FAKULTEIT INGENIEURSWESE
FACULTY OF ENGINEERING
Elektriese Aandryfstelsels 324 / Electrical Drive Systems 324
Tutoriaal 4 / Tutorial 4
27 en 28 Februarie 2014 / 27 and 28 February 2014
Die volgende probleme uit Chapman 5uit moet gedoen word en aan die dosent aan diens gewys word voor
16:00 om vir die Tutoriaal afgemerk te word.
The following problems from Chapman 5ed shall be done and shown to the lecturer on duty before 16:00
in order to be signed off for the Tutorial.
P8-15’ A 440 V shunt dc motor with a 560 A rated armature current and 7,52 A rated field current, has been
tested, and the following data were taken:
Blocked-rotor test (i.e. n = 0 rpm):
VT
IA
=
=
16,9 V (inclusive of brushes)
500 A
VF
IF
VT
IA
=
=
440 V (inclusive of brushes)
23,1 A
IF
n
=
=
440 V
7,52 A
No-load operation:
=
=
7,52 A
863 rpm
[Note: Assume that (1) the brush voltage drop is 2 V; (2) the rotational losses can be approximated by a
viscous frictional coefficient; and (3) stray load losses can be ignored.]
(a) Calculate the armature resistance of this motor.
RA =
VT,br −Vbrush 16,9 − 2
=
= 0,029 8 Ω
IA,br
500
(b) Calculate the rated Kφ value of this machine (i.e. at IF = 7,50 A)
EA,nl = VT,nl −Vbrush − IA,nl RA
= 440 − 2 − 23,1 · 0,029 8
= 437,3V
∴
= Kφ ωm nl
EA,nl
Kφ =
ωm nl
437,3
=
863 · 2π
60
= 4,839 rpm
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(c) Calculate the viscous frictional coefficient, b, that can be used to approximate the machine’s rotational losses.
Pnl = EA,nl IA,nl = 437,3 · 23,1
= 10,102 kW
2
= bωm,nl
2π 2
= b nm,nl ·
60
2π 2
= b 863 ·
60
10,102 × 103
b=
2
863 · 2π
60
∴
= 1,237 Nm/rad/s
(d) Calculate the rated speed of the shunt dc motor in rpm.
EA,rated = VT,rated −Vbrush − IA,rated · RA
= 440 − 2 − 560 · 0,029 8
= 421,3V
= Kφ ωm,rated
EA,rated
ωm,rated =
Kφ
421,3
=
4,839
= 87,066 rad/s
60
nm,rated = ωm,rated ·
2π
= 831,4 rpm
∴
∴
(e) Calculate the rated output power of the shunt dc motor.
Pout,rated = Pconv,rated − Protational,rated
2
= EA,rated IA,rated − bωm,rated
= 421,3 · 560 − 1,237 · 87,0662
= 226,55 kW
(f) Calculate the rated efficiency of the shunt dc motor.
Pin,rated = VT,rated IA,rated +VF,rated IF,rated
= 440 · 560 + 440 · 7,52
∴
= 249,71 kW
Pout,rated
ηrated =
× 100%
Pin,rated
226550
=
× 100%
249710
= 90,725%
P8-16’ A 240 V shunt dc motor with a rated armature current of 100 A has the following characteristics:
RA
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=
0,14 Ω
RF
=
200 Ω
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This motor has compensating windings and interpoles. The magnetization curve for this motor at 3000
rpm is shown in Figure P8-6.
[Note: Assume that the brush voltage drop, the rotational losses and the stray load losses can be ignored.]
(a) What is the no-load speed of this motor when Radj (i.e. the external resistance in series with the field
winding) is adjusted to 120 Ω.
IF =
⇒
VT
240
=
= 0,75 A
RF + Radj 200 + 120
EA,oc = 245 V
@
noc = 3 000 rpm
With:
Protational = 0 W
⇒
∴
∴
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IA = 0 A
EA = VT = 240 V
EA,nl
nnl =
· noc
EA,oc
240
=
· 3 000
245
= 2 939 rpm
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(b) What is its full-load (rated) speed?
IA,fl = 100 A
EA,fl = VT − IA,fl RA
∴
= 240 − 100 · 0,14
= 226 V
EA,fl
· noc
nfl =
EA,oc
226
=
· 3 000
245
= 2 767 rpm
∴
(c) What is its speed regulation?
nnl − nfl
× 100%
nfl
2 939 − 2 767
=
× 100%
2 767
= 6,22%
SR =
P8-20 An automatic starter circuit is to be designed for a shunt motor rated at 15 kW, 240 V, and a rated
armature current of 69 A. The armature resistance of the motor is 0,12 Ω, and the shunt field resistance
is 40 Ω. The motor is to start with no more than 250 percent of its rated armature current, and as soon
as the current falls to rated value, a starting resistor stage is to be cut out. How many stages of starting
resistance are needed, and how big should each one be?
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page 4 of 5
P9-21’ A 7,5 kW, 120 V shunt dc motor has a full-load armature current of 70 A when operating at rated
conditions. The armature resistance of the motor is RA = 0,12 Ω, and the field resistance RF is 50 Ω. The
adjustable resistance in the field circuit Radj may be varied over the range from 0 to 200 Ωand currently
set to 100 Ω. Armature reaction may be ignored in this machine. The magnetization curve for this motor,
taken at a speed of 1000 rpm, is given in following table:
EA [V]
IF [A]
5
0,00
78
0,80
95
1,00
112
1,28
118
1,44
126
2,88
130
4,00
[Note: Assume that the brush voltage drop and the stray load losses can be ignored.]
(a) What is the rated speed of this motor when it is running at the rated conditions specified above? (i.e.
when connected to a 120 V supply and drawing an armature current of 70 A)
IF =
⇒
120
VT
=
= 0,8 A
RF + Radj 50 + 100
EA,oc = 78 V
@
noc = 1 000 rpm
EA,rated = VT − IA,rated RA
∴
= 120 − 70 · 0,12
∴
= 111,6 V
EA,rated
nrated =
· noc
EA,oc
111,6
=
· 1 000
78
= 1 430 rpm
(b) The output power from the motor is specified as 7,5 kW at rated conditions. What is the rated output
torque of the motor?
τrated =
Prated
7 500
=
= 50,08 Nm
ωm,rated 1 430 · 2π
60
(c) What is the rated developed torque of the motor?
∴
Pconv, rated = EA,rated IA,rated = 111,6 · 70 = 7,812 kW
Pconv, rated
7 812
τind,rated =
=
= 52,167 Nm
ωm,rated
1 430 · 2π
60
(d) What are the rotational losses in the motor at full load?
∴
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τrotational = τind,rated − τrated = 52,167 − 50,08 = 2,087 Nm
2π
= 312,55W
Protational = τrotational ωm,rated = 2,087 · 1 430 ·
60
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