3-PHASE INDUCTION MOTORS
EXAMPLES OF MOTOR OPERATION
At the base of equivalent circuit showing one phase of multi-phase motor
we can consider different possibilities of motor operation, depending on
its speed n, slip s and, therefore, the values of:
resistance R2'
2
1− s
s
and the power I2' ⋅ R2' ⋅
1− s
s
a) n = n1 (s = 0 ) - ideal no-load operation (zero torque)
R2'
1− s
= ∞ (electrical equivalent circuit is opened – no-load).
s
Output power shown by means of this resistance is zero (I2’=0).
b) n = 0 (s = 1) (the shaft is blocked – is locked – locked rotor state)
R2'
1− s
=0
s
(electrical eq. circuit is short – circuited:
short-circuit state of the motor).
Output power shown by means of this resistance is zero.
IM2- 1
IM2- 2
c) Operation of induction machine with 0 < n < n1 (motor operation)
Sankey’s diagram – energy (power) balance diagram:
P1 = m1 I 1U 1 cos ϕ
∆PCu1 = m1I12R1 - stator copper loss
( m1 - number of phases)
∆PFe - stator core loss (can be
determined from the no-load test)
The remainder: Pe = P1 − ∆PCu1 − ∆PFe
is called “electromechanical power”
or “ideal power”. Physically it is power
transferred from the stator to rotor by
means of magnetic field (through the
air gap). It is a power delivered to the
rotor!
From the equivalent circuit:
R'
Pe = m1I2'2 2 = m1I2'2R2'
s
+
rotor copper
loss ∆PCu 2
∆PCu 2 = sPe
 1− s 
m1I2'2R2' 

 s 
“mechanical power” Pm
or “rotor developed power”
Pm = Pe − ∆PCu 2
From the first relation:→ the value of slip s should be as small as
possible to limit the value of rotor copper loss.
In practice s is at the range of several %, therefore f 2 = sf 1 is very
small and rotor iron (core) loss is small and can be neglected.
After subtraction mechanical loss (power loss due to friction & windage
of rotating rotor) from Pm we get:
Pm − ∆Pm = P2
(or Pout or P ) – shaft output power
Efficiency of energy (power) conversion:
η=
P2 Pout
=
P1
Pin
For further consideration let’s take the mechanical power which
corresponds to the power appearing across the resistance R2'
1− s
. This
s
power is connected with the torque- electromechanical torque –
rad
driving the rotor with the speed n (or Ω in
):
s
1− s
Pm = m1I2'2R2'
s
Electromechanical torque
Te =
Pm
Ω
=
60m1
1− s
I2'2R2'
2πn1(1 − s )
s
in [N⋅m] or [Nm]
Rotor core loss is not even
shown at Sankey’s diagram.
IM2- 3
From simplified equivalent circuit:
With two following
simplifications:
‘
R2'
- and can
s
a) R1 <<
be neglected,
b) parallel branch is
neglected,
I2' =
U1
(X
1+
)
2
X 2'
 R' 
+ 2
 s 


2
Substitution I 2' to Te relation yields:
Te =
60m1
⋅
2πn1
(
R2' U12
s X1 +
)
2
X 2'
In general Te = f (s, U1, f , R2 )
R '2
+ 2
s
The torque developed in the machine is proportional to the voltage
squared.
Testing of the function Te = f (s ) for U, f , R2 = const yields the following
results:
torque–slip curve
There are two critical points
60m1
U12
Tmax = ±
⋅
2πn1 2 X1 + X 2'
breakdown torque – maximum
torque that can be developed in
IM for given voltage U 1 and
frequency f .
(
)
Tmax doesn’t depend upon R2 !
smax = ±
R2'
X1 + X 2'
breakdown slip smax ∼ R2 !
The same result in other coordinates
Kloss formula
torque–speed curve
T
Tmax
=
2
smax
s
+
s
smax
Ts - starting torque.
Usually smax is at the range of
about 10%.
Notice that the maximum
(breakdown) torque is
proportional to the voltage
squared and inversely
proportional to frequency
squared (n1 and X ∼ f).
IM2- 4
STABILITY OF THE MOTOR OPERATION
The motor operates with
constant speed when torque
balance is zero:
Te − Tload = 0
From two possible points A & B
only point B corresponds to
stable operation.
Stable region of T-n curve
Stable region is for 0 < s < smax
N - point of nominal operation
( TN ; n N - nominal torque,
nominal speed).
Adequate surplus of stability is required. Usually minimum value of
torque ratio required by users is 2:
Tmax
≥2
TN
stability margin
For motor driving the load having fan-type torque-speed characteristic it
can happen that stable operation occurs for very low speed:
Such operation is very
dangerous for normal
construction motor:
n≈
1
s ≈ 0,5
n1
2
∆PCu 2 ≈ 0,5Pe
what means:
- very high rotor copper loss!,
- very low efficiency!.
IM2- 5
NUMERICAL EXAMPLE
Consider a 3-phase induction machine (motor) of rating:
PN = 75 kW
UN = 220/380 V
cosϕN = 0.85
ηN = 0.92
f = 50 Hz
nN = 975 rev/min
∆Pm = 0.5%PN
R1 = 0.033 Ω
rated power (always must be understood as output one)
rated voltage (for two possible connections)
rated power factor
rated efficiency
rated speed
mechanical loss at rated speed
stator winding (phase) resistance
Calculate:
1) Nominal stator current (line) for star and delta connections of stator
winding,
2) Apparent nominal power SN (power drawn by the stator from the
line),
3) Active and reactive power absorbed from the mains for nominal load,
4) Nominal torque and nominal slip,
5) Iron core loss.
1.
I1NY =
I1N∆ =
P1N
3UN cosϕ N
PN
3UNηN cosϕ N
2.
S1N =
3.
P1N =
PN
=
3UNηNcosϕ N
=
=
75 × 103
3 × 380 × 0.92 × 0.85
75 × 103
3 × 220 × 0.92 × 0.85
= 145.7 A
= 251.7 A
P1N
75
=
= 95.9 kVA
cosϕ N 0.92 × 0.85
PN
ηN
=
75
= 81.52 kW
0.92
Q1N = S1N sinϕ N = 95.9 1 − 0.852 = 50.5 kvar
75 × 103
PN
= 9.55 ×
= 734.6 N ⋅ m
975
nN
n − nN 1000 − 975
sN = 1
=
= 0.025
sN = 2.5%
n1
1000
4.
TN = 9.55
5.
2
∆PCu1N = 3INph
R1 = 3 × 145.72 × 0.033 = 2.102 kW
 Pe = P1N − ∆PFeN − ∆PCu1N

PN = Pm − ∆Pm = Pe (1 − s ) − ∆Pm
from this set of equations the unknown ∆PFeN is
∆PFeN = P1N − ∆PCu1N −
PN + ∆Pm
75(1 + 0.005 )
= 81.52 − 2.102 −
= 2.11 kW
1− s
1 − 0.025
BASIC CONSTRUCTIONS OF INDUCTION MACHINE
Slip-ring induction machine
Idea of construction
Example of connections
Basic characteristics
IM2- 6
Squirrel-cage machine (squirrel-cage motor, cage motor)
Different possible profiles of rotor slots and cage bars
Double-cage motor
Deep-bar rotor (deep-bar squirrel-cage induction motor)
IM2- 7
IM2- 8
STARTING OF INDUCTION MOTORS
Slip-ring motor starting
Cage motor starting
Star-delta starting (most popular for the motors of rated power above 5 kW).
NUMERICAL EXAMPLE
About induction motor of rating as in previous numerical example:
PN = 75 kW
UN = 220/380 V
cosϕN = 0.85
ηN = 0.92
f = 50 Hz
nN = 975 rev/min
∆Pm = 0.5%PN
R1 = 0.033 Ω
we know that it is a squirrel cage motor having the breakdown torque
ratio Tmax/TN=Tb/TN=2.2, starting torque ratio Ts/TN=1.3 and rated starting
current ratio IsN/IN=6. Calculate:
1. Values of starting torque and currents for direct switch-on starting (at
nominal conditions of supply),
2. Starting torque and current for Y/∆ starting.
1.
a) for Y connection (stator connected in Y, U=UN=380 V)
TsN=1.3×TN=1.3×734.6=954.98 N⋅m
IsNY=6×INY=6×145.7=874.2 A
b) for ∆ connection (stator connected in ∆, U=UN=220 V)
TsN=1.3×TN=1.3×734.6=954.98 N⋅m
IsN∆=6×IN∆=6×251.7=1510.2 A
2.
stator circuit connected in Y and supplied from the mains 3×220 V:
1
IsY =
3
× 6 × INY =
6
3
× 145.7 = 504.72 A
1
× 6 × INY
IsY
1
3
=
=
IsN∆
3
6 × 3 × INY
TsY
 U
= 
 UN
2
2

 1 
1
1
 TsN = 
 TsN = TsN = × 954.98 = 318.33 N ⋅ m
3
3
 3

IM2- 9
SPEED CONTROL OF INDUCTION MOTOR
By means of rotor resistance
By means of line voltage
By means of slip-power recovery
IM2- 10
By means of frequency
SPECIAL APPLICATION OF INDUCTION MACHINE
INDUCTION REGULATOR
Rotor position is controlled by means of worm gear (worm/pinion
device).
Controlled voltage (AC) source 0 ÷ 2V1 (laboratory application).
IM2- 11