ENERGY CONVERSION ONE (Course 25741)

1
ENERGY CONVERSION ONE (Course 25741)

• CHAPTER SIX
• SYNCHRONOUS MOTORS

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Synchronous Motors Steady-state Operation

• for field current less than value related to
  IA,min armature current is lagging, consuming
  reactive power, while when field current is
  greater than value related to IA,min armature
  current is leading, and supplying Q to power
  system
• Therefore, by adjusting the field current,
  reactive power supplied to (or consumed by) power
  system can be controlled

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Steady-state OperationUnder-excitation
Over-excitation

• If EA cosd projection of EA onto Vf lt Vf
• syn. motor has a lagging current consumes Q
  since If is small, is said to be underexcited
• If EA cosd projection of EA onto Vf gt Vf it has
  a leading current supply Q (since If is large,
  motor named overexcited

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Syn MotorSteady-state Operation2nd Example

• The 208, 45 kVA, 0.8 PF leading, ? connected, 60
  Hz syn. motor of last example is supplying a 15
  hp load with an initial PF of 0.85 PF lagging. If
  at these conditions is 4.0 A
• (a) sketch initial phasor diagram of this motor,
  find IA EA
• (b) motors flux increased by 25, sketch new
  phasor diagram of motor. what are EA, IA PF ?
• (c) assume flux in motor varies linearly with
  If, make a plot of IA versus If for a 15 hp load

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Syn MotorSteady-state Operation.Example

• Solution
• Pin13.69 kW,
• IAPin/3Vf cos?13.69/3x208x0.8525.8 A
• ?arc cos 0.8531.8? A
• IA 25.8 /_-31.8? A
• EAVf-j XS IA 208 (j2.5)(25.8/_-31.8?)
• 208 64.5/_58.2? 182 /_-17.5 V
• Related phasor diagram shown next

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Syn MotorSteady-state Operation.Example

• (b) if flux f increased by 25, EAKf? will
  increase
• by 25 too
• EA21.25 EA1 1.25(182)227.5 V
• since EA sind1 is proportional to Power, it
  remains constant when varying f to a new level,
  so
• EA sind1 EA2 sind2

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Syn MotorSteady-state Operation.Example

• d2arcsin(EA1/EA2 sind1)
• arcsin182/227.5 sin(-17.5)-13.9?
• armature current
• IA2Vf-EA2/ (jXS) 208-227.5/_-13.9/j2.5
  
• 56.2/_103.2/(j2.5) 22.5 /_13.2 A
• motor PF is PFcos(13.2)0.974 leading
• (c) assuming flux vary linearly with If, EA also
  vary linearly with If, since EA182 for If4.0A?
• EA2/182If2 /4.0 A or EA245.5 If2

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Syn MotorSteady-state Operation.Example PF
Correction

• Torque angle d for a given If found as follows
• EA sind1 EA2 sind2 ? d2 arcsin(EA1/EA2
  sind1)
• These two present the phasor voltage of EA
  then new armature current determined
• IA2 Vf-EA2/(jXS)
• Using a MATLAB M-file IA determined versus If
  and present the V shape (text book)
• SYNCHRONOUS MOTOR PF CORRECTION
• In next figure an infinite bus supply an
  industrial plant containing several motors (as
  load), through a transmission line
• Two of loads are induction motors with lagging
  PF, and third load is a synchronous motor with
  variable PF
• What does the ability to set PF of one load, do
  for power system? This studied in following
  Example

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Syn MotorSteady-state Operation3 rd Example

• Simple power system, infinite bus has a 480 V
• Load 1, an induction motor consuming 100 kW at
  0.78 PF lagging, load 2 is an induction motor
  consuming 200 kW at 0.8 PF lagging. Load 3 is
  synchronous motor whose real power consumption is
  150 kW

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Syn MotorSteady-state Operation3 rd Example

• (a) syn. Motor adjusted to operate at 0.85 PF
  lagging, what is transmission line current
• (b) if syn. Motor adjusted to operate at 0.85 PF
  lagging, what is transmission line current in
  this system
• (c) assume the transmission line losses are
  given by PLL3IL2 RL line loss
• Where LL stands for line losses. How do
  transmission losses compare in the two cases?

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Syn MotorSteady-state Operation3 rd
Example-Solution

• (a) Real power of load is 100 kW, its Q is
• Q1P1 tan? 100 tan (acos 0.78)
• 100xtan(38.7)80.2 kVAr
• Real power of load 2, is 200 kW, and Q of load 2
  is
• Q2P2 tan? 200 tan (acos 0.8)
• 200xtan36.87150 kVAr
• Real power of load 3 is 150 kW, Q of load 3
  is Q3P3 tan? 150 tan(acos0.85)150xtan 31.8
  93 kVAr
• Thus total real load is PtotP1P2P310020015
  0450 kW
• QtotQ1Q2Q380.215093323.2 kVAr

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Syn MotorSteady-state Operation3 rd
Example-Solution

• PFcos?cos(atanQ/P)cos(atan323.2/450)
• cos35.70.812 lagging
• ILPtot/v3 VL cos?450/v3x480x0.812
  667A
• (b) only Q3 changed in sign
• Q3P3 tan? 150 x tan(-acos0.85)150xtan(-31.8)
  -93 kVAr
• Ptot100200150450 kW
• Qtot80.2150-93137.2 kVAr
• PFcos(atanQ/P) cos(atan 137.2/450)
• cos(16.96)0.957 lagging
• ILPtot/v3 VL cos?450/v3x480x0.957566 A

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Syn MotorSteady-state Operation3 rd
Example-Solution

• (c) transmission losses in (a)
• PLL3 IL2 RL3x6672 RL1344700 RL
• in (b) PLL3x5562 RL961070 RL
• Note in (b) transmission losses reduced by 28
  while power supplied to loads is the same
• The ability to adjust PF of loads in a power
  system affects operating efficiency significantly
• Most loads in typical power system are induction
  motors ? lagging PF
• Having leading loads (overexcited syn. motor) is
  useful due to following reasons

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Syn MotorSteady-state Operation3 rd
Example-Solution

• 1-leading load, supply reactive power Q for
  nearby lagging loads, instead of coming from Gen.
  transmission losses reduced
• 2- since less current pass transmission lines,
  lower rating and investment is required for a
  rated power flow
• 3- requiring a synchronous motor to operate
  with leading PF means it must be overexcited
• this mode increase motors maximum torque
  reduces chance of accidentally pullout torque
• Application of syn. Motor or other equipment to
  improve PF is called power-factor correction
• Many loads that accept constant speed motor are
  driven by syn. Motors due to PF correction
  capability save money in industrial plants
• Any syn. Motor exists in a plant run overexcited
  to improve PF to increase its pullout torque,
  which need high If f heat

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Synchronous Capacitor or Synchronous Condenser

• A syn. Motor can operate overexcited to supply Q,
  some times in past syn. motor employed to run
  without a load simply for power-correction
• This mode of operation shown in below phasor
  diagram
• VfEAj XS IA since no power being drawn from
  motor ? EA sind IA cos? are zero

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Syn. capacitor or Syn. Condenser

• jXS IA points to left ? IA points straight up
• If Vf and IA examined voltage-current
  relationship is similar to a capacitor
• Some syn. Motors are used specifically for PF
  correction can not be connected to any load,
  these are called syn. Condensers or syn.
  Capacitors
• Today, conventional capacitors are more
  economical to be used than syn. Capacitors,
  however some syn. Capacitors may still be used in
  older industrial plants

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Syn. capacitor or Syn. Condenser

• V curve of Synchronous Capacitor corresponding
  phasor Diagram
• Since real power supplied to machine zero (except
  for losses) at unity PF, IA0

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STRATING SYNCHRONOUS MOTORS

• Sofar, in study of syn. motor, it is assumed that
  motor is initially turning at syn. Speed
• However, how did motor get to synchronous speed?
• realizing that for a 60 Hz motor at moment power
  applied to stator, rotor is stationary, and BR is
  stationary
• Stator magnetic field BS is starting to sweep
  around at syn. speed
• At t0 BR and BS are exactly lined up.
• Tind k BR x BS would be zero

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STRATING SYNCHRONOUS MOTORS

• Torque alternates rapidly in magnitude
  direction, ? net starting torque is zero

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STRATING SYNCHRONOUS MOTORS

• The average torque over complete cycle cycle was
  zero
• Motor vibrates heavily with each electrical cycle
  overheats
• This approach to syn. Motor starting is not
  satisfactory, burning up the expensive equipment
• The 3 basic approach to safely start the syn.
  Motor
• 1- Reduce speed of stator magnetic field to low
  enough value that rotor can accelerate lock in
  with it during half-cycle of magnetic field
  rotation, by reducing frequency of applied
  electric power

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STRATING SYNCHRONOUS MOTORS

• 2- use an external prime mover to accelerate
  motor up to syn. Speed, follo the paralleling
  procedure and bring machine on the line as a
  generator, turning off the prime mover will make
  syn. Machine a motor
• 3- use damper windings or amortisseur windings.
  Its function described next