DC Drives - K13 - DC motor model .pdf
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Transcript of DC Drives - K13 - DC motor model .pdf
04/01/2013
1
SEPARATELY ECXITED DC MOTOR
Applied Newtonian mechanics to find the differential equations for mechanical systems.
Using Newton’s second law:
Electromagnetic torque developed by separately excited DC motor:
Viscous torque :
Load torque : TL
dt
dJJT
J : equivalent moment
of inertia
afafe iiLT
rmviscous BT
Equivalent circuit for separately excited DC motors
VOLTAGE
SUPPLY
LOAD
rfafa iLE
+
-
er T,LT
+
-
ar
ai
arr
aL
frr
fi
fr
fu
fL
auaxisquadrature
axisdirect
armature
field
SEPARATELY EXCITED DC MOTORS
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SEPARATELY ECXITED DC MOTOR
LrmfaafLviscouse
r TBiiLJ
TTTJdt
d
11
a
a
rf
a
af
a
a
aa uL
iL
Li
L
r
dt
di 1
f
f
f
f
ffu
Li
L
r
dt
di 1
J
T
J
Bii
J
L
dt
d Lr
mfa
afr
From Newton’s Second Law, Torsional-Mechanical equation is given as
The nonlinear differential equation for separately excited DC motor which is
found using Kirchhoff’s Voltage Law
SEPARATELY ECXITED DC MOTOR
Using Newton’s second law :
Dynamics of rotor angular displacement :
The derived three first order differential equations are rewritten in the
s-domain
LrmfaafLviscouse
r TBiiLJ
TTTJdt
d
11
rr
dt
d
)()()(
1)( sussiL
rsLsi arfaf
aa
a
)(
1)( su
rsLsi f
ff
f
Lfaaf
m
r TsisiLBJs
s
)()(1
)(
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SEPARATELY ECXITED DC MOTOR
x
x
aa rsL
1
ff rsL
1
mBsJ
1
afL
afL
fu
auai
eT
LT
fi
r
SEPARATELY ECXITED DC
GENERATOR
pmrmfaafpmviscouse
r TBiiLJ
TTTJdt
d
11
a
a
rf
a
af
a
a
aa uL
iL
Li
L
r
dt
di 1
f
f
f
f
ffu
Li
L
r
dt
di 1
J
T
J
Bii
J
L
dt
d Lr
mfa
afr
From Newton’s Second Law, Torsional-Mechanical equation is given as
The nonlinear differential equation for separately excited DC generator which is
found using Kirchhoff’s Voltage Law
The expression for the voltage at the load terminal must be used.
For the resistive load LR
au
aLa iRu
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4
Analysis of eqn(3) indicates that the angular velocity of the separately excited motor can be regulated by changing the applied voltages to the armature and field windings.
The flux is a function of the field current in the stator winding, and higher angular velocity can be achieved by field weakening by reducing the stator current [eqn(3)]
However, there exists a mechanical limit imposed on the maximum angular velocity. The maximum allowed (rated) armature current is specified as well, one concludes that the electromagnetic torque is bounded.
afafe iiLT
fi
fuau
)3(
2
e
faf
a
faf
a
faf
aaar T
iL
r
iL
u
iL
iru
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SEPARATELY ECXITED DC MOTOR
A separately excited, 2 kW DC motor with rated armature current 20 A and angular velocity 200 rad/s operates at the constant voltages and . The motor parameters are: , , , and .
Calculate:
The steady state angular velocity at the minimum and maximum load conditions, Nm and Nm.
The armature current at the minimum and maximum load conditions, Nm and Nm.
Vua 100 Vu f 20 18.0ar 5.3fr 1.0afL
radNmsBm /007.0
0min LT
0min LT
10max LT
10max LT
Example
Steady state condition Le TT
f
f
fr
ui
)3(
2
e
faf
a
faf
a
faf
aaar T
iL
r
iL
u
iL
iru
NmTL 0min
NmTL 10max
rr 007.07.51.0
18.0
7.51.0
1002
rr 007.010
7.51.0
18.0
7.51.0
1002
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Steady state condition Le TT
faf
rmL
faf
ea
iL
BT
iL
Ti
NmTL 0min
NmTL 10max
faafe iiLT
7.51.0
007.0 min
r
faf
ea
iL
Ti
7.51.0
007.010 min
r
faf
ea
iL
Ti
Example
Plot the torque-speed characteristic curves for a
separately excited, 2-kW DC motor if the
rated (maximum) armature voltage is
and the field voltage is . The
motor parameters are: , ,
, and
The load characteristic if
Vua 100max
Vu f 20
rmLL BTT 0
18.0ar 5.3fr
1.0afL radNmsBm /007.0
NmTL 50
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% parameters of separately-exited motor
ra=0.18; Laf=0.1; Bm=0.007; If=5.7; Tl0=5;
Te=0:1:10;
for ua=11:10:100;
wr=ua/(Laf*If)-(ra/((Laf*If)^2))*Te;
wrl=0:1:200; Tl=Tl0+Bm*wrl;
plot(Te,wr,'-',Tl,wrl,'-');hold on;
axis([0, 10, 0, 160]);
end; disp('End')
SEPARATELY ECXITED DC MOTOR
(cont~)
%transient dynamics of a separately excited dc motor
function yprime=difer(t,y);
ra=0.18; rf=3.5; La=0.0062; Lf=0.0095; Laf=0.1; J=0.04; Bm=0.007;
T1=0;
%T1=10;
ua=100; uf=20;
yprime=[(-ra*y(1,:)-Laf*y(2,:)*y(3,:)+ua)/La;...
(-rf*y(2,:)+uf)/Lf;...
(Laf*y(1,:)*y(2,:)-Bm*y(3,:)-T1)/J];
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SEPARATELY ECXITED DC MOTOR
(cont~)
%transient dynamics of a separately excited dc motor
clc
t0=0; tfinal=0.4; tol=1e-7; trace=1e-7; y0=[0 0 0]';
[t,y]=ode45('CHP5_1mdno',t0,tfinal,y0,tol,trace);
subplot(2,2,1); plot(t,y(:,1),'r-');
xlabel('Time (seconds)'); title('Armature Current ia, [A]');
subplot(2,2,2); plot(t,y(:,2),'g-.');
xlabel('Time (seconds)'); title('Field Current if, [A]');
subplot(2,2,3); plot(t,y(:,3),'b-');
xlabel('Time (seconds)'); title('Angular Velocity wr, [rad/s]');
subplot(2,2,4);plot(t,y(:,1),'r-',t,y(:,2),'g-.',t,y(:,3),'b-')
xlabel('Time (seconds)'); title('LAB 1');
SEPARATELY ECXITED DC MOTOR
(cont~)
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4-100
0
100
200
300
X: 0.03529Y: 270.5
Time (seconds)
Armature Current ia, [A]
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40
1
2
3
4
5
6
Time (seconds)
Field Current if, [A]
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40
50
100
150
200
250
Time (seconds)
Angular Velocity wr, [rad/s]
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4-100
0
100
200
300
Time (seconds)
LAB 1
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SEPARATELY ECXITED DC MOTOR
(cont~)
tl and te
field current
combine
armature current
angular velocity
1
0.04s+0.007
Transfer Fcn2
1
0.0095s+3.5
Transfer Fcn1
1
0.0062s+0.18
Transfer Fcn
combine
To Workspace
Step1
Step
SignalGenerator
Product1
Product
1
Gain5
1
Gain4
1
Gain3
1
Gain2
0.1
Gain1
0.1
Gain
SHUNT CONNECTED DC MOTOR
The armature and field windings are connected in parallel
VOLTAGE
SUPPLY
LOAD
rfafa iLE
+
-
er T,LT
+
-
ar
ai
arr
aL
frr
fifr
fu
fL
auaxisquadrature
axisdirect
armature
field
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SHUNT CONNECTED DC MOTOR
LrmfaafLviscouse
r TBiiLJ
TTTJdt
d
11
a
a
rf
a
af
a
a
aa uL
iL
Li
L
r
dt
di 1
;1
f
f
f
f
ffu
Li
L
r
dt
di
J
T
J
Bii
J
L
dt
d Lr
mfa
afr
From Newton’s Second Law, Torsional-Mechanical equation is given as
The nonlinear differential equation for separately excited DC motor which is
found using Kirchhoff’s Voltage Law
fa uu
Steady state condition fa uu
f
af
r
ui
a
rfafa
ar
iLui
Substituting the currents equation into torque equation, gives
faafe iiLT
21 a
f
raf
fa
af
e ur
L
rr
LT
It shows that
The electromagnetic torque is a linear function of the angular velocity
The electromagnetic torque varies as the square of the armature voltage applied
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SHUNT CONNECTED DC
MOTOR (Example)
A shunt connected motor, drives a fan.
Given
When one applies the angular velocity is 150rad/s. For steady state operating condition and assuming the viscous friction is negligibly small, find the developed electromagnetic torque and the currents in the armature and field windings
,12.0,23,0,15.0 affraraf rrrrL
Vua 100
SHUNT CONNECTED DC
MOTOR (cont~)
21 a
f
raf
fa
af
e ur
L
rr
LT
mNTe .8.1110023
15015.01
2312.0
15.0 2
Ar
ui
f
f
f 35.423
100
AiL
Ti
faf
ea 1.18
35.415.0
8.11
faafe iiLT
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SERIES CONNECTED DC MOTOR
The armature and field windings are connected in series
VOLTAGE
SUPPLY
LOAD
rfafa iLE
+
-
er T,LT
+
-
ar
fa ii
arr
aL
fr
fL
auaxisquadrature
axisdirect
armature
field
Steady state condition 0dt
dia
dt
diLLirriLu a
faafaraafa
Then, currents equation
2
aafe iLT
21 a
f
raf
fa
af
e ur
L
rr
LT
It shows that
The developed electromagnetic torque is proportional to the square of the current
Saturation effect should be taken into account
The nonlinear differential equation for series connected DC motor which is
found using Kirchhoff’s Voltage Law
faraf
aa
rrL
ui
Substituting the currents equation into torque equation, gives
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SERIES CONNECTED DC
MOTOR
LrmfaafLviscouse
r TBiiLJ
TTTJdt
d
11
a
fa
ra
fa
af
a
fa
faa uLL
iLL
Li
LL
rr
dt
di
1
J
T
J
Bi
J
L
dt
d Lr
ma
afr 2
From Newton’s Second Law, Torsional-Mechanical equation is given as
The nonlinear differential equation for series connected DC motor which is
found using Kirchhoff’s Voltage Law
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