Download AC-DC MOTOR DRIVES (Converters)

AC-DC MOTOR DRIVES (Converters)
Basic Characteristics of DC Motors
Consider :
The equations describing the characteristics of a separately excited motor can be written as:
Vf = R f if + Lf
dif
dt
&
The motor back emf is expressed as:
eg = Kvωif
The torque developed by the motor is:
TD = Kv if ia
Va = R a ia + La
dia
+e
dt
The torque developed must be usual to the load torque:
d
+ B  + TL
dt
developed torque, Nm
moment of inertia Kgm2
motor speed, rad/s
viscous friction constant, Nm/rad/s
motor voltage, u/A - rad/s
motor torque constant
armature circuit inductance, H
armature circuit resistance, Ω
field circuit inductance, H
field circuit resistance, Ω
load torque, Nm
Td = j
Where
TD
J
ω
B
Kv
Kt
La
Ra
Lf
Rf
TL
=
=
=
=
=
=
=
=
=
=
=
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
Page 1
Under steady-state conditions, the time derivatives in these equations are zero and the
steady-state average quantities are:
Vf = Rf If
Eg = KvωIf
Va = Eg + Ia Ra = Kv ωIf + Ra Ia
TD = Kt If Ia = Bω + TL
PD = TD ω
The speed of a separately excited motor can be found from:
 = V a R a Ia = V a R a Ia
K v If
Kv
Vf
Rf
Motor Control
Without electronic control, it would be necessary to add a resistance in series with the
armature circuit.
Va = Ia (Rc + Ra) + Eg
 =
=
Ia (Rc + Ra) + Kv ωIf
Va - Ia R c + R a 
K u If
Variation of RC would vary ω ( N)
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
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EXAMPLE
Calculate the efficiency and Rc required of a 15-hp 220V,2000rpm separately excited d.c.
motor, controlling a load requiring a torque of TL = 45 Nm at a speed of:
(a)1800 rpm
(b) 400 rpm
The field at resistance is 147Ω, the armature resistance is 0.25Ω and the voltage constant of
the motor kv = 0.7302 v/A-rad/s. The field voltage is 220V.
Assume (TL = TD)
Solution:
(a)
Va = Ia (Ra + RC) + E
N Vf
x
60 R f
1800 220
= 0.7302 x 2 x
x
60 147
E
= Kv ωIf
E1800 = KV x 2 x
TD = K V Ia If = > Ia =
Ra + Rc =
TD
=
K V If

= 206V
E1800
45
TD =
= 41.18 A
220
Vf
KV
0.7302 x
147
Rf
Va - E 220 - 206
=
= > R a + R c = 0.34 
41.18
Ia

Rc=0.34-0.25=0.09 Ω
41.18 x 220 - 41.182 0.34
 = Po = P1 P1055 =
x 100 = 93%
Pi
41.18 x 220
Pi
(b)
E400 = 7302 x 2 x
Ra + Rc =
400 220
x
= 45. 78v
60 147
220 - 45.78
= 4.23 
41.18
 R c = 3.61
220 x 41.18 - 41.182 4.23
=
= 20%
220 x 41.18
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
Page 3
EXAMPLE
A 15-hp, 220-v, 2000-rpm separately excited dc motor controls a load requiring
a torque of TL = 45 Nm at a speed of 1200 rpm. The field circuit resistance
is Rf = 147 Ω, the armature resistance Ra = 0.25 Ω and the voltage constant
of the motor Kv = 0.7302 V/A-rad/s. The field voltage Vf = 220V. The viscous
frictions are negligible. The armature current may be assumed continuous and
ripple-free. Determine (a) E (b) Ia (c) Va (d) the rated armature current and efficiency if
machine losses are 600W
Solution:
(a)
E1200 = KV x 2 x
 E = 0.7032 x
N Vf
x
60 R f
2 x 1200 220
x
= 132.25 V
60
147
(b)
TD = K V Ia If = > Ia =
TD
=
K V If
45
TD =
= 41.18 A
220
Vf
KV
0.7302 x
147
Rf
Ia = 41.18 A
 Va= 132.25V + 41.18A x 0.25Ω
(c)
Va = E1200 + Ia Ra
(d)
Prated = I rated V rated = > I rated =
=
Va = 142.54V
Prated 15 x746
=
= 50.87 A
Vrated
220
output input - losses 142.54 x 41.18 - 600 - Ia2 R a
=
=
x 100 = 83.2%
input
input
142.54 x 41.17
PROBLEM
Referring to the example above, suppose we wish to drop the speed from 1200, to 750 rpm,
while maintaining the same torque (ie. the same Ia), without electronic control and by adding
a series resistor in armature circuit. (a) Calculate Rc
(b) Efficiency
Ans : (a) 1.13 Ω
(b) η = 26.16%
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
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Single Phase AC-DC MOTOR DRIVES
(i)
Single-Phase Half-wave Semi-Converter Drives [1/2 kW power]
Va =

1
 Vm sin td (t )  Vm  cos t   Vm 1  cos  
2 
2
2
EXAMPLE : The speed of a separately excited D.C. motor is controlled by a single pahse
half-wave semi-converter. The field current supplied is 1.274A and the A.C. supply is 240V
@ 50Hz. The armature resistance Ra = 0.25Ω and the motor voltage constant Kt = Kv =
0.7032 V/A-rads/S. The load torque TL = 45 Nm @ 1000 rpm. The viscous and no load
losses are negligible. The inductance of the armature is sufficient enough to make armature
current continous and ripple-free. Determine the delay angle of the converter of the armature
circuit.
Solution:
α = ?, Va 
Vm
1  cos   ,
2
Va = Eg + Ia Ra , Eg =Kv ωIf , TL =TD = Kt If Ia
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
Page 5
TD
45
=
= 50.23 A
KV If 0.7032 x 1.274
N
2 x 1000
= KV x 2 x
x I f  0.7032 x
x 1.274 = 93.82 V
60
60
TD = KV Ia If = > Ia =
E1000
Va = E1000 + Ia Ra
Va 
Vm
1  cos   
2
 Va= 93.82V + 50.23A x 0.25Ω
Va = 106.38V
2 x240
1  cos    106.38    14.23o
2
PROBLEM : The speed of a separately excited D.C. motor is controlled by a single pahse
half-wave semi-converter. The field current supplied is 1.2A and the A.C. supply is 240V @
50Hz. The armature resistance Ra =0 .25Ω and the motor voltage constant Kt = Kv = 0.7032
V/A-rads/S. Firing angle of the converter for the armature circuit is 25 o degrees. Determine
the speed N in revs/min if the torque required is 32 Nm.
Ans :???
(ii)
Single-Phase full-wave Semi-Convertor Drives [15 kW power]
Va =
1



V sin td (t ) 
 m
Vm

 cos t 

Vm

1  cos   
2Vs

1  cos  
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
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EXAMPLE : The speed of a separately excited D.C. motor is controlled by a single pahse
semi-converter. The field current is also controlled by a full-wave semi-converter, which is
set to the maximum possible value. The a.c. supply voltage to armature and field converter is
single-phase 240V, 50Hz. The armature resistance Ra = 0.25Ω, the field resistance is Rf =
147Ω, and the motor voltage constant is Kv = 0.7302 V/(A-rad/s). The load torque TL = 45
Nm @ 1200 rpm. The losses are negligible. The inductances of the armature and field are
large enough to make their currents continuous and ripple-free.
Determine (a) If, (b) The delay angle of the converter in the armature at a
Solution:
(a)
2 . Vs .
Vf = If R f =
If =

1 + cos f  =
2 2 . Vs

2 2. Vs 2 2. 240V
=
1.47 A
147 . 
R f .
(b)
Va =
2.240

1 + cos a 
TD  K v I a I f  I a =
E1200 = KV x 2 x
and Va = IaRa + Eg but Ia = ? and Eg = ?
TD
45
=
= 41.92 A
Kv I f
0.7302 x 1.47
N
2 x 1200
x I f  0.7302 x
x 1.274 = 91.76 V
60
60
Va = IaRa + Eg = 41.92Ax0.25Ω + 91.76V = 102.24V
102.24 =
2.240

1 + cos a 
=> 1 + cos a  = 0.946
a = cos-1 (-0.054)= 1.625radians => 93
Problem : The speed of a separately excited motor is controlled by a single-phase
semiconverter. The field current is also controlled by a semi-converter, is set to 2A. The a.c.
supply voltage to armature and field converter is one-phase 240V, 50Hz. The armature
resistance Ra = 0.20Ω, the field resistance is Rf = 120Ω, and the motor voltage constant is KV
= 0.8032 V/(A-rad/s). The losses are negligible. The inductances of the armature and field
are large enough to make their currents continuous and ripple-free. For a firing angle of 45 ,
for a torque of 40 Nm
Determine (a) E and
(b) The speed in rev/mins.
Ans : (a) ????
(b) ????
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
Page 7
(iii)
Va =
Single-Phase full wave Full convertor Drives [15 kW power]
1

 
 
V sin td (t ) 
m
Vm

 cos t  

2 2Vs

cos 
Note :cos(A ± B) = cosAcosB ∓ sinAsinB
The application of this drive is limited to up to 15KW. The current waveforms for highly
inductive load are shown above. During regeneration for reversing the direction of power
flow, the back emf of the motor can be reversed by reversing the field excitations.
EXAMPLE: A single phase full converter is used to control the speed of a separately excited
dc motor with Ra=0.25 Ω, Rf= 175 Ω, kv=0.7302 V/A-rad/s. The field is also controlled with
a full converter and the field current is set to the maximum value. The armature current
corresponding to the load demand is Ia=50 A. The supply is one-single phase of 240 V, 50
Hz. The vicous friction and no load losses are negligible and the inductance of the armature
and filed are sufficient to make to make armature and field current ripple free. If the delay
angle of the armature converter is set to αa = 30o, determine (a) the torque developed TD by
the motor, (b) the speed ω in revolution per minute, (c) the input power factor of the drive.
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
Page 8
Solution:
(a)
TD = ?, TD = KvIfIa, If = ? and If=Vf/Rf,
Vf 
2 2Vs

cos  f 
2 2 x 240V

cos 0 o  216V If =
216V
 1.235 A
147 
TD = KvIfIa= 0.7302 x 1.235A x 50A = 45 N.m
(b)
ω = ?, Eg =Kv ωIf , Eg = ?, Va = IaRa + Eg , Va = ?
Va 
2 2Vs

cos  a 
2 2 x240V

cos 30 o  187.12V
187.12V = 50Ax0.25Ω + Eg => Eg = 174.62V
V
E g = KV I f  174.62  KV x 2 x
N
2 x N
x I f  0.7302 x
x 1.235 A
60
60
N = 1849 RPM
(c)
Power Factor (PF) = ?, PF = Pi/Ps, Pi=VaIa + VfIf, Ps = VsIs, Is = (Isa2 +Isf2)1/2
Pi=VaIa + VfIf = 50Ax187.12V + 1.235Ax216V = 9622.76W
Ps = VsIs = Vs (Isa2 +Isf2)1/2 = 240V(502A +1.2352A)0.5 = 12003.66W
PF = Pi/Ps = 9622.76/12003.66 = 0.8 (lagging)
Problem : For the example above, The polarity of the back e.m.f. is reversed say by reversing
the field excitation. Determine (a) the delay angle αa of the armature converter circuit to
maintain Ia of 50 A, and (b) the power feedback Pg.
Ans : (a) 138.61o
(b) 8106W
(c) i?? ii??
(Note : Va = IaRa + Eg , When the If is reveresed, Eg = - Eg , Pg= [Ea-ER]Ia)
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
Page 9
(iv)
Three-Phase Halfwave Converter Drives [40 kW power]
When T1 is fired at ωt= π/6 + α, the phase voltage appears across the motor until T2 is fired
at ωt= 5π/6 + α. When T2 is fired, T1 is reverse biased, because the line to line voltage
vab(van-vab), is negative and T1 is turned off. The phase voltage vbn appears across the load
until T3 is fired at ωt= 3π/2 + α. When T3 is fired, then T2 is turned off and vcn appears
across the load until T1 is fired again at the beginning of the next cycle.
Vo =
3
2
5

6

6

Vm sin td (t ) 
5

3 3Vm
3 3Vm
cos 
  cos t  6 
2
2

6
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
Page 10
Example : The speed of a 20-hp 300V 1800 rpm separately excited DC motor is controlled
by a three phase half-wave converter. The input is a three-phase Y(wye)-connected 50 Hz VL
of 415V. Ra = 0.25 ohms, the motor constant Kv = 1.2 V/A-rad/S. The armature and field
currents are assumed to be continuous and ripple free and the viscous friction is negligible.
Determine
(a) The delay angle of the converter, if the motor supplies the rated power at the rated speed,
and field current is 0.5A.
(b) The no-load speed, if delay angles are the same as in part (a) and the armature current
at no-load is 20% of the rated value.
(a)
αa = ?, Irated=Ia =P/V, P=20x746, Irated=Ia =(20x746)/300=49.73 A
Va = IaRa + Eg , Eg = ?, Eg =Kv ωIf = 1.2x [(1800 x 2π)/60] x 0.5 = 113.1 v
Va = 49.73x0.25 + 113.1 = 125.53 v
Va 
3 3Vphase
2
cos  
3x415
cos   125.53    50.65o
2
(b) N = ?, Eg =Kv ωIf , Ia= 20% of 49.73 = 9.95 A
Eg =Va - IaRa = 125.53V – 9.95Ax0.25Ω = 123 V

Eg
Kv I f

123
 205
1.2 x0.5
rad/s
N
60 60 x205

 1957 rev/min
2
2
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
Page 11
(v)
Three-Phase full wave Semi-convertor Drives [115 kW power]
Vo =
3
2
7

6

6

Vm sin td (t ) 
5

3 3Vm
3 3Vm
(1  cos  )
  cos t  6 
2
2

6
Problem : Repeat the example above for the 3 phase semiconverter shown below.
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
Page 12
(vi)
Three-Phase full wave Full-convertor Drives [1500 kW power]

Vo =
3

2

6





3 3Vm 
3 3Vm
  2


3Vm sin  t   d (t ) 
 cos  t      
cos 

6
 
6  6



Problem : Repeat the example above for the 3 phase full converter shown below.
ENG732S1 – Machines and Power Drives : Hassan Parchizadeh
Page 13