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# Chapter 4 synchronous machine

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### Chapter 4 synchronous machine

1. 1. CHAPTER 4 SYNCHRONOUS MACHINE4.1 INTRODUCTION• Synchronous machine is designed to be operating at synchronous speed, nsync.• The rate rotation of the magnetic fields in the synchronous machine is given as: 120 f e nm = P Where, nm = rate rotation of synchronous machine’s magnetic field, rpm fe = electrical frequency/frequency supply,Hz P = number of poles in the machine.• The synchronous machine can be used to operate as: [a] Synchronous generator [b] Synchronous motor• Used principally in large power applications because of their - high operating efficiency, - reliability and - controllable power factor .• Rotates at constant speed in the steady state.• The rotating air gap field and the rotor rotate at the same speed.• Applications: - Used for pumps in generating stations - Electric clock - Timers - Mills - Refineries - Assist in power factor correction and etc• It can draw either leading or lagging reactive current from the ac supply system.• It is a doubly excited machine: - Rotor poles are excited by a DC current - Stator are connected to the ac supply• The air gap flux is the resultant of the fluxes due to both rotor and stator. 88
2. 2. Figure 4.14.2 SYNCHRONOUS GENERATORConstruction• Synchronous generator is also known as alternator. This machine consists of two main parts: i. Field winding (rotor winding) - winding that produce the main magnetic field in the machine. ii Armature winding (stator winding) - winding where the main voltage is induced.• Main construction: - Rotor - Stator• Stator - has a 3 phase distributed windings(AC supply)– similar to induction machine. - Stator winding is sometimes called the armature winding.• Rotor - has a winding(DC winding) called the field winding. - Field winding is normally fed from an external dc source through slip rings andbrushes. - Rotor can be divided into two groups: 89
3. 3. • High speed machines with cylindrical (or non salient pole) rotors • Low speed machines with salient pole rotors Figure 4.2• Two types of rotor: [a] Salient pole rotor – magnetic pole stick out from the surface of the rotor • Its rotor poles projecting out from the rotor core. • Is use for low-speed hydroelectric generator. • Need large number of poles to accumulate in projecting on a rotor (large diameter but small length). • Almost universal adapt. • Has non uniform air-gap. Figure 5.3: Two poles salient pole rotor [b] Non salient or cylindrical pole rotor – magnetic pole constructed flush with the surface of the rotor. 90
4. 4. • Has its rotor in cylindrical form with dc field winding embedded in the rotor slots. • Has uniform air-gap. • Provide greater mechanical strength. • Per-unit more accurate dynamic balancing. • For use in high speed turbo generator. • 2 / most 4 poles machine use. • Simple to model & analyze. Figure 4.4: Two poles round (cylindrical) rotor• Two type of armature winding: [a] Single layer winding. [b] Double layer winding.Generated voltage• The magnitude of the voltage induced in the given stator phase is given as: E A = 2πN C φf or E A = Kφω where φ = flux in the machine f = frequency of the machine ω = speed rotation of the machine K = Constant representing the construction of the machine• From the equation, it can concluded that (i) EA proportional to flux and speed (ii) Flux proportional to field current, If (iii) Thus EA also proportional to If 91
5. 5. The internal generated voltage EA Vs field current If plot is shown below: EA I Figure 4.5: Magnetization F curveEquivalent Circuit of a Synchronous Generator IF RF IA jXS RA + + + VF LF EA VΦ V (dc) - - - Per-phase equivalent circuit of synchronous generator E A∠δ = Vφ ∠00 + I A∠ ± θ 0 ( RA + jX S )EA = internal generated or generated emf per phaseIA = armature currentVθ = per phase terminal voltageXs = synchronous reactance 92
6. 6. Zs Ra Xs Direction Ia out from Ia the generator because generator - EG 1.0k Vt 1.0m load supply power to the load +Phasor Diagram of The Synchronous GeneratorThe load for synchronous machine may be of three types: (i) Pure resistive load (unity power factor) (ii) Inductive load (lagging power factor) (iii) Capacitive load (leading power factor)(i) Synchronous generator with pure resistive load (unity power factor) : Ra Xs I - 1.0k EG 1.0k 1.0m load + EA jXSIA Unity pf IA Vφ IARAThe equation: E A = Vφ ∠0 0 + I A ∠0 0 ( R A + jX S ) 93
7. 7. (ii) Synchronous generator with inductive load (lagging power factor): Ra Xs 1.0k 1.0m I - EG 1.0k 1.0m load + EA Lagging pf θ jXSIA Vφ IA IARAThe equation: E A = Vφ ∠0 0 + I A ∠ − θ ( R A + jX S )(iii) Synchronous generator with capacitive load (leading power factor): Ra Xs 1.0k 1.0u I- EG load 1.0k 1.0m + EA jXSIA IA Leading pf IARA VφThe equation: E A = Vφ ∠0 0 + I A ∠ + θ ( R A + jX S )NOTE:AT NO LOAD CONDITION, IA = 0 A, THUS EA = Vφ∟00 VOLT.Example 1 94
8. 8. A 3-phase Y-connected synchronous generator supplies a load of 10MW at power factor0.85 lagging and the terminal voltage is 11kV. The armature resistance is 0.1 ohm/phaseand synchronous reactance of 0.66 ohm/phase. Calculate the armature current, theinternal generated voltage and voltage regulation. Draw the phasor diagram.Solution Ia 0.1Ω 1, 0k 0 Vt j0.66Ω load - Eg + Power factor = 0.85 (lagging) θ = cos-1 PF = cos-1 0.85 = 31.790 Pout = 10MW = √3 VLIL cos θ IL = IA = Pout / (√3 VL cos θ) = 10M / (√3 x 11k x 0.85) = 617.5 A EA = Vφ∟00 + (IA∟-θ)(RA + jXS) = (11k/√3 )∟00 + (IA∟-31.790) (0.1 + j0.66) = 6624.07∟2.720 V EA =6624.07V Lagging pf θ 2.72º jXSIA 31.79º Vφ IA IR A APower flow diagram and Torque in Synchronous Generator 95
9. 9. Pin=Input torque x gen. speed in r/sec Pin=Pout + Total Losses Pout=√ 3VTILcosθ Pin=τ appω m 2 Cop.Loss=3(IA)2RA I R losses Core (copper losses) Friction Stray losses and losses windage losses Can be calculated Usually given in when RA is given If not given in the the question question, Pstray = 0 and IA is known If the question don’t mentioned about RA, the copper losses = 0 Synch Generator Pin = Pout + P iron loss + P copper loss = Tmwr P copper loss(3Ia2Ra)P in Pm Pout P iron loss, Pμ (stray+friction+windage+core+etc)For delta connection, 96
10. 10. Vφ = VLIA = IL / √3= Pout / (3 Vφ cosθ)= S / 3 VφFor Y connection,Vφ = VL / √3IA = IL= Pout / (√3 VL cosθ) or Pout / (3 Vφ cosθ)= S / √3 VL or S / 3 Vφ● Real output power can be determined using the following equation: Pout = 3VT I L cos θ OR P =3Vφ I A cos θ out● Since XS >> RA, then RA can be ignored. The new phasor diagram will be: EA c θ EAsinδ = jXSIA δ XSIAcosθ Vφ O γ θ a b IA● From the pahsor diagram, it can be seen that IA cosθ can be represented as: E A sin δ I A cos θ = XS● Insert IA cosθ in the real output power equation will give: 3Vφ E A sin δ P= XS 97
11. 11. ● At maximum condition, δ = 900, thus the equation will be: 3Vφ E A Pmax = XS● Other equation for induced torque for synchronous generator: 3Vφ E A sin δ τ ind = ωm X S4.2.6 Efficiency and Voltage Regulation● The voltage regulation of synchronous generator is given as: E A − VPHASE VR = × 100% VPHASE● The efficiency of the synchronous generator is given as: Pout η= × 100% Pin Pout η= × 100% Pout + PlossesExample 2A 480V, 60Hz, delta-connected, 4 pole synchronous generator has the OCC shown infigure 1. This generator has a synchronous reactance of 0.1 ohm per phase, and armatureresistance of 0.015 ohm per phase. At full-load, the machine supplies 1200 A at 0.8 PFlagging. Under full load conditions, the friction and windage losses are 40kW and thecore losses are 30 kWi) What is the speed rotation of the magnetic field in rpm?ii) How much is the field current must be supplied to the generator to make the terminalvoltage 480V at no load?iii) If the generator is now connected to the load and the load draws 1200 A at 0.8 PFlagging, how much field current will be required to keep the terminal voltage to 480V?What is the voltage regulation of this generator?iv) How much power is now generator is supplying? How much power is supplied to thegenerator by the prime mover? What is the generator’s efficiency? 98
12. 12. v) If the generator is now connected to a load drawing 1200 A at 0.8 PF leading, howmuch field current will be required to keep VT = 480 V and what is generator’s voltageregulation? Figure 1SolutionFor delta connection:Vφ = VLIA = IL / √3i) The speed rotation of magnetic field: nm = (120fe )/ P = (120 x 60) / 4 = 1800 rpmii) If at no load condition.At no load, IA = 0 A. Thus, 99
13. 13. EA = Vφ∟00 Volt. = 480 ∟00 V.Refer to OCC, when EA = 480 V, If = 4.5 A.iii) If at load (0.8 power factor lagging). IL = 1200 A. PF= cos θ = 0.8 (lagging) θ = cos-1 0.8 = 36.870 IA = IL / √3 = 1200 / √3 = 692.82 A. Vφ = VL = 480 VFor lagging load, the internal generated voltage is given as: EA = Vφ∟00 + (IA∟-θ) (RA + jXS) = 480∟00 + (692.82∟-36.870) (0.015 + j0.1) = 529.88 + j49.19 = 532.16 ∟5.300 VFrom OCC, when EA = 532.16 V, If = 5.7 A.The voltage regulation, VR: VR = [ (EA - Vφ) / Vφ ]x 100% = [(532.16 – 480) / 480] x 100% = 10.83%iv) The output power: PF= cos θ = 0.8 (lagging) Pout = √3 VL IL cosθ = √3 (480) (1200) (0.8) = 798.129kWThe input power: PIN = Pout + Pstray + Pf&w + Pcore + Pelect 100
14. 14. Pelect = 3 IA2RA = 3 (692.82)2(0.015) = 21.6kW PIN = 798.129k + 0 + 40k + 30k + 21.6k = 889.729kWThe generator’s efficiency: η = (Pout / PIN ) x 100% = (798.129k / 889.729k) x 100% = 89.7%v) If at load 0.8 leading. IL = 1200 A IA = IL/√3 = 1200 / √3 = 692.82 A. PF = cos θ = 0.8 (leading) θ = cos-1 0.8 = 36.870 EA = Vφ∟00 + (IA ∟+θ)(RA + jXS) = 480∟00 + (692.82∟+36.870)(0.015 + j0.1) = 446.74 + j61.66 = 450.98∟7.860Refer to OCC, when EA = 450.98 V, If = 4AThe generator’s voltage regulation: VR = [(EA - Vφ) / Vφ] x 100% = [(450.98 – 480) / 480] x 100% = 6.06 %Measuring Synchronous Generator Model Parameters 101
15. 15. ● 3 quantities that to be determined,  Relationship between field current, If and EA  Synchronous reactance, XS  Armature resistance, RA● 2 tests to be conducted,  open circuit test – terminal of generator is open-circuited, generator run at rated speed, If is gradually increased in steps, and terminal voltage is measured. Produced open-circuit characteristic (OCC) graph. Air-gap line VT (V) Open-circuit characteristic (OCC) If (A)  short circuit test – terminal of generator is short-circuited through an ammeter, IA or IL is measured as If is increased. Produced short-circuit characteristic (SCC) graph. IA (A) Short-circuit characteristic (SCC) If (A)● From the short circuit test, the armature current, IA is given as: IA = (EA/ (RA + jXS)● The magnitude of IA is given by: |IA| = |EA| / √(RA2 + jXs2)● The internal machine impedance is given as: ZS = √(RA2 + jXS 2 ) = EA / IA 102