Effect of change in mechanical power

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-39

2. 2
Learning Outcomes: - (Previous Lecture_38)
 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 V-Curves and inverted-V curves.

3. 3
Learning Outcomes: - (Today’s Lecture_39)
 To analyse the effect of varying load torque on a synchronous motor.
 To solve numerical on Synchronous Motor.

4. 4
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

5. 5
 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

6. 6
 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

7. 7
 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
 

8. 8
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.
Eb1
VIa1
jIa1XsjIa2XsjIa3Xs
Ia2
1δ
2δ
3δ
4φ
0
1φ = 0
2φ
Ia3
Eb2
Eb3
Ia4
3φ
Eb4
sin cos
( cos ) sin
b
i a
s
i b a
s
VE
P VI
X
V
Q V E VI
X
 
 
 
  

9. 9
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
 
 
 
  

10. 10
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
 
 
 
  

11. 11
Plotted by taking V=1 pu, Xs=0.5 pu, Eb = 1.2 pu and varying the load angle from 50 pu to 900.

12. 12
Plotted by taking V=1 pu, Xs=0.5 pu, Eb = 1.2 pu and varying the load angle from 50 pu to 900.

13. 13
Plotted by taking V=1 pu, Xs=0.5 pu, Eb = 1.2 pu and varying the load angle from 50 pu to 900.

14. 14
Plotted by taking V=1 pu, Xs=0.5 pu, Eb = 1.2 pu and varying the load angle from 50 pu to 900.

15. 15
Plotted by taking V=1 pu, Xs=0.5 pu, Eb = 1.2 pu and varying the load angle from 50 pu to 900.

16. 16
Numerical on Synchronous Motor
1. A 2000 V, 3-phase, star connected synchronous motor has synchronous impedance of
(0.5 + j5) Ω/phase. For an excitation voltage of 3000 V, the motor takes an input of 900
kW at rated voltage. Compute the line current and power factor.
0
2
3
0
2000
/ , 1154.7 .
3
3000
/ , 1732.05 .
3
, 0.5 5 5.025 84.3
, 900 .
900 10 cos( ) cos( )
49.14
b
s
i
b
s s
Supply voltahe phase V volt
Ecxitationemf phase E volt
Synchronousimpedance Z j
Input power drawnbythemotor P kW
VEV
Z Z
N
  

 
 
   

    
 
0 0
0
0
1154.7 0 1732.05 49.14
, , 260.71 4.77 .
0.5 5
cos(4.77 ) 0.997 .
b
a
s
V E
ow atmaturecurrent I A
Z j
Power factor leading
 
     
   

 

17. 17
2. A 2300 V, 3-phase, star connected synchronous motor operating at normal voltage is
excited to give excitation voltage of 2400 V. Determine the maximum power developed,
armature current and power factor under this excitation. Per phase synchronous
impedance is (1.5 + j21) Ω.
0
2
2300
/ , 1327.91 .
3
2400
/ , 1385.64 .
3
, 1.5 21 5.025 85.91
,
cos( ) cos( )
b
s
b b
g
s s
Supply voltage phase V volt
Ecxitationemf phase E volt
Synchronousimpedance Z j
Power developed ina synchronous motor
VE E
P
Z Z
Condition for m
  
 
 
   
  
0
85.91 .aximum power input is  

18. 18
2
0 0
0
0
, cos( ) 255.69 .
1327.91 0 1385.64 85.91
, 87.85 37.56 .
1.5 21
cos(37.56 ) 0.793 .
b b
g
s s
b
a
s
VE E
Maximum power developed P kW
Z Z
V E
Armaturecurrent I A
Z j
Power factor lagging

 

  
    
    

 

19. 19
3. A 3-phase, 400 V, 6-pole, 50 Hz, star connected synchronous motor has synchronous
impedance of (0.5 + j8) Ω/phase. Its input current is 10 A at unity power factor for a
certain field current. With the field current remaining constant, the load torque is
increased until the motor takes an input current of 35 A. Calculate the developed torque
and the new power factor.
0
1
0
1
1 1
0 0
400
/ , 230.94 .
3
, 0.5 8 8.016 86.42
, 10 , .
, 10 0 .
, ,
230.94 0 10 0
s
a
a
b a s
Supply voltage phase V volt
Synchronousimpedance Z j
Initial armaturecurrent I A and initial power factor isunity
So I A
Now back emf E V I Z

  
 
    

 
 
     0
0
1
(0.5 8) 239.69 19.5 .
, , 19.5 .
j volt
So load angle 
   

Eb1
Ia1
Ia1Ra
jIa1Xs
V
Ia1Zs

O
A
1
B

20. 20
2
2 2 2
2
2
, ,
35 .
, .
,
2 cos( )
cos
a
Now keeping theexcitationconstant mechanical torqueis
increased until thearmaturecurrent becomes I A
So theload angleincreases and power factor becomeslagging
Fromthe leOAB
AB OA OB OA OB 
 


     
 
2 2 2
1
2 1 2
0
2
0 0
2 0
2
2
239.69 , 230.94 , 35 8.016 280.55
, 73.16 .
230.94 0 239.69 73.16
, 35 31.57 .
0.5 8
b b a s
b
a
s
OA OB AB
OA OB
OA E E volt OB V volt AB I Z volt
So
V E
Now I A
Z j

 

  
    
        

    
    

Eb1=Eb2
Ia2Ra
jIa2Xs
V
Ia2Zs

O
A
2
B
2
Ia2

21. 21
2
2
3
, cos(31.57) 0.852 .
,
3 3
cos( ) cos( ) 188.23 .
188.23 10
, 179.75
2 / 60 2 1000 / 60
b b
g
s s
g
g
s
So thenewoperating power factor lagging
Total active power developed inthemotor
VE E
P kW
Z Z
P
Corresponding torque T Nm
N
  
 
 
   

  


22. 22
Thank you