A separately excited DC motor has the following data P=20.5kW (rated power); Va=440V (rated voltage); Ia=55A (rated current); n=1000 rpm (rated speed), the load torque is 90% rated torquea.Determine the original responses? Determine the artifical responses when the armature voltage is reduced to 90% rated voltage? b.The armature voltage is reduced to 270V. Draw the artifical responses and determine the rotor speed at that time. Explain the process of braking in this case. c.To stop the DC motor, the armature voltage is reversed and the resistance of 1.75Ω is added to armature circuit. Calculate the rotor speed and the torque at the first time of braking?
Trang 1HCMC UNIVERSITY OF
TECHNOLOGY AND EDUCATION
Faculty for High Quality Training
Program name: Mechatronics
FINAL EXAMINATION SEMESTER 2 – ACADEMIC YEAR 2018-2019 Course name: Electric drives
Course ID: ETDR336429E Exam code: 01 Number of pages: 01
Duration: 60 minutes
Open-book.
Question 1: (6/10)
A separately excited DC motor has the following data P=20.5kW (rated power); Va=440V (rated voltage); Ia=55A (rated current); n=1000 rpm (rated speed), the load torque is 90% rated torque
a Determine the original responses? Determine the artifical responses when the
armature voltage is reduced to 90% rated voltage?
b The armature voltage is reduced to 270V Draw the artifical responses and
determine the rotor speed at that time Explain the process of braking in this case
c To stop the DC motor, the armature voltage is reversed and the resistance of 1.75Ω
is added to armature circuit Calculate the rotor speed and the torque at the first time of braking?
a)
R u=U dm. I dm−P dm.
2 I dm2 =440 55−20 , 5 103
2 552 =0 612 Ω
kϕϕ dm=U dm−Idm R u
ω dm == 9 55
U dm−Idm R u
n dm =9 55
440−55∗0 612
1000 =3 88
I ST=U đm
R ư =
440
0.612=718.95( A)
T ST=U đm
R ư =2789.53 (Nm)
ω 0=44 0
3.88=113.4(rad /s )
The original responses
ω=113.4− 0.612
(3.88)2M =113.4−0.04 M 1.0d
The artifical responses when the armature voltage is reduced to 90% rated voltage
ω=102.06− 0.612
(3.88)2M=102.06−0.04 M 1.0d
b)
ω=270−55∗0.61 2∗0.9
3.88 =61 78(rad / s)1.0d
¿>n=581.71(v / p)
Trang 2the artifical responses
Regenerative Braking AB D A’
Regenerative Braking: BD 1.0d
c)
ω A=ω B=1000
9.55=104.71(rad /s)
the torque at the first time of braking
ω B=−440
3.88 −
0.612+1.75
(3.88)2 M B=104.71
¿>M B=−1390.16(Nm) 1.0d
ω A '=−440
3.88 −
0.612+1.75 (3.88)2 0.9.55=−143.53 rad /s 1.0d
Question 2: (4/10)
A 208-V, six pole, 60 Hz, Y-connected induction motor, R1=0.6Ω; R2’=0.4; Xeq=5Ω
a Calculate the motor speed if the load torque is 30 Nm?
b Calculate the smallest voltage source so that the motor could handle the load
torque 25 Nm?
c Calculate the starting current, starting torque, maximum torque,
d Calculate the resistance should be added to rotor circuit to reduce starting current a
half
e Calculate the resistance should be added to rotor circuit to keep the starting torque
equal to the maximum torque
a)
n s=60 f
p =
60 x 60
3 =1200(v / p)
V s=208
√3 =120.1(V )
the motor speed when the load torque is 30 Nm
Trang 330= 3 x 120.12x 0.4
sx1200
9.55 x[ (0.6+0.4
s )2+52
]=¿{
s=0.097 s=0.065(accept )1.0d n=n s (1−s )=1200 (1−0.065)=1122¿
b
2
sx1200
9.55 x[0.6+√0.62
+52❑]=¿V =±108.641 V =¿V =188.172V 0.5 d
c the starting current, starting torque, maximum torque
d
Ist’=11.776A
e
smax=1
Note: Proctors are not allowed to give any unauthorised explaination.
[LO 1.1]: Describe structure of electric motors; symbols
and roles of electric motors in mechatronic systems
[LO 2.1]: Create mathematic models of
electrical-mechanical responses of typical electric motors
Question 1
[LO 2.2]: Describe structure and working principles of
typical electric motors
Question 2
20/05/2019
Approved by program chair