Eet3082 binod kumar sahu lecture_38
- 1. Electrical Machines-II 6th Semester, EE and EEE By Dr. Binod Kumar Sahu Associate Professor, Electrical Engg. Siksha ‘O’ Anusandhan, Deemed to be University, Bhubaneswar, Odisha, India Lecture-38
- 2. 2 Learning Outcomes: - (Previous Lecture_37) To solve numerical on power equation of a Salient Pole Synchronous Motor. To analyse the effect of varying excitation on a synchronous motor. To analyse the effect of varying excitation on a synchronous motor.
- 3. 3 Learning Outcomes: - (Today’s Lecture_38) To analyse the V-Curves and inverted-V curves. To analyse the effect of varying load torque on a synchronous motor.
- 4. 4 Effect of varying excitation on a Synchronous Motor: - Infinite Bus Synchronous Motor XS V = Constant f = Constant Ia Eb Field Excitation Mechanical Load Te Tm If
- 5. 5 Effect of varying excitation under loaded condition:- It is known to us that, change in excitation changes the back emf ‘Eb’ but cannot change the active power input to the synchronous motor. Active power input can only be altered by changing the mechanical power output of the motor. Expression for active power and reactive power input/phase are: In the above power expression, V, and Xs are constant. So, change in excitation changes the values of Eb, Ia, δ and φ, but maintains the products Ebsinδ and Iacosφ constant, so that active power does not change before and after the change in excitation. This can be easily understood from the phasor diagram. , ( )b a b a s s VE V P sin VI cos Q V E cos VI sin X X
- 6. 6 Eb1 VIa1 jIa1Xs Eb2 jIa2Xs Ia2 jIa3Xs Eb3 Ia3 1δ 2δ 3δ 3φ 0 1φ = 02φ a1 1 a2 2 a3 3 I cosφ = I cosφ = I cosφ b1 1 b2 2 b3 3 E sinδ = E sinδ = E sinδ Phasor Diagram by increasing the excitation from its normal value, i.e. from unity power factor. From the phasor diagram it is clear that, increase in excitation from normal value (corresponding to unity power factor), a. Increases the back emf, ‘Eb’. b. Decreases the load angle, ‘δ’. c. Increases the armature current, ‘Ia’. d. Increases the power factor angle, ‘φ’, i.e. decreases the power factor. e. But the products ‘Ebsinδ’ and ‘Iacosφ’ remains constant as change in excitation cannot change the active power drawn by the motor. f. Increases ‘Ebcosδ’, and makes the motor to supply reactive power i.e. to operate at leading pf. , ( ) b a s b a s VE P sin VI cos X V Q V E cos VI sin X
- 7. 7 Eb1 VIa1 jIa1Xs Eb2 jIa2Xs Ia2 jIa3Xs Eb3 Ia3 1δ 2δ 3δ3φ 0 1φ = 0 2φ a1 1 a2 2 a3 3 I cosφ = I cosφ = I cosφ b1 1 b2 2 b3 3 E sinδ = E sinδ = E sinδ Phasor Diagram by decreasing the excitation from its normal value, i.e. from unity power factor. From the phasor diagram it is clear that, decrease in excitation from normal value (corresponding to unity power factor), a. Decreases the back emf, ‘Eb’. b. Increases the load angle, ‘δ’. c. Increases the armature current, ‘Ia’. d. Increases the power factor angle, ‘φ’, i.e. decreases the power factor. e. But the products ‘Eb x sinδ’ and ‘Ia x cosφ’ remains constant as change in excitation cannot change the active power drawn by the motor. f. Makes the motor to draw reactive power, i.e. to operate at lagging pf. , ( ) b a s b a s VE P sin VI cos X V Q V E cos VI sin X
- 8. 8Plotted by taking V=1 pu, Xs=0.2 pu, P = 1 pu and varying the excitation from 0.3 pu to 2 pu
- 9. 9Plotted by taking V=1 pu, Xs=0.2 pu, P = 1 pu and varying the excitation from 0.3 pu to 2 pu V-Curves
- 10. 10Plotted by taking V=1 pu, Xs=0.2 pu, P = 1 pu and varying the excitation from 0.3 pu to 2 pu
- 11. 11Plotted by taking V=1 pu, Xs=0.2 pu, P = 1 pu and varying the excitation from 0.3 pu to 2 pu
- 12. 12 Plotted by taking V=1 pu, Xs=0.2 pu, P = 1 pu and varying the excitation from 0.3 pu to 2 pu
- 13. 13 Plotted by taking V=1 pu, Xs=0.2 pu, P = 1 pu and varying the excitation from 0.3 pu to 2 pu
- 14. 14 Effect of varying mechanical load on a Synchronous Motor: - Infinite Bus Synchronous Motor XS V = Constant f = Constant Ia Eb Field Excitation Mechanical Load Te Tm If
- 15. 15 Initially, assume that |Eb| = |V| and are in phase opposition in the local circuit formed by interconnection of synchronous motor and infinite bus (i.e. load angle δ = 00 at no load). Expression for active and reactive power input/phase by the alternator are ( ) , , 0 0, ( ) 0, . i i s s i i s EV V P sin and Q V Ecos X X V So At noload P and Q V E as E V X S1 N2 Axis of Rotor Field Axis of Stator Field Direction of Movement bE V bE V f
- 16. 16 Voltage equation of the synchronous motor is: As δ = 0, and |E| = |V| => Ia = 0. So, the active and reactive power received by the synchronous motor: So at no load, no power is delivered or, received from the infinite bus. Therefore the synchronous motor is said to be in floating condition. In synchronous motor, increase in mechanical load momentarily (for a small duration) decreases the rotor speed there by making the rotor poles to fall slightly behind the stator poles. 0 0i a i aP VI cos and Q VI sin b a sV E j I X The angular displacement between stator and rotor poles (by torque angle or load angle, ‘δ’) causes the phase of back emf ‘Eb’ to change with respect to supply voltage ‘V’. r a s E jIX aI V bE bE V
- 17. 17 Increase in load angle increases the armature current drawn by the motor from the 3- phase supply and is given by: So, increase in load angle causes the armature current to increase and the increased electrical power input further accelerates the rotor and makes it to rotate at synchronous speed but behind the stator pole by the angle ‘δ’, (called load angle). b a s V E I Z & | | .bsV Z areconstants and E is alsoconstant if theexcitationisunaltered
- 18. 18 Phasor Diagram by increasing the mechanical load from normal excitation, i.e. from unity power factor From the phasor diagram it is clear that, increase in mechanical load from normal excitation (corresponding to unity power factor), a. Keeps the back emf ‘Eb’ constant. b. Increases the load angle, ‘δ’. c. Increases the armature current, ‘Ia’. d. Increases the power factor angle, ‘φ’, i.e. decreases the power factor. e. Operates at lagging power factor. f. Increases the active electrical power drawn from the supply as ‘sinδ’increases. g. Decreases the reactive power drawn by the motor as ‘Ebcosδ’ decreases. sin cos ( cos ) sin b i a s i b a s VE P VI X V Q V E VI X Eb1 VIa1 jIa1XsjIa2XsjIa3Xs Ia2 1δ 2δ 3δ 4φ 0 1φ = 0 2φ Ia3 Eb2 Eb3 Ia4 3φ Eb4
- 19. 19 Eb1 V Ia1 jIa1Xs jIa2Xs jIa3Xs 1δ 2δ 3δ 0 1φ = 02φ Eb2 Eb3 3φ Ia2 Ia3 Phasor Diagram by increasing the mechanical load from normal excitation, i.e. from unity power factor sin cos ( cos ) sin b i a s i b a s VE P VI X V Q V E VI X
- 20. 20 Phasor Diagram by increasing the mechanical load from normal excitation, i.e. from unity power factor From the phasor diagram it is clear that, increase in mechanical load from normal excitation (corresponding to unity power factor), a. Keeps the back emf ‘Eb’ constant. b. Decreases the load angle, ‘δ’. c. Decreases the armature current, ‘Ia’. d. Increases the power factor angle, ‘φ’, i.e. decreases the power factor. e. Operates at leading power factor. f. Decreases the active electrical power drawn from the supply as ‘sinδ’ decreases. g. Increases the reactive power supplied by the motor as ‘Ebcosδ’ increases. sin cos ( cos ) sin b i a s i b a s VE P VI X V Q V E VI X
- 21. 21 Plotted by taking V=1 pu, Xs=0.5 pu, Eb = 1.2 pu and varying the load angle from 50 pu to 900.
- 22. 22 Plotted by taking V=1 pu, Xs=0.5 pu, Eb = 1.2 pu and varying the load angle from 50 pu to 900.
- 23. 23 Plotted by taking V=1 pu, Xs=0.5 pu, Eb = 1.2 pu and varying the load angle from 50 pu to 900.
- 24. 24 Plotted by taking V=1 pu, Xs=0.5 pu, Eb = 1.2 pu and varying the load angle from 50 pu to 900.
- 25. 25 Plotted by taking V=1 pu, Xs=0.5 pu, Eb = 1.2 pu and varying the load angle from 50 pu to 900.
- 26. 26 Thank you