Complete Equivalent Circuit
for the Shunt Motor
Racir = resistance of
the armature circuit
Interpoles and Compensating Windings
ECE 441
1
General Speed Equation For a DC Motor
E  n k
a
p
G
E
n 
k
V I R
n 
k
a
p
G
T
a
p
ECE 441
acir
G
2
Example 10.8
• A 25-hp, 240-V shunt motor operating at
850 r/min draws a line current of 91 A
when operating at rated conditions. A
2.14-Ω resistor inserted in series with the
armature causes the speed to drop to 634
r/min. The respective armature-circuit
resistance and field-circuit resistance are
0.221-Ω and 120-Ω. Determine the new
armature current.
ECE 441
3
Solution
240V
I 
 2A
120
I  I  I  91  2  89A
f
a
T
f
ECE 441
4
V I R
n 
k
T
a
p
acir
G
[V  I R ]
k
[V  I R ]


[V  I (R  R )] [V  I (R  R )]
k
n
[V   (V  I R )]
n

(R  R )
634
[240 
 (240  89  0.221)]
850

 32.05A
0.221  2.14
T
n
n
1
2
a1
p
T
a2
acir
G
T
acir
p
x
T
a1
a2
acir
acir
x
G
2
T
I
T
acir
1
a2
acir
I
a1
a2
x
ECE 441
5
Example 10.9
• A shunt motor rated at 10hp, 240-V, 2500 r/min,
draws 37.5 A when operating at rated conditions.
• Ra=0.213Ω, RCW=0.065Ω, RIP=0.092Ω, Rf=160Ω
ECE 441
6
Example 10.9 (continued)
• Determine the steady-state armature
current if a rheostat in the shunt field
reduces the flux in the air gap to 75% of its
rated value, a 1.0-Ω resistor is placed in
series with the armature, and the load
torque on the shaft is reduced to 50%
rated
ECE 441
7
At rated conditions
V
240
I 

 1.5A
R
160
f
f
f
I  I  I  37.5  1.5  36A
a
T
f
ECE 441
8
At the new conditions
T BIk
D
p
a
M
[B  I ] [ I ]
T


T
[B  I ] [ I ]

T 
0.5T
I I
 36 

 24A
T 
T
0.75  
1
p
p
a
1
p
a
1
2
p
p
a
2
p
a
2
a2
2
p1
1
p2
1
p1
a1
1
ECE 441
p1
9
T BIk
D
p
a
M
[B  I ] [ I ]
T


T
[B  I ] [ I ]

T 
0.5T
I I
 36 

 24A
T 
T
0.75  
1
p
p
a
1
p
a
1
2
p
p
a
2
p
a
2
a2
2
p1
1
p2
1
p1
a1
1
ECE 441
p1
10
Determine the new steady-state speed
V I R
n 
k
T
a
p
acir
Check page 421 for = sign
G
k
n
V I R
[
] [
]
n
k
V I R
2
T
a
p
acir
G
2
1
p
G
1
T
a
acir
k
240  24  (1  0.370)
n  2500  [
] [
]
0.75 k
240  36  0.370
p
G
2
p
G
n  3046r / min
2
ECE 441
11
Speed Control of DC Motors
• Armature Control
– Insert a resistor or
rheostat in series with
the armature
– Reduce speed below
the base speed
• Shunt Field Control
– Insert a resistor or
rheostat in series with
the shunt field
– Increase speed above
the base speed
ECE 441
12
For speed reduction
Armature current decreases
Armature current increases
Torque decreases
Torque recovers (increases)
Speed decreases
Speed settles to new value
Counter-emf decreases
ECE 441
13
For speed increase
Decreasing field current reduces the
flux
Reduction in flux causes cemf to
decrease
Motor accelerates, cemf increases
Armature current decreases
Armature current increases
Torque increases, machine
accelerates
Torque settles down
Motor runs at new speed
Speed increases
ECE 441
14
Mechanical Power and Developed Torque
Pmech = Total Power Input to the Armature – Copper Losses in the Armature
Pmech = VTIa – Ia2Racir
Racir = Ra + RIP + RCW
Pmech = EaIa
ECE 441
15
Power-flow Diagram – DC Motor
Ploss = Pacir + Pb + Pcore + Pfcl + Pf,w + Pstray
Pb = VbIa
ECE 441
16
Power-flow Diagram – DC Generator
Ploss = Pacir + Pb + Pcore + Pfcl + Pf,w + Pstray
Pb = VbIa
ECE 441
17
Starting a DC Motor
• At “locked-rotor”, or “blocked-rotor”
V E
I 
R
V 0 V
I 

R
R
T
a
a
cir
T
T
a ,lr
acir
ECE 441
acir
18
Manually-Operated DC Motor Starter
“Start” with all of the
rheostat resistance
in the circuit
All resistance is
cut out when
motor reaches
full-speed
“Off” position
“Holds” the lever
in “Run” position
“Run” position
(A “break” in the
field circuit deenergizes the
coil, shutting the
motor down)
ECE 441
19
Example 10.12
Motor Starting
• A 15-hp, 230-V, 1750 r/min shunt motor with a
compensating winding draws 56.2A when
operating at rated conditions. The motor
parameters are Racir = 0.280 Ω and Rf = 137 Ω.
ECE 441
20
Example 10.12 (continued)
• Determine
• (a) the rated torque
Tn
P  5252
P 
T 
5252
n
15  5252
T 
 45.0lb  ft
1750
rated
rated
ECE 441
21
Example 10.12 (continued)
• (b) the armature current at locked-rotor if no
starting resistance is used
I
V E
230  0


 821.4A
R
0.280
T
a ,lr
a
cir
ECE 441
22
Example 10.12 (continued)
• (c) the external resistance required in the
armature circuit that would limit the current and
develop 200% rated torque when starting
ECE 441
23
V
230
I 

 1.68A
R
137
T
f
f
I
a ,rated
I
T
I

T
I
1
a1
2
a2
T ,rated
 I  56.2  1.68  54.52A
f
T
2T
I  I   54.52 
 109.0A
T
T
2
a2
1
a1
1
1
ECE 441
24
V  E  I (R  R )
T
a
a
acir
x
V E
R 
R
I
230  0
R 
 2.80  1.83
109.0
T
a
x
acir
a
x
ECE 441
25
Example 10.12 (continued)
• (d) Assuming that the system voltage drops to
215 V, determine the locked-rotor torque using
the external resistor in (c).
– Assume that the flux density is proportional to the
field current
T BIk
D
p
B I
p
a
M
f
T I I
f
ECE 441
a
26
Example 10.12 (continued)
T
T
D1
D2
[I I ]

[I I ]
f
a
1
f
a
2
[I I ]
T T 
[I I ]
1.57  101.9
T  45.0 
 78.6lb  ft
1.68  54.52
D2
f
a
2
f
a
1
D1
D2
ECE 441
27