Unit-4 Induction Motors Objectives: After completing this Unit-4, you will be able to: • Explain why an induction motor cannot run at synchronous speed. • Describe the construction of two types of rotors for an induction motor. • Determine slip for a given speed of the motor. • Derive an expression for the emf induced in the rotor. • State how the induction motor differs from a transformer. • Draw and explain the power flow diagram for an induction motor. • Draw the equivalent circuit of an induction motor. • Draw and explain the torque-slip characteristic of an induction motor. • Explain how the starting torque can be increased in a phase-wound i nduction motor by inserting more resistance. • Derive the condition for obtaining maximum torque.
•
Derive an expression for the ratio of torque
τ for any s to the maximum torque τ m .
4.1 INTRODUCTION The ac synchronous motor has limited practical applications. An alternative is the asynchronous motor which we usually call induction motor. motor. Most motors that we meet in the home and in industry are induction motors. These motors are more rugged, they need less maintenance, and they are less expensive than the synchronous motors or dc motors 1. Induction motors are available both for three-phase and single-phase operation. Three-phase induction motors are used for high power and industrial applications such as lifts, cranes, pumps, exhaust fans, lathes, etc. Single-phase induction motors 2 find use in domestic electric appliances such as fans, refrigerators, washing machines, pumps, hair-driers, etc.
4.2 PRINCIPLE OF WORKING The stator of an induction motor (Fig. 4.1 a) is similar to that of a synchronous machine. It has three-phase windings, P poles, sinusoidal mmf and flux distribution. For simplicity, the stator slots and windings are omitted in the figure. When three-phase currents flow through the stator windings, a magnetic flux is produced that rotates at synchronous speed given by Eq. 14.8, repeated here for convenience,
N s =
120 f P
...(4.1) The rotor is an iron laminated cylinder with large embedded conductors in the form of copper or aluminium bars in th e semiclosed slots. The slots are usually not made parallel to the axis, but are given a slight twist. The rotor is then known as skewed 3 rotor. The bars are short-circuited at each end by a conducting ring or plate (Fig. 4.1 b). The bars and the shorting rings look like a squirrel cage, cage, as shown in Fig. 4.1c. The air-gap between the rotor and the stator is uniform and is made as s mall as possible mechanically.
(a) Front view.
(b) Side view. Fig. 4.1
(c) Squirrel-cage rotor. Induction motor
Suppose that the stator is wound for two poles. Let the distribution of magnetic flux due to stator currents at a particular instant be as shown in Fig. 4.1 a. Assume that the stator flux rotates anticlockwise. With respect to this flux, the rotor conductors move in clockwise direction. The emfs are thus induced 4 in the rotor conductors, whose directions can be determined by Fleming’s right hand 1
We shall discuss dc motors in the next Chapter. We shall discuss single-phase induction motors in Chapter 17. 3 The skewing of the rotor helps in reducing noise, in increasing the starting torque and in eliminating clogging. This is why this type of motor is called an induction motor . In fact, an induction motor is like a transformer with its secondary winding short-circuited. short-circuited. The o nly difference is that in an induction motor, the secondary winding is free to rotate. 2
1
rule, as indicated by crosses and dots in Fig. 4.1 a. The emf generated in the rotor conductors is a maximum in the region of maximum flux density. N F1 A
B F2 S Fig. 4.2 Torque generated on the rotor. A Now, consider conductors and B of the rotor which face the N -pole -pole and S-pole of the stator, as shown in Fig. 4.2. The emf generated in these conductors circulates a current, which in turn produces their own flux. The resultant of these two fluxes is such as to strengthen the flux density on the right-side and weaken that on the left side for the conductor A. Consequently, the conductor A experiences a force F 1 left-ward. Similar action takes place for the conductor B, so that it experiences a force F 2 right-ward. These two forces create a torque
τ e that tends to rotate the rotor in the direction of the rotating flux.
Once the rotor starts rotating, the relative movement between the stator’s rotating field and the rotor-conductors is reduced. As a result, the induced emf, the current, and its frequency are all reduced. If the motor shaft is not loaded, the machine has to rotate to meet the mechanical losses. The rotor speed can approach very close to the synchronous speed. But, it can never be the same as the synchronous speed. If it does, the induced emf in the rotor-conductors would become zero and there would be no torque produced. Hence, the rotor speed always remains slightly less than the synchronous speed. Now, suppose we put a mechanical load on the shaft. Its immediate reaction is to slow down the rotor. As a result, the relative speed with respect to the rotating field increases. The induced emf in the rotor-conductors increases and hence the torque τ exerted on the rotor increases. Ultimately, an equilibrium state is attained. The rotor speed adjusts itself to make the torque τ sufficient to balance the mechanical-loss torque and the load torque. Obviously, the speed of the motor running under full-load is less than the no-load spe Note that unlike a synchronous machine, the induction motor has field on the stator and armature on the rotor.
Slip of Induction Motor As discussed above, the rotor speed must always remain less than the synchronous speed N s given by Eq. 4.1. The difference between the synchronous speed N s and the actual speed N of the rotor is known as slip speed , N∆
= Ns − N . This term is descriptive of the
speed , or simply the slip manner the rotor slips back from the exact synchronous speed. The normalized The normalized slip speed , the slip s is usually expressed as per-unit or fraction of the synchronous speed,
s=
N ∆
=
N s − N
N s
N s ...(4.2)
For a given slip s, the rotor speed is given as
N = Ns (1 − s) ...(4.3) The slip can also be expressed as a percentage of the synchronous speed, as
s=
N s − N N s
×100%
...(4.4) When the motor is at standstill (that is, it is not running), the rotor speed N is zero, and hence s = 1. The value of s can never be zero. Because this would mean that the rotor is rotating at synchronous speed which is impossible. In practice, the value of slip is very small. At no load, the slip is around 1% only and at full load, it is around 3 %. For large size, efficient motors, motors, the slip at full load may be around 1 % only. Thus, an induction motor almost has a constant speed. Thus, we find that 0 ≤ s < 1 . Is it possible to make the slip s have a negative value? Yes, if the rotor is made to rotate by a primemover at a speed higher t han the synchronous speed. The negative slip corresponds to the generator action . Frequency of Rotor Currents: When the induction motor is at standstill, the frequency of the currents induced in the rotor winding is the same as the supply frequency. However, when the motor runs, the frequency of rotor currents depends upon the relative speed or slip-speed. If the rotor-speed N and the synchronous-speed N s are expressed in rpm (revolutions per minute), the frequency f r r of the rotor currents is given by an expression similar to that in Eq. 4.1, as
N s − N =
120 f r P
Dividing the above equation by Eq. 4.1, we get
2
N s − N N
s
∴
=
fr f
f r
s=
or
f
fr = s⋅ f
...(4.5) Speed of Rotation of Rotor-Field: The rotor currents produce their own rotating magnetic field. Since the frequency f r of the rotor currents is s ⋅ f , the speed of this rotating field is s ⋅ N s with respect to the rotor winding. However, the rotor itself is running at a speed N with respect to the stator. Hence, The speed of rotor field in space = Speed of rotor field relative to the rotor + Speed of rotor relative to stator = sN s
+ N = sN s + N s (1 − s) = N s
Thus, we find that even though the rotor is not rotating at synchronous speed, the rotor field rotates at the synchronous speed. In fact, the rotor field remains locked with the stator field, irrespective of the rotor speed.
4.3 CONSTRUCTION OF INDUCTION MOTOR Like any other rotating machine, an induction motor has a stator and a rotor. The stator has three-phase windings which receive energy form three-phase ac supply. The rotor carries windings in which the working currents are induced.
Stator The stator core is a hollow cylindrical structure. It is made of sheet-steel laminations, each about 0.4 mm thick, slotted on its inner surface. The slots in large size motors are open type to facilitate the insertion of form-wound coils, which are well insulated before they are slipped into the slots. In small size motors, the slots are partially closed type. This helps in reducing the effective length of the air gap between the stator and the rotor. The coils are externally wound and are inserted through the narrow openings one wire at a time.
Rotor The rotor is an inner cylindrical core. It may be either squirrel-cage type or wire-wound type. (1) Squirrel Cage Rotor: About 90 % of the motors in use have this type of rotor (see Fig. 4.1 c). This rotor has two main advantages. First, it is adaptable to any number of poles. Secondly, it is simple in construction, has no slip-rings and brushes, is very rugged, and is very economical in manufacturing. The only disadvantage it has is that its resistance is low (and fixed) and hence it has low starting torque. (2) Wire- or Phase-Wound Rotor: It has three-phase double layer distributed windings placed in the slots of the rotor core. It is wound for the same number of poles as the stator. The windings are usually connected in star, though they may be connected in delta. The three ends of the windings are brought out and are soldered to the slip rings mounted on the shaft. Carbon brushes, fixed with the stator, make contact with these moving slip rings. This arrangement makes it possible to connect additional resistances in the rotor windings to give high starting torque (see Fig. 4.16). The external resistances are gradually reduced to zero as the motor picks up speed. Under normal running conditions, the wound rotor is short-circuited like a squirrel cage rotor.
Example 4.1
A 3-phase, 6-pole, 50-Hz induction motor has a slip of 1 % at no load and of 3 % at full load. Find ( a) the synchronous speed, (b) the no-load speed, ( c) the full-load speed, ( d ) the frequency of rotor-currents at standstill, and ( e) the frequency of rotor-currents at full-load.
(c) The full-load speed, N = Ns (1 − s)
=
120 f
=
120 × 50
= 1000 rpm 6 (b) The no-load speed, N = Ns (1 − s) = 1000(1 − 0.01) = 990 rpm Solution: (a) The synchronous speed, N s
P
= 1000(1 − 0.03) = 970 rpm
(d ) At standstill, s = 1. Hence, the frequency of rotor-currents,
fr = s⋅ f = 1× 50 = 50 Hz (e) At full-load, s = 0.03. Therefore, the frequency of rotor-currents,
fr = s⋅ f = 0.03× 50 = 1.5 Hz Example 4.2
A 12-pole, 50-Hz, 3-phase induction motor runs at 485 rpm. What is the frequency of rotor-current?
Solution: The synchronous speed, N s
∴
=
120 f P
120 × 50
12 500 − 485
= 500 rpm
= 0.03 500 ∴ Rotor currents frequency, fr = s⋅ f = 0.03 × 50 = 1.5 Hz Example 4.3
Slip, s =
=
A 6-pole induction motor is fed from 50-Hz supply. If the frequency of rotor emf at full-load is 2 Hz, find the full-
load slip and speed.
3
Solution: The synchronous speed, N s The slip at full load, s
f r
=
The full-load speed, N
f
=
2 50
=
120 f
=
120 × 50 6
P
= 1000 rpm
= 0.04 = 4%
= N s (1 − s) = 1000(1 − 0.04) = 960 rpm
Example 4.4
A three-phase induction motor is wound for four poles and is supplied from a 50-Hz supply. Calculate ( a) the synchronous speed, (b) the speed of the rotor when the slip is 4 %, and ( c) the rotor frequency when the speed of the rotor is 600 rpm. Solution: (a) The synchronous speed, N s
=
120 f
=
120 × 50 4
P
= 1500 rpm
(b) The speed of the rotor when the slip is 4 %,
N = Ns (1 − s) = 1500(1 − 0.04) = 1440 rpm (c) When the speed of the rotor is 600 rpm, the slip is
s=
1500 − 600
= 0.6 1500 ∴ The rotor frequency, fr = s⋅ f = 0.6 × 50 = 30 Hz 4.4 ROTOR EMF, CURRENT AND POWER FACTOR The analysis of induction motor performance is done under the assumption that the applied voltage V 1 per phase is constant and purely sinusoidal of constant frequency f . It is further assumed that the flux per pole Ф is sinusoidally distributed in the space around the air gap and is rotating at synchronous speed N s. We define following quantities: V 1 = applied voltage to stator per phase (V) N 1 = number of turns in series per phase of the stator N 2 = number of turns in series per phase of the rotor Ф = flux per pole (Wb) f = frequency of the supply (Hz) E 1 = emf induced per phase in the stator (V) E 2 = emf induced per phase in the rotor (V) (when the motor is running) E 20 = emf induced per phase in the rotor (V) (when the motor is standstill) R2 = resistance of the rotor winding per phase (Ω) X 2 = reactance of the rotor winding per phase (Ω) (when the motor is running) X 20 = reactance of the rotor winding per phase (Ω) (when the motor is standstill) L20 = inductance of the rotor winding per phase (H) (when the motor is standstill) k d1 , k d2 = distribution factors of stator and rotor windings, respectively k p1, k p2 = pitch factors of stator and rotor windings, respectively N s = synchronous speed (rpm) When the rotor is standstill, an emf is induced in both the stator winding and rotor winding. The rotating field produced by stator three-phase currents cuts the stator and rotor conductors at synchronous speed N s. Since the rotating field makes N s /60 rps (revolutions per second) and the stator (or rotor) conductors cut total flux PФ per revolution, Flux cut per revolution
=
PΦN s
60
This is also the average value of the emf induced in each conductor. Therefore, if K f (= 1.11 for sinusoidal waveshape) is the form factor, the rms value of induced emf in each conductor of the stator is
e1 = K f ×
PΦN s
60
= K f × 2Φ ×
PN s
120
= 1.11× 2Φf = 2.22Φf
In the stator, there are N 1 turns per phase or 2 N 1 conductors in series per phase. Therefore, the induced emf per phase in the stator is given as
E1 = e1 × 2 N1 = 4.44 f Φ N1 On taking the distribution factor and pitch factor into account, above expression is modified as
E1 = e1 × 2 N1 = 4.44 f Φ N1 kd 1 kp1 ...(4.6)
4
Similarly, the induced emf E 20 per phase in the rotor at standstill is given as
E20 = 4.44 f Φ N2 kd 2 kp 2
...(4.7) When the rotor rotates at a slip s, the induced emf also reduces by a factor s. Thus, the induced emf E 2 per phase in the rotor running at a slip s is given as
E2 = sE20 = 4.44 sf Φ N2 kd 2 kp 2 = 4.44( sf )Φ N2 kd 2 kp 2 = 4.44 fr Φ N2 kd 2 kp 2 ...(4.8) The induced emf E 2 is zero when the rotor revolves at synchronous speed (that is, N = N s or s = 0). It increases in proportion to the slip speed because the speed of the rotor-conductors relative to the stator flux increases. Rotor Impedance: The rotor reactance X 2, when it is rotating at a slip s, is given as
X2 = 2π fr L20 = 2π ( s⋅ f ) L20 = s⋅ 2π fL20 = sX20 Therefore, the rotor impedance per phase, under running condition, is given as
Z 2 = R2 + jX2 = R2 + jsX20 = Thus,
Z2 =
2 2 R 2 + ( sX 20 )
R2 + ( sX20 ) ∠ tan 2
2
θ 2 = ∠ tan −1
and
−1
sX 20 R2
sX 20 R2
...(4.9) The impedance of the rotor circuit increases with slip speed because of the increase in the frequency of the induced emf E 2. At very small slip speeds, the impedance is largely resistive. But at larger slip speeds, the inductive reactance of the rotor dominates. Rotor Current: The induced emf per phase in the rotor circuit, under running condition, is given by Eq.4.8. The magnitude of the rotor current is therefore given as
I 2 =
E 2 Z 2
=
sE 20 R22 + ( sX 20 )2
...(4.10) For small values of s, on increasing slip speed, the induced emf E 2 increases at a faster rate than the rotor impedance Z 2. As a result, on increasing slip speed s, the rotor current I 2 initially increases, and then tends to approach a maximum value, where the increase in E 2 is offset by the corresponding increase in Z 2. Power Factor: The phase angle between the induced emf E 2 and the rotor current I 2 is same as the impedance angle θ 2, as given by Eq. 4.9. Thus, the po wer factor of the rotor circuit, under running condition, is given as
pf = cos θ 2 =
R2 2 2 R 2 + ( sX 20 )
(lagging)
...(4.11) As the slip speed increases, the rotor circuit becomes more inductive and the po wer factor becomes poorer.
Induction Motor as Transformer The working of an induction motor resembles in many respect with that of a transformer. The stator winding works as primary. The rotor winding works as secondary. Under standstill condition, when a 3-phase supply is connected to the stator, emfs are induced in both the stator and the rotor, as given by Eqs. 4.6 and 4.7, respectively. These equations are similar to Eqs. 13.4 and 13.5 for a transformer. In a transformer, the primary and secondary coils are concentrated. But in an induction motor, the stator and rotor windings are distributed. Hence, the distribution factor k d and pitch factor k p are included in Eqs. 4.6 and 4. 7 for an induction motor. When the rotor circuit is closed, a current flows in the rotor winding. This current creates an mmf. The rotor mmf and stator mmf can be combined just in the same way as in a transformer. In a transformer, the primary and secondary fields remain stationary. But in an induction motor, the stator and rotor fields keep rotating in space. However, this does not make any difference; these two fields, though rotating, remain stationary with respect to each other. In a transformer, a greater secondary current causes an increased primary current. Similarly, in an induction motor, a greater load on the shaft causes an increased stator current to balance the rotor mmf. If the shaft is held stationary with rotor circuit closed (a condition known as blocked rotor or locked rotor), the situation becomes same as in a transformer with its secondary short-circuit. Under such blocked rotor condition, excessive heating of the induction motor occurs as it draws a very heavy current from the ac supply. Differences from a Transformer: An induction motor differs from a transformer in following ways: 1. Magnetic leakage and hence the leakage reactances of stator and rotor are much higher than those in a transformer. 2. Because of the presence of the air gap, the magnetizing current required is much greater than that in a transformer. 3. Because of the distributed windings, the ratio o f the stator and the rotor currents is not equal to the ratio of turns. 4. The losses are much greater, and hence the efficiency is lower than that of a transformer. 5. The no-load current is about 25 % to 40 % of the rated current, whereas it is only about 2 % to 5 % of the rated current in a transformer.
5
Example 4.5 The induced emf between the slip-ring terminals of a three-phase induction motor, when the rotor is standstill, is 100 V. The rotor windings are star-connected and have resistance and standstill reactance of 0.05 Ω and 0.1 Ω per phase, respectively. Calculate the rotor current and phase difference between the rotor voltage and rotor current at ( a) 4 % slip, and ( b) 100 % slip. Solution: For star-connected rotor windings, the induced emf per phase is given as
E 20 =
E L
=
100
3
3
= 57.7 V
(a) At s = 4 % = 0.04:
I 2 =
cos θ 2 =
∴
E 2 Z 2
sE 20
=
=
R22 + ( sX 20 )2
R2 R + ( sX20 ) 2 2
= 2
0.04 × 57.7 (0.05) 2 + (0.04 × 0.1) 2
0.05 (0.05) + (0.04 × 0.1) 2
2
= 46 A
= 0.99
θ 2 = cos −1 0.99 = 8.1°
(b) At s = 100 % = 1.0:
I 2 =
cos θ 2 =
∴
E 2 Z 2
sE 20
=
R + ( sX 20 ) 2 2
R2 R 2 + ( sX 20 ) 2
= 2
= 2
1.0 × 57.5 (0.05) + (1.0 × 0.1) 2
0.05 (0.05) 2 + (1.0 × 0.1) 2
2
= 514 A
= 0.447
θ 2 = cos −1 0.449 = 63.4°
4.5 POWER RELATIONS FOR AN INDUCTION MOTOR The entire power Pin supplied by the three-phase source to the stator of the induction motor is not converted into the mechanical power at the shaft. As shown in Fig. 4.3, a portion of the input power Pin is lost in the stator as copper and iron loss, which heats the stator. The remaining power Pg is transferred to the rotor via the air-gap magnetic field, similar to the power transferred from primary to the secondary in a transformer. This rotor input power Pg less the rotor losses P R (both the copper loss and iron loss) is the mechanical power Pd developed by the rotor. Since the frequency of the rotor currents is very small (say, around 1.5 Hz), the rotor iron loss is negligibly small. Thus, we can say that the power developed is given as
Pd = Pg − PR = Pg − Rotor copper loss ...(4.12) Out of the mechanical power Pd developed by the rotor, some power loss occurs due to the friction at bearings and slip-rings, and some due to windage (i.e., due to the air resistance experienced by the rotating shaft). The remaining power Po is the net mechanical output power available at the shaft to meet the external mechanical load. Electrical
Mechanical Rotor
Stator
o
Pin
P
PS Stator Cu and iron loss
Fig. 4.3
Pd = (1 − s )
P R Rotor Cu and iron loss
Po
Output power at shaft
Pm Friction and windage loss
Power flow diagram for an induction motor.
Let τ (in Nm) be the electromagnetic torque exerted on the rotor by the rotating magnetic field at synchronous speed N s (in rpm). Then the air-gap power Pg (in W) transferred from the stator to the rotor is given as
Pg =
2πτ N s 60
...(4.13) This is the input power to the rotor. If the rotor rotates at a speed N (in rpm), the total mechanical power developed by the rotor is given as
6
Pd =
2πτ N 60 ...(4.14)
Putting Eqs. 4.13 and 4.14 in Eq. 4.12, we get
P R = P g− P d =
2πτ 60
( N s− N ) ...(4.15)
Dividing the above equation by Eq. 4.13, we get
P R Pg or
=
N s − N N s
=s
(according to the definition of slip)
= s × Pg
The rotor copper loss, P R
...(4.16) Therefore, the power Pd developed by the rotor, using Eq. 4.12, is then given as The power developed by the rotor, Pd
= Pg − PR = Pg − sPg = (1 − s ) Pg ...(4.17)
We can put the air-gap power as
Pg = Pg − sPg + sPg = (1 − s) Pg + sPg
Pd
⇒
Pd
P R
P R
=
1− s s ...(4.18)
This shows that the air-gap power divides between the developed power and rotor-copper loss in a ratio that depends only on the slip speed. Example 4.6 A three-phase, four-pole, 50-Hz induction motor has a full-load output power of 5 hp at 1470 rpm. The efficiency of the motor at full load is 87.5 %. The mechanical losses are 5 % of the total losses. Determine the developed power, air-gap power, rotor copper loss, and stator loss.
Po = 5 hp = 5 × 746 = 3730 W
Solution: The output power,
∴
The input power,
Pin =
Po
η
=
3730 0.875
= 4263 W
Total losses,
Pl = 4263 − 3730 = 533 W
∴
Mechanical losses,
Pm = 0.05× 533 = 26.65 W
and
Electrical losses,
Pe = 533 − 26.65 = 506.35 W
Developed power,
Pd = Po + Pm = 3730 + 26.65 = 3756.65 W
To determine the air-gap power Pg from the developed power Pd , we need the slip s. First we calculate synchronous speed,
N s = s=
Therefore, the slip,
120 f
=
120 × 50
= 1500 rpm
P 4 N s − N 1500 − 1470
=
N s
1500
= 0.02
From Eq. 4.17, the air-gap power,
Pg =
Pd
1− s
=
3756.65 1 − 0.02
= 3833.3 W
From Eq. 4.16, the rotor copper loss,
P R = s × Pg = 0.02 × 3833.3 = 76.7 W Hence, the stator loss,
PS = 506.35 − 76.7 = 429.65 W
Note that the stator loss could also be determined by subtracting air-gap power from the input power,
PS = 4263 − 3833.3 = 429.7 W 4.6 EQUIVALENT CIRCUIT OF AN INDUCTION MOTOR Since an induction motor is similar to a transformer, except that its secondary (i.e., the rotor) is not stationary but rotating at a speed N . Also, the load on the secondary is not electrical but mechanical.
7
Stator Equivalent circuit In the per-phase stator circuit in Fig. 4.4, R1 accounts for the stator copper loss, X 1 the stator leakage magnetic flux, and E 1 the induced emf in the stator winding due to rotating air-gap magnetic flux Ф. V1 is the per-phase stator voltage, and I1 the per-phase current drawn by the stator from the three-phase power supply. Obviously, the frequency of stator voltage and current is same as that of the supply. *
Three times the complex power E1I1 into the stator emf represents the power and magnetic energy leaving the stator and passing into the air gap. The real power accounts for the rotor copper loss and developed mechanical power. The reactive power accounts for the stored energy in the air gap and stray magnetic field in the rotor.
I1 R1
jX 1
jX 2
V1
E1
Ф
E2
(a) Stator
R2
Short circuit
I2
(b) Rotor (Frequency = fr
(Frequency = f ) Fig. 4.4
= s⋅ f )
Per-phase equivalent circuit for stator and rotor.
Rotor Equivalent circuit The per-phase rotor circuit shown in Fig. 4.4 shows an emf E2 induced in the rotor circuit due to the rotating air- gap flux Ф, a resistance R2 to account for rotor copper loss, a reactance X 2 to account for rotor leakage magnetic flux, and a short circuit. The electrical frequency f r of the rotor emf E2 and current I2 is s times the stator frequency f . The reactance X 2 is also proportional to slip s, that is, X 2 = sX 20. The reactance X 20 is the blocked rotor reactance because the slip is unity when the rotor is stationary. Also, the emf E 2 = sE 20, where E 20 is the blocked rotor emf induced . The total rotor copper loss is
P R = 3I 2 R2 2
...(4.19) where the factor 3 accounts for the three phases. The rotor iron loss, being very small, has been ignored. The equivalent circuit shown in Fig. 4.4 at present is unable to account for the stored energy in the air gap or for the developed mechanical power. Let N 1 and N 2 represent the equivalent turns for the stator and rotor, respectively, and let X ag represent a reactance that accounts for the stored energy in the air gap. Then, for the currents only, the coupling between the stator and rotor can be '
represented by the ideal transformer shown in Fig. 4.5 a. Here, I1 is the rotor current referred to the stator side and is given as N 2 I2 N 1
I1' =
...(4.20) The current I ag is the current that accounts for the stored magnetic energy in the air gap, and is given as I ag =
E1 jX ag
...(4.21) According to KCL, we must have
N 2 E1 I2 + jX ag N1
I1 = I1 + I ag = '
...(4.22) Next, we consider coupling between the stator and the rotor, for the voltages only . Since the rotor is rotating with a slip s in the same direction as the magnetic flux, the rotor emf E2 has to be reduced by a factor s. This can be achieved by imagining that the rotor turns N 2 are reduced by the same factor s, so that
E1 E2
=
N 1 sN 2 ...(4.23)
This suggests the ideal transformer shown in Fig. 4.5 b.
8
I1'
I1 Ia
E
jX a
E
(a) Current relationship Fig. 4.5
E2
( b) Voltage relationship. Per-phase equivalent circuits.
As per Fig. 4.4b, the rotor current and rotor emf are related through
I2 =
E2 Z2
E2
=
R jsX2 2 +
...(4.24) Note that the ideal transformer in Fig. 4.5 b has different turns-ratio from the one shown in Fig. 4.5 a. Such a strange transformer would not obey the basic principle of conservation of energy, and o f course, it should not because electrical energy is not conserved in this device due to the mechanical output. 5 The strange transformer can be eliminated by scaling up the rotor emf to “impose” the conservation of electrical energy on the circuit. Thus, dividing both the numerator and the denominator on the right side of Eq. 4.24 by s, we get
I2 =
E 2 / s Z2 / s
=
E 2 / s
(R2 / s ) + jX 2
...(4.25) We may now use a normal ideal transformer to make the equivalent circuit shown in Fig. 4.6. It gives relationships between the stator and the scaled rotor circuits.
I1'
I1
E1
jX ag
Stator
Fig. 4.6
I2
N 1: N 2
E1
Ideal transformer
jX 2
E2
R2
s
s
Rotor + Mechanical
Equivalent circuit with rotor voltage and impedance scaled by a factor 1/ s.
The secondary circuit in Fig. 4.6 now accounts for the entire power passing from the stator to the rotor, including that transferred to mechanical power . The scaled up resistor R2 / s accounts for both the rotor-copper loss power ( P R) and the developed mechanical power (Pd ). We can therefore determine the developed power by subtracting P R from the total air-gap power,
2 1 2 R 2 Pd = Pag − PR = 3 I 2 2 − I 2 R2 = 3 I 2 R2 − 1 s s or
1 − s s
Pd = 3 I 22 R2 ×
...(4.26) Here,
[(1 − s) / s ] × R2 is an equivalent resistance accounting for the developed mechanical power per phase. Thus, we can modify
the equivalent circuit of Fig. 4.6 by dividing resistor R2 / s into two resistors: (1) R2 which accounts for the rotor electrical loss , and (2)
[(1 − s) / s] × R2 which accounts for the developed mechanical power . The result is shown in Fig. 4.7. The figure also shows resistor R1 to account for the stator copper loss, Ri for iron loss, and X 1 as the leakage-flux reactance.
5
This idea of scaling up the rotor emf was first suggested by Carl Steinmetz (1865-1923).
9
P R /3
Pag /3 I1 R1
I1' N 1: N 2
jX 1
Ideal transformer
Stator
Fig. 4.7
jX 2 R2
1− s s R2
E2
jX ag E1
Ri
V1
I2
Pd /3
Rotor
Mechanical
Equivalent circuit accounting for developed mechanical power.
We can further simplify the circuit of Fig. 4.7 using the properties of an ideal transformer. If K
= ( N 2 / N 1 ) is
transformation ratio, the equivalent impedances as referred to the primary (stator) are given by Eq. 13.8 as '
and X2
the
R2 = R2 / K '
2
= X2 / K2 . Thus, we get the equivalent circuit of Fig. 4.8. I1
R1
jX 1
A
'
R2
1− s ' R2 s
R L' =
jX ag
Ri
V1
jX 2'
Air gap
Stator
B Rotor
Mechanical
Fig. 4.8 Equivalent circuit with rotor impedance transformed to stator-side. As for a transformer, we could further simplify the circuit of Fig. 4.8 by shifting the shunt branches Ri and X ag to the right of rotor impedance. However, this simplification cannot be done in an induction motor without incurring considerable error, as the no-load current is about 25 % to 40 % of the rated current. We can make use of the equivalent circuit of Fig. 4.8 by finding Thevenin’s equivalent of the circuit on the left of terminals A-B. In practice, the core losses in an induction motor are quite small compared to the other powers. We can therefore ignore the resistance Ri. Thevenin’s voltage is given as
VTh = Voc = VAB = V1 ×
jX ag R1 + j( X1 + Xag ) ...(4.27)
Thevenin’s equivalent impedance is given as
ZTh =
jXag ( R 1 + jX 1 ) R1 + j( X1 + Xag ) ...(4.28)
Thus, the equivalent circuit can be represented as in Fig. 4.9
ZTh
A
jX 2'
1− s ' R2 s
VTh
R L' =
B Fig. 4.9
Simplified equivalent circuit using Thevenin’s theorem.
10
Example 4.7 A three-phase, 25 hp, 400-V, 50-Hz, four-pole, star-connected induction motor has following impedances per phase as referred to the stator side: ' ' R X R Ω ag = 26.3 1 = 0.641Ω , 1 = 1.106 Ω , 2 = 0.332 Ω , X 2 = 0.464Ω and X
Assume that the core losses are negligible and the rotational losses are constant at 0.34 kW. If the slip is 2 % at the rated voltage and frequency, determine ( a) the rotor speed, ( b) the stator current, ( c) the power factor, ( d ) the output and input power, and ( e) the efficiency of the motor.
N= Ns (1 − s) =
(a) The rotor speed,
400
=
Solution: The per phase applied voltage, V 1
3
120 f p
= 231 V
(1 − s) =
120 × 50 × (1 − 0.02) 4
= 1470 rpm
(b) Referring to Fig. 4.9 and taking V1 as the reference phasor, Thevenin’s equivalent voltage and impedance are given by Eqs. 4.27 and 4.28, respectively, as
jX ag
VTh = V1 ×
and
R1 + j( X1 + Xag )
= (231∠0°) ×
j 26.3
0.641 + j(1.106 + 26.3)
= 221∠1.34° V jXag ( R1 + jX1 ) (26.3∠90°)(0.641 + j1.106) ZTh = = = (059 + j1.08) Ω R1 + j( X1 + Xag ) 0.641 + j(1.106 + 26.3)
The equivalent resistance representing mechanical load is
1 − s ' 1 − 0.02 R2 = 0.02 × 0.332 = 16.268 Ω s
R L' =
Thus, the stator current is Ι1
=
VTh ZTh + ( R+ ' 2
(c) Power factor, pf
jX) + R ' 2
' L
=
221.6∠1.34° j j (0.59 + 1.08) + (0.332 + 0.464) + 16.268
= 12.84∠ − 3.79
A
= cos φ 1 = cos(−3.79° ) = 0.998lagging
(d ) Power output,
Po = 3I12 RL' − Rotational losses = 3 × (12.84) 2 × 16.268 − 340 = 7706.1 W Power input, Pin
= 3V1I 1 cos φ 1 = 3 × 231× 12.82 × 0.998 = 8866.5 W
(e) Efficiency of the motor,
η =
Po Pin
=
7706.1 8866.5
= 86.91pu = 86.91 %
4.7 TORQUE-SLIP CHARACTERISTICS The input power to the rotor transferred from the air gap is given by Eq. 4.13, repeated here for convenience,
Pg =
2πτ N s 60 ...(4.29)
The rotor copper loss is given as
P R = 3E2 I 2 cos θ 2 = 3 × ( sE20 ) × =
2 3s 2 E20 R2
R2 + ( sX20 ) 2
2
=
× 2 2 R 2 + ( sX 20 ) sE20
R2 2 2 R 2 + ( sX 20 )
2 3s 2 E20 R2
R2 + s X20 2
2
2
...(4.30) where, we have made use of Eqs. 4.10 and 4.11. According to Eq. 4.16, we have
P R = s × Pg
or
2 R2 3s 2 E20 2 R22 + s 2 X 20
=s⋅
2πτ N s 60
Thus, the electromechanical torque exerted on the rotor is given as
11
τ =
60 2π Ns
2 3sE20 R2
×
R2 + s X20 2
2
2
...(4.31) The induced emf E 20 is proportional to the air-gap flux Ф and in turn, the flux Ф is approximately proportional to the voltage V 1 applied to the stator. That is,
E ∝ Φ ∝ 1V
E=
or
20
20
kV 1
Equation 4.31 can then be modified to
τ =
or
60 2π Ns
τ =
3s( kV1 ) 2 R2
×
2 R22 + s2 X20
2
= K
V1 sR2 2 R22 + s2 X20
KV12sR2 2 R22 + s2 X20
...(4.32) where, constant K is given as
K=
60
× (3k ) = 2
2π N s
90k 2 π N s
...(4.33) Thus, the torque for a given machine is seen to depend on two factors: (i) the applied voltage V 1, and (ii) the slip, s. Starting Torque At starting, the rotor is stationary. Hence, s = 1, and Eq. 4.32 reduces to 2
τ st =
KV1 sR2 2 R22 + s2 X20
2
=
KV1 R2 2 R22 + X20
...(4.34) However, in practice, the value of X 20 is much greater than that of R2 (typically, X 20 ≈ 1.5 Ω and R2 ≈ 0.2 Ω). We can therefore ignore 2 2 R2 compared to X 20 , so that the starting torque is given as 2
τ st =
KV1 R2 2 X 20
...(4.35) For a given machine, the reactance X 20 is constant. Hence, we find that
τ st ∝ V and τ st ∝ R2 . Thus, to obtain large starting torque 2 1
we should have large rotor-resistance R2 as well as large applied voltage V 1.
Torque-Slip Characteristic Curve From Eq. 4.32, we can predict the general shape of the torque-slip characteristic curve. If we keep the applied voltage V 1 constant, Eq. 4.32 can be written as
τ = K 1
sR2 2 R + s2 X20 2 2
...(4.36) where, K 1 is another constant. 2
2
For Small Values of Slip: For very small values of slip (say, from 0 to 0.1), the term s X 20 is negligibly small compared to R2. We can therefore rewrite Eq. 4.36 as
τ = K1 Thus,
τ ∝ s
sR2 2 2
R
= K 1
s R2
[ R2 being constant]
Hence, for small values of s, torque is seen to be directly proportional to slip s. The torque-slip curve should be a straight line, as shown in Fig. 4.10. 2
2
2
For large value of slip: When s is large (say, from 0.2 to 1), the term R2 becomes negligibly small as compared to s X 20 , so that Eq. 4.36 can be rewritten as
τ = K1
sR2 2
s X
2 20
= K 1
R2 2
sX 20
12
Thus,
τ ∝
1
[ R2 and X 20 being constant]
s
Hence, for large values of s, torque is seen to be inversely proportional to slip s. The torque-slip curve should be a rectangular hyperbola, as shown in Fig. 4.10.
Fig. 4.10
Torque-slip characteristic curve for an induction motor.
The overall torque-slip characteristic curve has a shape as shown in Fig. 4.10. We can interpret the values of s in terms of the rotor speed N . When s = 0, the rotor speed N is same as the synchronous speed N s. When s = 1, the rotor speed N becomes zero (i.e., the motor is standstill).
Three Modes of Operation Depending on the value of slip s, there can be following three modes of operation of an induction motor (Fig. 4.11): (1) Motor action (0 < s ≤ 1) , (2) Brake action ( s > 1) , and (3) Generator action ( s < 0) .
τ -1.0 0
Generator
(1) Motor Action (0 <
0.5
Motor
1.0
s
Brake
Fig. 4.11 Three modes of operation of a n induction motor. s ≤ 1) : In this mode, the rotor rotates in the same direction as the stator field. The speed is less than the
synchronous speed. When s = 1, the rotor speed N is zero corresponding to point C on the curve (Fig. 4.10). The torque at zero speed is called starting torque τ st . Point B on the characteristic curve corresponds to a value of slip s for which the torque developed by the motor is maximum torque τ m . The shaded portion (for 0.01 < s
< 0.06 ) shows the normal working-range of the induction motor. Obviously, if the motor is to
start running, the load torque at the shaft must be less than the starting torque
τ st . The motor will accelerate from standstill, until the
torque developed and the load torque comes to equality at a speed close to but less than the synchronous speed. In the region AB, the machine has a stable motor action . If the load torque increases (but still remains below the value τ m ), the motor develops increased torque at a slightly reduced speed. If the load torque is increased beyond τ m , the speed decreases and the point of operation goes beyond B. The motor further decelerates and ultimately comes to a standstill. Thus, the r egion BC represents an unstable motor action. (2) Brake Action ( s > 1) : There are two ways of making s greater than unity. First, the rotor can be driven by a prime mover in a direction opposite to the rotating magnetic field. Second, we can reverse any two of the phase supplies while operating the machine as a motor. The effect of reversing two supply-phases is to make the stator field rotate in the opposite direction. Thus, at the time of
13
switch-over, the rotor is rotating almost at synchronous speed in one direction and the stator field is rotating at synchronous speed in the opposite direction. The difference is almost twice the synchronous speed and hence the slip s is almost 2. The effect is that the rotor now attempts to reverse its direction of rotation. This amounts to braking of the rotor in order to bring it to a standstill prior to commencing rotation in the opposite direction. This braking effect is known as plugging. As soon as the machine stops, the power supply is switched off and the machine remains at standstill. (3) Generator Action ( s < 0) : The slip s can be made negative if with the help of a prime mover the rotor is made to rotate at a speed higher than the synchronous speed. In such cases, the machine works as a generator. However, induction generators are rarely used as the most significant generators are synchronous machines.
Condition for Maximum Torque We have seen that the torque developed depends on the value of slip s. To determine the value of s that gives maximum torque, we differentiate the expression for τ (Eq. 4.36) with respect to s and equate the differential to zero.
∴ For
d τ ds
= K1 ⋅
2 2 ( R22 + s 2 X 20 ) R2 − sR2 ⋅ 2 sX 20 2 2 ( R22 + s 2 X 20 )
τ to have a maximum value, we should have 2 2 2 ( R or 2 − s X 20 ) = 0
R2 ( R2 − s X 20 ) 2
= K 1 ⋅
2
2
2 2 ( R22 + s 2 X 20 )
R 2 = sX 20 ...(4.37)
This means that the torque is maximum when the rotor resistance is equal to the rotor reactance under running condition. For a given machine (i.e., for given R2 and X 20), the value of slip s at which the torque is maximum is given as
sm =
R2 X 20 ...(4.38)
Maximum Torque Putting the value of s from Eq. 4.37 in Eq. 4.36, we get the maximum value of the torque as
τ m = K1
( R2 / X 20 ) R2 2 R22 + ( R2 / X 20 ) 2 X 20
= K 1
R22
2 R22 X 20
=
K1
2
⋅
1 X 20
...(4.39) Thus, we find that the maximum torque of an induction motor is inversely proportional to the leakage reactance at standstill X 20 and is independent of the value of the rotor resistance R2. The ratio of torque
τ τ m
K 1
=
τ for any slip s to the maximum torque τ m can be determined by dividing Eq. 4.36 by Eq.4.39, sR2
2 R + s2 X20 2 sR X 2 s( R2 / X20 ) 2 s⋅ sm = 2 2 2 202 = = K 1 1 R2 + s X 20 ( R2 / X 20 )2 + s2 sm2 + s 2 ⋅ 2 X 20 2 2
...(4.40) Since s = 1 at the starting, the ratio of starting torque
τ st τ m
=
τ st to the maximum torque τ m is given by putting s = 1 in Eq. 4.40,
2 sm sm + 1 2
...(4.41)
Effect of Rotor Resistance on the Starting Torque For a squirrel-cage induction motor, the rotor resistance R2 is fixed. Such motors are designed to have torque-slip characteristic of the type shown in Fig. 4.10. For such motors, the starting torque is small and hence they cannot start with heavy loads connected. However, in case of phase-wound induction motors it is possible to include suitable value of resistance in the rotor circuit (see Fig. 4.13) so as to give desired starting torque. If the impedance of the stator winding is assumed negligible, then for a given supply voltage, the torque is given by Eq. 4. 36,
τ = K 1 ⋅
sR2 R2 + s X20 2
2
2
The value of X 20 is far greater than that of R2. For simplicity, assume X 20 = 1.0 Ω. Using above equation, we can plot torque-slip characteristics for four different values of R2 as 0.1 Ω, 0.2 Ω, 0.6 Ω and 1.0 Ω. The results are shown in Fig. 4.12. Let us see the effect of doubling the resistance R2 from 0.1 Ω to 0.2 Ω. In the normal working range (say, s = 0.05) the torque reduces by about 50 %. But, the starting torque (for s = 1), almost doubles. Hence, if a large starting torque is required, the rotor must have relatively high resistance. From the characteristic curves given in Fig. 4.1 2, following observations are made:
14
τ st increases on increasing the rotor resistance.
(i)
The starting torque
(ii)
The maximum torque
(iii) (iv)
The slip corresponding to maximum torque is greater for higher values of rotor resistance. Maximum starting torque is obtained when R2 = sX 20.
τ m is constant for all the curves.
Fig. 4.12 Effect of rotor resistance on torque-slip characteristic of an induction motor.
Example 4.8 A three-phase, 400-V, 50-Hz, six-pole, star-connected induction motor develops maximum torque at a speed of 940 rpm. If the rotor resistance per phase is 0.1 Ω, determine the standstill rotor reactance. Solution: The synchronous speed, N s
∴
Slip, s =
N s − N N s
=
=
120 f
=
P 1000 − 940
1000
120 × 50 6
= 1000rpm
= 0.06
Maximum torque occurs at a slip so as to satisfy the condition: R2 = sX 20. Therefore,
X 20 =
R2 s
=
0.1 0.06
= 1.66Ω
4.8 STARTING OF INDUCTION MOTORS There are two problems in starting an induction motor: (1) Low starting torque, and (2) Heavy starting current, though for a short duration. As can be seen from Fig. 4.13, an induction motor having a low-resistance rotor, such as the usual type of squirrel-cage rotor, the starting torque is small compared to the maximum torque available. On the other hand, if the bars of the cage rotor were made with sufficiently high resistance to give high starting torque, the slip for full-load torque would be quite large. This would have two adverse effects. First , the I 2 R loss in the rotor would be high causing excessive heating and reduced efficiency of the motor. Secondly, the variation of speed with load would be quite large. Another problem is that the motor draws about four to seven times the full-load current, if it is directly switched to the power supply. At the instant of starting, the slip is unity, the frequency of induced emf in the rotor is
f = sf = 50Hz , and hence a large
r
emf is induced. The induction motor works as a transformer with its secondary short-circuited. This results in a large circulating current in the rotor winding. Such a large current causes a large voltage drops in the lines and thereby produces an objectionable dimming of the lamps in the vicinity. Therefore, the direct-on-line starting of induction motors is not desirable. Rules laid down by Electric Supply Companies do not permit the direct-on-line starting of 3-phase induction motors above 5 hp. For such heavy duty motors, we use some method by which either more resistance is included in the rotor circuit in the starting, or a reduced voltage is applied at the starting.
Starting of Wound-Rotor Induction Motor The best (but costly) solution is to use a phase-wound rotor induction motor, with a starting arrangement shown in Fig. 4.13. The motor is designed to have low rotor resistance. External variable resistance is connected across the slip rings. The motor is started with all the resistance included in the rotor circuit. This gives a high starting torque. As the motor picks up speed, the resistance is slowly reduced. When the motor attains full speed, all the resistance is cut out.
15
Fig. 4.13 Starting of wound-rotor induction motor Large motors are often fitted with a short-circuiting and brush lifting device. On attaining full speed, first the three-rings are short-circuited and then the brushes are lifted off the rings. This eliminates the losses due to the brush-contact resistance and the brush friction. Also, the wear of the brushes and slip-rings is reduced.
Starting of Cage-Rotor Induction Motor Most induction motors have squirrel-cage type rotor. It is usual to start cage-rotor motors –except small machines—with a reduced voltage, using one of the methods given below. Star-Delta Starter: This starter can be used only for those motors whose stator winding is designed for delta-connection during its normal operation. All the six terminals of the three-phase stator windings are brought out and connected as shown in Fig. 4.14 a. Using the double-throw triple-pole switch, in the starting, the stator windings are connected in star, so that the voltage across each phase is
1/ 3 times the normal value. As the motor picks up speed, the changeover switch disconnected the winding terminals and then reconnected them in delta across the supply terminals. Each phase now gets the normal voltage. This method reduces the current drawn by the motor to one-third the current it would have drawn if it was directly connected in delta. However, the starting torque too is reduced to one-third. This method is cheap but is limited to applications where high starting torque is not necessary, e.g., machine tools, pumps, etc.
(a) Star-delta starter. ( b) Auto-transformer starter Fig. 4.14 Two ways of starting a cage-rotor induction motor. Auto-Transformer Starter: We use a three-phase auto-transformer to supply a reduced voltage to the motor at the starting (Fig. 4.14b). Two or three voltage-steps (in the auto-transformer) are used during the starting process. The auto-transformer is completely cut out once the motor picks up speed to its rated value.
ADDITIONAL SOLVED EXAMPLES Example 4.9 A 3-phase, 6-pole induction motor runs at 960 rpm on full-load. It is supplied from a 4-pole alternator running at 1500 rpm. Calculate the full-load slip of the motor. Solution: The frequency of generated emf by the alternator is
f =
NP
120
=
1500 × 4 120
= 50Hz
Therefore, the synchronous speed of the motor is
N s =
120 f P
=
120 × 50 6
= 1000rpm
16
∴
Slip, s =
N s − N N s
=
1000 − 960
= 0.04 = 4%
1000
Example 4.10 A 3-phase, 4-pole induction motor supplies a useful torque of 160 Nm at 5 % slip. The stator losses are 1000 W and the friction and windage losses are 500 W. Calculate ( a) the rotor input, ( b) the motor input, and ( c) the efficiency.
∴
120 f
The output power, Po
2πτ sh N
=
=
2π ×160 × 1425
=
60
=
120 × 50
= 1500rpm P 4 Motor speed, N = (1 − s) Ns = (1 − 0 .05) × 1500 = 1425 rpm
Solution: (a) The synchronous speed, N s
60
= 23876 W
Thus, the power developed by the rotor is given as
Pd = Po + Pm = 23876 + 500 = 24376 W Pd
(c) The efficiency,
η =
Po
23876
=
Pin
=
24376
= 25659 W = 25.659kW (1 − s ) (1 − 0.05) (b) The input power, Pin = Pg + PS = 25659 + 1000 = 26659W Therefore, the rotor input power, Pg
26659
=
= 0.8956 = 89.56%
Example 4.11 An induction motor has an efficiency of 0.9 when load is 50 kW. At this load, the stator copper and rotor copper loss each equals the iron loss. The mechanical losses are one-third of the no-load loss. Calculate the slip. Solution: The power input, Pin
=
50
= 55.56kW
0.9
= Pin − Po = 55.56 − 50 = 5.56 kW
Neglecting the rotor iron loss, the total losses, Ptl
The no-load loss consists of the stator iron loss ( Pi) and mechanical losses ( Pm) (since the stator and rotor copper losses are negligible). These two losses are independent of the load.
=
Given, the mechanical loss, Pm
no-load loss 3
Pi + Pm
=
⇒
3
Pm =
Pi
2
We know that Total loss = Stator copper loss + Stator iron loss + Rotor copper loss + Mechanical loss or or
Ptl = Pi + Pi + Pi + Pi / 2
5.56 = 3Pi +
Pi
⇒
2
Pi = 1.59 kW
Now, the power developed by the rotor is given as
Pd = Po + Pm = 50 +
1.59 2
= 50.795 kW
The rotor input power is given as
Pg = Power developed by the rotor + Rotor copper loss
= Pd + PR = 50.795 + 1.59 = 52.385kW Therefore, the slip, s
=
P R Pg
=
1.59 52.385
= 0.03 = 3 %
Example 4.12 The induced emf between the slip-ring terminals of an induction motor at standstill is 100 V. The rotor windings are star-connected and have a resistance of 0.4 Ω per phase. Calculate the rotor current when the slip-ring terminals are short-circuited and the rotor is rotating at a slip of 4 %. Solution: For star-connected rotor windings, the induced emf per phase when the rotor is at standstill is given as
E 20 =
V 2
=
100
3
3
V
The rotor reactance X 2 = sX 20 is negligible for small values of s, and hence can be ignored. The rotor current is therefore given as
I 2 =
E 2 R + sX 2 2
2 20
=
sE 20 R2
=
0.04 × 100/ 3 0.4
= 5.77A 17
Example 4.13 A slip ring induction motor runs at 285 rpm on full load when fed from a 50-Hz supply. Calculate ( a) the number of poles, (b) the slip, ( c) the slip for full-load torque if the rotor resistance is doubled, and ( d ) the rotor copper losses with added rotor resistance if the original value of the rotor losses were 250 W. Solution: (a) Approximate number of poles,
P=
120 f N
=
120 × 50 285
= 21.05 .
There has to be an even number of poles, such that N s > N. Thus, the actual number of poles is 20. (b) With 20 poles, the synchronous speed is given as
N s =
120 f
120 × 50
= 300 P 20 N s − N 300 − 285
=
Therefore, the slip, s
=
=
N s
300
= 0.05
(c) For small values of s, the reactance sX 20 is much smaller than the resistance R2, and hence we can write Eq. 4.31 as
τ ∝
sR2 R +s X 2 2
2
2 20
∝
sR2 2 2
R
∝
s R2
R2 ratio should remain the same. If R2 is doubled, s also has to be doubled. Thus, new It means that to keep the torque same, the s / value of s = 2 × 0.05 = 0.1 (d ) Since the full-load current remains the same, on doubling the rotor resistance, the copper loss ( I 2 R) is also doubled. Hence, the new
copper loss P R
= 2 × 250 = 500 W .
Example 4.14 The full-load slip of a 500-hp, 50-Hz, 3-phase induction motor is 0.02. The rotor winding has a resistance of 0.25 Ω/phase. Calculate the slip and the power output, if external resistances of 2 ohm each are inserted in each rotor phase. Assume the torque to remain unaltered. Solution: Given: R2
= 0.25 Ω; s = 0.02; R2' = 2 + 0.25 = 2.25 Ω .
If the torque remains the same, the ratio s / R2 must also
remain same. Hence,
s' =
s R2
× R2' =
0.02 0.25
× 2.25 = 0.18
Now, if N s is the synchronous speed of the motor, the speed of the rotor before inserting the external resistance,
N = (1 − s) Ns = (1 − 0.02) Ns = 0.98 Ns
and the speed of the rotor after inserting the external resistance,
N' = (1 − s') Ns = (1 − 0.18) Ns = 0.82 Ns
Since the torque remains the same, the output is directly proportional to the speed. Hence, the new output of the motor is
Po' =
0.82 N s 0.98 N s
× 500 hp = 418.4 hp
Example 4.15 A 3-phase, 4-pole, 50-Hz, 7.46-kW induction motor, while working at rated voltage and frequency has a starting torque of 160 % and a maximum torque of 200 % of the full-load torque. Determine ( a) the full-load speed, and (b) the speed at maximum torque. Solution: (a) Given: The ratio
τ st τm
=
τ st = 1.6τ fl ; τ m = 2.0τ fl . 1.6τ fl 2.0τ fl
= 0.8
Therefore, using Eq. 4.41, we have
0.8 =
2 sm
sm2 + 1 = 2.5sm
or
s +1 2 m
⇒
Obviously, sm = 2 is not possible. Hence, sm = 0.5. Now, using Eq. 4.40,
2s ⋅ s fl
τ τ m
= fl
m
sm2 + s 2fl
s − s fl + 0.125 =0 fl 2
or
or
⇒
0.5 =
sm = 2, 0.5
2 × s × 0.5 fl (0.5)2 + s 2fl
s = 0.8535, 0.1465 fl
Since, for stable motor action, s fl must be less than sm, we have s fl = 0.1465. The synchronous speed is
N s =
120 f P
=
120 × 50 4
= 1500 rpm 18
= N (1 s − s )fl = 1500 × (1 − 0 .1465) = 1280.25 rpm
Therefore, full-load speed, N
fl
= N s (1 − sm ) = 1500 × (1 − 0.5) = 750 rpm
(b) The speed at maximum torque, N m
Example 4.16 A 3-phase, 4-pole, 50-Hz, 18.65-kW induction motor has friction and windage losses of 2.5 % of the output, and fullload slip of 4 %. Determine ( a) the rotor copper loss, ( b) the rotor input, ( c) the output torque, and ( d ) the gross torque.
= 0.025 × 18650 = 466.25 W . Pd = Po + Pm = 18650 + 466.25 = 19 116.25 W
Solution: The friction and windage losses, Pm The rotor gross output,
(a) Using Eq. 4.18, the rotor copper loss is
s 0.04 = 19116.25 = 796.5 W 1− s 1 − 0.04
P R = Pd (b) The rotor input, Pg
=
P R s
796.5
=
= 19 912.5 W
0.04 120 f
=
The rotor speed,
=
120 × 50
= 1500 rpm P 4 N = Ns (1 − s) = 1500 × (1 − 0.04) = 1440 rpm
(c) The synchronous speed, N s
Po
τ sh =
18 650
= 123.7 Nm 2π ( N / 60) 2π × (1440 / 60) Pd 19 116.25 τ d = = = 126.8 Nm 2π ( N / 60) 2π × (1440 / 60)
The shaft torque,
(d ) The gross torque,
=
Example 4.17 A 3-phase, 4-pole, 1100-V, 50-Hz, delta-connected induction motor has a star-connected rotor with a phase transformation ratio of 3.8: 1. The rotor resistance and standstill reactance are 0.012 Ω and 0.25 Ω per phase, respectively. The motor runs at 1440 rpm at full load. Neglecting the stator impedance and magnetizing current, determine ( a) the rotor current at starting with slip-rings shorted, ( b) the rotor power factor at starting with slip-rings shorted, ( c) the rotor current while running at full load with sliprings shorted, ( d ) the rotor power factor while running at full load with slip-rings shorted, and ( e) the external resistance per phase required to limit the starting current to 100 A in the stator supply lines.
=
Solution: The synchronous speed, N s The slip at full load, s
=
1500 − 1440 1500 N 2
The transformation ratio, K =
N 1
120 f P
=
120 × 50 4
= 1500rpm .
= 0.04
1
=
3.8
The induced voltage in rotor per phase at standstill, E 20
= KE 1 =
1 3.8
× 1100 = 289.5 V
= R22 + X 202 = (0.012) 2 + (0.25) 2 ≈ 0.25 Ω
The rotor impedance at standstill, Z 20 The rotor impedance at full load,
Z2 =
2 2 R (0.012)2 + (0.04 × 0.25)2 = 0.0156 Ω 2 + ( sX 20 ) =
(a) The rotor current at starting with slip-rings shorted,
I 20 =
E 20 Z 20
289.5
=
0.25
= 1158 A
(b) The rotor power factor at starting with slip-rings shorted,
pf 20 = cos tan
−1
0.25 = cos87.25° = 0.048(lagging) 0.012
(c) The rotor current while running at full load with slip-rings shorted,
I 2 =
E 2 Z 2
=
sE 20 Z 2
=
0.04 × 289.5 0.25
= 742.3 A
(d ) The rotor power factor while running at full load with slip-rings shorted,
19
pf =
R2
=
Z 2
0.012
= 0.769(lagging)
0.0156
(e) The stator phase current corresponding to starting line-current of 100 A is
I 1 =
100 3
= 57.73 A
∴
Rotor current per phase, I 2 =
∴
Rotor total impedance, Z 2' =
I 1 K E 20
= 57.73 × 3.8 = 219.4A =
I 20
289.5 219.4
= 1.32 Ω
Thus, the rotor total resistance required is
R2 + r =
2 Z2'2 − X20 = (1.32)2 − (0.25)2 = 1.296 Ω
Therefore, the external rotor resistance required is
r = 1.296 − 0.012 = 1.284Ω SUMMARY
1.
An induction motor has a distributed three-phase field winding on the stator and armature on the rotor.
2.
The synchronous speed in rpm is given as N s
3.
The speed of the rotor of an induction motor always remains slightly less than the synchronous speed.
4.
The slip, s
=
120 f P
.
= ( N s − N ) / N s . At no load s is around 1 % and at full load it is around 3 %. Thus, an induction motor operates at
6.
almost constant speed. Induction motor action is, in many respects, similar to that of a transformer. However, due to the air gap, it has much greater magnetizing current. The phasor diagram and the e quivalent circuit of a three-phase induction motor are always drawn per phase basis.
7.
The frequency of the rotor current,
5.
fr = s⋅ f . Since s is very small, the frequency f r is very small. Therefore, the rotor iron
losses are negligibly small. 8.
The rotor current is given as I 2
9.
The rotor copper loss , P R
=
E 2
sE 20
=
Z 2
R2 + ( sX 20 ) 2
2
= s × Pg , where Pg is the air-gap power.
10. The power developed by the rotor, Pd
= Pg − PR = Pg − sPg = (1 − s ) Pg .
11. The power developed Pd (in W) and torque developed τ d (in Nm) are related as
Pd = τ d ω = τ d (2π n) = 12.
2πτ d N
60 P τ d = d 2π N
or
60
1 hp = 746 W.
13. The mechanical load on the shaft of the motor is represented by an equivalent (fictitious) electrical load R L
= R2 (1 − s) / s .
2
14. The torque exerted on the rotor is given as
15. The starting torque ,
τ st =
KV12 R2 2
X 20
τ =
KV1 sR2 2 R22 + s 2 X 20
.
.
16. The motor has three modes of operations: ( i) Motor Action (0 < s
≤ 1) ,
(ii) Brake Action ( s
> 1) , and (iii) Generator Action
( s < 0) . 17. Maximum torque occurs for a value of s, such that 18. The maximum torque is given as τ m
=
K 1
⋅
1
2 X 20
R2 = sX20 .
.
20
19. The ratio of the torque for any s to the maximum torque is given as
τ τ m
=
2s ⋅ sm sm + s 2
2
20. An induction motor draws considerably high current at starting. To limit the starting current, star-delta starter or auto-transformer starter is used.
CHECK YOUR UNDERSTANDING Before you proceed to the next Chapter, take this Test. Give yourself two marks for each correct answer rand minus one for each wrong answer. If your score is 12 or more, go to the next Chapter; otherwise study this Chapter again.
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
No Statement The more poles an induction motor has, the faster it runs. For maximum torque, the rotor and the stator poles should be spatially orthogonal. The slip speed is defined as positive in the negative angular direction. The change in speed of a three-phase induction motor from no load to full load is about 98 %. When a squirrel cage rotor is used in an induction motor, its stator can have any number of poles. The current in the rotor winding of an induction motor is an induced current. The frequency of the rotor currents is independent of the speed of the induction motor. An induction motor has field on the rotor and armature on the stator . An induction motor develops maximum torque at a speed at which the rotor reactance has the value same as its resistance. The reason why most of the induction motors use squirrel cage rotor is that the starting torque is high.
True False
Your Score = Answers 1. False
2. True
3. True
4. False
5. True
6. True
7. False
8. False
9. True
10. False
REVIEW QUESTIONS 1. 2.
Explain the principle of working of a three-phase induction motor. What is meant by slip of an induction motor? Explain the importance of slip in the operation and performance of an induction motor. 3. Describe the constructional differences between a squirrel cage rotor and wound rotor of an induction motor. Discuss their relative advantages and disadvantages. 4. Draw and explain the power flow diagram of a three-phase induction motor. Show that the rotor copper-loss is slip times the power input to the rotor. 5. Derive the equation for the torque developed by an induction motor. Draw a typical torque-slip characteristic curve and deduce the condition for maximum torque. 6. Explain why the starting torque of a squirrel-case induction motor is low, although the current drawn form the line may be quite high in starting of the motor. 7. What is the effect of introducing extra resistance in the rotor circuit? Under what condition is it done and why? 8. Prove that the frequency of the induced emf in the rotor of an induction motor is slip times its stator supply frequency. 9. Explain why an induction motor cannot run at synchronous speed, and why a synchronous motor cannot run at any speed other than the synchronous speed. 10. Starting from first principles, develop the equivalent circuit of an induction motor and explain how mechanical load is accounted for in this equivalent circuit. 11. Explain why a starter is needed for starting an induction motor. What are different methods of starting an induction motor? 12. With the help of a circuit diagram explain how a manual auto-transformer is used in starting an induction motor. 13. Give the reasons for the following: (a) The speed of an induction motor can never be the same as the synchronous speed. (b) The reactance of the rotor varies greatly between starting and running conditions. (c) The induction motor can be called a generalized rotating transformer. (d ) The rotor core-loss in a three-phase induction motor is negligible. (e) The induction motor is the most commonly used motor.
MULTIPLE CHOICE QUESTIONS Here are some incomplete statements. Four alternatives are provided below each. Tick the alternative that completes the statement correctly: 1. The main reason why three-phase induction motors are widely used in industries is that (a) they are rugged in construction, require less maintenance and are less expensive than other motors (b) their operating characteristics are superior over other electrical motors
21
2.
3. 4. 5.
6.
(c) their speed can be controlled very smoothly over a wide range (d ) they can be manufactured easily for any hp-rating In a three-phase induction motor, (a) three-phase supply is connected to the stator winding and a dc supply is connected to the rotor winding (b) three-phase supply is connected to both the stator and rotor windings (c) three-phase supply is connected to rotor winding only (d ) three-phase supply is connected to stator winding only The speed of a 50-Hz, three-phase induction motor under full-load condition is 720 rpm. The number of pole in the motor is ( a) 4 (b) 6 (c) 8 (d ) 12 In a 50-Hz, three-phase induction motor, the frequency of the currents induced in the rotor is about (a) 50 Hz (b) 10 Hz (c) 2 Hz (d ) zero In a wound-rotor type induction motor, the rotor winding terminals are brought out through slip rings and brushes. This is done (a) to enable us to connect the rotor windings either in star or in delta as per requirement (b) to enable us to close the rotor circuit externally (c) to enable us to connect 3-phase supply to the rotor windings (d ) to enable us to connect extra resistances across them while starting the motor The relation between the synchronous speed N s (in rpm), stator supply frequency f , and the number of poles P of a three-phase induction motor is given as (a) N s
7.
8. 9.
10.
11.
12. 13.
14.
15.
16.
17.
1. a 11. a
=
P
120 f
(b) N s
=
120 P f
(c) f
=
PN s
(d ) f
120
=
120 N s P
The synchronous speed of an induction motor can be increased by (a) reducing mechanical friction ( b) increasing supply voltage (c) increasing number of poles ( d ) increasing frequency of supply When an induction motor is standstill, its slip is (a) zero (b) 0.5 (c) 1 (d ) infinity A three-phase, 400-V induction motor develops a torque of 200 Nm. If the supply voltage is reduced to 200 V, the torque developed will be (a) 50 Nm (b) 100 Nm (c) 200 Nm (d ) 400 Nm If an induction motor is made to run at synchronous speed, then (a) the rotor emf would be zero (b) the rotor current would be zero (c) the torque developed would be zero (d ) all the above The rotor circuit of a three-phase induction motor under running condition (a) is always closed (b) is always open (c) is sometimes closed and sometimes open (d ) is neither completely closed nor completely open The number of slip rings in a 3-phase wound rotor induction motor is ( a) 3 (b) 4 (c) 9 (d ) 12 A 3-phase, 50-Hz induction motor at standstill, when energized, has a voltage of 50 V between its slip rings. While running at full load with a slip of 0.04, the voltage between its slip rings would be (a) 50 V (b) 20 V (c) 4 V (d ) 2 V At standstill, the power factor of the rotor circuit of an induction motor is found to be 0.2 lagging. While running at full load, the power factor of the rotor circuit will be (a) 0.2 lagging (b) about 0.8 lagging (c) almost unity (d ) any of the above If the rotor circuit resistance of a three-phase induction motor is increased, (a) both the starting torque and the maximum value of the torque developed increase (b) both the starting torque and the maximum value of the torque developed remain unchanged (c) the starting torque increases but the maximum value of the torque developed decreases (d ) the starting torque increases but the maximum value of the torque developed remains unchanged An induction motor has a starting torque of 600 Nm when started by direct switching. If it is started through an autotransformer with 50 % tapping, the starting torque will be (a) 1200 Nm (b) 600 Nm (c) 300 Nm (d ) 150 Nm An induction motor has a starting torque of 600 Nm when started by direct switching. If a star-delta starter is used for starting this motor, the starting torque will be (a) 1200 Nm (b) 600 Nm (c) 300 Nm (d ) 200 Nm
2. d 12. a
3. c 13. d
4. c 14. c
5. d 15. d
6. c 16. d
Answers 7. d 17. d
8. c
9. a
10. d
22
PROBLEMS (A) Simple Problems 1. A 6-pole induction motor is fed from a 50-Hz supply. If the frequency of the rotor emf at full load is 2 Hz, find the full-load slip and speed. [Ans. 0.04, 960 rpm] 2. A 3-phase, 6-pole, 50-Hz induction motor has a slip of 1 % at no load and 3 % at full load. Find ( a) the synchronous speed, ( b) the no-load speed, (c) the full-load speed, ( d ) the frequency of rotor current at standstill, and ( e) the frequency of rotor current at full load. [Ans. (a) 1000 rpm; (b) 990 rpm; ( c) 970 rpm; ( d ) 50 Hz; (e) 1.5 Hz] 3. A 10-pole induction motor is supplied by a 6-pole alternator which is driven by a prime mover at 1200 rpm. If the motor runs at slip of 3 %, what is its speed? [Ans. 698.4 rpm] 4. A 4-pole induction motor is energized from a 50-Hz supply system. If the machine runs on full-load at 3 % slip, determine the running speed and the frequency of the rotor currents. [Ans. 1455 rpm, 1.5 Hz] 5. If the electromotive force in the stator of an 8-pole induction motor has a frequency of 50 Hz, and that in the rotor is 1.5 Hz, at what speed is the motor running? What is the slip? [Ans. 727.5 rpm, 0.03] 6. A 3-phase, 440-V, 50-Hz, 4-pole induction motor the standstill rotor emf per phase of 115 V. If the motor is running at 1440 rpm, calculate for this speed ( a) the slip, (b) the frequency of the rotor induced emf, and ( c) the value of the rotor induced emf per phase. [Ans. (a) 4 %; (b) 2 Hz; (c) 4.6 V] 7. A 3-phase, 440-V, 50-Hz, 4-pole, phase-wound induction motor at standstill has the induced emf of 157 V between the slip-ring terminals. The rotor windings are star- connected and have a resistance of 0.6 Ω per phase. Calculate the rotor current when the slip-ring terminals are short-circuited and the rotor is rotating at a speed of 1455 rpm. [ Ans. 4.53 A] 8. A 6-pole, 3-phase, 50-Hz motor with star-connected rotor h as the rotor resistance of 0.3 Ω per phase and the rotor standstill reactance of 1.5 Ω per phase. When the motor is at standstill, the emf between the slip-rings on open circuit is 175 V. If the motor runs at a speed of 950 rpm, find ( a) the slip, ( b) the rotor emf per phase, and ( c) the rotor frequency and reactance. [Ans. (a) 0.05; (b) 5.05 V; (c) 2.5 Hz, 0.075 Ω per phase] 9. A 500-V, 3-phase, 6-pole, 50-Hz induction motor, working at a power factor of 0.87, develops 20 hp inclusive of windage and friction losses when running at 975 rpm. Calculate ( a) the slip, ( b) the rotor copper loss, ( c) the total input if the stator losses are 1500 W, (d ) the current drawn by the motor from the supply, and ( e) the frequency of rotor current. [Ans. (a) 2.5 %; (b) 382.6 W; (c) 16.8 kW; (d ) 22.3 A; (e) 1.25 Hz] (B) Tricky Problems 10. A 3-phase, 12-pole alternator is driven by an engine running at 500 rpm. The alternator supplies an induction motor which has a full-load speed of 1455 rpm. Find the slip and the number of poles of the motor. [Ans. 3 %, 4 poles] 11. A 3-phase, 400-V, 50-Hz, 30-hp, 2-pole induction motor operates at an efficiency of 85 % with a power factor of 0.75 lag. Calculate the current drawn by the motor from the power mains. [Ans. 50.67 A] 12. The power input to a 3-phase induction motor is 60 kW. The stator losses total to 1 kW. Find the total mechanical power developed and the rotor copper losses per phase, if the motor is running with a slip of 3 %. [Ans. 57.23 kW, 1.77 kW] 13. A 3-phase, 50-Hz, 4-pole, induction motor has a rotor resistance of 0.024 ohm per phase and standstill reactance of 0.6 ohm per phase. Determine the speed at which maximum torque is developed. [Ans. 1440 rpm] 14. A 3-phase, 4-pole induction motor runs at a speed of 1440 rpm on 500-V, 50-Hz mains. The mechanical power available at the shaft is 20.3 hp. The mechanical losses are 2.23 hp. The stator copper loss and iron loss add up to 1 kW. Calculate ( a) the slip, (b) the rotor copper losses, and ( c) the efficiency. [Ans. (a) 4 %; (b) 701 W; (c) 81.8 %] 15. The power input to a three-phase induction motor is 50 kW and the corresponding stator losses are 2 kW. Calculate ( a) the total 2 mechanical power developed and the rotor I R loss when the slip is 3 %, (b) the output horse power of the motor, if the friction and windage losses are 1.0 kW, and ( c) the efficiency of the motor. [Ans. (a) 46.56 kW; (b) 61 hp; (c) 91.1 %] 16. The rotor of a 6-pole, 50-Hz, slip-ring induction motor has a resistance of 0.2 ohm per phase. It is running at 960 rpm. Calculate the resistance to be added to the rotor circuit to reduce the speed to 600 rpm, the torque remaining the same. [ Ans. 1.8 Ω] 17. A 3-phase, 50-Hz, 6-pole, 18.65-kW, slip-ring induction motor runs at 960 rpm on full-load with a rotor current of 35 A per phase. Calculate the resistance per phase of the rotor winding, allowing 1 kW for mechanical losses. [ Ans. 0.223 Ω/phase] 18. The power supplied to a 3-phase induction motor is 40 kW and the corresponding stator losses are 1.5 kW. Calculate ( a) the total mechanical power developed and the rotor copper loss when the slip is 0.04 per unit, and ( b) the efficiency of the motor. Neglect the rotor iron losses. [Ans. (a) 36.96 kW, 1.54 kW; (b) 92.4 %] (C) Challenging Problems 19. A three-phase, 415-V, phase-wound, 50-Hz, 4-pole induction motor has a delta-connected stator with 240 conductors per phase and a star-connected rotor with 48 conductors per phase. The rotor winding has a resistance of 0.013 Ω per phase and a leakage reactance of 0.048 Ω per phase at standstill. Calculate ( a) the rotor emf per phase at standstill with the rotor on opoen circuit, ( b) the rotor emf and current per phase at 4 % slip, and ( c) the phase difference between the rotor emf and current for a slip of 4 %. [Ans. (a) 83 V; (b) 3.32 V, 252.7 A; (c) φ 2
= 8.4° ]
20. A 50-Hz, 8-pole induction motor has a full-load slip of 2 %. The rotor resistance is 0.001 Ω/phase and the standstill reactance is 0.005 Ω/phase. Find the ratio of the maximum to full-load torque, and the speed at which the maximum torque occurs. [Ans. 5.05, 600 rpm] 21. A 3-kV, 24-pole, 50-Hz, 3-phase, star-connected induction motor has a slip-ring rotor of resistance 0.016 ohm per phase and standstill reactance of 0.265 ohm per phase. Full-load torque is obtained at a speed of 247 rpm. Neglecting the stator impedance, calculate ( a) the ratio of maximum to full-load torque, and ( b) the speed of maximum torque. [Ans. (a) 2.6; (b) 235 rpm]
23
22. A 6-pole, 3-phase induction motor develops 30 hp including 2 hp mechanical losses at a speed of 950 rpm on 550-V, 50-Hz mains. The power factor is 0.88 lagging. Calculate ( a) the slip, ( b) the rotor copper loss, (c) total input, f stator losses are 2 kW, (d ) the efficiency, and (e) the line current. [Ans. (a) 0.05; (b) 1083.9 W; ( c) 23 678 W; (d ) 93.18 %; ( e) 28.24 A]
EXPERIMENTAL EXERCISE 15 .1 Load Test on a Three-Phase Induction Motor OBJECTIVES: For different settings of the load, To plot the efficiency versus output power characteristic curve. To plot the torque versus output power characteristic curve. To plot the line current versus output power characteristic curve. To plot the power factor versus output power characteristic curve. To plot the slip versus output power characteristic curve. To plot the torque versus speed characteristic curve. APPARATUS: A 400-V, Three-phase supply; One 15-A, Three-phase variac; Two single-phase wattmeters, (0 - 10 A), 400 V; One voltmeter, MI, (0 - 400 V); One ammeter, MI, (0-10 A); One three-phase, squirrel-cage, induction motor with brake and pulley arrangement. CIRCUIT DIAGRAM: The circuit diagram is shown in Fig. 4.15. BRIEF THEORY: The load test on an induction motor helps us to determine its complete performance characteristics. Normal rated voltage at normal frequency is supplied to the motor and a variable load is applied to its shaft. Load Torque: The mechanical load of the shaft can be varied by using brake and pulley arrangement. A pulley or a drum (called brake drum) is attached to the shaft. A belt of suitable length rides over this drum. Two spring balances S1 and S2 are attached at each end of the belt. The other end of the spring balance is fixed to a nut and bolt. If the nut is tightened, the belt presses harder on to the rotating drum thereby increasing the mechanical load. If S1 and S2 are the readings of the two spring balances, the net restraining force applied to the drum is 1. 2. 3. 4. 5. 6.
F = ( S1 − S2 ) kgf = ( S1 − S 2 ) g = (S1 − S 2 ) × 9.8 N If D is the effective diameter of the brake drum, the torque applied to the shaft is given as
τ = F ×
D
2
= ( S1 − S2 ) × 9.8×
D
2
Nm
…(i)
Slip: The speed at which the stator magnetic field rotates is called synchronous speed and is given by
N s =
120 f P
Here, the frequency f of the 3-phase ac supply is 50 Hz, and the number of poles P can be seen on the name plate of the induction motor. The actual speed N of the rotor is always less than N s. The difference between the two is called slip , which can be defined in percentage as
s=
N s − N N s
×100 %
…(ii)
The slip at full-load normally is 2 to 5 %.
Fig. 4.15 Circuit arrangement to conduct load test on a 3-phase, squirrel cage induction motor.
24
Output Power: The output power Po of the motor is given as
Po =
2π N τ 60
watts
…(iii)
Input Power: The input power Pin to the motor can be determined from the readings of the two wattmeters connected in the input circuit (see Fig. 4.15). If P1 and P2 are the readings of the two wattmeters, the input power is given as
Pin = P1 + P2
…(iv)
Note that at low power factors, one of the wattmeters may give negative deflection. If it happens, the connections of one of the coils (either current coil or pressure coil ) should be reversed and the readings should be treated negative. Input Power Factor: Using the readings P1 and P2 of the two wattmeters, the input power factor angle can be calculated from following:
φ = tan −1 3
P1 − P2 P1 + P2
The power factor is then given as
pf = cos φ
…(v)
Efficiency: The efficiency of the induction motor is given as
η =
Po Pin
× 100 %
…(vi)
PROCEDURE: 1. 2. 3. 4. 5.
Make the connections as shown in Fig. 4.15. Keep the three-phase variac at its minimum position. Release the brake and pulley system so that the shaft is not loaded. Switch on the three-phase supply. Gradually increase the position of the variac towards higher voltage. Note the direction of deflection in the wattmeters. If the deflection is found negative in a wattmeter, bring the variac back to its minimum and reverse the connections of either the current coil or pressure coil of that wattmeter. The reading of this wattmeter should be recorded negative. 6. Slowly increase variac position to apply rated voltage to the induction motor. 7. Note the readings of all the meters. Measure the speed of the rotor by attaching the tachometer to the shaft. These are no-load readings. 8. Increase the mechanical load by tightening the nut of the springs and note the readings. Repeat this for 6 – 8 settings of the load. 9. At some stage of the load, one wattmeter may again start reading negative. If so, again repeat step 5. 10. Switch off the three-phase supply.
OBSERVATIONS: See Table 4.1a, (in Excel) attached separately.
CALCULATIONS: See Table 4.1b, (in Excel) attached separately
RESULTS: For eight different observations, different characteristic curves can be plotted.
PRECAUTIONS: 1. 2. 3. 4.
Before switching on the supply, the zero of the ammeter, voltmeter, and wattmeters should be checked. The readings in the ammeter should not exceed the current ratings of the wattmeter. Special care should be taken a bout the sign of the readings of wattmeters. During the experiment, the mechanical power developed by the motor is dissipated as frictional loss (heat) at the brake and pulley arrangement. Cooling of the brake pulley should be done properly; otherwise, the belt may wear away in short time.
VIVA-VOCE: 1.
2. 3. 4.
5.
Q.: What do you mean by the slip? Ans.: The difference between the synchronous speed and the actual speed of the rotor, expressed as percentage, is known as slip . Q.: Usually, what is the order of slip for an induction motor, while running at no load and running at full load? Ans.: At no load, it is about 1 - 2 %, and at full-load it is about 2 - 5 %. Q.: At what voltage should the load test on an induction motor be performed? Ans.: At its rated voltage. Q.: Suppose that an induction motor is running on no load. What do you think its power factor will be? Ans.: At no load, the motor draws very small active power. Therefore, its power factor is very low, of the order of 0.1 to 0.3 lagging. Q.: And when it is running at full load, what would be the power factor? Ans.: At full load, the power factor is about 0.85 lagging.
25
6.
Q.: Under what conditions, one of the wattmeters may give negative reading? Ans.: If the reading of one wattmeter, say P2, is negative, the value of ratio
( P1 − P2 ) /( P1 + P2 ) will always be greater than
unity. Thus,
tan φ = 3 × (greater than unity) cos φ < 0.5 7. 8. 9.
10.
11.
12. 13.
14.
⇒
tan φ > 3
⇒
φ > 60°
or
Hence, if one of the wattmeters reads negative, the power factor has to be less than 0.5. Q.: Approximately, how much is the efficiency of an induction motor while running at full load? Ans.: It is about 0.75 to 0.85. Q.: What do you mean by ‘ pull-out torque ’ of an induction motor? Ans.: The maximum torque the motor can develop is also called pull-out torque. Q.: Why should it be called pull-out torque? Ans.: If the load on an induction motor is increased beyond the maximum torque it can develop, the motor slows down, the slip increases and the operation becomes unstable. Ultimately, the machine stops. It pulls out of the motion. Q.: What is meant by plugging of an induction motor? Ans.: In case of a running motor, if suddenly two of the three-phase supply connections to the stator are inter-changed, the stator magnetic field reverses its direction of rotation. As a result, the machine comes to a stop quickly. This amounts to braking or pluggingof the machine. Q.: What do you mean by single-phasing of a 3-phase induction motor? Ans.: If one of the lines of the three-phase supply gets open due to some fault, the 3-phase motor continues running as a single-phase motor. This phenomenon is called single phasing. When this happens, the current drawn by the two remaining lines increases and the motor gets overheated, and it may even get damaged. Q.: Do you know any other machine whose output-speed characteristic is similar to that of a three-phase induction motor? Ans.: Yes, dc shunt motor. Q.: In what way, an induction motor is similar to a transformer? Ans.: (i) Both work on the principle of electromagnetic induction. (ii) The stator of the induction motor is analogous to the primary of the transformer and the rotor to the secondary of the transformer. (iii) Both can be represented by similar equivalent circuits. (iv) Both machines can be tested through similar tests. The no-load test on an induction motor is similar to the open-circuit test on a transformer. Q.: In what way, an induction motor differs from a transformer? Ans.: (i) The windings of a transformer are cylindrical, whereas those of an induction motor are distributed. (ii) The induction motor has an air gap in its magnetic path and hence the magnetizing current of an induction motor is much higher than that of a transformer. (iii) Due to mechanical losses (frictional and windage losses), the efficiency of an induction motor is lower than that of a transformer. (iv) An induction motor converts electrical energy into mechanical energy, whereas a transformer transforms the electrical energy from one voltage level to another.
26
