Download Induction Machine

4. INDUCTION “ASYNCHRONOUS” MACHINES
4.1. Introduction
The induction machine is the most commonly used electric machine. Most industrial
motors of one horsepower or larger are three-phase induction machines. The construction of an
induction machine is similar to a synchronous machine, except that induction machines require
no electrical connection to the rotor windings. Instead, the rotor windings are short circuited.
Magnetic flux flowing across the air-gap links these closed-rotor circuits. As the rotor moves
relative to the air-gap flux, voltages are induced by Faraday’s Law in the shorted rotor windings,
causing currents to flow in them. The fact that the rotor current arises from induction, rather
than conduction is the basis for the name of this class of machines. They are also called
“asynchronous” machines because their operating speed is slightly less than synchronous speed
in the motor mode, and slightly greater than synchronous speed in generator mode.
When the induction machine operates as a generator, their particular characteristics make
them suitable for only a few special applications.
As motors, the have many advantages. They are rugged, relatively inexpensive, and
require very little maintenance. They range in size from a few watts to about 10,000 hp. The
speed of an induction motor is nearly but not quite constant, dropping only a few percent in
going from no load to full load. The main disadvantages of induction motors are:
•
The speed is not easily controlled.
•
The starting current may be five to eight times full-load current.
•
The power factor is low and lagging when the machine is lightly loaded.
For most applications, their advantages far outweigh their disadvantages.
4.2. Construction Details of an Induction Machine
The stator core and stator windings of a three-phase induction machine are exactly like
those of a synchronous machine. The only difference in construction between the two machines
is in the rotor. Induction machine rotors are of two types, wound and squirrel-cage.
In the wound rotor type, the rotor windings are contained in slots in a laminated iron
case, which is mounted on the shaft. The winding of a wound rotor is a polyphase winding. It is
quite similar to stator winding of a synchronous machine. It is almost always three phase, and
wye-connected. The three terminal leads are brought to slip rings, mounted on the shaft. Carbon
brushes riding on these slip rings are shorted together for normal operation. Wound rotors are
normally used only in large machines. External resistances are inserted into the rotor circuit to
improve starting characteristics. As the motor accelerates, the external resistances are gradually
reduced to zero.
Squirrel-cage rotor windings consist of solid bars of conducting material in the rotor
slots. These rotor bars are shorted together at the two ends of the rotor by end rings.
1
4.3. Operation of a Three-Phase Induction Motor
When three-phase voltages are applied to the terminals of the stator winding, balanced,
three-phase currents flow in the phase windings. As a result, a rotating mmf field is produced in
the air-gap of the machine. This field is exactly like that of the stator of a synchronous machine.
Its speed of rotation is given by:
nS =
120 ⋅ f e
P
(rpm)
(4.1)
ωS =
4π ⋅ f e
P
(rad / S )
(4.2)
or
where P is the number of magnetic poles designed into the machine, and fe is the power line
frequency of the stator voltages and currents. The speed of rotation of the mmf field is called the
“synchronous speed.” The strength of this field is proportional to the rms stator current.
This rotating magnetic field induces voltages in the rotor conductors, similar to the
transformer action in a three-phase transformer. The induced voltages give rise to rotor currents,
which interact with the air-gap to produce a torque. The torque is maintained as long as the
rotating magnetic field and the induced rotor current exist. Consequently, the rotor starts rotating
in the direction of the rotating field. The rotor will achieve a steady-state speed, n, such that n <
ns. Clearly, when n = ns, the rotor is moving exactly with the stator’s rotating magnetic field, and
there will be no induced currents, and hence no torque. The condition where n > ns corresponds
to the generator mode, and will be discussed later.
An alternative approach to explaining the operation of the polyphase induction motor is
by considering the interaction of the “excited” stator magnetic field with the “induced” rotor
magnetic field. The stator excitation produces a rotating magnetic field, which rotates in the air
gap at a synchronous speed. The field induces polyphase currents in the rotor, thereby giving
rise to another rotating magnetic field, which also rotates at the same synchronous speed as that
of the stator, and with respect to the stator. Thus two rotating magnetic fields exist, rotating as a
synchronous speed with respect to the stator but stationary with respect to each other.
Consequently, according to the principle of alignment of magnetic fields, the rotor experiences a
torque. The rotor rotates in the direction of the rotating field of the stator.
4.4. The Concept of Slip
The actual mechanical speed, n, of the rotor is often expressed as a fraction of the
synchronous speed, ns, as related by slip, s, defined as
s≡
ns − n ω s − ω
=
ns
ωs
(4.3)
where ns and ωs are given by Equation 4.1. The slip may also be expressed as percent slip,
defined as follows:
percent slip ≡
ns − n
× 100%
ns
(4.4)
2
At standstill, the rotating magnetic field produced by the stator has the same relative
speed with respect to the stator windings. Thus, the frequency of the rotor currents, f2, is the
same as the frequency of the stator currents, f1 = fe. At synchronous speed, there is no relative
motion between the rotating field and the rotor. Thus, the frequency of the rotor current is zero.
At other speed, the rotor frequency is proportional to the slip; that is
f 2 = s f1
(4.5)
where f2 is the frequency of the rotor currents, and f1 is the frequency of the stator input current
(or voltage).
EXAMPLE 4.1
A six-pole, three-phase, 60 Hz induction motor runs at 4% slip at a certain load.
Determine: (a) the synchronous speed, (b) the rotor speed, (c) the frequency of the rotor currents,
(d) the speed of the rotor’s rotating field with respect to the stator, and (e) the speed of the rotor
rotating field with respect to the stator rotating field.
Solution
a) The synchronous speed is
ns =
120 f e 120 (60 Hz )
=
= 1200 rpm
P
6
b) The rotor speed is
n = (1 − s )n s = (1 − 0.04)(1200 rpm) = 1152 rpm
c) The frequency of the rotor current is
f 2 = s f 1 = (0.04)(60 Hz ) = 2.4 Hz
d) First, the speed of the rotor field with respect to the rotor, nmmf2, is
nmmf 2 =
120 f 2 120 (s f )1
=
= s nmmf 1 = s n s
P2
P
where P2 is the number of poles on the rotor, and P2 = P.
Then the speed of the rotor field with respect to the stator, nRF, is
n R F = n mmf 2 + nrotor = s n s + (1 − s )n s = n s = 1200 rpm
e) The speed of the rotor field with respect to the stator field is
n R F − n s = (1200 rpm) − (1200 rpm) = 0 rpm
4.5. Circuit Model of the Induction Machine
In developing the equivalent circuit of the induction machine, let us consider the
similarities between a transformer and an induction machine on a per-phase basis.
3
The primary of a transformer can be considered to be similar to the stator of the
induction machine. The rotor corresponds to the secondary of the transformer. It follows that
the stator and rotor each have their respective resistances and leakage reactances. Because the
stator and the rotor are magnetically coupled, we must have a magnetizing reactance, Xm, just as
in a transformer. The air gap in an induction machine makes its magnetic circuit relatively poor
and the corresponding magnetizing reactance is relatively smaller, compared to that of a
transformer. A shunt resistance, Rc, represents the hysteresis and eddy-current losses in an
induction machine as was done for the transformer. This is where the similarities between the
transformer and induction machine end; the major difference is introduced because of the
rotation of the rotor. Consequently, the frequency of the rotor current is different from the
frequency of the stator currents. Keeping these facts in mind, we can now proceed to represent a
three-phase induction motor by a stationary equivalent circuit.
Considering the rotor first, and recognizing that the frequency of the rotor currents is the
slip frequency, the rotor leakage reactance, x2, may be expressed in terms of a standstill reactance
X2 and the slip, s, on a per phase basis:
x2 = s X 2
(4.6)
For the next step, it is observed that the magnitude of the voltage induced in the rotor
circuit is also proportional to the slip. From transformer theory, we may view the induction
machine’s rotor at standstill as a transformer with an air gap. For the transformer, the induced
voltage on the secondary is given by
E 2 = 4.44 f N 2 φ max
(4.7)
where f is the line frequency and is the same as fe or f1. But in the case of an induction machine,
the frequency becomes f2, which is equal to s f1. Substituting this value of the frequency into
Equation 4.7 yields the induced voltage at a slip s:
e2 = 4.44 s f 1 N 2 φ max = s E 2
(4.8)
where E2 is the per phase voltage induced in the rotor at standstill.
Using Equations 4.6 and 4.8, the rotor equivalent circuit, shown in Figure 4.1(a), is
obtained. The rotor current, I2, is given by
I2 =
s E2
E2
=
R 2 + j s X 2 R2 s + j X 2
(4.9)
resulting in the alternative form of the equivalent circuit shown in Figure 4.1(b). Notice that
these circuits are drawn on a per phase basis. To this circuit, the per-phase stator equivalent
circuit is added to obtain the complete equivalent circuit of the induction machine.
I2
s X2
R2
I2
s E2
X2
R2 / s
E2
(a)
(b)
Figure 4.1. Two Forms of the Rotor Equivalent Circuit.
4
In an induction machine, only the stator is connected to the ac source, and the rotor’s
voltage and current are produced by induction. In this regard, as mentioned earlier, the induction
machine may be viewed as a transformer with an air gap, having a variable resistance in the
secondary. Thus we may consider that the primary of the transformer corresponds to the stator
of the machine, whereas the secondary corresponds to the rotor on a per phase basis. Because of
the air gap, the value of the magnetizing reactance, Xm, tends to be relatively low compared with
that of a transformer. As in a transformer, a mutual flux links both the stator and rotor, and is
represented by the magnetizing reactance and various leakage fluxes. The total stator leakage
flux is denoted by X1 and the total rotor leakage flux is denoted by X2 as illustrated in Figures 4.1
and 4.2. We may consider that the rotor is coupled to the stator like the secondary of a
transformer is coupled to its primary winding. Hence, the induction machine circuit may be
drawn as shown in Figure 4.2.
I1
R1
X1
V1
X2
a:1
E1
E2
I2
R2 / s
Figure 4.2. Stator and Rotor as Coupled Circuits.
To develop this circuit further, the rotor quantities should be express with reference to
the stator, as was done for the transformer. The pertinent details are cumbersome and are not
considered here. However, having referred the rotor quantities to the stator, the per-phase exact
equivalent circuit is obtained and shown in Figure 4.3(a) with equivalent rotor impedance values
R'2 and X'2. For reasons that will be clear later in the chapter, it is convenient to split the
secondary resistance as
R2′
R′
= R2′ + 2 (1 − s )
s
s
(4.10)
which results in the circuit shown in Figure 4.3(b). Here R'2 is simply the per phase standstill
rotor resistance referred to the stator, and R'2 (1 - s) / s is a dynamic resistance that depends on
the rotor speed and corresponds to the load on the motor. Notice that all the parameters shown in
Figure 5.3(b) are standstill values and the circuit is the per-phase exact equivalent circuit model
referred to the stator.
Note that the value of Rc is very large compared to Xm and it can often be neglected.
5
I1
R1
X1
X'2
I2
I0
R'2 / s
V1
Xm
Rc
(a)
I1
R1
X1
X'2
R'2
I2
I0
V1
Xm
Rc
R'2 (1 - s)
s
(b)
Figure 4.3. Two Forms of the Exact Equivalent Circuits of an Induction Motor.
4.6. Losses, Power Flow, and Efficiency of Induction Motors
The electrical power input to an induction motor is given by
Pin = 3 VLL I L cos φ = 3 V p I p cos φ
(4.11)
There are several losses in an induction motor. They are:
1. Stator copper loss, SCL
SCL = 3 I 12 R1
(4.12)
2. Core loss, Pc
3. Rotor copper loss, RCL
RCL = 3 I 22 R2′
(4.13)
4. Mechanical losses or friction and windage losses, Pfw
5. Stray load loss, Psll, consisting of all losses not covered above, such as losses due to
harmonic fields. Sometimes it can be neglected.
The equivalent circuit of an induction motor without Rc is shown in Figure 4.4(a). Also
illustrated if Figure 4.4(b) is the power flow with the different losses.
6
I1
R1
X1
X'2
R'2
I2
I0
Pg / 3
V1
R'2 (1 - s)
s
Xm
(a)
Pin
Pg
SCL
Pd
RCL
Pout
Prot
(b)
Figure 4.4. (a) An Equivalent Circuit of an Induction Motor with Rc Neglected.
(b) The Power Flow and Associated Losses of an Induction Motor.
There is no general agreement as to how to treat core losses in the model. The core loss
resulting from stator leakage flux is not negligible, as it is in a transformer. The rotor-core loss
varies with rotor frequency, and hence with the slip. Under running condition, the slip is on the
order of 0.03; the rotor frequency is only about 2 Hz, and the rotor-core loss is negligible. At
starting, and during acceleration, however, the rotor-core loss is high and decreases while the
friction and windage losses start at zero and increase. As a result, the sum of the friction,
windage, and core losses is roughly constant. This leads to the term “rotational losses”:
Prot ≡ Pfw + Pc + Psll
(4.14)
A very important quantity in the analysis of induction machines is the power transferred
by the air-gap magnetic field from the stator windings to the rotor. The quantity will be given the
symbol, Pg, and is called the “air gap power.” Neglecting the stator core losses at this point, the
air gap power is given by subtracting the stator copper loss from the input to the stator from the
electrical source:
Pg = Pin − SCL
(4.15)
= 3 VLL I L cos φ − 3 I 12 R1
The air gap power is the power input to the rotor and it is partly consumed by the rotor
copper loss, RCL, with the remainder available to develop mechanical power. The “Developed
Mechanical Power” is denoted by Pd, and is obtained by subtracting the copper losses.
7
Pd = Pg − RCL
(4.16)
= Pg − 3 I 22 R2′
Having determined the developed mechanical power, Pd, the net mechanical output
power is found by subtracting the rotational losses from Pd.
Pout = Pd − Prot
(4.17)
= Pd − Pfw − Pc − Psll
Note that the power transferred from one of the three stator-phase circuits of a threephase machine is shown in Figure 4.4(a), and is one-third of the total air-gap power. It is evident
that this power must be consumed by the total resistance R'2 / s:
Pg
3
= I 22
R2′
s
→ Pg = 3 I 22
R2′
s
(4.18)
From Equation 4.13, the rotor copper loss is given by
RCL = 3 I 22 R2′
(4.19)
= Pg s
By comparison with Equation 4.18, the developed mechanical power is
Pd = Pg − RCL
 R′

= 3 I 22  2 − R2′ 
 s

1− s 
2 R 2′
(1 − s )
= 3 I 22 R2′ 
 = 3 I2
s
 s 
Pd = Pg (1 − s )
(4.20)
The developed mechanical torque is given by
τd =
Pd
ω
Nm
(4.21)
where ω is the angular velocity of the rotor and it is equal to ωs (1 - s). Then
τd =
Pg (1 − s ) Pg
Pd
=
=
ω s (1 − s ) ω s (1 − s ) ω s
Nm
(4.22)
Once the air gap power is determined, three quantities (RCL, Pd, and τd) may be found from the
slip and synchronous speed, as shown in the above equations.
EXAMPLE 4.2
An induction motor draws 25 A from a 460 V, three-phase line at a power factor of 0.85
lagging. The stator copper loss is 1000 W and the rotor copper loss is 500 W. The “rotational”
losses include 250 W of friction and windage, 800 W of core, and 200 W of stray load losses.
Calculate (a) the air-gap power, (b) the developed mechanical power, Pd, (c) the output horse
power, and (d) the efficiency.
8
Solution
a)
Pg = Pin − SCL
= 3 (460V )(25 A)(0.85 pf ) − 1000W
= 15.93kW
b)
Pd = Pg − RCL
= 15.93kW − 500W
= 15.43kW
c)
Pout = Pd − Prot = Pd − ( Pc + Pfw + Psll )
= 15.43kW − (800W + 250W + 200W )
= 14.18kW
Pout =
14.18kW
= 19.0hp
746hp / kW
d)
η=
Pout 14.18kW
=
= 83.3%
Pin 16.93kW
EXAMPLE 4.3
If the frequency of the source in Example 4.2 is 60 Hz, and the machine has four poles,
fine (a) the slip, (b) the operating speed, (c) the developed torque, and (d) the output torque.
Solution
a) By applying Equation 5.18
s=
500W
RCL
=
= 0.0314
15.93kW
Pg
b)
4π f e 4π (60 Hz )
=
= 188.5 rad / s
4
P
120 f e 120 (60 Hz )
ns =
=
= 1800 rpm
4
P
ω = ω s (1 − s ) = 188.5 (1 − 0.0314) = 182.6 rad / s
ωs =
n = n s (1 − s ) = 1800 (1 − 0.0314) = 1744 rpm
9
c)
τd =
Pd 15.43kW
=
= 84.5 N ⋅ m
ω
182.6
d)
τ out =
Pout 14.18kW
=
= 77.7
ω
182.6
N ⋅m
Note that the output torque is 8% less than the developed torque.
4.7. Performance Calculations Using the Circuit Model
If the impedance elements of the model of a given motor are known, the performance of
the motor at any speed may be calculated on the basis of relationships previously derived. The
steps are as follows:
1. Calculate the synchronous speed.
ωs =
4π f 1
120 f 1
or n s =
P
P
(4.1)
2. For the desired speed, calculate the slip.
s=
ω s − ω ns − n
=
ωs
ns
(4.3)
3. Calculate the rotor impedance.
z2 =
R2′
+ jX 2′
s
(4.23)
4. Calculate the field impedance, Zf. This is the effect of the resultant magnetic field on
the impedance of one phase of the stator. It is the parallel combination of the rotor
impedance, z2, and the core’s magnetizing reactance, j Xm, referring to Figure 4.5(a).
′
j X m  R2 + j X 2′ 
 s

Z f = Rf + j X f ≡
R2′
+ j (X m + X 2′ )
s
(4.24)
5. Calculate the impedance looking in from the phase terminals of the machine, as
shown in Figure 4.5(b).
Z in = z1 + Z f = (r1 + R f ) + j (x1 + X f
)
(4.25)
6. Calculate the phase current.
I1 =
V1 ∠0°
(4.26)
Z in
10
I1
I1
X1
R1
X'2
Zf
Xm
E1
Xf
Pg / 3
V1
Zin
Zf
R'2 / s
Rf
(a)
(b)
Figure 4.5. The Field Impedance Concept in the Induction Machine.
(a) The Field Impedance, Zf. (b) The Model Employing Zf.
7. Calculate the power factor.
pf = cos φ
8. Calculate the input power.
Pin = 3 VLL I L cos φ
= 3 V1 I 1 cos φ
9. Calculate the stator copper loss.
SCL = 3 I 12 r1
(4.27)
10. Calculate the air gap power.
Pg = 3 I 12 R f
(4.28)
11. Calculate the rotor copper loss.
RCL = s Pg
12. Calculate the developed mechanical power.
Pd = (1 − s ) Pg
13. Calculate the developed torque.
τd =
Pg
ωs
N m or 7.04
Pg
ns
ft lb
14. Calculate the output power.
Pout = Pd − Prot
11
15. Calculate the output torque.
τ out =
Pout
ω
N m or 7,04
Pout
n
ft lb
16. Calculate the efficiency.
η=
Pout
Pin
EXAMPLE 4.4
A 220 V line-to-line, three-phase, wye-connected, 10 hp, 60 Hz 4-pole induction motor
has the following constants in ohms per phase with reference to the stator:
r1 = 0.39 Ω
R'2 = 0.14 Ω
x1 = 0.35 Ω
X'2 = 0.35 Ω
Xm = 16.0 Ω
Rotational losses are 350 W.
(a) Calculate the performance of this motor for a speed of 1746 rpm.
(b) Find the starting current and torque.
Solution:
(a)
120 (60 Hz )
= 1800 rpm
4
1800 rpm − 1746 rpm
s=
= 0.03
1800 rpm
R′
0.14 Ω
z 2 = 2 + j X 2′ =
+ j 0.35 Ω = 4.68 Ω∠4.29°
0.03
s
ns =
Note how the resistance part dominates this impedance.
Zf =
( j X m )(z 2 )
j X m + z2
=
( j 0.35 Ω )(4.68 Ω∠4.29°) = 74.9 Ω∠94.3°
( j 0.35 Ω ) + (4.67 + j 0.35)Ω 17.0 Ω∠74.1°
= 4.40 Ω∠20.2° = 4.13 Ω + j 1.52 Ω
R f = 4.13 Ω
Z in = (0.39 Ω + 4.13 Ω ) + j (0.35 Ω + 1.52 Ω ) = (4.52 + j 1.87 )Ω = 4.89 Ω∠22.5°
I1 =
V1
V
1
= LL ∠0°
=
Z in
Z in
3
220V∠0°
3 (4.89 Ω∠22.5°)
pf = cos − 22.5° = 0.924
12
= 26.0 A∠ − 22.5°
Pin = 3 VLL I L cos φ = 3 (220V )(26.0 A)(0.924) = 9150 W
Pg = 3 I 12 R f = 3 (26.0 A) 2 (4.13 Ω) = 8380 W
Pd = (1 − s ) Pg = (1 − 0.03)(8380 W ) = 8120 W
Pout = Pd − Prot = 8120 W − 350W = 7770 W
=
1 hp
7770W = 10.4 hp
746 W
τ d = 7.04
Pg
ns
τ out = 7.04
η=
= 7.04
8380 W
= 32.8 ft lb
1800 rpm
Pout
7770W
= 7.04
= 31.3 ft lb
1746 rpm
n
Pout 7770 W
=
= 84.9%
9150 W
Pin
(b) At starting, the machine is at standstill.
s=
ns − 0
=1
ns
+ j x 2 = 0.14 Ω + j 0.35 Ω = 0.377 Ω∠68.2°
1
s
( j 16 Ω)(0.377 Ω∠68.2°) 6.03 Ω∠158.2°
Zf =
=
= 0.369 Ω∠68.7°
0.14 Ω + j (0.35 + 16) Ω 16.35 Ω∠89.5°
= (0.134 + j 0.344) Ω
Z in = (0.39 + 0.134) Ω + j (0.35 + 0.344) Ω = (0.524 + j 0.694) Ω
z 2 = r2
= 0.870 Ω∠52.9°
I 1− Starting =
220V 3
= 146 A
0.870 Ω
Note that the starting current is nearly six times the current for a slip of 0.03.
Rotational losses may be neglected at zero speed (s = 1), and the starting torque is equal
to the developed torque.
Pg = 3 I 12 R f = 3 (146 A) 2 (0.134 Ω) = 8570W
Td = 7.04
Pg
ns
= 7.04
(8570W )
= 33.5 ft lb
(1800 rpm)
Note that, in spite of the high starting current, the starting torque is only slightly higher than that
at 1746 rpm.
13
4.8. Torque-Speed Characteristics of Induction Machines
An expression for the torque of an induction machine as a function of its slip may be
obtained by application of Thevenin’s theorem to the exact equivalent circuit model of an
induction motor. According to Thevenin’s theorem, the equivalent source voltage VTh is the
voltage that would appear across the terminals a and b of Figure 4.6 with the rotor circuit open.
Hence,
VTh =
Vin ( j X m )
r1 + j (x1 + X m )
I1
R1
(4.29)
X1
X'2
I2
a
I0
Air Gap
V1
R'2 / s
Xm
Stator
Rotor
b
(a)
ITh
RTh
XTh
X'2
I2
a
VTh
R'2 / s
Stator
b
Rotor
(b)
Figure 4.6. Application of Thevenin’s Theorem to the Circuit Model of an Induction Motor.
The Thevenin equivalent stator impedance RTh + j XTh is the impedance between the
terminals a and b of Figure 4.6, viewed towards the source with the source voltage Vin shortcircuited. The impedance is calculated as:
Z Th = RTh + j X Th ≡ (r1 + j x1 ) // ( j X m )
=
(r1 + j x1 ) ( j X m )
(r1 + j x1 ) + ( j X m )
(4.30)
From the Thevenin equivalent circuit as shown in Figure 4.6, and the torque expression of
Equation 4.22, the develop torque can be described as
3 I 22 R2′
τd =
=
ωs
ωs s
Pg
(4.31)
where
14
I2 =
VTh
 R + R2′  + j (X + X ′ )
 Th
2
Th
s 

By combining Equations 4.31 and 4.32, we find
τd =
3
ωs 
R′
 RTh + 2 s

VTh2
2
 + (X + X ′ )2

2
Th

R2′
s
(4.33)
The general shape of the torque-speed or the torque-slip curve is that shown in Figure 4.7. Both
the motor region (s > 0) and the generator region (s < 0) are shown in the Figure.
The internal torque is at maximum when the power delivered to R'2 / s is at maximum, as
illustrated in Figure 4.6(b). By the familiar impedance-matching principle in circuit theory, this
power will be the greatest when the impedance of R'2 / s equals the magnitude of the source
impedance. In other words, maximum torque and power happen at the matching of the source
impedance and a slip value, smax, for which
R2′
2
= Z Th + j X 2′ = RTh2 + (X Th + X 2′ )
s
(4.34)
The slip at maximum torque, smax, is therefore:
s max =
R2′
R + (X Th
2
Th
(4.35)
2
+ X 2′ )
The strength of the maximum torque may be found by inserting the expression for smax,
Equation 4.35, into the equation for the developed torque, Equation 4.33. Note that in Equation
4.33, the slip, s, always occurs in terms of R'2 / s, and that R'2 / smax by Equation 4.34, is
independent of the value of R'2. Therefore, the maximum torque, τmax, is independent of the rotor
resistance, R'2:
τ max =
VTh2
3
⋅
ω s 2  R + R 2 + (X + X ′ )2 
2
Th
Th
 Th

(4.36)
Note that the maximum torque is independent of the rotor resistance while the slip at which it
occurs is proportional to the rotor resistance.
15
M ax im um
1
Torque
0 .5
to rq u e
M otor M ode
B rak e M ode
0
-2
-1
0
G enerator M ode
1
2
-0 .5
-1
M inim um
Torque
-1 .5
s lip
(a) Torque as a function of slip
1
S tarting
F ull load torque
torque
to rq u e
0 .5
0
−1 ω
No load
M otor m ode
B rak e m ode
0 ω
1 ω
G enerator
m ode
2 ω
3 ω
-0 .5
-1
-1 .5
sp eed
(b) Torque as a function of speed
Figure 4.7. Torque of an Induction Machine. (a) Developed Torque as a Function of Slip.
(b) Developed Torque as a Function of Speed.
In motor operation, the resistance of the rotor may be reduced to maintain high
accelerating torque. For a “soft” start, the machine may be connected to the line with the rotor
open-circuited. Relatively high resistance would then be connected to the slip rings and
gradually reduced until sufficient torque was produced to start the load.
16
EXAMPLE 4.5
A 220 V line-to-line, three-phase, wye-connected, 10 hp, 60 Hz 6-pole induction motor
has the following constants in ohms per phase with reference to the stator:
r1 = 0.294 Ω
R'2 = 0.144 Ω
x1 = 0.503 Ω
X'2 = 0.209 Ω
Xm = 13.25 Ω
(a) For a slip s = 0.03 determine the load current, I2, the developed torque, τd, and the developed
power Pd. (b) Find the maximum torque and the corresponding speed. (c) Calculate the starting
torque and the corresponding current I2.
Solution:
Using the circuit in Figure 4.6 and Equations 4.29 and 4.30,
VTh = 122.3V
Z Th = RTh + j X Th = 0.273 Ω + j 0.490 Ω
(a) At s = 0.03,
R2′ 0.144 Ω
=
= 4.8 Ω
s
0.03
VTh
I2 =
=
2
′
2
R
 R + 2  + (X + X ′ )
 Th
2
Th
s 

122.3V
(5.07 Ω) 2 + (0.699 Ω) 2
= 23.9 A
3 I 22 R2′ 3 (23.9 A) 2 (4.8 Ω)
τd =
=
= 65.5 N m
ωs s
125.6
R′
Pd = 3 I 22 2 (1 − s ) = 3 (23.9 A) 2 (4.8 Ω) (1 − 0.03) = 7970 W
s
(b) At the maximum torque point, the slip is calculated from Equation 4.35.
s max =
nτ − max
R2′
RTh2 + (X Th + X 2′ )
0.144 Ω
=
(0.273 Ω )2 + (0.699 Ω) 2
= (1 − s max ) ns = (1 − 0.192) 1200 rpm = 970 rpm
2
VTh2
3
⋅
2 ω s  R + R 2 + (X + X ′ )2 
2
Th
Th
 Th

2
(122.3V )
3
=
⋅
= 175 N m
2 (125.6) [(0.273 Ω) + (0.750 Ω)]
τ max =
17
=
0.144
= 0.192
0.750
(c) At starting, s = 1, and R'2 will be assumed constant. Therefore,
R2′
= R2′ = 0.144 Ω
s
R′
RTh + 2 = 0.417 Ω
s
I 2 − starting =
VTh
R′ 

2
 RTh + 2  + (X Th + X 2′ )
s 

2
(122.3V )
(0.417 Ω )2 + (0.699)2
=
= 150.5 A
From Equation 4.31,
τ starting
3 I 22 R2′ 3 150.5 A 2
=
=
(0.144 Ω ) = 78.0 N m
ωs s
125.6
4.9. The Approximate Equivalent Circuit
The preceding examples illustrate the usefulness of equivalent circuits of induction
motors. For most purposes, an approximate equivalent circuit is adequate. One such circuit is
shown in Figure 4.8. Obviously, in order to use this circuit for calculations, its parameters must
be known. The parameters of the circuit shown in the figure can be obtained from the following
test: (a) no-load test, (b) dc resistance test, and (c) the blocked rotor test.
I1
R1
X1
X'2
R'2
I2
I0
R'2 (1 - s)
s
V1
Xm
Rc
Figure 4.8. Approximate Equivalent Circuit of the Induction Motor.
4.9.1. The No-Load Test
In the no-load test or open-circuit test, the motor runs at no-load, or in other words, no
mechanical loading on the shaft. A wattmeter measures the core, friction and windage losses.
The mechanical load per phase, represented by R'2 (1 - s) / s, is simply the internal windage and
friction of the machine. This amounts to only a few percent of the rated mechanical load, and the
slip is very small, say less than 0.001, and R'2 (1 - s) / s is very large. Therefore, the circuit
shown in Figure 4.8 can be considered as open-circuit for the branch of I2, and the magnetizing
current, I0, is equal to the input current and the open-circuit test current, I1 = Ioc. The no-load or
open-circuit test input-power, Poc, may be taken to be the sum of the friction and windage loss,
Pfw, and the core loss, Pc. The core loss can be determined by
Pc = 13 ( Poc − Pfw )
(4.37)
where Poc is the three-phase no-load input-power to the machine at rated voltage and frequency.
18
Note that under no-load conditions, the power factor is quite low, 0.2 or less, so the circuit is
essentially reactive. This input impedance at no load is thus approximately Rc in parallel with
j Xm and they can be obtained as:
Rc =
Voc2 − LL
Poc
(4.38)
where Voc-LL is the line-to-line voltage during the open circuit test. Next, Xm is obtained. First, Ioc
is the vectorial sum of Im and Ic. Also
Ic =
Voc −φ
Rc
Poc
=
(4.39)
3 Voc − LL
and
I m = I oc2 − I c2
(4.40)
Then
Xm =
Voc −φ
(4.41)
Im
The above equations show that the core parameters (Rc and Xm) can be easily obtained from the
no load test.
4.9.2. The DC Resistance Test
In the case of a transformer, it was only necessary to obtain Req1 or Req2 to solve practical
problems. Knowledge of the individual winding resistances was not essential. In the induction
machine, however, R2 plays a crucial role, and R1 must be known to evaluate the stator copper
loss.
With a steady direct current flowing in the stator windings, the stator magnetic field is
fixed in space. If no external torque turns the shaft, the flux linkage with the rotor circuits are
constant in time. As a result, no voltages are induced in the rotor, no rotor currents flow and no
torque is developed. There is no rotor mmf, and the rotor circuits have no electrical effect on
those of the stator. Thus the dc stator winding resistance may be measured independently of the
rotor impedance. In this test, a rheostat is connected in series with the stator winding and it must
be capable of handling the rated current of the motor. The current is adjusted to approximately
rated value, so that the temperature of the winding approximates that of running conditions.
Applying Ohm’s law to the circuit of Figure 4.9 results in the following equations:
2 R1 =
Vdc
I dc
or
R1 =
Vdc
2 I dc
(4.42)
19
c
Stator
R1
R1
R1
a
b
Idc
Vdc
Figure 4.9. DC Measurement of the Stator Winding Resistance.
4.9.3. The Blocked Rotor Test
This test, which corresponds to the short-circuit test of transformer, is also called
“blocked-rotor test.” The shaft is clamped, so that it cannot turn; thus ω = 0 and s = 1 during this
test. If full voltage at rated frequency were applied, the current would be five to eight or more
times rated value. For this reason, the blocked-rotor tests are not done at full voltage, except for
small motors. Instead, the applied voltage is usually adjusted upward from zero until the stator
current is at rated value. Blocked-rotor test measurements are thus taken at rated current rather
than at rated voltage.
Under running conditions the frequency of the stator currents is line frequency (f1), but
the frequency of the rotor currents is only a few hertz. However in the blocked rotor test, the slip
is unity and the rotor and stator currents have the same frequency. It is usually the practice to use
a lower test frequency for the input voltage and current for the blocked rotor test.
The effect of blocking the rotor on the model is shown in Figure 4.10. The voltage
appearing across one phase of the stator winding is VBR-φ, and the magnitude of the impedance
looking into the terminal of one phase of the machine is:
Z BR =
VBR −φ
(4.43)
I BR −φ
where IBR-φ is the average of the three line currents, and should roughly equal the rated current.
IBR-φ
R1
X1
X'2
V BR-φ
Figure 4.10. Circuit Model at Blocked Rotor.
20
R'2
I2
With unity slip, (R'2/s) = R'2, the applied voltage is quite low and the iron core of the
machine is unsaturated. As a result, Xm is larger than normal. Even at rated voltage, Xm is
typically 25 times the magnitude of R'2 + j X'2. The effect of Xm may thus be neglected, and the
assumption made that
′ = R1 + R2′ + j X BR
′
Z BR = RBR + j X BR
(4.44)
where X'BR is the blocked-rotor reactance at the test frequency, ft. The real part of ZBR is given by
the power per phase divided by the phase current squared, as
RBR =
PBR
= R1 + R2′
2
3 I BR
(4.45)
or
RBR = Z BR cos θ BR
(4.46)
where
θ BR =
PBR
(4.47)
3 VBR I BR
Now, R1 was found by the dc test (Equation 4.42). Then
R2′ = RBR − R1 =
PBR
− R1
2
3 I BR
(4.48)
The imaginary part of ZBR is given by
2
′ = Z BR
X BR
sin θ BR
(4.49)
2
2
′ = Z BR
X BR
− RBR
(4.50)
or
If rated frequency is f1 and the frequency of the blocked-rotor test is ft, then let
X BR ≡
f1
′ = X 1 + X 2′
X BR
ft
(4.51)
The designer of the machine knows, on the basis of the design calculations, how much of
XBR is X1 and how much is X2. There is no way to determine this ratio from tests on a squirrelcage motor.
The above equations show that the stator and the rotor parameters can be determined
from the dc test and the blocked rotor test.
EXAMPLE 4.6
The results of a no-load test and a block rotor test on a three-phase wye-connected
induction motor are as follows:
no-load test:
line-to-line voltage
input power
frequency
400 V
1770 W
60 Hz
21
input current
friction and winding loss
18.5 A
600 W
blocked rotor test:
line-to-line voltage
input power
45 V
2700 W
input current
test frequency, ft
Determine the parameters of the approximate equivalent circuit of Figure 4.8.
Solution:
a) no-load test:
V0 =
400V
= 231V
3
P0 = 13 (1770W − 600W ) = 390W
I 0 = 18.5 A
V02 (231V )2
Rc =
=
= 136.8Ω
P0
390W
Ic =
P0 390W
=
= 1.69 A
V0
231V
(18.5 A)2 − (1.69 A)2
I m = I 02 − I c2 =
Xm =
= 18.42 A
V0
231V
=
= 12.5Ω
I m 18.42 A
b) blocked rotor test:
VB =
PBR
45V
= 25.198V
3
= 13 (2700 W ) = 900W
I BR = 63 A
RBR = R1 + R2′ =
Z BR =
PBR
900 W
=
= 0.23 Ω
2
VBR (63 A) 2
VBR (25.198V )
=
= 0.412 Ω
I BR
(63 A)
2
2
′ = Z BR
X BR
− RBR
= (0.412 Ω) 2 − (0.23 Ω) 2 = 0.34 Ω
X BR =
where
f1
f test
60 Hz
(0.34 Ω) = 1.36 Ω
15 Hz
= X 1 + X 2′
′ =
X BR
X BR
22
63 A
15 Hz