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ENERGY CONVERSION ONE (Course 25741)

ENERGY CONVERSION ONE (Course 25741). CHAPTER SIX ……SYNCHRONOUS MOTORS . Synchronous Motors …Steady-state Operation.

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ENERGY CONVERSION ONE (Course 25741)

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  1. ENERGY CONVERSION ONE(Course25741) CHAPTER SIX ……SYNCHRONOUS MOTORS

  2. Synchronous Motors …Steady-state Operation • for field current less than value related to IA,minarmature current is lagging, consuming reactive power, while when field current is greater than value related to IA,minarmature 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

  3. …Steady-state OperationUnder-excitation & Over-excitation • If EA cosδ; projection of EA onto Vφ; < Vφ syn. motor has a lagging current & consumes Q& since If is small, is said to be:underexcited • If EA cosδ; projection of EA onto Vφ; > Vφit has a leading current & supply Q (since If is large, motor named overexcited

  4. Syn Motor…Steady-state Operation2nd Example • The 208, 45 kVA, 0.8 PF leading, Δ 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) motor’s 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

  5. Syn Motor…Steady-state Operation….Example • Solution: • Pin=13.69 kW, IA=Pin/[3Vφ cosθ]=13.69/[3x208x0.85]=25.8 A θ=arc cos 0.85=31.8◦ A IA = 25.8 /_-31.8◦ A EA=Vφ-j XS IA =208 – (j2.5)(25.8/_-31.8◦)= = 208 – 64.5/_58.2◦ = 182 /_-17.5 V Related phasor diagram shown next

  6. Syn Motor…Steady-state Operation….Example • (b) if flux φ increased by 25%, EA=Kφω will increase by 25% too: EA2=1.25 EA1 =1.25(182)=227.5 V since EA sinδ1 is proportional to Power, it remains constant when varying φ to a new level, so: EA sinδ1 =EA2 sinδ2

  7. Syn Motor…Steady-state Operation….Example • δ2=arcsin(EA1/EA2 sinδ1) = arcsin[182/227.5 sin(-17.5)]=-13.9◦ • armature current : IA2=[Vφ-EA2]/ (jXS) = [208-227.5/_-13.9]/[j2.5]= 56.2/_103.2/(j2.5) =22.5 /_13.2 A motor PF is : PF=cos(13.2)=0.974 leading (c) assuming flux vary linearly with If, EA also vary linearly with If, since EA=182 for If=4.0A EA2/182=If2 /4.0 A or EA2=45.5 If2

  8. Syn Motor…Steady-state Operation….Example & PF Correction • Torque angle δ for a given If found as follows: EA sinδ1 =EA2 sinδ2  δ2 =arcsin(EA1/EA2 sinδ1) These two present the phasor voltage of EA & then new armature current determined: IA2= [Vφ-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

  9. Syn Motor…Steady-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

  10. Syn Motor…Steady-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: PLL=3IL^2 RL line loss Where LL stands for line losses. How do transmission losses compare in the two cases?

  11. Syn Motor…Steady-state Operation3 rd Example-Solution (a) Real power of load is 100 kW, & its Q is: Q1=P1 tanθ =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 Q2=P2 tanθ =200 tan (acos 0.8) =200xtan36.87=150 kVAr • Real power of load 3 is 150 kW, & Q of load 3 is: Q3=P3 tanθ =150 tan(acos0.85)=150xtan 31.8 =93 kVAr • Thus total real load is: Ptot=P1+P2+P3=100+200+150=450 kW • Qtot=Q1+Q2+Q3=80.2+150+93=323.2 kVAr

  12. Syn Motor…Steady-state Operation3 rd Example-Solution • PF=cosθ=cos(atanQ/P)=cos(atan323.2/450)= cos35.7=0.812 lagging IL=Ptot/[√3 VL cosθ]=450/[√3x480x0.812]=667A (b) only Q3 changed in sign: Q3=P3 tanθ =150 x tan(-acos0.85)=150xtan(-31.8)=-93 kVAr Ptot=100+200+150=450 kW Qtot=80.2+150-93=137.2 kVAr PF=cos(atanQ/P)= cos(atan 137.2/450) =cos(16.96)=0.957 lagging IL=Ptot/[√3 VL cosθ]=450/[√3x480x0.957]=566 A

  13. Syn Motor…Steady-state Operation3 rd Example-Solution • (c) transmission losses in (a): PLL=3 IL^2 RL=3x667^2 RL=1344700 RL in (b); PLL=3x556^2 RL=961070 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:

  14. Syn Motor…Steady-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 motor’s 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 & φ & heat

  15. 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 • Vφ=EA+j XS IAsince no power being drawn from motor EA sinδ & IA cosθ are zero

  16. Syn. capacitor or Syn. Condenser • jXSIA points to left  IA points straight up • If Vφ 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

  17. 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, IA=0

  18. 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 t=0 BR and BS are exactly lined up. • Tind= k BR x BS would be zero

  19. STRATING SYNCHRONOUS MOTORS • Torque alternates rapidly in magnitude & direction,  net starting torque is zero

  20. 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

  21. 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

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