ECE 3600
DC Motor Examples
Ex.1 A 3-hp dc motor has the following nameplate information: 150 V, 1400 rpm, 18 A, RA = 0.8 Ω, and RF = 300 Ω.
The field is shunt connected and the 18 A includes the field current. Assume rotational losses are constant.
1 . hp = 745.7 W
a) Find the efficiency of the motor at nameplate operation. (Include the field in your calculations)
VT
150. V
3 . hp.
P out
P in
1400. rpm
n FL
18. A
I FL
745.7. W
0.8 . Ω
RA
A.Stolp
11/29/11
rev 12/1/14
300. Ω
RF
P out = 2.237 kW
hp
V T. I FL
b) Find the rotational losses at nameplate operation.
VT
field current: I F
RF
armature full-load current: I AFL
I AFL = 17.5 A
I AFL. R A
VT
P conv
E AFL. I AFL
P conv
= 82.86 %
P in
I F = 0.5 A
I FL I F
E AFL
P rot
P out
η =
P in = 2.7 kW
E AFL = 136 V
P conv = 2.38 kW
P out
P rot = 142.9 W
P rot = 0.192 hp
either answer
c) Find the required current for a developed power of 1.5 hp with V T = 150 V.
P conv
1.5 . hp
VT = EA
P conv = 1.119 kW
I A. R A =
P conv
IA
1
Solving for
2.R A
IA =
1
2.R A
= E A. I A
I A. R A
. V
T
VT
2
0 = R A. I A
Rearrange
2
4 . R A. P conv
=
. V
T
VT
2
V T. I A
4 . R A. P conv
179.72
A
7.78
P conv
IA
7.78. A
IS
IA
IF
I S = 8.28 A
d) Find the output power if the developed power is 1.5 hp with V T = 150 V.
P out
P conv
P rot
P out = 975.7 W
P out = 1.308 hp
e) Find the shaft speed if the developed power is 1.5 hp with V T = 150 V.
P conv
EA
.n
EA
E A = 143.773 V
n
FL
IA
E AFL
n = 1480 rpm
ω
f) A deranged Mouse chews through part of the field winding so that the field
current drops and the field flux drops to 40% of its former value. Find the shaft
speed if the developed power is still 1.5 hp with V T = 150 V.
2 . π. rad
n.
sec .
60.
rev
min
EA
E A = K. φ. ω
so...
n
n FL
=
ω
ω orig
=
φ new
EA
=
E A φ orig
.
E A φ new
φ orig
g) Find the load torque if the developed power is still 1.5 hp with V T = 150 V.
ECE 3600
DC Motor Examples
p1
n new
E A 100. %
.
.n
E A 40. %
either answer
ω = 155
rad
sec
either
answer
n new = 3700 rpm
P out
τ
n new.
2 . π. rad
sec .
60.
rev
min
τ = 2.518 N . m
min
ECE 3600
DC Motor Examples
p2
Ex.2 A dc motor has the following nameplate information: 5-hp, 1200 rpm, Armature: 200 V, 22 A, 1 Ω; Field: 200 V, 1 A
a) Find the rotational losses at nameplate operation.
VT
200. V
P outFL
E AFL
n FL
RA
1.Ω
P outFL = 3.728 kW
E AFL = 178 V
E AFL. I FL
P convFL
22. A
I FL
745.7. W
5 . hp.
hp
V T I FL. R A
P convFL
P rotFL
1200. rpm
P convFL = 3.916 kW
P outFL
P rotFL = 187.5 W
P rotFL = 0.251 hp
b) The load torque is constant (not dependent on speed) and the rotational losses are proportional to speed, find
the shaft speed at half the rated armature voltage and full rated field voltage.
P = τ. ω m
so, if a power is proportional to speed, then the torque is constant.
OR, conversely, if the torque is constant, the power is proportional to speed.
In this case, ALL the power converted is proportional to speed and ALL the the induced torque is constant.
n new
.P
P conv =
convFL
n FL
One way: τ ind = K. φ. I A
VT
so, if τind and the field current are constant, then IA is constant.
200. V
EA
2
I A. R A
VT
Another solution: recognize that: E A =
VT = EA
EA
200. V
2
I A. R A = E A
P convFL
.R
E AFL
n new
P conv
EA
.E
n FL
.R
A
n new =
E A = 78 V
EA
.n
E AFL
because the field is constant
n new
.P
convFL
n FL
.R
= EA
= EA
A
n new
.E
AFL
n FL
I FL
FL = 525.8 rpm
AFL
n new =
E A = 78 V
A
IA
P convFL
.R
E AFL
EA
.n
E AFL
A
FL = 525.8 rpm
c) The torque is constant (like part b)) , find the shaft speed at half the rated voltage for both the armature and field.
You may assume that the flux is proportional to the field current.
One way: τ ind = K. φ. I A
so, if τind is constant and the field current (and flux) is halved,
then IA is constant at twice the value it used to be.
VT
200. V
2
EA
VT
I A. R A
E A = 56 V
2 . I FL
IA
n new =
EA
E AFL
.n
FL = 377.5 rpm
1 E AFL .
E A = K. φ. ω m is halved because the flux is halved
EA = .
n new
2
n
FL
n new
.P
convFL
P conv
n FL
P convFL
.R
.R
.R
V T = E A I A. R A = E A
= E A 2.
A = EA
A
A
EA
n
E
1 . new .
AFL
E AFL
2 n FL
P convFL
EA
200. V
.R
.n
EA
2.
E A = 56 V
n new =
A
FL = 377.5 rpm
2
E AFL
E AFL
Another solution: recognize that:
ECE 3600
DC Motor Examples
p2
ECE 3600
DC Motor Examples
p3
d) If the load torque and rotational loss torque are both proportional to speed (most like real life), find the shaft
speed when the motor armature is hooked to half the rated voltage. The field is at rated voltage.
τ ind =
n new
n FL
.τ
P = τ. ω m
indFL
which leads to:
n new
P conv =
2
.P
n FL
convFL
One way: τ ind = K. φ. I A
V T= E A
so, if τind is proportional to speed and the field current is constant,
n new
.I
IA
then IA is also proportional to speed.
FL
n FL
n new
n new
n new
n new
. I .R
. I .R
.E
.I .R
I A. R A = E A
=
FL A = E A
FL A
AFL
FL A
n FL
n FL
n FL
n FL
n new
. E
.
=
AFL I FL R A
n FL
\___ V TFL ___/
n new
. ( 200. V )
=
n FL
100. V .
n FL = 600 rpm
200. V
n new =
Another solution:
V T= E A
EA =
n new
.E
n FL
I A. R A = E A
AFL
P conv
EA
because the field is constant
n new
.R
A
= EA
2
.P
n FL
n new
n FL
n new
convFL
.R
.E
A
=
n new
n FL
.E
n FL
AFL
.P
convFL
.R
E AFL
A
AFL
=
n new
n FL
same as above
. E
AFL
\___
P convFL
E AFL
.R
A
V TFL ___/
d) If the load torque is proportional to speed and rotational loss torque is constant, find the shaft speed when the
motor armature is hooked to half the rated voltage. The field is at rated voltage.
n new
P rotFL
.τ
τ load =
τ
τ
=
=
loadFL
loss
lossFL
n FL
ω mFL
n new
.τ
τ ind =
= K. φ. I A
loadFL τ lossFL
n FL
n new τ loadFL τ lossFL
n new
.
.I
IA =
=
the current can be thought
AloadFL I AlossFL
n FL
n FL
K. φ
K. φ
of as composed of two parts
P outFL
P rotFL
I AloadFL
I AloadFL = 20.947 A
I AlossFL
I AlossFL = 1.053 A
E AFL
E AFL
n new
n new
.E
.I
.
V T = E A I A. R A
=
AFL
AloadFL I AlossFL R A
n FL
n FL
n new
n new
.E
.I
.
.
=
AFL
AloadFL R A I AlossFL R A
n FL
n FL
n new
V T I AlossFL. R A
. E
.R
.n
V T I AlossFL. R A =
I
n
=
AFL
AloadFL A
new
FL = 596.8 rpm
n FL
E AFL I AloadFL. R A
ECE 3600
DC Motor Examples
p3