Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this document? Why not share!

- Synchronous machines by Anu71 12236 views
- DETERMINATION OF VOLTAGE REGULATION... by vishalgohel12195 4381 views
- Lectures synchronous machines (1) by Nakul Surana 3026 views
- Chapter 3 induction machine by mkazree 48325 views
- 14+synchronous+motors by huseyin28 8427 views
- Synchronous motor drive by Guru Moorthi 13907 views

57,111 views

56,720 views

56,720 views

Published on

No Downloads

Total views

57,111

On SlideShare

0

From Embeds

0

Number of Embeds

35

Shares

0

Downloads

2,124

Comments

7

Likes

27

No embeds

No notes for slide

- 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. 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. • 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. • 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. 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. 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. (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. 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. 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. 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. ● 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. 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. 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. 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. ● 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
- 16. ● Since Xs >>RA, the equation reduce to: Xs = Vφoc / IA = EA/IA5.2.8 The Synchronous Generator Operating Alone E’A E’A EA jXSI’A jXSIA δ δ’ EA θ δ δ’ V’φ Vφ IA I’A IA I’A V’φ Vφ Unity PF +P or resistive load added, Vφ Lagging PF and VT ↓ VR = small +ve+Q or inductive load added,Vφ and VT ↓↓VR = large +ve E’A jXSIA Leading PF I’A -Q or capacitive loadIA EA jXSI’A added, Vφ and VT ↑↑ δ δ’ Vφ V’φ VR = -veGeneral conclusions from synchronous generator behavior are – If lagging loads (+Q or inductive load) are added, Vφ and VT decrease significantly but voltage regulation VR is positive large. – If unity power factor loads (no reactive load) are added to a generator, there is a slight decrease in Vφ and VT and VR is positive small – If leading loads (-Q or capacitive power loads) are added, Vφ and VT will rise and VR is negative.5.3 SYNCHRONOUS MOTOR 103
- 17. ● Synchronous motor converts electrical power to mechanical power.● The equivalent circuit of synchronous motor: IF RF IA jXS RA + + + VF LF EA V - VΦ (dc) - -● The difference between synchronous generator equivalent circuits and the synchronous generator equivalent circuit is the direction of IA. Vφ ∠0 0 = E A ∠ + I A ∠±θ ( R A + jX S ) δThe phasor diagram of synchronous motor with various power factor are shown below:i) Unity Power Factor E A = Vφ ∠0 0 − I A ∠0 0 ( R A + jX S )ii) Lagging Power Factor 104
- 18. E A = Vφ ∠0 0 − I A ∠ − θ 0 ( R A + jX S )iii) Leading Power Factor E A = Vφ ∠0 0 − I A ∠ + θ 0 ( R A + jX S )Power flow for synch motor is reverse from synch motor Synch Motor Pin = Pout + P iron loss + P copper loss = Tmwr P iron loss, Pμ (stray+friction+windage+core+etc)P in Pm Pout P copper loss(3Ia2Ra)Efficiency and Voltage RegulationThe voltage regulation of synchronous motor is given as: VPHASE − E A VR = ×100% EA 105
- 19. The efficiency of the synchronous motor is given as: Pout η= × 100% Pin Pout η= × 100% Pout + PlossesExample 3A three phase, Y connected synchronous generator is rated 120 kVA, 1.5 kV, 50 Hz, 0.75pf lagging. Its synchronous inductance is 2.0mH and effective resistance is 2.5 Ω.(i) Determine the voltage regulation at this frequency.(ii) Determine the rated voltage and apparent power if the supply frequency is going to be twice.(iii) Determine the voltage regulation if the frequency is increased to 120% of the original frequency.Solution(i) 2.0 mH 2.5 ohm + DC M1 Ea 1,0m 1,0k 120kVA 1.5kV - 50HZZ S = RS + jX S X S = 2πfLS = 2π (50)(2m) = 0.628ΩTherefore 106
- 20. S = 3V L I L SZ S = 2.5 + j 0.628 IL = 3VL= 2.578∠14.1° 120k = 3 x1.5k cos φ = 0.75 = 46.188∠ − φ φ = 41.41°I L = 46.188∠ − 41.41°E g = V∠0° + I a Z S 1.5k= ∠0° + 46.188∠ − 41.41°(2.578∠14.1°) 3= 971.825 − j 54.63= 973.36∠ − 3.22 0 A E g − VT 973.36 − 866.025VR = = = 12.39% VT 866.025(ii) f = 100 H zRated Voltage:50 H Z → 1.5k 100100 H Z → 1.5k ⇒ x1.5k = 3kV 50Apparent Power50 H Z → 120kVA100 H Z → 240kVA(iii) f = 120%of 50 H Zf = 60 H Z New X S = 2πfLS = 2π (60)(2m) = 0.754Z S = 2.5 + j 0.754 = 2.61∠16.78°E g = VT = I a Z sVT = ? at 60Hz 107
- 21. Rated Voltage:50 H Z → 1.5k 6060 H Z → 1.8k ⇒ x1.5k = 1.8kV 50Apparent Power50 H Z → 120kVA 144k IL = 60 3 x1.8k60 H Z → x120 = 144kVA 50 = 46.188 AE g = VT + I a Z s 1.8k = + ( 46.188∠ − 41.41°)(2.61∠16.78°) 3 = 1148.81 − j 50.24V = 1149.91∠ − 2.5 0 V 1.8k 1149.91 − E g − VT 3VR = = = 10.65% VT 1.8k 3Example 4A 2300V, 120hp, 50Hz, eight poles, Y-connected synchronous motor has a synchronousinductance of 6.63mH/phase and armature resistance of 1Ω/phase at rated power factor of0.85 leading. At full load, the efficiency is 90 percent. Find the following quantities forthis machine when it is operating at full load.(i) Draw a phasor diagram to represent back emf, supply voltage and armature current.(ii) What is its voltage regulation?(iii) Output power.(iv) Input power.(v) Developed mechanical power.(vi). Draw the power flow diagram. 108
- 22. Solution1hp = 746W2300V, 120hp=89.52kW, 50Hz, 8 poles, efficiency = 90%Y connected(IΦ=IL, VΦ=VL/√3) Xs = 2πfL=2π(50)(6.63m)=2.08Ω Z = Ra+jXs = 1 + j2.08 Ω(i) Pout Pout 89.52kW η= , Pin = = = 99.47kW Pin η 0.9 Pin = 3VφIφ cos θ Pin 99.47 k Iφ = = = 29.38∠31.79° A 3Vφ cos θ 3(2300 / 3 )(0.85) Iφ = 24.97 + j15.48 A 2300 E = V − IZ = − (24.97 + j15.48)(1 + j 2.08) = 1335.13 − j 67.42V = 1336.83∠ − 2.89°V 3 θ = 31.79° δ = −2.89° Vt − E 2300 / 3 − 1336.83(ii) VR = = = −0.0067 x100% = −0.67% E 1336.83(iii) Pout = 120hp = 89.52kW Pout 89.52kW(iv) Pin = = = 99.47kW η 0 .9(v) Pmech = Pin-Pcl = 99.47kW - 3Ia2R = 99.47kW - 3(29.38)2(1) = 96.88kW(vi) P iron loss, Pμ (stray+friction+windage+core+etc)P in Pm Pout P copper loss 109
- 23. Steady-state Synchronous Motor OperationTorque Speed Characteristic Curve Maximum torque τind when δ=90° τpullout nnl − n fl SR = X 100% = 0% n fl τrated nsync nm1) Speed of the motor will be constant (because it’s locked to the electrical frequency). The result torque speed characteristic is shown in the figure above.2) The speed is constant from no load torque until max torque, thus SR=0%.Effect Load Changes on Synchronous Motor 3Vφ E A sin δ P = 3Vφ I A cos θ = XS 110
- 24. IA1 I A2 IA3 Vφ IA4 ∝P1 ∝P2 EA1 ∝P3 ∝P4 EA2 EA3 EA4When the load is increased: 1) θ is change from leading to lagging. 2) jXSIA is increased, thus IA is increased too. 3) Torque angle, δ, is increased 4) |EA| is constantEffect Field Current Changes on Synchronous Generator ∝P (=constant) IA4 IA3 IA2 Vφ IA1 ∝P (=constant) EA1 EA2 EA3 EA4When the field current is increased: 111
- 25. I V V δI δ E E 1) |EA| increased 2) Torque angle, δ, is decreased 3) IA first is decreased and then increased. 4) θ changed from lagging to leading. 5) Real power supply is constant 6) VL is constant. Synchronous V Curve IA P = P2 P = P1 Lagging power Leading factor power factor PF = 1.0 IF Starting Synchronous Motor 3 methods to start a synchronous motor: i) Reduce electrical frequency, fe ii) Use external prime mover iii) Use damper windings or amortisseur windings. 4.4 RELATIONSHIP BETWEEN SYNCHRONOUS MOTOR AND SYNCHRONOUS GENERATOR P Q Supply Q EAcosδ > Vφ Consume Q EAcosδ < Vφ Supply P E E Generator I δ V δ V EA leads Vφ I Consume P I V V Motor δ δ 112 EA lags Vφ E I E
- 26. 4.5 INDUSTRIAL APPLICATIONThe three phase synchronous motor is used when a prime mover having a constant speedfrom a no-load condition to full load is required. Such as fans, air compressors andpumps.Also used to drive mechanical load and also to correct the power factor.Only used as a correct power factor of an industrial power system, as a bank capacitorused for power factor correction, also called a synchronous capacitor.Rating up to 10hp are usually started directly across the rated three-phase voltage.Synchronous motor of larger sizes are started through a starting compensator or anautomatic starter.Tutorial 41. A 90kVar, 445 V, 60 Hz 3-phase synchronous motor is delta connected. The motor has a synchronous inductance 1.3 mH/phase and armature resistance of 4 ohm/phase. Calculate the back emf and power angle if the motor is operating at 0.85 lagging power factor.2. A 2400 V, 60 kW, 50 Hz, 6 poles, delta-connected synchronous motor has a synchronous reactance of 4 Ω/phase and armature resistance of 2 Ω/phase. At full load, the efficiency is 92 %. Find the followings for this machine when it is operating at full load at rated power factor 0.85 lagging. (i) Phasor diagram to represent back emf, supply voltage and armature current. (ii) Voltage regulation. 113
- 27. (iii) Input power (iv)Developed mechanical power (v) Power flow diagram3. A 3-phase, 50 Hz, Y-connected synchronous generator supplies a load of 10MW. The terminal voltage is 22 kV at power factor 0.85 lagging . The armature resistance is 0.2 ohm/phase and synchronous inductance is 0.3 mH/phase. Calculate the line value of emf generated.4. A hydraulic turbine turning at 200 rpm is connected to a synchronous generator. If the induced voltage has a frequency of 60 Hz, how many poles does the rotor have.5. A 2300 V 3-phase, star connected synchronous motor has an armature resistance of 0.2 ohm/phase and a synchronous reactance of 2.2 ohm/phase. The motor is operating on 0.5 power factor leading with a line current of 200A. Determine the value of generated or counter emf per phase. Also draw the phasor diagram. From the phasor diagram, discuss what happen to the counter emf if the power factor is increased to 0.8 leading. No need to recalculate the new emf being generated.6. A 200kVA, 600 V, 50 Hz three-phase synchronous generator is Y- connected. The generator has a synchronous reactance 0.1 ohm/phase and armature resistance of 2 ohm/phase. Calculate the voltage regulation if the generator is operating at 0.75 leading power factor. Draw the phasor diagram.7. A 2000V, 500hp, 3-phase Y-connected synchronous motor has a resistance of 0.3 ohm/phase and synchronous reactance of 3 ohm/phase respectively. Determine the induced emf per phase if the motor works on full-load with an efficiency of 92% and p.f = 0.8 leading. 114

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

10Q!