EE 560 – Electric Machines and Drives Homework #3 solved

Description

Analysis of AC Induction Motors
You are the controls engineer for XYZ Company and your boss has asked you to analyze the
performance of an induction motor. The induction motor has the following parameters:
𝑟𝑠 = 0.4 Ω stator resistance
𝑟𝑟 = 0.8 Ω rotor resistance
𝐿𝑚 = 70.0 𝑚𝐻 magnetizing inductance
𝐿𝑙𝑠 = 2.0 𝑚𝐻 magnetizing inductance
𝐿𝑙𝑟 = 2.0 𝑚𝐻 magnetizing inductance
𝑉𝜑𝑟𝑚𝑠 =
230
√3
𝑉 phase voltage
𝜔𝑒 = 2𝜋60 𝑟𝑎𝑑
𝑠𝑒𝑐
excitation frequency
𝐽 = 0.1 𝑘𝑔 ∙ 𝑚2
rotor inertia
𝑏 = 1.0 ∙ 10−3 𝑁𝑚
(
𝑟𝑎𝑑
𝑠𝑒𝑐 )
magnetizing inductance
𝐿𝑠 = 𝐿𝑚 + 𝐿𝑙𝑠 stator self inductance
𝐿𝑟 = 𝐿𝑚 + 𝐿𝑙𝑟 rotor self inductance

You will develop a transient model of the induction machine using the stator referred differential
equations from class (repeated below) in PLECS. Use native blocks in PLECS, do not use the
induction motor model provided (this will be used for comparison).
To do:
a. Turn in your PLECS model. Each of the variables should be clearly labeled.
b. Plot the following start-up transient variables:
a. Stator phase currents (abc)
b. Rotor phase currents (dq)
c. Rotor speed
d. Electromagnetic torque
c. Apply a step change in load torque from 0 to 10 Nm after 0.5 seconds. Plot this with the
variables of part b. (Parts b and c should be plotted on the same plot.)
d. Compare your model with the PLECS induction motor model results.
EE 56 – Electric Machines and Drives Prof. N.J. Nagel
Homework #3 – Due November 30th, 2020
Page 2
ia
if
field winding
armature winding
stator
Equation Summary for Simulation:
The following is a listing of state equations which can be used in simulating an AC machine:
p
s
qs = v
s
qs – rsi
s
qs
p
s
ds = v
s
ds – rsi
s
ds
p
s
qr = v
s
qr – rri
s
qr + r 
s
dr
p
s
dr = v
s
dr – rri
s
dr – r 
s
qr
Te =
3
2
P
2
Lm (i s
qs i
s
dr – i
s
qr i
s
ds)
where

s
qs = Lsi
s
qs + Lm i
s
qr

s
ds = Lsi
s
ds + Lm i
s
dr


s
qr = Lr i
s
qr + Lmi
s
qs

s
dr = Lr i
s
dr + Lmi
s
ds
so
i
s
qs =
1
Ls






s
qs –
Lm
Lr

s
qr
i
s
ds =
1
Ls






s
ds –
Lm
Lr

s
dr
i
s
qr =
1
Lr





1 +
L
2
m
LrLs

s
qr –
Lm
LrLs

s
qs
i
s
dr =
1
Lr





1 +
L
2
m
LrLs

s
dr –
Lm
LrLs

s
ds
 = 1 –
L
2
m
LsLr
(coupling factor)