AC Induction Motor Developing Torque • Torque of an AC induction motor dependent upon strength of interacting rotor and stator fields and phase relationship between them • Equation shows mathematical expression for AC motor torque: 𝑇 = 𝐾𝛷𝐼𝑅 cos 𝜃𝑅 • Where: T = torque (lb-ft or N-m) K = constant Φ = stator magnetic flux IR = rotor current (A) cos 𝜃𝑅 = power factor of rotor © Copyright 2016 ELO 2.3 Operator Generic Fundamentals 39 Developing Torque in AC Induction Motor • During normal operation, K and cos 𝜃R are essentially constant, so torque is directly proportional to rotor current • Rotor current increases in almost direct proportion to slip, so maximum current will occur at maximum slip, which is at motor startup © Copyright 2016 ELO 2.3 Operator Generic Fundamentals 40 Developing Torque in AC Induction Motor • Rotating field induces a voltage in the rotor, causing current to flow C • Current flow builds rotor magnetic field • Rotor field chases stator • Always turns slower • Called “slip” • More load = More slip • Rotor current induces Counter ElectroMotive Force (CEMF) in stator C • CEMF controls current © Copyright 2016 ELO 2.3 Operator Generic Fundamentals 41 Starting Current on AC Induction Motor ELO 2.4 – Explain differences between starting and running current for an AC induction motor and the operating limits used to mitigate the effects of starting current. • There are several reasons for high starting current: – Power is required to build up rotating magnetic field in stator – Extra energy is required to overcome inertia of rotor – Interactions occur between rotor currents and stator’s magnetic field, result in motor drawing high currents – Rotor speed is too low to generate sufficient counter electromotive force in stator © Copyright 2016 ELO 2.4 Operator Generic Fundamentals 42 Starting Current and CEMF • Three induced fields C – Stator field from electric source – Rotor field induced from stator field – Counter field from rotor back into stator - limits current flow in stator (CEMF) C © Copyright 2016 ELO 2.4 Operator Generic Fundamentals 43 Starting Current and CEMF • Stator resistance is low at start • When breaker is closed, stator voltage applied to rotor – Causing high current on pump start (5 – 7 times full load) • CEMF builds in as rotor comes to speed – Limits stator current • Current quickly stabilizes at no load running current – With pump discharge valve closed or throttled © Copyright 2016 ELO 2.4 Operator Generic Fundamentals 44 Starting Current and CEMF • Terminal voltage applied is EMF EMF – (index finger of right hand) • As rotor cuts line of force, a current flows in the rotor CEMF • Rotor current generated a field that opposes the EMF Net Field – Called Counter Electro Motive Force (CEMF) – (index finger of left hand) • Results in lower “net” field – No load running current © Copyright 2016 ELO 2.4 Operator Generic Fundamentals 45 Starting Current on AC Induction Motor Starting Current Indications • Amps immediately increase (normally five to seven times) the normal running current value, usually off scale high • After a few seconds, current comes back on scale, decreases to normal running current • If current does not drop back into normal band within a few seconds, this is indication that equipment not functioning properly and is appropriate to shut it down and investigate • Due to the large starting current and heat produced on motor startup on large pumps, plant operating procedures will often limit the number of pump starts allowed within a given time period to prevent overheating the motor windings. © Copyright 2016 ELO 2.4 Operator Generic Fundamentals 46 Starting Current on AC Induction Motor Knowledge Check – NRC Question The starting current in an AC motor is significantly higher than the fullload running current because... A. little counter electromotive force is induced in the rotor windings during motor start. B. motor torque production is highest during motor start. C. little counter electromotive force is induced in the stator windings during motor start. D. work performed by the motor is highest during motor start. Correct answer is C. © Copyright 2016 ELO 2.4 Operator Generic Fundamentals 47 Starting Current on AC Induction Motor Knowledge Check – NRC Question If the discharge valve of a large motor-driven centrifugal pump remains closed during a normal pump start, the current indication for the AC induction motor will rise to... A. several times the full-load current value, and then decrease to the no-load current value. B. approximately the full-load current value, and then decrease to the no-load current value. C. approximately the full-load current value, and then stabilize at the full-load current value. D. several times the full-load current value, and then decrease to the full-load value. Correct answer is A. © Copyright 2016 ELO 2.4 Operator Generic Fundamentals 48 AC Synchronous Motor Operation ELO 2.5 – Describe how an AC synchronous motor is started and its operating characteristics. • Synchronous motors are like induction motors in that they both have stator windings that produce rotating magnetic field • Unlike an induction motor, an external DC source excites the synchronous motor rotor and therefore requires slip rings and brushes Figure: AC Synchronous Motor © Copyright 2016 ELO 2.5 Operator Generic Fundamentals 49 AC Synchronous Motor Operation • Synchronous motors use wound rotor • Slip rings and brushes needed • Rotor is much more expensive to manufacture than squirrel cage rotor associated with AC induction motors Figure: Wound Rotor © Copyright 2016 ELO 2.5 Operator Generic Fundamentals 50 AC Synchronous Motor Operation Starting a Synchronous Motor • Not self-starting, torque only developed when running at synchronous speed • Needs some type of device to bring rotor to synchronous speed • Two ways of starting a synchronous motor: – Use a separate DC motor – Embed squirrel-cage windings on the face of the rotor • If starting a synchronous motor with a separate DC motor, both motors may share a common shaft • Upon bringing DC motor to synchronous speed, stator windings receive AC current, DC motor then acts as a DC generator and supplies DC field excitation to rotor of synchronous motor • Once operating at synchronous speed, synchronous motor is ready for load © Copyright 2016 ELO 2.5 Operator Generic Fundamentals 51 AC Synchronous Motor Starting a Synchronous Motor • More commonly, a squirrel-cage winding is embedded in face of rotor poles to start motor • Starts as an induction motor, speed increases to ≈95% of synchronous speed, then direct current is applied and motor begins to pull into synchronism • Some plants use synchronous motors for Circ Water pumps – Improves power factor by being able to adjust excitation © Copyright 2016 ELO 2.5 Operator Generic Fundamentals 52 AC Motor and Generator Operation TLO 3 – Explain operating characteristics and limitations on AC motors. 3.1 State the applications of the following types of AC motors: a. Induction b. Single-phase c. Synchronous 3.2 Describe the indications resulting from a locked motor rotor, sheared shaft, or miswired phases. 3.3 Describe the consequences of overheated motor windings or motor bearings. 3.4 Describe the causes of excessive current in motors and generators resulting from conditions such as low voltage, overloading, and mechanical binding. 3.5 Describe the relationship between pump motor current and the following parameters: a. Pump flow b. Pressure c. Speed d. Motor stator temperature © Copyright 2016 TLO 3 Operator Generic Fundamentals 53 Motor Applications ELO 3.1 – State the applications of the following types of AC motors: induction, single-phase, and synchronous. Induction Motors • Most commonly used motor in industrial applications – less expensive to manufacture and maintain and operates reliably • Pumps, fans, and compressors are examples of induction motors © Copyright 2016 ELO 3.1 Operator Generic Fundamentals 54 Motor Applications Induction Motors • Named from rotating magnetic field of stator – induces a voltage into the rotor • Most commonly used AC motor in industrial applications – simplicity, rugged construction, and relatively low manufacturing costs – no external connections • Most AC induction motors use a “squirrel-cage” rotor © Copyright 2016 ELO 3.1 Figure: Squirrel Cage Rotor Operator Generic Fundamentals 55 Motor Applications Squirrel-Cage Rotor • Made of a laminated cylinder with slots in its surface • Made of heavy copper bars that connect at each end by a metal ring made of copper or brass • No insulation required because of low voltages induced into rotor bars • For maximum field strength, size of air gap between rotor bars and stator windings is small © Copyright 2016 Figure: Squirrel-Cage Induction Rotor ELO 3.1 Operator Generic Fundamentals 56 Motor Applications Single-Phase Motors • Very small commercial applications • Mostly used in household appliances Synchronous Motors • More complex design and higher manufacturing cost • Rarely used in commercial applications • Sometimes used for Circ Water pumps – Excitation can be adjusted to minimize impact of inductive loading © Copyright 2016 ELO 3.1 Operator Generic Fundamentals 57 Indications of Motor Malfunctions ELO 3.2 – Describe the indications resulting from a locked motor rotor or shaft, sheared shaft, or miswired phases. Indications of Motor Malfunctions • Motors or their connected loads are subject to several different types of mechanical failures, including: – Locked (seized) rotor – Sheared rotor – Miswired phases Figure: Damage From Locked Rotor © Copyright 2016 ELO 3.2 Operator Generic Fundamentals 58 Indications of Motor Malfunctions Locked (Seized) Rotor • Typical indications of a locked rotor: – Rotor speed decreases – Immediate increase in current to essentially starting current o No rotor speed, no CEMF – Immediate reduction in system flow rate – Immediate reduction in component discharge pressure – Immediate rise in motor winding temperatures resulting from higher current flow – Eventual timed delay tripping of circuit breaker due to high current o Therefore, current drops to ZERO after breaker trip © Copyright 2016 ELO 3.2 Operator Generic Fundamentals 59 Indications of Motor Malfunctions Sheared Rotor • Allows motor to operate freely (spinning) with no load mechanically attached • Following indications are typical of a sheared rotor: – Rotor speed initially increases due to loss of load (pump impeller) – Load has no (low) running current (as indicated on ampere meter) – Immediate reduction in system flow rate – Immediate reduction in component discharge pressure • Reduction in flow and pressure similar to locked rotor indication • Current indication differs from locked rotor – Sheared shaft goes to no load running current – Locked rotor goes to starting current then 0 after breaker trip © Copyright 2016 ELO 3.2 Operator Generic Fundamentals 60 Indications of Motor Malfunctions Miswired Phases • Phases can be miswired after work in which motor leads were disconnected • If two motor leads (phases) reversed, motor will turn backwards • Motor will work fine, but component it is driving generally will not © Copyright 2016 ELO 3.2 Operator Generic Fundamentals 61 Indications of Motor Malfunctions Knowledge Check – NRC Question A motor-driven cooling water pump is operating normally. How will pump motor current respond if the pump experiences a locked rotor? A. Decreases immediately to zero due to breaker trip. B. Decreases immediately to no-load motor amps. C. Increases immediately to many times running current, then decreases to no-load motor amps. D. Increases immediately to many times running current, then decreases to zero upon breaker trip. Correct answer is D. © Copyright 2016 ELO 3.2 Operator Generic Fundamentals 62 Consequences of Motor Overheating ELO 3.3 – Describe the consequences of overheated motor windings or bearings. Loss of Motor Cooling • Continuous operation of a motor at rated load with a loss of required cooling to motor windings will result in breakdown of motor insulation due to overheating • Causes increased temperature and current flow due to a decrease in insulation resistance – Possible short circuit • Thermal overload device protects many large motors from high current © Copyright 2016 ELO 3.3 Operator Generic Fundamentals 63 Consequences of Motor Overheating Loss of Bearing Cooling or Lubrication • Various means of cooling and lubricating motor bearings, depending on size and application of motor • Two needs are common to all bearings: – Must be properly lubricated to minimize friction – Heat generated in the bearing must be removed • High bearing friction results in additional load that causes motor current and motor winding temperature to rise • Bearing wear can increase runout, permitting destructive contact between rotor and stator © Copyright 2016 ELO 3.3 Operator Generic Fundamentals 64 Consequences of Motor Overheating Loss of Bearing Cooling or Lubrication • Increase in bearing friction will cause bearing to degrade until it fails, resulting in component trip and possible damage to component – If increase is significant, can lead to motor damage • If a bearing is running with higher temperature, but friction is not excessive, it may run for some time without causing damage – Most bearing lubricants are effective within a specified temperature range © Copyright 2016 ELO 3.3 Operator Generic Fundamentals 65 Consequences of Motor Overheating Knowledge Check – NRC Question Which one of the following will result from prolonged operation of an AC induction motor with excessively high stator temperatures? A. Decreased electrical current demand due to reduced counter electromotive force B. Decreased electrical resistance to ground due to breakdown of winding insulation C. Increased electrical current demand due to reduced counter electromotive force D. Increased electrical resistance to ground due to breakdown of winding insulation Correct answer is B. © Copyright 2016 ELO 3.3 Operator Generic Fundamentals 66 Excessive Current in Motors ELO 3.4 – Describe the causes of excessive current in motors resulting from conditions such as low voltage, overloading, and mechanical binding. Motor Overloading • Overloading or just increased loading can result in increased current • Overloading can be caused by: – Gradual bearing failure – Undervoltage – Locked rotor – Shorted windings – One open phase © Copyright 2016 ELO 3.4 Operator Generic Fundamentals 67 Excessive Current in Motors Malfunction Likely Result Gradual motor bearing failure Increased friction, current and temperature could be high enough to cause a thermal overload trip Packing on a motor-operated valve tightened excessively Motor current increases due to increased torque on motor caused by additional friction associated with packing © Copyright 2016 High currents can result in excessive heat within motor, causing breakdown of motor winding insulation and damage to motor (grounds) ELO 3.4 Operator Generic Fundamentals 68 Excessive Current in Motors Malfunction Likely Result Locked or seized Overcurrent trip of the supply circuit breaker rotor of motor or component driven (pump, etc.) Reduced voltage Motor supplies power needed to drive the load, supplied to motor when voltage drops, current increases Operating for extended periods with reduced voltage can lead to winding damage Sheared shaft or Abnormally low or no running currents as indicated rotor on an amp meter, with low or no flow/pressure indicated © Copyright 2016 ELO 3.4 Operator Generic Fundamentals 69 Excessive Current in Motors Failed Bearings • Bearing failure can result from a number of circumstances, including: – Insufficient lubrication – Poor bearing maintenance practices – Improper loading of component – Motor undervoltage situation • Any condition leading to overheating of bearings can cause bearing failure • Excessive work can cause heat buildup in machine windings © Copyright 2016 ELO 3.4 Operator Generic Fundamentals 70 Excessive Current In Motors Knowledge Check Which of the following will cause an increase in motor current? (Select all that are true.) A. An increase in bearing friction due to inadequate maintenance B. Increasing the load on the component driven, such as opening a pump discharge valve to provide more flow C. Increased voltage to the motor D. Reduced voltage to the motor Correct answers are A, B, and D. © Copyright 2016 ELO 3.4 Operator Generic Fundamentals 71 Excessive Current In Motors Knowledge Check – NRC Question Excessive current will be drawn by an AC induction motor that is operating _____________. A. completely unloaded B. at full load C. with open circuited stator windings D. with short circuited stator windings Correct answer is D. © Copyright 2016 ELO 3.4 Operator Generic Fundamentals 72 Pump Motor Current Relationships ELO 3.5 – Describe the relationship between pump motor current and the following parameters: pump flow, pressure, speed, and motor stator temperature. Operating Limits to Mitigate Effect of Starting Current • Recall from the pumps module that centrifugal pumps operate within a known set of laws, known as pump laws • Applying pump laws, we can determine how changes in pump parameters will affect motor power and current © Copyright 2016 ELO 3.5 Operator Generic Fundamentals 73 Pump Motor Current Relationships Pump Law Review • Flow rate of a pump is directly proportional to pump speed 𝑉∝𝑛 • Head of a pump is directly proportional to pump speed squared 𝐻𝑝 ∝ 𝑛2 • Power of a pump is directly proportional to pump speed cubed 𝑝 ∝ 𝑛3 • Where: n = speed of pump impeller (rpm) 𝑉 = volumetric flow rate of pump (gpm or ft3/hr) Hp = head developed by pump (feet) p = pump power (kW) © Copyright 2016 ELO 3.5 Operator Generic Fundamentals 74 Pump Motor Current Relationships • These relationships are useful in determining the effects on pump performance when flow is increased or decreased. Changes in power can affect the amount of current an electric motor will draw. Action Equation Determine change in flow by using this relationship. 𝑉2 = 𝑉1 𝑛2 𝑛1 Determine change in head by using this relationship. Notice that head changes by the square of change in speed. 𝐻𝑃2 = 𝐻𝑃1 Determine change in power by using this relationship. Notice that power changes by the cube of change in speed. 𝑃2 = 𝑃1 © Copyright 2016 ELO 3.5 𝑛2 𝑛1 𝑛2 𝑛1 2 3 Operator Generic Fundamentals 75 Pump Motor Current Relationships Flow Example • A cooling water pump is operating at a speed of 1,800 rpm. Its flow rate is 400 gpm at a head of 48 ft. The pump’s motor is drawing 45 kW. Determine the new pump flow rate if the pump speed increases to 3,600 rpm. • Solution: 𝑉2 = 𝑉1 𝑛2 𝑛1 𝑉2 = 400 𝑔𝑝𝑚 3,600 𝑟𝑝𝑚 1,800 𝑟𝑝𝑚 𝑉2 = 800 𝑔𝑝𝑚 © Copyright 2016 ELO 3.5 Operator Generic Fundamentals 76 Pump Motor Current Relationships Head Example • A cooling water pump is operating at a speed of 1,800 rpm. Its flow rate is 400 gpm at a head of 48 ft. The pump's motor is drawing 45 kW. Determine the new pump head if the pump speed increases to 3,600 rpm. • Solution: 𝐻𝑃2 = 𝐻𝑃1 𝐻𝑃2 𝑛2 𝑛1 2 3,600 𝑟𝑝𝑚 = 48 𝑓𝑡 1,800 𝑟𝑝𝑚 2 𝐻𝑃2 = 192 𝑓𝑡 © Copyright 2016 ELO 3.5 Operator Generic Fundamentals 77 Pump Motor Current Relationships Power Example • A cooling water pump is operating at a speed of 1,800 rpm. Its flow rate is 400 gpm at a head of 48 ft. The pump's motor is drawing 45 kW. Determine the new pump power requirements if the pump speed increases to 3,600 rpm. • Solution: 𝑃2 = 𝑃1 𝑛2 𝑛1 3 3,600 𝑟𝑝𝑚 𝑃2 = 45 𝑘𝑊 1,800 𝑟𝑝𝑚 3 𝑃2 = 360 𝑘𝑊 © Copyright 2016 ELO 3.5 Operator Generic Fundamentals 78 Pump Motor Current Relationships • Relationship of these parameters can be calculated and plotted on a pump performance curve for specific pumps and operating conditions (see graph) • NRC GFE questions use these curves frequently Figure: Pump Flow, Head, and Brake Horsepower © Copyright 2016 ELO 3.5 Operator Generic Fundamentals 79 Pump Motor Current Relationships • Brake horsepower is input power delivered to pump by motor (kW) • If flow rate is increased, brake horsepower will increase, and pump head will decrease • If flow rate decreases, brake horsepower decreases and pump head will increase • Changes in brake horsepower are proportional to changes in motor amps • Increases in motor amps will result in increases in motor winding temperature © Copyright 2016 ELO 3.5 Operator Generic Fundamentals 80 Pump Motor Current Relationships Example: Consider a pump driven by a single-speed AC induction motor. A throttled discharge flow control valve controls the pump flow rate. • The following initial pump conditions exist: – Pump motor current = 50 amps – Pump flow rate = 400 gpm What will be the approximate value of pump motor current if the flow control valve is repositioned such that pump flow rate is 800 gpm? a. Less than 100 amps b. 200 amps c. 400 amps d. More than 500 amps © Copyright 2016 Figure: Pump Flow, Head, and Brake Horsepower ELO 3.5 Operator Generic Fundamentals 81 Pump Motor Current Relationships • Using the pump curve, 400 gpm is equivalent to ~ 220 BHP; – 800 gpm is ~ 270 BHP – 220/270 = 50 amps/x • Solving for x: – x = 50/(220/270) – x = 50/0.82 = ~ 61 amps, • 61 amps is less than 100 amps, so the correct answer is A © Copyright 2016 Figure: Pump Flow, Head, and Brake Horsepower ELO 3.5 Operator Generic Fundamentals 82 Pump Motor Current Relationships Knowledge Check – NRC Question A multispeed centrifugal pump is operating with a flow rate of 1,800 gpm at a speed of 3,600 rpm. Which one of the following approximates the new flow rate if the pump speed is decreased to 2,400 rpm? A. 900 gpm B. 1,050 gpm C. 1,200 gpm D. 1,350 gpm Correct answer is C. © Copyright 2016 ELO 3.5 Operator Generic Fundamentals 83 AC Generator Operation TLO 4 – Analyze operation of generators in parallel and predict system response to changes in frequency, voltage, and load. 4.1 Define apparent, true, and reactive power and power factor using a power triangle and their effect on generator operation. 4.2 Describe the purpose of a voltage regulator and function of each of the following typical components: a. Sensing circuit b. Reference circuit c. Comparison circuit d. Amplification circuit(s) e. Signal output circuit f. Feedback circuit 4.3 Describe the conditions that must be met prior to paralleling two generators, including consequences of not meeting these conditions. 4.4 Describe the consequences of over-excitation and under-excitation when load sharing. © Copyright 2016 TLO 4 Operator Generic Fundamentals 84 Power Factor and Generators ELO 4.1 – Define apparent, true, and reactive power and power factor using a power triangle and their effect on generator operation. Power Triangle • In AC circuits, current and voltage are normally out of phase due to effects of inductive and capacitive reactance • As a result, not all power produced by a generator in an AC application is available to accomplish work • Similarly, power calculations in AC circuits vary from those in DC circuits © Copyright 2016 ELO 4.1 Operator Generic Fundamentals 85 Power Triangle • Can be used to determine a generator’s efficiency • The power output of a generator is referred to as apparent power (S) ‒ Measured in volt-amperes (VA) • True power (P) is the amount of usable power ‒ Measured in watts • Wasted or stored power is referred to as reactive power (Q) ‒ Measured in volt-amperesreactive (VAR) • Phase angle (θ) represents inefficiency of circuit ‒ Corresponds total reactive impedance (Z) to current flow © Copyright 2016 ELO 4.1 Figure: Power Triangle Operator Generic Fundamentals 86 Power Factor and Generators • Apparent power is power delivered to an electrical circuit 𝑆 = 𝐼2 𝑍 = 𝐼𝐸 • Where: S = apparent power (VA) I = RMS current (A) E = RMS voltage (V) Z = impedance (Ω) • True power is power consumed by resistive loads in electrical circuit 𝑃 = 𝐼2 𝑅 = 𝐸𝐼 cos 𝜃 • Where: P = true power (watts) I = RMS current (A) E = RMS voltage (V) R = resistance (Ω) θ = angle between E and I sine waves © Copyright 2016 ELO 4.1 Operator Generic Fundamentals 87 Power Factor and Generators • Reactive power is power component in AC circuit (inductive) and (capacitive) fields, units are volt-amperes-reactive (VAR) 𝑄 = 𝐼 2 𝑋 = 𝐸𝐼 sin 𝜃 • Where: Q = reactive power (VAR) I = RMS current (A) X = net reactance (Ω) E = RMS voltage (V) θ = angle between the E and I sine waves • Reactive power is unusable power because it is stored in circuit by: – Inductors, expand and collapse magnetic fields, try to keep current constant – Capacitors, charge and discharge in an attempt to keep voltage constant – Inductance and capacitance alternately exchange reactive power (consume and give back) with circuit’s AC source © Copyright 2016 ELO 4.1 Operator Generic Fundamentals 88 Power Factor and Generators • Total power delivered by the source, same as apparent power • Part of this apparent power, called true power, dissipates by circuit resistance in form of heat • Rest of apparent power returns to source by circuit inductance and capacitance (reactive power) © Copyright 2016 ELO 4.1 Operator Generic Fundamentals 89 Power Factor and Generators • Power factor (PF) is represented by cos θ in an AC circuit • Ratio of true power to apparent power, where θ is phase angle between applied voltage and current sine waves and is angle between P and S on a power triangle 𝑃 cos 𝜃 = 𝑆 • Where: cos 𝜃 = power factor (pf) P = true power (watts) S = apparent power (VA) Figure: Power Triangle NOTE: The longer the hypotenuse, the greater the generator current, and higher the heat from that current. © Copyright 2016 ELO 4.1 Operator Generic Fundamentals 90 Power Factor and Generators Lagging Power Factor • Power factor determines what part of apparent power is true power • Can vary from 1, when the phase angle is 0°, to 0, when the phase angle is 90° • Inductive circuit, current lags voltage • This type of circuit has a lagging power factor ELI the ICE Man • ELI refers to an inductive circuit (L) – Where, current (I) lags voltage (E) © Copyright 2016 ELO 4.1 Operator Generic Fundamentals 91 Power Factor and Generators Leading Power Factor • In capacitive circuit, current leads voltage • Circuits with loads like motors will have lagging power factor • Circuits with loads like fluorescent lighting will have leading power factor • Most industrial electrical systems exhibit lagging power factor because inductive loads account for larger percentage of reactance in these circuits ELI the ICE Man • ICE refers to a capacitive circuit (C) where current (I) leads voltage (E) © Copyright 2016 ELO 4.1 Operator Generic Fundamentals 92 Sample Generator Capability Curve • Megawatts is the solid horizontal line (also, 0 VARs) • In a purely resistive circuit – Power triangle is a straight line – Power Factor = 1 – True Power = Apparent Power • As you raise (or lower) excitation (voltage), you become more (or less) “overexcited” – Move “up (down)” the Reactive KVA axis – Phase angle increases as you move away from “Megawatts” line o Generator current increases © Copyright 2016 Figure: Sample Generator Capability Curve ELO 4.1 Operator Generic Fundamentals 93 VARs Terminology • Consider the Generator Capability Curve: Increase excitation • Lagging VARs: – Overexcited Increase generator amps – Positive VARs – VARs OUT – Supplying VARs • Leading VARs: – Underexcited Increase generator amps – Negative VARs Decrease excitation © Copyright 2016 ELO 4.1 – VARs IN Operator Generic Fundamentals 94 VARs Terminology • Consider an increase in excitation from an initial point of some “leading VARs”: New Power Triangle Still LEADING VARs, just less leading. Power Factor closer to 1. (smaller phase angle) Increase excitation © Copyright 2016 ELO 4.1 Operator Generic Fundamentals 95 Voltage Regulation and Components ELO 4.2 – Describe the purpose of a voltage regulator and function of each of the following typical voltage regulator components: sensing circuit, reference circuit, comparison circuit, amplification circuit(s), signal output circuit, and feedback circuit. • Output voltage of an AC generator varies with both changes in load (i.e., current) and changes in power factor • Due to output variation, AC generators require voltage regulation • Voltage regulator compares output voltage of machine to a desired value – Varies excitation applied to generator’s field to maintain generator output voltage constant © Copyright 2016 ELO 4.2 Operator Generic Fundamentals 96 Voltage Regulation and Components Voltage Regulator Components • Below is a typical block diagram of a voltage regulator, which consists of six basic circuits that act together to regulate output voltage from no-load to full-load Figure: Voltage Regulator Block Diagram © Copyright 2016 ELO 4.2 Operator Generic Fundamentals 97 Voltage Regulation and Components Sensing Circuit • Senses output voltage of AC generator • As generator is loaded or unloaded, output voltage changes and sensing circuit provides a signal of voltage changes • Signal proportional to output voltage, sends signal to comparison circuit Reference Circuit • Maintains a constant output for reference, represents desired voltage output Comparison Circuit • Electrically compares reference voltage to received sensed voltage and provides error signal • Sends error signal to amplification circuit © Copyright 2016 ELO 4.2 Operator Generic Fundamentals 98 Voltage Regulation and Components Amplification Circuit • Magnetic amplifier or transistor amplifier, takes error signal from comparison circuit and amplifies milliamp input to an amp output then sends amplified signal to signal output or field circuit Signal Output Circuit • Controls field excitation of AC generator, increases or decreases field excitation to either raise or lower AC output voltage Feedback Circuit • Takes some output of signal output circuit and feeds it back to amplification circuit, prevents overshooting or undershooting desired voltage by slowing down circuit response Changing Output Voltage • As generator load changes, output voltage will change if excitation remains constant; voltage regulator senses change in output voltage and changes generator excitation to return output voltage to desired value © Copyright 2016 ELO 4.2 Operator Generic Fundamentals 99 Voltage Regulation and Components Generator Voltage Control (most plants are similar to this) • Steam applied to rotor of turbine to get it up to 1800 rpm • Permanent Magnet Generator (PMG) generates AC and sends to voltage regulator – Sometimes called Field Flashing • Voltage regulator rectifies the AC to DC then sends DC to the exciter – This is the “excitation” discussed earlier to change VARs • Exciter generates AC and sends it to the rectifier banks • Rectifier banks rectify AC to DC • DC applied to the rotor of the main generator and generates AC in the stator (generator output voltage) • This voltage also sent to voltage regulator for voltage control © Copyright 2016 ELO 4.2 Operator Generic Fundamentals 100 Voltage Regulation and Components Increasing Generator Load (Output Voltage Drop) • Increase in generator load results in a drop in generator output voltage – voltage regulator will respond to raise output voltage • Sensing circuit senses decrease in output voltage as compared to reference and lowers its input to comparison circuit • Comparison circuit develops an error signal • Error signal developed will be a positive value with magnitude of signal dependent on difference between sensed and reference voltage • Amplifier circuit amplifies this output and sends amplified signal to signal output circuit • Signal output circuit increases field excitation to AC generator, causing generated voltage to increase to desired output © Copyright 2016 ELO 4.2 Operator Generic Fundamentals 101 Voltage Regulation and Components Decreasing Generator Load (Output Voltage Increase) • If load on generator decreases, voltage output of machine would rise • Actions of voltage regulator for an increase in output voltage would be opposite of those for a lowering output voltage • Comparison circuit will develop a negative error signal whose magnitude is again dependent on difference between sensed voltage and reference voltage • Signal output circuit will decrease field excitation to AC generator, causing generated voltage to decrease to desired output © Copyright 2016 ELO 4.2 Operator Generic Fundamentals 102 Paralleling Generators ELO 4.3 – Describe the conditions that must be met prior to paralleling two generators, including consequences of not meeting these conditions. • Most power grids and distribution systems have more than one AC generator operating at one time • Three required conditions for paralleling (or synchronizing) AC generators: – Terminal voltages must be essentially equal o Minimizes VAR loading – Frequency of incoming generator slightly > grid frequency o Ensures generator picks up load instead of becoming a load (motorizing) – Output voltages must be in phase o Minimize current through generator output breaker © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 103 Paralleling Generators • During paralleling operations, voltmeters indicate voltages of generators • Output frequency meters facilitate frequency matching • A synchroscope, a device that senses two frequencies and indicates phase differences between generators, allows phase matching of two generators © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 104 Paralleling Generators • With two generators in parallel, real load carried by each generator can be determined by examining frequency-power characteristics for each generator • The “house curves” to the right graphically describes load sharing between two power sources • Note that power values for generators G1 increases to the left and G2 increases to the right © Copyright 2016 Figure: Two AC Generators Operating in Parallel ELO 4.3 Operator Generic Fundamentals 105 Paralleling Generators • Gen G1 – power output to left of the solid vertical centerline (frequency axis) • Gen G2 - power output to right of solid vertical centerline (frequency axis) • Since parallel generators must operate at same frequency, we can determine load carried by each generator by noting the intersection of generator characteristic for particular speed regulator setting in use and operating frequency © Copyright 2016 Figure: Two AC Generators Operating in Parallel ELO 4.3 Operator Generic Fundamentals 106 Paralleling Generators Real Power Sharing • For settings in figure, at frequency of 60 Hz, G1 would carry 350 kW and G2 would carry 50 kW • Points A (generator G1) and B (generator G2) are the points where each generator curve intersects with frequency • Total length of line A-B represents total load (400 kW) • If total load increased to 500kW with no changes in speed regulator settings, frequency would drop to 59.95 Hz and G1 would carry 400 kW and G2 100kW, line C-D © Copyright 2016 ELO 4.3 Figure: Two AC Generators Operating in Parallel Operator Generic Fundamentals 107 Paralleling Generators Real Load Sharing • Load initially unbalanced between generators and proper distribution of load between generators • Dashed lines show original condition (freq. = 60Hz, G1 load = 350 kW, and G2 load = 50 kW) • Increasing speed setting on G2 and decreasing speed setting on G1 until no-load frequencies are same, load on G2 increases from 50 kW to 200 kW • Load on G1 decreases from 350 kW to 200 kW © Copyright 2016 ELO 4.3 Figure: Kilowatt Load Balanced by Adjusting Speed Regulator Operator Generic Fundamentals 108 Paralleling Generators Reactive Load Sharing • Consider two generators operating in parallel, supplying a constant real power • The reactive loads carried by each generator determined by terminal voltage • Different curves represent different values of field excitation • Fixed value of field excitation, terminal voltage drops off as reactive load increases, exhibiting a drooping characteristic © Copyright 2016 Figure: Reactive Load Balance by Adjusting Voltage Regulator ELO 4.3 Operator Generic Fundamentals 109 Paralleling Generators Reactive Load Sharing • Generators typically use automatic voltage regulators to correct terminal voltage-reactive power characteristic to a linear droop • Parallel generators operate at same voltage • Reactive load determined by intersection of generator characteristic curve for voltage regulator setting and operating voltage © Copyright 2016 Figure: Reactive Load Balance by Adjusting Voltage Regulator ELO 4.3 Operator Generic Fundamentals 110 Paralleling Generators Reactive Load Sharing • For example, using figure, at an operating voltage of 450KV, G1 initially carries 600 KVAR and G2 carries 200 KVAR, shown by dotted lines; to balance reactive load while maintaining terminal voltage at 450KV, decrease noload voltage regulator setting on G1 to 460 KV while increasing voltage regulator setting on G2 to 460 KV (desirable to distribute load in proportion to generator ratings) © Copyright 2016 ELO 4.3 Figure: Reactive Load Balance by Adjusting Voltage Regulator Operator Generic Fundamentals 111 Paralleling Generators Operation in Parallel with a Large Grid • When paralleling a generator with a large power system (grid), output of generator is minimal in comparison • House curve shows system as a constant frequency, constant voltage power source • Power grid automatically supplies remainder of required load; when single generator supplies negative reactive power – decreases total reactive load, improves power factor, and requires less current for same total real power © Copyright 2016 ELO 4.3 Figure: Parallel Operation of a Single Generator with the Grid Operator Generic Fundamentals 112 Paralleling Generators Example An operator is connecting a main generator to an infinite power grid that is operating at 60 Hz. Generator output voltage is equal to the grid voltage, but generator frequency is at 57 Hz. Which one of the following generator conditions is most likely to occur if the operator closes the generator output breaker with the voltages in phase (synchronized) but with the existing frequency difference? (Assume no generator breaker protective trip occurs.) A. Reverse power B. Under-frequency C. Under-voltage D. Over-speed Answer A is the only credible choice. However, in practice, the generator would certainly be reverse powered and with a frequency difference this large it would almost certainly trip on reverse power if the synchronizing circuit allowed the breaker to close at all. © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 113 Paralleling Generators Re-Energizing a Dead Bus • Consider the process of re-energizing a dead bus with a generator – Could occur when using plant emergency diesel generators • Assuming none of the loads on the bus are running • If you connect generator onto bus without first stripping loads, all of the bus loads will start at once when generator energizes bus • Starting current is several times higher than running current, if all start at once, the bus will draw several times its normal full load current • Generator rating is sufficient to carry all bus loads • Not sufficient to carry all starting currents simultaneously • Starting all loads together will result in an over-current condition, which may trip the generator © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 114 Paralleling Generators Typical NRC Exam Question Closing the output breaker of a three-phase generator onto a deenergized bus can result in... A. An over-voltage condition on the bus B. An over-current condition on the generator if the bus was not first unloaded C. A reverse power trip of the generator circuit breaker if generator frequency is low D. A large reactive current in the generator Only correct choice is B. This is a real operating concern. If a dead bus is re-energized without stripping loads first, it will result in an overcurrent condition, possibly tripping the bus again and possibly damaging either the generator or the feeder breaker. © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 115 Paralleling Generators Explanation for NRC Exam Question Closing the output breaker of a three-phase generator onto a de-energized bus can result in... A. An over-voltage condition on the bus. Incorrect. If voltage was set correctly prior to energizing dead bus, voltage would tend to drop slightly when loaded. Voltage regulator would compensate and keep it within the band, but it would certainly not result in voltage going too high. B. An over-current condition on the generator if the bus was not first unloaded. Correct. All motors supplied by this bus would be accelerating to speed and drawing starting current at the same time. This will result in an overcurrent condition that will likely trip the generator and de-energize the bus again. C. A reverse power trip of the generator circuit breaker if generator frequency is low. Incorrect. Since the generator is only source of power to this bus, it cannot receive power from any other source; reverse power is not plausible. D. A large reactive current in the generator. Incorrect. Reactive power should not be higher than normal reactive load for running this bus. Real load from starting currents will be the problem in this circumstance. © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 116 Paralleling Generators Example A main generator is about to be connected to an infinite power grid. Closing the generator output breaker with generator voltage slightly lower than grid voltage and generator frequency slightly higher than grid frequency will initially result in… (assume no generator breaker protective trip occurs) A. The generator supplying reactive power to the grid B. The generator attaining a leading power factor C. The generator acting as a real load to the grid D. Motoring of the generator Requires you to consider both real load and reactive load. Since generator frequency is slightly higher than grid frequency (the synchroscope will be rotating slowly clockwise as it should be with parallel generators), the generator will pick up real load from the grid. This eliminates options C and D, since both of them imply the generator becomes a load on the grid. You should evaluate reactive load next. Since generator voltage is slightly lower than grid voltage, it will not supply reactive power to the grid, eliminating option A. It will have a leading power factor and appear as a capacitive load to the rest of the grid. © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 117 Paralleling Generators Knowledge Check – NRC Question A main generator is being paralleled to the power grid. Generator voltage has been properly adjusted and the synchroscope is rotating slowly clockwise. The generator breaker must be closed just as the synchroscope pointer reaches the 12 o'clock position to prevent... A. motoring of the generator due to unequal frequencies. B. excessive MW load transfer to the generator due to unequal frequencies. C. excessive MW load transfer to the generator due to out-of-phase voltages. D. excessive arcing within the generator output breaker due to outof-phase voltages. Correct answer is D. © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 118 Paralleling Generators Knowledge Check A main generator is about to be connected to an infinite power grid with the following conditions: Generator frequency: 59.5 Hz Grid frequency: 59.8 Hz Generator voltage: 115.1 kV Grid voltage: 114.8 kV When the generator output breaker is closed, the generator will... A. acquire real load and reactive load. B. acquire real load, but become a reactive load to the grid. C. become a real load to the grid, but acquire reactive load. D. become a real load and a reactive load to the grid. Correct answer is C. © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 119 Paralleling Generators Knowledge Check – Explanation • Generator Capability Curve only shown with “positive” MW loading – But “negative” MW loading can be shown for training purposes • Since generator frequency < grid frequency and generator voltage > grid voltage • Power triangle would actually look like this: VARs (Supplying/Acquiring) MW (Becoming) (Reverse Power) MW (Supplying/Acquiring) Therefore, correct answer is: C. become a real load to the grid, but acquire reactive load. VARs (Becoming, for lack of better word) © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 120 Paralleling Generators Knowledge Check A main generator is about to be connected to an infinite power grid. Closing the generator output breaker with generator and grid voltages matched but with generator frequency lower than grid frequency will initially result in the generator... A. picking up a portion of the grid real load. B. picking up a portion of the grid reactive load. C. experiencing reverse power conditions. D. experiencing over-speed conditions. Correct answer is C. © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 121 Generator Excitation ELO 4.4 – Describe the consequences of over-excitation and underexcitation when load sharing. Generator Excitation Guidelines • Terms commonly used to explain changes in generator field current: – over-excitation – under-excitation © Copyright 2016 ELO 4.4 Operator Generic Fundamentals 122 Generator Excitation Guidelines • If generator in under-excited condition, power factor angle is leading – Machine is absorbing reactive power from system • If generator in over-excited condition, power factor angle is lagging – Generator is supplying reactive power to system • Based on grid conditions and loads, power control/load dispatch may request change in generator excitation to maintain grid voltage stability • Generator will react differently to governor speed and field changes when operating un-paralleled, paralleled with a generator, or supplying power grid • During parallel operation, generator is a small power supply in comparison to entire grid network, not able to change grid voltage or frequency significantly © Copyright 2016 ELO 4.4 Operator Generic Fundamentals 123 Generator Excitation Guideline Terms • Unity power factor – (VARs at lowest point or zero) for every operating condition, one value of field current causes generator to deliver only real power • Under-excitation – when current falls below value that yields operation at unity power factor, the generator absorbs reactive power (VARs increase) – Generator acts as inductive coil – Not desired mode of operation, since generator is a reactive load • Over-excitation – increasing current causes generator to supply increased reactive power – Generator acts as capacitor (delivery of reactive power) – Generators normally provide VARs together with watts and nearly always operate in overexcited conditions © Copyright 2016 ELO 4.4 Operator Generic Fundamentals 124 Generator Excitation When generator is under-excited: • Reducing generator excitation will make the generator more underexcited • It will look like a larger inductive load to the grid and draw more reactive load from the grid • Increasing generator voltage will make the generator less underexcited • It will draw less reactive load and move closer to unity power factor • In under-excited condition, generator has a leading power factor © Copyright 2016 ELO 4.4 Operator Generic Fundamentals 125 Generator Excitation When generator is over-excited: • Increasing generator voltage will make the generator more overexcited – will supply more reactive load to the grid • In over-excited condition, generator has a lagging power factor • Power plants generally operate in over-excited condition because the grid has more inductive than capacitive loads Note: Generator voltage changes will only affect reactive load (VAR) and not real load (kW). © Copyright 2016 ELO 4.4 Operator Generic Fundamentals 126 Infinite Versus Isolated Bus Infinite Bus • Nuclear power plants have turbine generators that are tied to an infinite bus – Speed and voltage changes do not effect grid frequency or voltage, just real and reactive loading Isolated Bus • Large factories might supply their own power • This is considered an isolated or “finite” bus • There is a finite amount of real and reactive load being supplied • If two generators are operated in parallel – changes made by one or the other will effect frequency or voltage • Several bank questions in this chapter and 191008 test this concept © Copyright 2016 Operator Generic Fundamentals 127 Generator Excitation Example A main generator that connects to an infinite power grid has the following initial indications: 100 MW 0 MVAR 2,900 Amps 20,000 VAC If we reduce the main generator excitation slightly, amps will _________________ and MW will ______________. In order to evaluate this item, we need to consider real load and reactive load separately. © Copyright 2016 ELO 4.4 Operator Generic Fundamentals 128 Generator Excitation Example Answer • Main generator voltage excitation controls reactive load • Main generator real load is controlled by changing speed control – Speed will actually be unchanged due to the connection to an infinite grid, but the generator will take load from the grid in its attempt to raise speed – Since speed control has not been touched, real load will remain unchanged • The generator begins at 0 MVARS or no reactive load – An adjustment in either direction will increase current • Since excitation is “reduced slightly”, generator amps will increase • Therefore, in answer to the question, amps will increase and MW will remain the same © Copyright 2016 ELO 4.4 Operator Generic Fundamentals 129 Generator Excitation Knowledge Check Two identical 1,000 MW electrical generators are operating in parallel, supplying the same isolated electrical bus. The generator output breakers also provide identical protection for the generators. Generator A and B output indications are as follows: Generator A 22.5 kV 60.2 Hertz 750 MW 25 MVAR (out) © Copyright 2016 Generator B 22.5 kV 60.2 Hertz 750 MW 50 MVAR (out) ELO 4.3 Operator Generic Fundamentals 130 Generator Excitation Knowledge Check (continued) A malfunction causes the voltage regulator for generator B to slowly and continuously increase the terminal voltage for generator B. If the operator takes no action, which one of the following describes the electrical current indications for generator A? A. Current will decrease continuously, until the output breaker for generator A trips on reverse power. B. Current will decrease continuously, until the output breaker for generator B trips on reverse power. C. Current will initially decrease, then increase until the output breaker for generator A trips on overcurrent. D. Current will initially decrease, then increase until the output breaker for generator B trips on overcurrent. Correct answer is D. © Copyright 2016 ELO 4.3 Operator Generic Fundamentals 131 Generator Excitation - Explanation • Stem states it is an “isolated” bus (finite amount of load) – Total of: 1500 MW and 75 MVAR (Out) • Power triangles for both generators shown below (not to scale) • B regulator starts to fail “high” (increases excitation) – When “B” raises to 75 MVAR, “A” lowers to 0 MVAR – “B” generator current increases, “A” generator current decreases VARs (Supplying/Acquiring) MW (Becoming) (Reverse Power) B B A A MW (Supplying/Acquiring) VARs (Absorbing, for lack of better word) © Copyright 2016 ELO 4.4 Operator Generic Fundamentals 132 Generator Excitation - Explanation • As “B” continues to fail high (“B” to 100, “A” to -25) – “B” current continues to increase, while “A” now increases – “B” will reach an “upper” limit before “A” reaches a “lower” limit – Therefore, since the question asks what happens to “A”: o Current will initially decrease, then increase until the output breaker for generator B trips on overcurrent VARs (Supplying/Acquiring) MW (Becoming) (Reverse Power) B B A A MW (Supplying/Acquiring) VARs (Becoming, for lack of better word) © Copyright 2016 ELO 4.4 Operator Generic Fundamentals NRC KA to ELO Tie KA # KA Statement RO SRO ELO K1.01 Indication of a locked rotor 2.8 3.1 3.2 K1.02 Potential consequences of overheating insulation or bearings Causes of excessive current in motors and generators, such as low voltage, overloading, K1.03 and mechanical binding Relationship between pump motor current (ammeter reading) and the following: pump fluid K1.04 flow, head, speed, and stator temperature 2.8 2.9 3.3 2.7 2.8 3.4 2.7 2.8 3.5 K1.05 Explain the difference between starting current and operating (running) current in a motor 2.8 2.7 2.4 K1.06 Reason for limiting the number of motor starts in a given time period 3.0 3.1 2.4 K1.07 Electrical units: Volts, Amps, VARs, Watts, and Hertz 2.1 2.3 1.3 K1.08 Consequences of overexcited/under excited 2.1 2.3 K1.09 Interrelations of the following: VARs, Watts, Amps, Volts, Power factor 1.9 2.1 4.4 1.3, 4.1 K1.10 Load sharing with parallel generators 2.3 2.4 4.3 K1.11 Motor and generator protective devices * Covered in greater detail in 191008 - Breakers, Relays, and Disconnects for KA's K1.05 and K1.08 2.3 2.4 2.4* © Copyright 2016 Operator Generic Fundamentals