Presentation is loading. Please wait.

Presentation is loading. Please wait.

EE216 Electrical Engineering

Similar presentations


Presentation on theme: "EE216 Electrical Engineering"β€” Presentation transcript:

1 EE216 Electrical Engineering
Dr. Unnikrishnan P.C. Professor, EEE

2 Synchronous Generator-Alternator

3 Stator Construction Stator is identical to the induction motor
Laminated low silicon steel rings joined together Slots insulated with Mylar Example of 36 slot stator with 3 coil conductors per slot, 12 slots per phase

4 Stator Construction

5 Rotor Construction Two Types of Rotor Salient Pole Cylindrical

6 Rotor Construction

7 Salient-Pole Rotor with brushless excitation

8 Rotor Construction

9 Operation as a Synchronous Generator

10 Equations 𝑁 𝑠 = 120 𝑓 𝑃 𝑅𝑃𝑀 𝐸 𝑅𝑀𝑆 =2.22  𝑓 𝑧=4.44  𝑓 𝑇 π‘β„Ž Volts
Synchronous Speed 𝑁 𝑠 = 120 𝑓 𝑃 𝑅𝑃𝑀 where f = supply frequency required and P = Poles Induced EMF in an alternator 𝐸 𝑅𝑀𝑆 =2.22  𝑓 𝑧=4.44  𝑓 𝑇 π‘β„Ž Volts where  = Flux per pole set up by rotor current Z = Conductor in series per phase 𝑇 π‘β„Ž =π‘‡π‘’π‘Ÿπ‘›π‘  π‘π‘’π‘Ÿ π‘β„Žπ‘Žπ‘ π‘’= 𝑧 2

11 Synchronous Impedance
E = Induced Emf per phase V = Terminal voltage per phase 𝐼 π‘Ž = Armature current per phase 𝑅 π‘Ž =Armature resistance per phase 𝑋 𝑙 = Armature leakage reactance per phase 𝑋 π‘Ž = Reactance per phase representing Armature reaction Induced emf per phase 𝑬=𝑽+ 𝑰 𝒂 𝑹 𝒂 +j 𝑰 𝒂 𝑿 𝒍 + 𝒋𝑰 𝒂 𝑿 𝒂 𝑿 𝒔 = (𝑿 𝒍 + 𝑿 𝒂 ) is called synchronous reactance 𝒁 𝒔 = (𝑹 𝒂 +j 𝑿 𝒔 ) is called synchronous Impedance 𝑬=𝑽+ 𝑰 𝒂 𝑹 𝒂 +j 𝑰 𝒂 (𝑿 𝒍 + 𝑿 𝒂 ) =𝑽+ 𝑰 𝒂 ( 𝑹 𝒂 +j 𝑿 𝒔 ) =𝑽+ 𝑰 𝒂 𝒁 𝒔

12 Phasor Diagram of an Alternator at Lagging pf load
BE=VSinβˆ… 𝑂𝐷=π‘‰πΆπ‘œπ‘ βˆ… E = (𝑉 π‘π‘œπ‘ βˆ…+ 𝐼 π‘Ž 𝑅 π‘Ž ) 2 + (π‘‰π‘ π‘–π‘›βˆ…+ 𝐼 π‘Ž 𝑋 𝑠 ) βˆ’βˆ’βˆ’βˆ’βˆ’πΉπ‘œπ‘Ÿ πΏπ‘Žπ‘”π‘”π‘–π‘›π‘” 𝑝𝑓 E = (𝑉 π‘π‘œπ‘ βˆ…+ 𝐼 π‘Ž 𝑅 π‘Ž ) 2 + (π‘‰π‘ π‘–π‘›βˆ…βˆ’ 𝐼 π‘Ž 𝑋 𝑠 ) βˆ’βˆ’βˆ’βˆ’βˆ’πΉπ‘œπ‘Ÿ πΏπ‘’π‘Žπ‘‘π‘–π‘›π‘” 𝑝𝑓

13 Voltage Regulation % π‘‰π‘œπ‘™π‘‘π‘Žπ‘”π‘’ π‘…π‘’π‘”π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘›= 𝐸 βˆ’ 𝑉 𝑉 x 100
A convenient way to compare the voltage behaviour of two generators is by their voltage regulation (VR). % π‘‰π‘œπ‘™π‘‘π‘Žπ‘”π‘’ π‘…π‘’π‘”π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘›= 𝐸 βˆ’ 𝑉 𝑉 x 100 Pre-Determination of Voltage Regulation EMF Method or Synchronous Impedance Method MMF Method

14 EMF Method or Synchronous Impedance Method
Circuit diagram for open circuit and short circuit test on alternator

15 EMF Method …… The Armature resistance per phase 𝑅 π‘Ž
Open circuit characteristics (OCC)Β  (𝑉 π‘œπ‘ ) π‘β„Ž 𝑉 𝑠 𝐼 𝑓 Short Circuit Characteristics (SCC) (𝐼 π‘Ž ) 𝑠𝑐 𝑉 𝑠 𝐼 𝑓 The Armature resistance per phase 𝑅 π‘Ž From the Equivalent Circuit: 𝑍 𝑠 = 𝐸 π‘β„Ž 𝐼 π‘Ž 𝑠𝑐 = (𝑉 π‘œπ‘ ) π‘β„Ž ( 𝐼 π‘Ž ) 𝑠𝑐 π‘“π‘œπ‘Ÿ π‘ π‘Žπ‘šπ‘’ 𝐼 𝑓 𝑋 𝑠 = (𝑍 𝑠 ) 2 βˆ’ (𝑅 π‘Ž )  π‘β„Ž Equivalent circuit on short circuit

16 OCC & SCC of an Alternator

17 EMF Method …… No load induced e.m.f. per phase, EphΒ can be determined from the equation 𝐸 π‘β„Ž = (𝑉 π‘π‘œπ‘ βˆ…+ 𝐼 π‘Ž 𝑅 π‘Ž ) 2 + (π‘‰π‘ π‘–π‘›βˆ… ο‚± 𝐼 π‘Ž 𝑋 𝑠 ) 2 whereΒ Β Β Β  VphΒ = Phase value of rated voltage IaΒ = Phase value of current depending on the load condition cosΞ¦ = p.f. of load % π‘…π‘’π‘”π‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘›= 𝐸 π‘β„Ž βˆ’ 𝑉 π‘β„Ž 𝑉 π‘β„Ž x 100

18 EMF Method Advantages & Limitations
Advantage: Β synchronous impedance ZsΒ for any load condition can be calculated. Hence regulation of the alternator at any load condition and load power factor can be determined. Limitation: This method gives large values of synchronous reactance. This leads to high values of % regulation than the actual. Hence this method is calledΒ Pessimistic method.Β 

19 MMF Method or Ampere Turn Method
The effect of armature leakage reactance by an equivalent additional m.m.f so that this m.m.f may be combined with the armature reaction m.m.f. An alternator requires m.m.f. which is product of field current and turns of field winding-two components 1. An m.m.f. necessary to induce the rated terminal voltage on open circuit. 2. An m.m.f. equal and opposite to that of armature reaction m.m.f. The number of turns in the field winding is not known normally, so the m.m.f. is calculated in terms of the field current itself.

20 MMF Method …………. The field m.m.f. required to induce the rated terminal voltage on open circuit can be obtained from o.c.c. This is denoted as 𝐹 𝑂 . In s.c. test, field m.m.f. is necessary to overcome drop across armature resistance and leakage reactance and also to overcome effect of armature reaction But drop across armature resistance and leakage reactance is very small and can be neglected. So in s.c. test, field m.m.f. circulates full load current to balance the armature reaction effect. Ampere-turns required to circulate full load current can be obtained from s.c.c. Denoted asΒ  𝐹 𝐴𝑅 .

21 MMF Method ………….

22 MMF Method …………. At full load, the total field m.m.f. is the vector sum of its two componentsΒ  𝐹 𝑂 Β and 𝐹 𝐴𝑅 denoted by 𝐹 𝑅 This depends on the power factor of the load which alternator is resultant field m.m.f. is denoted as FR. Let us consider the various power factors and the resultant 𝐹 𝑅 .

23 MMF Method …………. Let us consider the various power factors and the resultant 𝐹 𝑅 . Zero lagging p.f. : The armature reaction is completely de-magnetizing. Hence the resultant 𝐹 𝑅 is the algebraic sum of 𝐹 𝑂 and 𝐹 𝐴𝑅

24 MMF Method …………. Zero leading p.f. : The armature reaction is completely magnetizing.Hence the resultant 𝐹 𝑅 is the algebraic difference of 𝐹 𝑂 and 𝐹 𝐴𝑅

25 MMF Method …………. Unity p.f. : The armature reaction is completely cross- magnetizing (Distorting). Resultant 𝐹 𝑅 is the vector sum of 𝐹 𝑂 and 𝐹 𝐴𝑅

26 MMF Method …………. Lagging p.f. :
The component 𝐹 𝑂 Β is at right angles to 𝑉 π‘β„Ž while 𝐹 𝐴𝑅 Β is in phase with the current ( 𝐼 π‘Ž ) π‘β„Ž . 𝐹 𝑅 Β is the vector sum of 𝐹 𝑂 Β and 𝐹 𝐴𝑅 (𝐹 𝑅 ) 2 = 𝐹 𝑂 + 𝐹 𝐴𝑅 π‘ π‘–π‘›βˆ… 𝐹 𝐴𝑅 π‘π‘œπ‘ βˆ… 2

27 MMF Method …………. Leading p.f. :
The component 𝐹 𝑂 Β is at right angles to 𝑉 π‘β„Ž while 𝐹 𝐴𝑅 Β is in phase with the current ( 𝐼 π‘Ž ) π‘β„Ž . 𝐹 𝑅 Β is the vector sum of 𝐹 𝑂 Β and 𝐹 𝐴𝑅 (𝐹 𝑅 ) 2 = 𝐹 𝑂 βˆ’ 𝐹 𝐴𝑅 π‘ π‘–π‘›βˆ… 𝐹 𝐴𝑅 π‘π‘œπ‘ βˆ… 2

28 MMF Method …………. Once 𝐹 𝑅 Β is known, obtain corresponding voltage which is induced e.m.f. 𝐸 π‘β„Ž , required to get rated terminal voltage 𝑉 π‘β„Ž is obtained from o.c.c.

29 Operation as a Synchronous Motor


Download ppt "EE216 Electrical Engineering"

Similar presentations


Ads by Google