ENERGY CONVERSION ONE (Course 25741) CHAPTER SIX ……SYNCHRONOUS MOTORS

Synchronous Motors …Steady-state Operation for field current less than value related to I A,min armature current is lagging, consuming reactive power, while when field current is greater than value related to I A,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

…Steady-state Operation Under-excitation & Over-excitation If E A cosδ; projection of E A onto V φ ; < V φ syn. motor has a lagging current & consumes Q & since I f is small, is said to be: underexcited If E A cosδ; projection of E A onto V φ ; > V φ it has a leading current & supply Q (since I f is large, motor named overexcited

Syn Motor…Steady-state Operation 2 nd 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. I f at these conditions is 4.0 A (a) sketch initial phasor diagram of this motor, & find I A & E A (b) motor’s flux increased by 25%, sketch new phasor diagram of motor. what are E A, I A & PF ? (c) assume flux in motor varies linearly with I f, make a plot of I A versus I f for a 15 hp load

Syn Motor…Steady-state Operation ….Example Solution: (a)P in =13.69 kW, I A =P in /[3V φ cosθ]=13.69/[3x208x0.85]=25.8 A θ=arc cos 0.85=31.8◦ A I A = 25.8 /_-31.8◦ A E A =V φ -j X S I A =208 – (j2.5)(25.8/_-31.8◦)= = 208 – 64.5/_58.2◦ = 182 /_-17.5 V Related phasor diagram shown next

Syn Motor…Steady-state Operation ….Example (b) if flux φ increased by 25%, E A =Kφω will increase by 25% too: E A2 =1.25 E A1 =1.25(182)=227.5 V since E A sinδ 1 is proportional to Power, it remains constant when varying φ to a new level, so: E A sinδ 1 =E A2 sinδ 2

Syn Motor…Steady-state Operation ….Example δ 2 =arcsin(E A1 /E A2 sinδ 1 ) = arcsin[182/227.5 sin(-17.5)]=-13.9◦ armature current : I A2 =[V φ -E A2 ]/ (jX S ) = [ /_-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 I f, E A also vary linearly with I f, since E A =182 for I f =4.0A  E A2 /182=I f2 /4.0 A or E A2 =45.5 I f2

Syn Motor…Steady-state Operation ….Example & PF Correction Torque angle δ for a given I f found as follows: E A sinδ 1 =E A2 sinδ 2  δ 2 =arcsin(E A1 /E A2 sinδ 1 ) These two present the phasor voltage of E A & then new armature current determined: I A2 = [V φ -E A2 ]/(jX S ) Using a MATLAB M-file I A determined versus I f 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

Syn Motor…Steady-state Operation 3 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

Syn Motor…Steady-state Operation 3 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: P LL =3I L ^2 R L line loss Where LL stands for line losses. How do transmission losses compare in the two cases?

Syn Motor…Steady-state Operation 3 rd Example-Solution (a) Real power of load is 100 kW, & its Q is: Q 1 =P 1 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 Q 2 =P 2 tanθ =200 tan (acos 0.8) =200xtan36.87=150 kVAr Real power of load 3 is 150 kW, & Q of load 3 is: Q 3 =P 3 tanθ =150 tan(acos0.85)=150xtan 31.8 =93 kVAr Thus total real load is: P tot =P 1 +P 2 +P 3 = =450 kW Q tot =Q 1 +Q 2 +Q 3 = =323.2 kVAr

Syn Motor…Steady-state Operation 3 rd Example-Solution PF=cosθ=cos(atanQ/P)=cos(atan323.2/450)= cos35.7=0.812 lagging I L =P tot /[√3 V L cosθ]=450/[√3x480x0.812]=667A (b) only Q 3 changed in sign: Q 3 =P 3 tanθ =150 x tan(-acos0.85)=150xtan(-31.8)=-93 kVAr P tot = =450 kW Q tot = =137.2 kVAr PF=cos(atanQ/P)= cos(atan 137.2/450) =cos(16.96)=0.957 lagging I L =P tot /[√3 V L cosθ]=450/[√3x480x0.957]=566 A

Syn Motor…Steady-state Operation 3 rd Example-Solution (c) transmission losses in (a): P LL =3 I L ^2 R L =3x667^2 R L = R L in (b); P LL =3x556^2 R L = R L 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:

Syn Motor…Steady-state Operation 3 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 I f & φ & heat

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 V φ =E A +j X S I A since no power being drawn from motor  E A sinδ & I A cosθ are zero

Syn. capacitor or Syn. Condenser jX S I A points to left  I A points straight up If V φ and I A 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

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, I A =0

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 B R is stationary Stator magnetic field B S is starting to sweep around at syn. speed At t=0 B R and B S are exactly lined up. T ind = k B R x B S would be zero

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

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

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