Introduction to 3 phase induction motors Definition: Induction motor is an a.c. motor in which currents in the stator winding (which is connected to the supply ) set up a flux which causes currents to be induced in the rotor winding ; these currents interact with the flux to produce rotation . Also called asynchronous motor. The rotor receives electrical power in exactly the same way as the secondary of a two winding transformer receiving its power from primary. That is why an induction motor can be called as a rotating transformer i.e., in which primary winding is stationary but the secondary is free to rotate.
Types of induction motors: Depending on rotor there are two types of induction motors :1- Squirrel cage induction motor : Rotors is very simple and consist of bars of aluminum (or copper) with shorting rings at the ends. 2- wound rotor induction motor : Three phase windings (star connected) with terminals brought out to slip rings for external connections. The first type is more common used compared to second one due to: 1- Robust : No brushes . No contacts on rotor shaft. 2- Easy to manufacture. 3- Almost maintenance-free , except for bearing and other mechanical parts. 4- Since the rotor has very low resistance, the copper loss is low and efficiency is high .
Construction: There are two main types of components which are used in induction motor manufacturing as follows: 1- Active components : which are classified into two categories:a- Magnetic materials (0.5 mm electrical steel) b- Electrical materials ( copper wires ,insulations ,bars , end rings ,slip rings ,brushes , and lead wires)
1
2- Constructional components: like frame , end shields , shaft , bearings , and fan . These components are shown in Figures 1 & 2 .
Figure- 1 parts of squirrel cage induction motor
Figure- 2 Axial view of squirrel cage induction motor
2
Stator construction : The stator is made up of several thin laminations (0.5 mm )of electrical steel (silicon steel) , they are punched and clamped together to form a hollow cylinder ( stator core ) with slots , as shown in Figure- 3. Coils of insulated wires are inserted into these slots . Each grouping of coils , together with the core it surrounds , forms an electromagnet (a pair of poles).The number of poles of an induction motor depends on the internal connection of the stator windings .
Rotor construction: The squirrel cage rotor is made up of several thin electrical steel lamination (0.5mm) with evenly spaced bars , which are made up of aluminum or copper , along the periphery .In the most popular type of rotor (squirrel cage rotor) , these bars are connected at ends mechanically and electrically by the use of end rings as shown in Fig.4. Almost 90 % of induction motors have squirrel cage rotors . This is because the squirrel cage rotors has a simple and rugged construction .The rotor slots are not exactly parallel to the shaft. Instead , they are given a skew for two main reasons :The first reason is to make the motor run quietly by reducing magnetic hum and to decrease slot harmonics. The second reason is to help reduce the locking tendency of the rotor (the rotor teeth tend to remain locked under the stator teeth due to direct magnetic attraction between the two). The rotor is mounted on the shaft using bearings on each end ; one end of the shaft is normally kept longer than the other for driving the load . Between the stator and the rotor, there exist an air gap , through which due to induction , the energy is transferred from the stator to the rotor.
3
The wound rotor has a set of windings on the rotor slots which are not short circuited , but are terminated to a set of slip rings . These are helpful in adding external resistors and contactors , as shown in Figure 5.
4
Typical name plate of induction motor A typical name plate of induction motor is shown in Figure 6, and Table 1 .
Motor standards NEMA : National Electrical Manufactures Association IEC : International Electrotechnical Commission
5
Motor insulation class Insulations have been standardized and graded by their resistance to thermal aging and failure. Four insulation classes are in common use , they have been designated by the letters A , B , F, and H . The temperature capabilities of these classes are separated from each other by 25 °C increments. The temperature capabilities of each insulation class is defined as being the maximum temperature at which the insulation can be operated to yield an average life of 20,000 hours , as in Table below. Insulation Class
Temperature Rating
A
105° C
B
130° C
F
155° C
H
180° C
Motor degree of protection I P : International Protection , I P * # * 0 1 2
3 4
5 6
Protection against ingress of bodies Non protected
#
Protected against ingress of foreign solid bodies of 50 mm or greater. Protected against ingress of foreign solid bodies of 12 mm or greater.
1
Protected against ingress of dripping water .
2
Protected against ingress of foreign solid bodies of 2.5 mm or greater. Protected against ingress of foreign solid bodies of 1 mm or greater. Partially protected against ingress of dust . Totally protected against ingress of dust.
3
Protection against ingress of dripping water at maximum angle of 150 degrees from the vertical. Protection against water falling like rain.
0
Protection against ingress of water Non protected
4
Protection against splashing water.
5
Protection against water jets . Protection against special conditions on ship's board.
6 7 8
6
Protection against immersion in water . Protection against prolonged immersion in water .
Chapter (8)
Three Phase Induction Motors Introduction The three-phase induction motors are the most widely used electric motors in industry. They run at essentially constant speed from no-load to full-load. However, the speed is frequency dependent and consequently these motors are not easily adapted to speed control. We usually prefer d.c. motors when large speed variations are required. Nevertheless, the 3-phase induction motors are simple, rugged, low-priced, easy to maintain and can be manufactured with characteristics to suit most industrial requirements. In this chapter, we shall focus our attention on the general principles of 3-phase induction motors.
8.1 Three-Phase Induction Motor Like any electric motor, a 3-phase induction motor has a stator and a rotor. The stator carries a 3-phase winding (called stator winding) while the rotor carries a short-circuited winding (called rotor winding). Only the stator winding is fed from 3-phase supply. The rotor winding derives its voltage and power from the externally energized stator winding through electromagnetic induction and hence the name. The induction motor may be considered to be a transformer with a rotating secondary and it can, therefore, be described as a “transformertype” a.c. machine in which electrical energy is converted into mechanical energy.
Advantages (i) (ii) (iii) (iv) (v)
It has simple and rugged construction. It is relatively cheap. It requires little maintenance. It has high efficiency and reasonably good power factor. It has self starting torque.
Disadvantages (i)
It is essentially a constant speed motor and its speed cannot be changed easily. (ii) Its starting torque is inferior to d.c. shunt motor.
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8.2 Construction A 3-phase induction motor has two main parts (i) stator and (ii) rotor. The rotor is separated from the stator by a small air-gap which ranges from 0.4 mm to 4 mm, depending on the power of the motor.
1.
Stator It consists of a steel frame which encloses a hollow, cylindrical core made up of thin laminations of silicon steel to reduce hysteresis and eddy current losses. A number of evenly spaced slots are provided on the inner periphery of the laminations [See Fig. (8.1)]. The insulated connected to form a Fig.(8.1) balanced 3-phase star or delta connected circuit. The 3-phase stator winding is wound for a definite number of poles as per requirement of speed. Greater the number of poles, lesser is the speed of the motor and vice-versa. When 3-phase supply is given to the stator winding, a rotating magnetic field (See Sec. 8.3) of constant magnitude is produced. This rotating field induces currents in the rotor by electromagnetic induction.
2.
Rotor The rotor, mounted on a shaft, is a hollow laminated core having slots on its outer periphery. The winding placed in these slots (called rotor winding) may be one of the following two types: (i) Squirrel cage type (ii) Wound type (i)
Squirrel cage rotor. It consists of a laminated cylindrical core having parallel slots on its outer periphery. One copper or aluminum bar is placed in each slot. All these bars are joined at each end by metal rings called end rings [See Fig. (8.2)]. This forms a permanently short-circuited winding which is indestructible. The entire construction (bars and end rings) resembles a squirrel cage and hence the name. The rotor is not connected electrically to the supply but has current induced in it by transformer action from the stator. Those induction motors which employ squirrel cage rotor are called squirrel cage induction motors. Most of 3-phase induction motors use squirrel cage rotor as it has a remarkably simple and robust construction enabling it to operate in the most adverse circumstances. However, it suffers from the disadvantage of a low starting torque. It is because the rotor bars are permanently short-circuited and it is not possible to add any external resistance to the rotor circuit to have a large starting torque.
182
Fig.(8.2)
Fig.(8.3)
(ii) Wound rotor. It consists of a laminated cylindrical core and carries a 3phase winding, similar to the one on the stator [See Fig. (8.3)]. The rotor winding is uniformly distributed in the slots and is usually star-connected. The open ends of the rotor winding are brought out and joined to three insulated slip rings mounted on the rotor shaft with one brush resting on each slip ring. The three brushes are connected to a 3-phase star-connected rheostat as shown in Fig. (8.4). At starting, the external resistances are included in the rotor circuit to give a large starting torque. These resistances are gradually reduced to zero as the motor runs up to speed.
Fig.(8.4) The external resistances are used during starting period only. When the motor attains normal speed, the three brushes are short-circuited so that the wound rotor runs like a squirrel cage rotor.
8.3 Rotating Magnetic Field Due to 3-Phase Currents When a 3-phase winding is energized from a 3-phase supply, a rotating magnetic field is produced. This field is such that its poles do no remain in a fixed position on the stator but go on shifting their positions around the stator. For this reason, it is called a rotating Held. It can be shown that magnitude of this rotating field is constant and is equal to 1.5 φm where φm is the maximum flux due to any phase.
183
To see how rotating field is produced, consider a 2-pole, 3i-phase winding as shown in Fig. (8.6 (i)). The three phases X, Y and Z are energized from a 3-phase source and currents in these phases are indicated as Ix, Iy and Iz [See Fig. (8.6 (ii))]. Referring to Fig. (8.6 (ii)), the fluxes produced by these currents are given by:
φ x = φ m sin ωt
Fig.(8.5)
φ y = φ m sin (ωt − 120°) φ z = φ m sin (ωt − 240°) Here φm is the maximum flux due to any phase. Fig. (8.5) shows the phasor diagram of the three fluxes. We shall now prove that this 3-phase supply produces a rotating field of constant magnitude equal to 1.5 φm.
Fig.(8.6) 184
(i)
At instant 1 [See Fig. (8.6 (ii)) and Fig. (8.6 (iii))], the current in phase X is zero and currents in phases Y and Z are equal and opposite. The currents are flowing outward in the top conductors and inward in the bottom conductors. This establishes a resultant flux towards right. The magnitude of the resultant flux is constant and is equal to 1.5 φm as proved under: At instant 1, ωt = 0°. Therefore, the three fluxes are given by;
φ x = 0;
φ y = φ m sin (− 120°) = −
φz = φ m sin (− 240°) =
Fig.(8.7)
3 φ ; 2 m
3 φ 2 m
The phasor sum of − φy and φz is the resultant flux φr [See Fig. (8.7)]. It is clear that: Resultant flux, φr = 2 ×
3 60° 3 3 φ m cos = 2× φm × = 1.5 φ m 2 2 2 2
(ii) At instant 2, the current is maximum (negative) in φy phase Y and 0.5 maximum (positive) in phases X and Y. The magnitude of resultant flux is 1.5 φm as proved under: At instant 2, ωt = 30°. Therefore, the three fluxes are given by;
φm 2 φ y = φ m sin (−90°) = −φ m
Fig.(8.8)
φ x = φ m sin 30° =
φ z = φ m sin (−210°) =
φm 2
The phasor sum of φx, − φy and φz is the resultant flux φr φ 120° φ m Phasor sum of φx and φz, φ'r = 2 × m cos = 2 2 2 φ Phasor sum of φ'r and − φy, φ r = m + φ m = 1.5 φ m 2 Note that resultant flux is displaced 30° clockwise from position 1.
185
(iii) At instant 3, current in phase Z is zero and the currents in phases X and Y are equal and opposite (currents in phases X and Y arc 0.866 × max. value). The magnitude of resultant flux is 1.5 φm as proved under: At instant 3, ωt = 60°. Therefore, the three fluxes are given by;
3 φ x = φ m sin 60° = φ ; 2 m 3 φ y = φ m sin (− 60°) = − φ ; 2 m φz = φ m sin (− 180°) = 0
Fig.(8.9)
The resultant flux φr is the phasor sum of φx and − φy (Q φ z = 0 ) .
φr = 2 ×
3 60° φ m cos = 1.5 φ m 2 2
Note that resultant flux is displaced 60° clockwise from position 1. (iv) At instant 4, the current in phase X is maximum (positive) and the currents in phases V and Z are equal and negative (currents in phases V and Z are 0.5 × max. value). This establishes a resultant flux downward as shown under: At instant 4, ωt = 90°. Therefore, the three fluxes are given by;
Fig.(7.10)
φ x = φ m sin 90° = φ m φm 2 φ φz = φ m sin (−150°) = − m 2 φ y = φ m sin (−30°) = −
The phasor sum of φx, − φy and − φz is the resultant flux φr φ 120° φ m Phasor sum of − φz and − φy, φ'r = 2 × m cos = 2 2 2 φ Phasor sum of φ'r and φx, φ r = m + φ m = 1.5 φ m 2 Note that the resultant flux is downward i.e., it is displaced 90° clockwise from position 1. 186
It follows from the above discussion that a 3-phase supply produces a rotating field of constant value (= 1.5 φm, where φm is the maximum flux due to any phase).
Speed of rotating magnetic field The speed at which the rotating magnetic field revolves is called the synchronous speed (Ns). Referring to Fig. (8.6 (ii)), the time instant 4 represents the completion of one-quarter cycle of alternating current Ix from the time instant 1. During this one quarter cycle, the field has rotated through 90°. At a time instant represented by 13 or one complete cycle of current Ix from the origin, the field has completed one revolution. Therefore, for a 2-pole stator winding, the field makes one revolution in one cycle of current. In a 4-pole stator winding, it can be shown that the rotating field makes one revolution in two cycles of current. In general, fur P poles, the rotating field makes one revolution in P/2 cycles of current. ∴ or
Cycles of current =
Cycles of current per second =
P × revolutions of field 2 P × revolutions of field per second 2
Since revolutions per second is equal to the revolutions per minute (Ns) divided by 60 and the number of cycles per second is the frequency f,
∴ or
f=
P Ns Ns P × = 2 60 120
Ns =
120 f P
The speed of the rotating magnetic field is the same as the speed of the alternator that is supplying power to the motor if the two have the same number of poles. Hence the magnetic flux is said to rotate at synchronous speed.
Direction of rotating magnetic field The phase sequence of the three-phase voltage applied to the stator winding in Fig. (8.6 (ii)) is X-Y-Z. If this sequence is changed to X-Z-Y, it is observed that direction of rotation of the field is reversed i.e., the field rotates counterclockwise rather than clockwise. However, the number of poles and the speed at which the magnetic field rotates remain unchanged. Thus it is necessary only to change the phase sequence in order to change the direction of rotation of the magnetic field. For a three-phase supply, this can be done by interchanging any two of the three lines. As we shall see, the rotor in a 3-phase induction motor runs in the same direction as the rotating magnetic field. Therefore, the
187
direction of rotation of a 3-phase induction motor can be reversed by interchanging any two of the three motor supply lines.
8.4 Alternate Mathematical Analysis for Rotating Magnetic Field We shall now use another useful method to find the magnitude and speed of the resultant flux due to three-phase currents. The three-phase sinusoidal currents produce fluxes φ1, φ2 and φ3 which vary sinusoidally. The resultant flux at any instant will be the vector sum of all the three at that instant. The fluxes are represented by three variable Fig.(8.11) magnitude vectors [See Fig. (8.11)]. In Fig. (8.11), the individual flux directions arc fixed but their magnitudes vary sinusoidally as does the current that produces them. To find the magnitude of the resultant flux, resolve each flux into horizontal and vertical components and then find their vector sum.
φh = φ m cos t ωt − φ m cos(ωt − 120°) cos 60° − φ m cos(ωt − 240°) cos 60° 3 = φm cos ωt 2 3 φ v = 0 − φm cos(ωt − 120°) cos 60° + φ m cos(ωt − 240°) cos 60° = φ m sin ωt 2 The resultant flux is given by;
[
]
1/ 2 3 3 φr = φ 2h + φ 2v = φ m cos 2 ωt + (− sin ωt ) 2 = φ m = 1.5 φ m = Constant 2 2
Thus the resultant flux has constant magnitude (= 1.5 φm) and does not change with time. The angular displacement of φr relative to the OX axis is
∴
3 φ v 2 φ m sin ωt = tan ωt = tan θ = φh 3 φ cos ωt 2 m θ = ωt
Fig.(8.12)
Thus the resultant magnetic field rotates at constant angular velocity ω( = 2 πf) rad/sec. For a P-pole machine, the rotation speed (ωm) is
ωm =
2 ω rad / sec P
188
2 π Ns 2 = × 2πf 60 P
or
∴
Ns =
... N s is in r.p.m.
120 f P
Thus the resultant flux due to three-phase currents is of constant value (= 1.5 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a synchronous speed of 120 f/P r.p.m. For example, for a 6-pole, 50 Hz, 3-phase induction motor, N, = 120 × 50/6 = 1000 r.p.m. It means that flux rotates around the stator at a speed of 1000 r.p.m.
8.5 Principle of Operation Consider a portion of 3-phase induction motor as shown in Fig. (8.13). The operation of the motor can be explained as under: (i) When 3-phase stator winding is energized from a 3-phase supply, a rotating magnetic field is set up which rotates round the stator at synchronous speed Ns (= 120 f/P). (ii) The rotating field passes through the air gap and cuts the rotor conductors, Fig.(1-) which as yet, are stationary. Due to the relative speed between the rotating flux and the stationary rotor, e.m.f.s are induced in the rotor conductors. Since the rotor circuit is short-circuited, currents start flowing in the rotor conductors. (iii) The current-carrying rotor conductors are placed in the magnetic field produced by the stator. Consequently, mechanical force acts on the rotor conductors. The sum of the mechanical forces on all the rotor conductors produces a torque which tends to move the rotor in the same direction as the rotating field. (iv) The fact that rotor is urged to follow the stator field (i.e., rotor moves in the direction of stator field) can be explained by Lenz’s law. According to this law, the direction of rotor currents will be such that they tend to oppose the cause producing them. Now, the cause producing the rotor currents is the relative speed between the rotating field and the stationary rotor conductors. Hence to reduce this relative speed, the rotor starts running in the same direction as that of stator field and tries to catch it. 189
8.6 Slip We have seen above that rotor rapidly accelerates in the direction of rotating field. In practice, the rotor can never reach the speed of stator flux. If it did, there would be no relative speed between the stator field and rotor conductors, no induced rotor currents and, therefore, no torque to drive the rotor. The friction and windage would immediately cause the rotor to slow down. Hence, the rotor speed (N) is always less than the suitor field speed (Ns). This difference in speed depends upon load on the motor. The difference between the synchronous speed Ns of the rotating stator field and the actual rotor speed N is called slip. It is usually expressed as a percentage of synchronous speed i.e., % age slip, s =
Ns − N × 100 Ns
(i) The quantity Ns − N is sometimes called slip speed. (ii) When the rotor is stationary (i.e., N = 0), slip, s = 1 or 100 %. (iii) In an induction motor, the change in slip from no-load to full-load is hardly 0.1% to 3% so that it is essentially a constant-speed motor.
8.7 Rotor Current Frequency The frequency of a voltage or current induced due to the relative speed between a vending and a magnetic field is given by the general formula; Frequency =
NP 120
where N = Relative speed between magnetic field and the winding P = Number of poles For a rotor speed N, the relative speed between the rotating flux and the rotor is Ns − N. Consequently, the rotor current frequency f' is given by;
( N s − N)P 120 s Ns P = 120
f '=
N −N Q s = s Ns N P Q f = s 120
= sf
i.e., Rotor current frequency = Fractional slip x Supply frequency (i) When the rotor is at standstill or stationary (i.e., s = 1), the frequency of rotor current is the same as that of supply frequency (f' = sf = 1× f = f). 190
(ii) As the rotor picks up speed, the relative speed between the rotating flux and the rotor decreases. Consequently, the slip s and hence rotor current frequency decreases. Note. The relative speed between the rotating field and stator winding is Ns − 0 = Ns. Therefore, the frequency of induced current or voltage in the stator winding is f = Ns P/120—the supply frequency.
8.8 Effect of Slip on The Rotor Circuit When the rotor is stationary, s = 1. Under these conditions, the per phase rotor e.m.f. E2 has a frequency equal to that of supply frequency f. At any slip s, the relative speed between stator field and the rotor is decreased. Consequently, the rotor e.m.f. and frequency are reduced proportionally to sEs and sf respectively. At the same time, per phase rotor reactance X2, being frequency dependent, is reduced to sX2. Consider a 6-pole, 3-phase, 50 Hz induction motor. It has synchronous speed Ns = 120 f/P = 120 × 50/6 = 1000 r.p.m. At standsill, the relative speed between stator flux and rotor is 1000 r.p.m. and rotor e.m.f./phase = E2(say). If the fullload speed of the motor is 960 r.p.m., then,
s= (i)
1000 − 960 = 0.04 1000
The relative speed between stator flux and the rotor is now only 40 r.p.m. Consequently, rotor e.m.f./phase is reduced to:
E2 ×
40 = 0.04 E 2 1000
or
sE 2
(ii) The frequency is also reduced in the same ratio to:
50 ×
40 = 50 × 0.04 1000
or
sf
(iii) The per phase rotor reactance X2 is likewise reduced to:
X2 ×
40 = 0.04X 2 1000
or
sX 2
Thus at any slip s, Rotor e.m.f./phase = sE2 Rotor reactance/phase = sX2 Rotor frequency = sf where E2,X2 and f are the corresponding values at standstill.
191
8.9 Rotor Current Fig. (8.14) shows the circuit of a 3-phase induction motor at any slip s. The rotor is assumed to be of wound type and star connected. Note that rotor e.m.f./phase and rotor reactance/phase are s E2 and sX2 respectively. The rotor resistance/phase is R2 and is independent of frequency and, therefore, does not depend upon slip. Likewise, stator winding values R1 and X1 do not depend upon slip.
Fig.(8.14) Since the motor represents a balanced 3-phase load, we need consider one phase only; the conditions in the other two phases being similar. At standstill. Fig. (8.15 (i)) shows one phase of the rotor circuit at standstill. Rotor current/phase, I 2 = Rotor p.f., cos φ 2 =
E2 E2 = Z2 R 22 + X 22
R2 R2 = Z2 R 22 + X 22
Fig.(8.15) When running at slip s. Fig. (8.15 (ii)) shows one phase of the rotor circuit when the motor is running at slip s. Rotor current, I'2 =
sE 2 sE 2 = Z' 2 R 22 + (sX 2 )2 192
Rotor p.f., cos φ'2 =
R2 R2 = Z'2 R 22 + (sX 2 )2
8.10 Rotor Torque The torque T developed by the rotor is directly proportional to: (i) rotor current (ii) rotor e.m.f. (iii) power factor of the rotor circuit
∴
T ∝ E 2 I 2 cos φ 2 T = K E 2 I 2 cos φ 2
or where
I2 = rotor current at standstill E2 = rotor e.m.f. at standstill cos φ2 = rotor p.f. at standstill
Note. The values of rotor e.m.f., rotor current and rotor power factor are taken for the given conditions.
8.11 Starting Torque (Ts) Let
E2 = rotor e.m.f. per phase at standstill X2 = rotor reactance per phase at standstill R2 = rotor resistance per phase Rotor impedance/phase, Z 2 = R 22 + X 22 Rotor current/phase, I 2 = Rotor p.f., cos φ 2 = ∴
...at standstill
E2 E2 = Z2 R 22 + X 22
...at standstill
R2 R2 = Z2 R 22 + X 22
...at standstill
Starting torque, Ts = K E 2 I 2 cos φ 2
= K E2 × =
K E 22 R 2 R 22 + X 22
193
E2 R 22 + X 22
×
R2 R 22 + X 22
Generally, the stator supply voltage V is constant so that flux per pole φ set up by the stator is also fixed. This in turn means that e.m.f. E2 induced in the rotor will be constant.
∴
Ts =
K1 R 2 R 22 + X 22
=
K1 R 2 Z 22
where K1 is another constant. It is clear that the magnitude of starting torque would depend upon the relative values of R2 and X2 i.e., rotor resistance/phase and standstill rotor reactance/phase. It can be shown that K = 3/2 π Ns.
∴
E 22 R 2 3 Ts = ⋅ 2π N s R 22 + X 22
Note that here Ns is in r.p.s.
8.12 Condition for Maximum Starting Torque It can be proved that starting torque will be maximum when rotor resistance/phase is equal to standstill rotor reactance/phase. Now
Ts =
K1 R 2
(i)
R 22 + X 22
Differentiating eq. (i) w.r.t. R2 and equating the result to zero, we get,
dTs R 2 (2 R 2 ) 1 =0 = K1 2 − dR 2 R 2 + X 22 R 2 + X 2 2 2 2
(
or
R 22 + X 22 = 2R 22
or
R 2 = X2
)
Hence starting torque will be maximum when: Rotor resistance/phase = Standstill rotor reactance/phase Under the condition of maximum starting torque, φ2 = 45° and rotor power factor is 0.707 lagging [See Fig. (8.16 (ii))]. Fig. (8.16 (i)) shows the variation of starting torque with rotor resistance. As the rotor resistance is increased from a relatively low value, the starting torque increases until it becomes maximum when R2 = X2. If the rotor resistance is increased beyond this optimum value, the starting torque will decrease. 194
Fig.(8.16)
8.13 Effect of Change of Supply Voltage Ts =
K E 22 R 2 R 22 + X 22
Since E2 ∝ Supply voltage V
∴
Ts =
K2 V2 R 2 R 22 + X 22
where K2 is another constant.
∴
Ts ∝ V 2
Therefore, the starting torque is very sensitive to changes in the value of supply voltage. For example, a drop of 10% in supply voltage will decrease the starting torque by about 20%. This could mean the motor failing to start if it cannot produce a torque greater than the load torque plus friction torque.
8.14 Starting Torque of 3-Phase Induction Motors The rotor circuit of an induction motor has low resistance and high inductance. At starting, the rotor frequency is equal to the stator frequency (i.e., 50 Hz) so that rotor reactance is large compared with rotor resistance. Therefore, rotor current lags the rotor e.m.f. by a large angle, the power factor is low and consequently the starting torque is small. When resistance is added to the rotor circuit, the rotor power factor is improved which results in improved starting torque. This, of course, increases the rotor impedance and, therefore, decreases the value of rotor current but the effect of improved power factor predominates and the starting torque is increased. (i)
Squirrel-cage motors. Since the rotor bars are permanently shortcircuited, it is not possible to add any external resistance in the rotor circuit at starting. Consequently, the stalling torque of such motors is low. Squirrel 195
cage motors have starting torque of 1.5 to 2 times the full-load value with starting current of 5 to 9 times the full-load current. (ii) Wound rotor motors. The resistance of the rotor circuit of such motors can be increased through the addition of external resistance. By inserting the proper value of external resistance (so that R2 = X2), maximum starting torque can be obtained. As the motor accelerates, the external resistance is gradually cut out until the rotor circuit is short-circuited on itself for running conditions.
8.15 Motor Under Load Let us now discuss the behaviour of 3-phase induction motor on load. (i) When we apply mechanical load to the shaft of the motor, it will begin to slow down and the rotating flux will cut the rotor conductors at a higher and higher rate. The induced voltage and resulting current in rotor conductors will increase progressively, producing greater and greater torque. (ii) The motor and mechanical load will soon reach a state of equilibrium when the motor torque is exactly equal to the load torque. When this state is reached, the speed will cease to drop any more and the motor will run at the new speed at a constant rate. (iii) The drop in speed of the induction motor on increased load is small. It is because the rotor impedance is low and a small decrease in speed produces a large rotor current. The increased rotor current produces a higher torque to meet the increased load on the motor. This is why induction motors are considered to be constant-speed machines. However, because they never actually turn at synchronous speed, they are sometimes called asynchronous machines. Note that change in load on the induction motor is met through the adjustment of slip. When load on the motor increases, the slip increases slightly (i.e., motor speed decreases slightly). This results in greater relative speed between the rotating flux and rotor conductors. Consequently, rotor current is increased, producing a higher torque to meet the increased load. Reverse happens should the load on the motor decrease. (iv) With increasing load, the increased load currents I'2 are in such a direction so as to decrease the stator flux (Lenz’s law), thereby decreasing the counter e.m.f. in the stator windings. The decreased counter e.m.f. allows motor stator current (I1) to increase, thereby increasing the power input to the motor. It may be noted that action of the induction motor in adjusting its stator or primary current with
196
changes of current in the rotor or secondary is very much similar to the changes occurring in transformer with changes in load.
Fig.(8.17)
8.16 Torque Under Running Conditions Let the rotor at standstill have per phase induced e.m.f. E2, reactance X2 and resistance R2. Then under running conditions at slip s, Rotor e.m.f./phase, E'2 = sE2 Rotor reactance/phase, X'2 = sX2 Rotor impedance/phase, Z'2 = R 22 + (sX 2 )2 Rotor current/phase, I'2 = Rotor p.f., cos φ'm =
E '2 sE 2 = Z' 2 R 22 + (sX 2 )2 R2 R 22 + (sX 2 )2
Fig.(8.18) Running Torque, Tr ∝ E'2 I'2 cos φ'2
∝ φ I'2 cos φ'2
197
(Q E '2 ∝ φ)
∝ φ×
s E2 R 22 + (s X 2 )2
×
R2 R 22 + (s X 2 )2
φ s E2 R 2
∝
R 22 + (s X 2 )2 K φ s E2 R 2
=
R 22 + (s X 2 )2 K1 s E 22 R 2
=
(Q E 2 ∝ φ)
R 22 + (s X 2 )2
If the stator supply voltage V is constant, then stator flux and hence E2 will be constant.
∴
Tr =
K2 s R2
R 22 + (s X 2 )2
where K2 is another constant. It may be seen that running torque is: (i) directly proportional to slip i.e., if slip increases (i.e., motor speed decreases), the torque will increase and vice-versa. (ii) directly proportional to square of supply voltage (Q E 2 ∝ V ) . It can be shown that value of K1 = 3/2 π Ns where Ns is in r.p.s.
∴
s E 22 R 2 s E 22 R 2 3 3 Tr = ⋅ = ⋅ 2π N s R 2 + (s X )2 2π N s (Z' )2 2 2 2
At starting, s = 1 so that starting torque is
E 22 R 2 3 Ts = ⋅ 2π N s R 22 + X 22
8.17 Maximum Torque under Running Conditions Tr =
K2 s R 2
(i)
R 22 + s 2 X 22
In order to find the value of rotor resistance that gives maximum torque under running conditions, differentiate exp. (i) w.r.t. s and equate the result to zero i.e.,
[ (
)
]
dTr K 2 R 2 R 22 + s 2 X 22 − 2s X 22 (s R 2 ) = =0 ds 2 2 2 2 R2 + s X2
(
198
)
or
(R
or
R 22 = s 2 X 22
or
R 2 = s X2
2 2 2+s
)
X 22 − 2s X 22 = 0
Thus for maximum torque (Tm) under running conditions : Rotor resistance/phase = Fractional slip × Standstill rotor reactance/phase Now
Tr ∝
s R2
… from exp. (i) above
R 22 + s 2 X 22
For maximum torque, R2 = s X2. Putting R2 = s X2 in the above expression, the maximum torque Tm is given by;
Tm ∝
1 2 X2
Slip corresponding to maximum torque, s = R2/X2. It can be shown that:
E 22 3 Tm = ⋅ N-m 2π N s 2 X 2 It is evident from the above equations that: (i) The value of rotor resistance does not alter the value of the maximum torque but only the value of the slip at which it occurs. (ii) The maximum torque varies inversely as the standstill reactance. Therefore, it should be kept as small as possible. (iii) The maximum torque varies directly with the square of the applied voltage. (iv) To obtain maximum torque at starting (s = 1), the rotor resistance must be made equal to rotor reactance at standstill.
8.18 Torque-Slip Characteristics As shown in Sec. 8.16, the motor torque under running conditions is given by;
T=
K2 s R 2 R 22 + s 2 X 22
If a curve is drawn between the torque and slip for a particular value of rotor resistance R2, the graph thus obtained is called torque-slip characteristic. Fig. (8.19) shows a family of torque-slip characteristics for a slip-range from s = 0 to s = 1 for various values of rotor resistance.
199
Fig.(8.19) The following points may be noted carefully: (i) At s = 0, T = 0 so that torque-slip curve starts from the origin. (ii) At normal speed, slip is small so that s X2 is negligible as compared to R2.
∴
T ∝ s/R2
∝s
... as R2 is constant
Hence torque slip curve is a straight line from zero slip to a slip that corresponds to full-load. (iii) As slip increases beyond full-load slip, the torque increases and becomes maximum at s = R2/X2. This maximum torque in an induction motor is called pull-out torque or break-down torque. Its value is at least twice the full-load value when the motor is operated at rated voltage and frequency. (iv) to maximum torque, the term 2 2 2 s X 2 increases very rapidly so that R 2 may be neglected as compared to s 2 X 22 .
∴
T ∝ s / s 2 X 22 ∝ 1/ s
... as X2 is constant
Thus the torque is now inversely proportional to slip. Hence torque-slip curve is a rectangular hyperbola. (v) The maximum torque remains the same and is independent of the value of rotor resistance. Therefore, the addition of resistance to the rotor circuit does not change the value of maximum torque but it only changes the value of slip at which maximum torque occurs.
8.19 Full-Load, Starting and Maximum Torques 200
Tf ∝ Ts ∝ Tm ∝
s R2
R 22
+ (s X 2 )2
R2 R 22 + X 22 1 2 X2
Note that s corresponds to full-load slip. (i)
∴
Tm R 22 + (s X 2 )2 = Tf 2s R 2 X 2
Dividing the numerator and denominator on R.H.S. by X 22 , we get,
Tm (R 2 / X 2 )2 + s 2 a 2 + s 2 = = Tf 2s(R 2 / X 2 ) 2a s R2 Rotor resistance/phase = X 2 Standstill rotor reactance/phase
where
a=
(ii)
Tm R 22 + X 22 = Ts 2 R 2 X 2
Dividing the numerator and denominator on R.H.S. by X 22 , we get,
Tm (R 2 / X 2 )2 + 1 a 2 + 1 = = Tf 2(R 2 / X 2 ) 2a where
a=
R2 Rotor resistance/phase = X 2 Standstill rotor reactance/phase
8.20 Induction Motor and Transformer Compared An induction motor may be considered to be a transformer with a rotating shortcircuited secondary. The stator winding corresponds to transformer primary and rotor winding to transformer secondary. However, the following differences between the two are worth noting: (i) Unlike a transformer, the magnetic circuit of a 3-phase induction motor has an air gap. Therefore, the magnetizing current in a 3-phase induction motor is much larger than that of the transformer. For example, in an induction motor, it may be as high as 30-50 % of rated current whereas it is only 15% of rated current in a transformer. (ii) In an induction motor, there is an air gap and the stator and rotor windings are distributed along the periphery of the air gap rather than concentrated 201
on a core as in a transformer. Therefore, the leakage reactances of stator and rotor windings are quite large compared to that of a transformer. (iii) In an induction motor, the inputs to the stator and rotor are electrical but the output from the rotor is mechanical. However, in a transformer, input as well as output is electrical. (iv) The main difference between the induction motor and transformer lies in the fact that the rotor voltage and its frequency are both proportional to slip s. If f is the stator frequency, E2 is the per phase rotor e.m.f. at standstill and X2 is the standstill rotor reactance/phase, then at any slip s, these values are: Rotor e.m.f./phase, E'2 = s E2 Rotor reactance/phase, X'2 = sX2 Rotor frequency, f' = sf
8.21 Speed Regulation of Induction Motors Like any other electrical motor, the speed regulation of an induction motor is given by: % age speed regulation = where
N 0 − N F.L. × 100 N F.L.
N0 = no-load speed of the motor NF.L. = full-load speed of the motor
If the no-load speed of the motor is 800 r.p.m. and its fall-load speed in 780 r.p.m., then change in speed is 800 − 780 = 20 r.p.m. and percentage speed regulation = 20 × 100/780 = 2.56%. At no load, only a small torque is required to overcome the small mechanical losses and hence motor slip is small i.e., about 1%. When the motor is fully loaded, the slip increases slightly i.e., motor speed decreases slightly. It is because rotor impedance is low and a small decrease in speed produces a large rotor current. The increased rotor current produces a high torque to meet the full load on the motor. For this reason, the change in speed of the motor from noload to full-load is small i.e., the speed regulation of an induction motor is low. The speed regulation of an induction motor is 3% to 5%. Although the motor speed does decrease slightly with increased load, the speed regulation is low enough that the induction motor is classed as a constant-speed motor.
8.22 Speed Control of 3-Phase Induction Motors
202
N = (1 − s) N s = (1 − s)
120 f P
(i)
An inspection of eq. (i) reveals that the speed N of an induction motor can be varied by changing (i) supply frequency f (ii) number of poles P on the stator and (iii) slip s. The change of frequency is generally not possible because the commercial supplies have constant frequency. Therefore, the practical methods of speed control are either to change the number of stator poles or the motor slip.
1.
Squirrel cage motors The speed of a squirrel cage motor is changed by changing the number of stator poles. Only two or four speeds are possible by this method. Two-speed motor has one stator winding that may be switched through suitable control equipment to provide two speeds, one of which is half of the other. For instance, the winding may be connected for either 4 or 8 poles, giving synchronous speeds of 1500 and 750 r.p.m. Four-speed motors are equipped with two separate stator windings each of which provides two speeds. The disadvantages of this method are: (i) It is not possible to obtain gradual continuous speed control. (ii) Because of the complications in the design and switching of the interconnections of the stator winding, this method can provide a maximum of four different synchronous speeds for any one motor.
2.
Wound rotor motors The speed of wound rotor motors is changed by changing the motor slip. This can be achieved by; (i) varying the stator line voltage (ii) varying the resistance of the rotor circuit (iii) inserting and varying a foreign voltage in the rotor circuit
8.23 Power Factor of Induction Motor Like any other a.c. machine, the power factor of an induction motor is given by; Power factor, cos φ =
Active component of current (I cos φ) Total current (I)
The presence of air-gap between the stator and rotor of an induction motor greatly increases the reluctance of the magnetic circuit. Consequently, an induction motor draws a large magnetizing current (Im) to produce the required flux in the air-gap. (i) At no load, an induction motor draws a large magnetizing current and a small active component to meet the no-load losses. Therefore, the induction motor takes a high no-load current lagging the applied voltage 203
by a large angle. Hence the power factor of an induction motor on no load is low i.e., about 0.1 lagging. (ii) When an induction motor is loaded, the active component of current increases while the magnetizing component remains about the same. Consequently, the power factor of the motor is increased. However, because of the large value of magnetizing current, which is present regardless of load, the power factor of an induction motor even at fullload seldom exceeds 0.9 lagging.
8.24 Power Stages in an Induction Motor The input electric power fed to the stator of the motor is converted into mechanical power at the shaft of the motor. The various losses during the energy conversion are:
1.
Fixed losses (i) Stator iron loss (ii) Friction and windage loss The rotor iron loss is negligible because the frequency of rotor currents under normal running condition is small.
2.
Variable losses (i) Stator copper loss (ii) Rotor copper loss Fig. (8.20) shows how electric power fed to the stator of an induction motor suffers losses and finally converted into mechanical power. The following points may be noted from the above diagram: (i) Stator input, Pi = Stator output + Stator losses = Stator output + Stator Iron loss + Stator Cu loss (ii) Rotor input, Pr = Stator output It is because stator output is entirely transferred to the rotor through airgap by electromagnetic induction. (iii) Mechanical power available, Pm = Pr − Rotor Cu loss This mechanical power available is the gross rotor output and will produce a gross torque Tg. (iv) Mechanical power at shaft, Pout = Pm − Friction and windage loss Mechanical power available at the shaft produces a shaft torque Tsh. Clearly, Pm − Pout = Friction and windage loss
204
Fig.(8.20)
8.25 Induction Motor Torque The mechanical power P available from any electric motor can be expressed as:
P= where
2π N T 60
watts
N = speed of the motor in r.p.m. T = torque developed in N-m
∴
T=
60 P P = 9.55 N-m 2π N N
If the gross output of the rotor of an induction motor is Pm and its speed is N r.p.m., then gross torque T developed is given by:
Tg = 9.55 Similarly,
Pm N-m N
Tsh = 9.55
Pout N-m N
Note. Since windage and friction loss is small, Tg = Tsh,. This assumption hardly leads to any significant error.
8.26 Rotor Output If Tg newton-metre is the gross torque developed and N r.p.m. is the speed of the rotor, then, Gross rotor output =
2π N Tg 60
watts
If there were no copper losses in the rotor, the output would equal rotor input and the rotor would run at synchronous speed Ns.
205
2π N s Tg 60
∴
Rotor input =
∴
Rotor Cu loss = Rotor input − Rotor output
= (i) ∴
watts
2π Tg ( N s − N) 60
Rotor Cu loss N s − N =s = Rotor input Ns Rotor Cu loss = s × Rotor input
(ii) Gross rotor output, Pm = Rotor input − Rotor Cu loss = Rotor input − s × Rotor input ∴ Pm = Rotor input (1 − s)
Gross rotor output N =1 − s = Rotor input Ns Rotor Cu loss s = Gross rotor output 1 − s
(iii) (iv)
It is clear that if the input power to rotor is Pr then s Pr is lost as rotor Cu loss and the remaining (1 − s)Pr is converted into mechanical power. Consequently, induction motor operating at high slip has poor efficiency. Note.
Gross rotor output =1 − s Rotor input If the stator losses as well as friction and windage losses arc neglected, then, Gross rotor output = Useful output Rotor input = Stator input
∴
Useful output = 1 − s = Efficiency Stator output
Hence the approximate efficiency of an induction motor is 1 − s. Thus if the slip of an induction motor is 0.125, then its approximate efficiency is = 1 − 0.125 = 0.875 or 87.5%.
8.27 Induction Motor Torque Equation The gross torque Tg developed by an induction motor is given by;
206
Tg = =
Rotor input 2π N s
... N s is r.p.s.
60 × Rotor input 2π N s
(See Sec. 8.26)
... N s is r.p.s.
2 Rotor Cu loss 3(I'2 ) R 2 = Rotor input = s s
Now
(i)
As shown in Sec. 8.16, under running conditions,
I'2 = where
s E2 R 22 + (s X 2 ) 2
=
K = Transformation ratio = ∴
Rotor input = 3 ×
s K E1 R 22 + (s X 2 ) 2
Rotor turn s/phase Stator tur ns/phase
s 2 E 22 R 2
3 s E 22 R 2 1 × = R 22 + (s X 2 ) 2 s R 22 + (s X 2 ) 2 (Putting me value of I'2 in eq.(i))
s 2 K 2 E12 R 2
2 2 1 3 s K E1 R 2 Rotor input = 3 × 2 × = R 2 + (s X 2 ) 2 s R 22 + (s X 2 ) 2
Also
(Putting me value of I'2 in eq.(i))
∴
s E 22 R 2 Rotor input 3 Tg = = × 2π N s 2π N s R 22 + (s X 2 ) 2 s K 2 E12 R 2 3 = × 2π N s R 22 + (s X 2 ) 2
...in terms of E2
...in terms of E1
Note that in the above expressions of Tg, the values E1, E2, R2 and X2 represent the phase values.
8.28 Performance Curves of Squirrel-Cage Motor The performance curves of a 3-phase induction motor indicate the variations of speed, power factor, efficiency, stator current and torque for different values of load. However, before giving the performance curves in one graph, it is desirable to discuss the variation of torque, and stator current with slip.
(i) Variation of torque and stator current with slip Fig. (8.21) shows the variation of torque and stator current with slip for a standard squirrel-cage motor. Generally, the rotor resistance is low so that full207
load current occurs at low slip. Then even at full-load f' (= sf) and. therefore, X'2 (= 2π f' L2) are low. Between zero and full-load, rotor power factor (= cos φ'2) and rotor impedance (= Z'2) remain practically constant. Therefore, rotor current I'2(E'2/Z'2) and, therefore, torque (Tr) increase directly with the slip. Now stator current I1 increases in proportion to I'2. This is shown in Fig. (8.21) where Tr and I1 are indicated as straight lines from no-load to full-load. As load and slip are increased beyond full-load, the increase in rotor reactance becomes appreciable. The increasing value of rotor impedance not only decreases the rotor power factor cos φ'2 (= R2/Z'2) but also lowers the rate of increase of rotor current. As a result, the torque Tr and stator current I1 do not increase directly with slip as indicated in Fig. (8.21). With the decreasing power factor and the lowered rate of increase in rotor current, the stator current I1 and torque Tr increase at a lower rate. Finally, torque Tr reaches the maximum value at about 25% slip in the standard squirrel cage motor. This maximum value of torque is called the pullout torque or breakdown torque. If the load is increased beyond the breakdown point, the decrease in rotor power factor is greater than the increase in rotor current, resulting in a decreasing torque. The result is that motor slows down quickly and comes to a stop.
Fig.(8.21) In Fig. (8.21), the value of torque at starting (i.e., s = 100%) is 1.5 times the fullload torque. The starting current is about five times the full-load current. The motor is essentially a constant-speed machine having speed characteristics about the same as a d.c. shunt motor.
(ii) Performance curves Fig. (8.22) shows the performance curves of 3-phase squirrel cage induction motor.
208
Fig.(8.22) The following points may be noted: (a) At no-load, the rotor lags behind the stator flux by only a small amount, since the only torque required is that needed to overcome the no-load losses. As mechanical load is added, the rotor speed decreases. A decrease in rotor speed allows the constant-speed rotating field to sweep across the rotor conductors at a faster rate, thereby inducing large rotor currents. This results in a larger torque output at a slightly reduced speed. This explains for speed-load curve in Fig. (8.22). (b) At no-load, the current drawn by an induction motor is largely a magnetizing current; the no-load current lagging the applied voltage by a large angle. Thus the power factor of a lightly loaded induction motor is very low. Because of the air gap, the reluctance of the magnetic circuit is high, resulting in a large value of no-load current as compared with a transformer. As load is added, the active or power component of current increases, resulting in a higher power factor. However, because of the large value of magnetizing current, which is present regardless of load, the power factor of an induction motor even at full-load seldom exceeds 90%. Fig. (8.22) shows the variation of power factor with load of a typical squirrelcage induction motor. Output (c) Efficiency = Output + Losses The losses occurring in a 3-phase induction motor are Cu losses in stator and rotor windings, iron losses in stator and rotor core and friction and windage losses. The iron losses and friction and windage losses are almost independent of load. Had I2R been constant, the efficiency of the motor would have increased with load. But I2R loss depends upon load. 209
Therefore, the efficiency of the motor increases with load but the curve is dropping at high loads. (d) At no-load, the only torque required is that needed to overcome no-load losses. Therefore, stator draws a small current from the supply. As mechanical load is added, the rotor speed decreases. A decrease in rotor speed allows the constant-speed rotating field to sweep across the rotor conductors at a faster rate, thereby inducing larger rotor currents. With increasing loads, the increased rotor currents are in such a direction so as to decrease the staler flux, thereby temporarily decreasing the counter e.m.f. in the stator winding. The decreased counter e.m.f. allows more stator current to flow. (e) Output = Torque × Speed Since the speed of the motor does not change appreciably with load, the torque increases with increase in load.
8.29 Equivalent Circuit of 3-Phase Induction Motor at Any Slip In a 3-phase induction motor, the stator winding is connected to 3-phase supply and the rotor winding is short-circuited. The energy is transferred magnetically from the stator winding to the short-circuited, rotor winding. Therefore, an induction motor may be considered to be a transformer with a rotating secondary (short-circuited). The stator winding corresponds to transformer primary and the rotor finding corresponds to transformer secondary. In view of the similarity of the flux and voltage conditions to those in a transformer, one can expect that the equivalent circuit of an induction motor will be similar to that of a transformer. Fig. (8.23) shows the equivalent circuit (though not the only one) per phase for an induction motor. Let us discuss the stator and rotor circuits separately.
Fig.(8.23) Stator circuit. In the stator, the events are very similar to those in the transformer primal y. The applied voltage per phase to the stator is V1 and R1 and X1 are the stator resistance and leakage reactance per phase respectively. The applied voltage V1 produces a magnetic flux which links the stator winding (i.e., primary) as well as the rotor winding (i.e., secondary). As a result, self-
210
induced e.m.f. E1 is induced in the stator winding and mutually induced e.m.f. E'2(= s E2 = s K E1 where K is transformation ratio) is induced in the rotor winding. The flow of stator current I1 causes voltage drops in R1 and X1.
∴
V1 = − E1 + I1 (R 1 + j X1 )
...phasor sum
When the motor is at no-load, the stator winding draws a current I0. It has two components viz., (i) which supplies the no-load motor losses and (ii) magnetizing component Im which sets up magnetic flux in the core and the airgap. The parallel combination of Rc and Xm, therefore, represents the no-load motor losses and the production of magnetic flux respectively.
I0 = Iw + Im Rotor circuit. Here R2 and X2 represent the rotor resistance and standstill rotor reactance per phase respectively. At any slip s, the rotor reactance will be s X2 The induced voltage/phase in the rotor is E'2 = s E2 = s K E1. Since the rotor winding is short-circuited, the whole of e.m.f. E'2 is used up in circulating the rotor current I'2.
∴
E ' 2 = I' 2 ( R 2 + j s X 2 )
The rotor current I'2 is reflected as I"2 (= K I'2) in the stator. The phasor sum of I"2 and I0 gives the stator current I1. It is important to note that input to the primary and output from the secondary of a transformer are electrical. However, in an induction motor, the inputs to the stator and rotor are electrical but the output from the rotor is mechanical. To facilitate calculations, it is desirable and necessary to replace the mechanical load by an equivalent electrical load. We then have the transformer equivalent circuit of the induction motor. It may be noted that even though the frequencies of stator and rotor currents are different, yet the magnetic fields due to them rotate at synchronous speed Ns. The stator currents produce a magnetic flux which rotates at a speed Ns. At slip s, the speed of rotation of the rotor field relative to the rotor surface in the direction of rotation of the rotor is
=
120 f ' 120 s f = = s Ns P P
But the rotor is revolving at a speed of N relative to the stator core. Therefore, the speed of rotor 211
Fig.(8.24)
field relative to stator core
= sN s + N = ( N s − N ) + N = N s Thus no matter what the value of slip s, the stator and rotor magnetic fields are synchronous with each other when seen by an observer stationed in space. Consequently, the 3-phase induction motor can be regarded as being equivalent to a transformer having an air-gap separating the iron portions of the magnetic circuit carrying the primary and secondary windings. Fig. (8.24) shows the phasor diagram of induction motor.
8.30 Equivalent Circuit of the Rotor We shall now see how mechanical load of the motor is replaced by the equivalent electrical load. Fig. (8.25 (i)) shows the equivalent circuit per phase of the rotor at slip s. The rotor phase current is given by;
I'2 =
s E2 R 22 + (s X 2 ) 2
Mathematically, this value is unaltered by writing it as:
I' 2 =
E2 ( R 2 / s) 2 + ( X 2 ) 2
As shown in Fig. (8.25 (ii)), we now have a rotor circuit that has a fixed reactance X2 connected in series with a variable resistance R2/s and supplied with constant voltage E2. Note that Fig. (8.25 (ii)) transfers the variable to the resistance without altering power or power factor conditions.
Fig.(8.25) The quantity R2/s is greater than R2 since s is a fraction. Therefore, R2/s can be divided into a fixed part R2 and a variable part (R2/s − R2) i.e.,
R2 1 = R 2 + R 2 − 1 s s
212
(i)
The first part R2 is the rotor resistance/phase, and represents the rotor Cu loss. 1 (ii) The second part R 2 −1 is a variable-resistance load. The power s delivered to this load represents the total mechanical power developed in the rotor. Thus mechanical load on the induction motor can be replaced by 1 a variable-resistance load of value R 2 −1 . This is s
∴
1 R L = R 2 − 1 s
Fig. (8.25 (iii)) shows the equivalent rotor circuit along with load resistance RL.
8.31 Transformer Equivalent Circuit of Induction Motor Fig. (8.26) shows the equivalent circuit per phase of a 3-phase induction motor. Note that mechanical load on the motor has been replaced by an equivalent electrical resistance RL given by;
1 R L = R 2 − 1 s
(i)
Fig.(8.26) Note that circuit shown in Fig. (8.26) is similar to the equivalent circuit of a transformer with secondary load equal to R2 given by eq. (i). The rotor e.m.f. in the equivalent circuit now depends only on the transformation ratio K (= E2/E1). Therefore; induction motor can be represented as an equivalent transformer connected to a variable-resistance load RL given by eq. (i). The power delivered to RL represents the total mechanical power developed in the rotor. Since the equivalent circuit of Fig. (8.26) is that of a transformer, the secondary (i.e., rotor) values can be transferred to primary (i.e., stator) through the appropriate use of transformation ratio K. Recall that when shifting resistance/reactance from secondary to primary, it should be divided by K2 whereas current should be multiplied by K. The equivalent circuit of an induction motor referred to primary is shown in Fig. (8.27). 213
Fig.(8.27) Note that the element (i.e., R'L) enclosed in the dotted box is the equivalent electrical resistance related to the mechanical load on the motor. The following points may be noted from the equivalent circuit of the induction motor: (i) At no-load, the slip is practically zero and the load R'L is infinite. This condition resembles that in a transformer whose secondary winding is open-circuited. (ii) At standstill, the slip is unity and the load R'L is zero. This condition resembles that in a transformer whose secondary winding is short-circuited. (iii) When the motor is running under load, the value of R'L will depend upon the value of the slip s. This condition resembles that in a transformer whose secondary is supplying variable and purely resistive load. (iv) The equivalent electrical resistance R'L related to mechanical load is slip or speed dependent. If the slip s increases, the load R'L decreases and the rotor current increases and motor will develop more mechanical power. This is expected because the slip of the motor increases with the increase of load on the motor shaft.
8.32 Power Relations The transformer equivalent circuit of an induction motor is quite helpful in analyzing the various power relations in the motor. Fig. (8.28) shows the equivalent circuit per phase of an induction motor where all values have been referred to primary (i.e., stator).
Fig.(8.28) 214
(i)
R' 1 Total electrical load = R '2 − 1 + R '2 = 2 s s Power input to stator = 3V1 I1 cos φ1
There will be stator core loss and stator Cu loss. The remaining power will be the power transferred across the air-gap i.e., input to the rotor. 3(I"2 )2 R '2 (ii) Rotor input = s 2 Rotor Cu loss = 3(I"2 ) R '2 Total mechanical power developed by the rotor is Pm = Rotor input − Rotor Cu loss
3(I"2 )2 R '2 1 = 3(I"2 )2 R '2 = 3(I"2 )2 R '2 − 1 s s This is quite apparent from the equivalent circuit shown in Fig. (8.28). (iii) If Tg is the gross torque developed by the rotor, then,
Pm =
2π N Tg 60
or
2π N Tg 1 3(I"2 )2 R '2 − 1 = 60 s
or
1 − s 2π N Tg 3(I"2 )2 R '2 = 60 s
or
1 − s 2π N s (1 − s) Tg 3(I"2 )2 R '2 = 60 s
∴ or
3(I"2 )2 R '2 s Tg = 2π N s 60
[Q N = N s (1 − s)]
N-m
3(I"2 )2 R '2 s Tg = 9.55 Ns
N-m
Note that shaft torque Tsh will be less than Tg by the torque required to meet windage and frictional losses.
215
8.33 Approximate Equivalent Circuit of Induction Motor As in case of a transformer, the approximate equivalent circuit of an induction motor is obtained by shifting the shunt branch (Rc − Xm) to the input terminals as shown in Fig. (8.29). This step has been taken on the assumption that voltage drop in R1 and X1 is small and the terminal voltage V1 does not appreciably differ from the induced voltage E1. Fig. (8.29) shows the approximate equivalent circuit per phase of an induction motor where all values have been referred to primary (i.e., stator).
Fig.(8.29) The above approximate circuit of induction motor is not so readily justified as with the transformer. This is due to the following reasons: (i) Unlike that of a power transformer, the magnetic circuit of the induction motor has an air-gap. Therefore, the exciting current of induction motor (30 to 40% of full-load current) is much higher than that of the power transformer. Consequently, the exact equivalent circuit must be used for accurate results. (ii) The relative values of X1 and X2 in an induction motor are larger than the corresponding ones to be found in the transformer. This fact does not justify the use of approximate equivalent circuit (iii) In a transformer, the windings are concentrated whereas in an induction motor, the windings are distributed. This affects the transformation ratio. In spite of the above drawbacks of approximate equivalent circuit, it yields results that are satisfactory for large motors. However, approximate equivalent circuit is not justified for small motors.
8.34 Starting of 3-Phase Induction Motors The induction motor is fundamentally a transformer in which the stator is the primary and the rotor is short-circuited secondary. At starting, the voltage induced in the induction motor rotor is maximum (Q s = 1). Since the rotor impedance is low, the rotor current is excessively large. This large rotor current is reflected in the stator because of transformer action. This results in high starting current (4 to 10 times the full-load current) in the stator at low power 216
factor and consequently the value of starting torque is low. Because of the short duration, this value of large current does not harm the motor if the motor accelerates normally. However, this large starting current will produce large line-voltage drop. This will adversely affect the operation of other electrical equipment connected to the same lines. Therefore, it is desirable and necessary to reduce the magnitude of stator current at starting and several methods are available for this purpose.
8.35 Methods of Starting 3-Phase Induction Motors The method to be employed in starting a given induction motor depends upon the size of the motor and the type of the motor. The common methods used to start induction motors are: (i) Direct-on-line starting (ii) Stator resistance starting (iii) Autotransformer starting (iv) Star-delta starting (v) Rotor resistance starting Methods (i) to (iv) are applicable to both squirrel-cage and slip ring motors. However, method (v) is applicable only to slip ring motors. In practice, any one of the first four methods is used for starting squirrel cage motors, depending upon ,the size of the motor. But slip ring motors are invariably started by rotor resistance starting.
8.36 Methods of Starting Squirrel-Cage Motors Except direct-on-line starting, all other methods of starting squirrel-cage motors employ reduced voltage across motor terminals at starting.
(i) Direct-on-line starting This method of starting in just what the name implies—the motor is started by connecting it directly to 3-phase supply. The impedance of the motor at standstill is relatively low and when it is directly connected to the supply system, the starting current will be high (4 to 10 times the full-load current) and at a low power factor. Consequently, this method of starting is suitable for relatively small (up to 7.5 kW) machines. Relation between starling and F.L. torques. We know that: Rotor input = 2π Ns T = kT Rotor Cu loss = s × Rotor input
But
∴ or
3(I'2 )2 R 2 = s × kT T ∝ (I'2 )2 s 217
(Q I'2 ∝ I1 )
T ∝ I12 s
or
If Ist is the starting current, then starting torque (Tst) is T ∝ I st2
(Q at starting s = 1)
If If is the full-load current and sf is the full-load slip, then, Tf ∝ I f2 s f 2
∴
Tst I st = × sf Tf I f
When the motor is started direct-on-line, the starting current is the short-circuit (blocked-rotor) current Isc. 2
∴
Tst I sc = × sf Tf I f
Let us illustrate the above relation with a numerical example. Suppose Isc = 5 If and full-load slip sf =0.04. Then, 2
∴
2
Tst I sc 5 I = × s f = f × 0.04 = (5) 2 × 0.04 = 1 Tf I f If Tst = Tf
Note that starting current is as large as five times the full-load current but starting torque is just equal to the full-load torque. Therefore, starting current is very high and the starting torque is comparatively low. If this large starting current flows for a long time, it may overheat the motor and damage the insulation.
(ii) Stator resistance starting In this method, external resistances are connected in series with each phase of stator winding during starting. This causes voltage drop across the resistances so that voltage available across motor terminals is reduced and hence the starting current. The starting resistances are gradually cut out in steps (two or more steps) from the stator circuit as the motor picks up speed. When the motor attains rated speed, the resistances are completely cut out and full line voltage is applied to the rotor. This method suffers from two drawbacks. First, the reduced voltage applied to the motor during the starting period lowers the starting torque and hence increases the accelerating time. Secondly, a lot of power is wasted in the starting resistances.
218
Fig.(8.30) Relation between starting and F.L. torques. Let V be the rated voltage/phase. If the voltage is reduced by a fraction x by the insertion of resistors in the line, then voltage applied to the motor per phase will be xV.
I st = x I sc 2
Now
Tst Ist = × sf Tf I f
or
Tst I = x 2 sc × s f Tf If
2
Thus while the starting current reduces by a fraction x of the rated-voltage starting current (Isc), the starting torque is reduced by a fraction x2 of that obtained by direct switching. The reduced voltage applied to the motor during the starting period lowers the starting current but at the same time increases the accelerating time because of the reduced value of the starting torque. Therefore, this method is used for starting small motors only.
(iii) Autotransformer starting This method also aims at connecting the induction motor to a reduced supply at starting and then connecting it to the full voltage as the motor picks up sufficient speed. Fig. (8.31) shows the circuit arrangement for autotransformer starting. The tapping on the autotransformer is so set that when it is in the circuit, 65% to 80% of line voltage is applied to the motor. At the instant of starting, the change-over switch is thrown to “start” position. This puts the autotransformer in the circuit and thus reduced voltage is applied to the circuit. Consequently, starting current is limited to safe value. When the motor attains about 80% of normal speed, the changeover switch is thrown to
219
“run” position. This takes out the autotransformer from the circuit and puts the motor to full line voltage. Autotransformer starting has several advantages viz low power loss, low starting current and less radiated heat. For large machines (over 25 H.P.), this method of starting is often used. This method can be used for both star and delta connected motors.
Fig.(8.31) Relation between starting And F.L. torques. Consider a star-connected squirrel-cage induction motor. If V is the line voltage, then voltage across motor phase on direct switching is V 3 and starting current is Ist = Isc. In case of autotransformer, if a tapping of transformation ratio K (a fraction) is used, then phase voltage across motor is KV 3 and Ist = K Isc, 2
2
2
Tst Ist KI I = × s f = sc × s f = K 2 sc × s f Tf I f If If
Now
2
∴
Tst I = K 2 sc × s f Tf If
Fig.(8.32)
220
The current taken from the supply or by autotransformer is I1 = KI2 = K2Isc. Note that motor current is K times, the supply line current is K2 times and the starting torque is K2 times the value it would have been on direct-on-line starting.
(iv) Star-delta starting The stator winding of the motor is designed for delta operation and is connected in star during the starting period. When the machine is up to speed, the connections are changed to delta. The circuit arrangement for star-delta starting is shown in Fig. (8.33). The six leads of the stator windings are connected to the changeover switch as shown. At the instant of starting, the changeover switch is thrown to “Start” position which connects the stator windings in star. Therefore, each stator phase gets V 3 volts where V is the line voltage. This reduces the starting current. When the motor picks up speed, the changeover switch is thrown to “Run” position which connects the stator windings in delta. Now each stator phase gets full line voltage V. The disadvantages of this method are: (a) With star-connection during starting, stator phase voltage is 1 3 times the
(
)2
line voltage. Consequently, starting torque is 1 3 or 1/3 times the value it would have with ∆-connection. This is rather a large reduction in starting torque. (b) The reduction in voltage is fixed. This method of starting is used for medium-size machines (upto about 25 H.P.). Relation between starting and F.L. torques. In direct delta starting, Starting current/phase, Isc = V/Zsc where V = line voltage Starting line current = In star starting, we have,
3 Isc
Starting current/phase, I st =
V 3 1 = I sc Z sc 3
Now
2 I sc Tst Ist = × s f = Tf I f 3 × If
or
Tst 1 Isc = × sf Tf 3 I f
2
× sf
2
where
Isc = starting phase current (delta) If = F.L. phase current (delta)
221
Fig.(8.33) Note that in star-delta starting, the starting line current is reduced to one-third as compared to starting with the winding delta connected. Further, starting torque is reduced to one-third of that obtainable by direct delta starting. This method is cheap but limited to applications where high starting torque is not necessary e.g., machine tools, pumps etc.
8.37 Starting of Slip-Ring Motors Slip-ring motors are invariably started by rotor resistance starting. In this method, a variable star-connected rheostat is connected in the rotor circuit through slip rings and full voltage is applied to the stator winding as shown in Fig. (8.34).
Fig.(8.34) (i)
At starting, the handle of rheostat is set in the OFF position so that maximum resistance is placed in each phase of the rotor circuit. This reduces the starting current and at the same time starting torque is increased. (ii) As the motor picks up speed, the handle of rheostat is gradually moved in clockwise direction and cuts out the external resistance in each phase of the rotor circuit. When the motor attains normal speed, the change-over switch is in the ON position and the whole external resistance is cut out from the rotor circuit.
222
8.38 Slip-Ring Motors Versus Squirrel Cage Motors The slip-ring induction motors have the following advantages over the squirrel cage motors: (i) High starting torque with low starting current. (ii) Smooth acceleration under heavy loads. (iii) No abnormal heating during starting. (iv) Good running characteristics after external rotor resistances are cut out. (v) Adjustable speed. The disadvantages of slip-ring motors are: (i) The initial and maintenance costs are greater than those of squirrel cage motors. (ii) The speed regulation is poor when run with resistance in the rotor circuit
8.39 Induction Motor Rating The nameplate of a 3-phase induction motor provides the following information: (i) Horsepower (ii) Line voltage (iii) Line current (iv) Speed (v) Frequency (vi) Temperature rise The horsepower rating is the mechanical output of the motor when it is operated at rated line voltage, rated frequency and rated speed. Under these conditions, the line current is that specified on the nameplate and the temperature rise does not exceed that specified. The speed given on the nameplate is the actual speed of the motor at rated fullload; it is not the synchronous speed. Thus, the nameplate speed of the induction motor might be 1710 r.p.m. It is the rated full-load speed.
8.40 Double Squirrel-Cage Motors One of the advantages of the slip-ring motor is that resistance may be inserted in the rotor circuit to obtain high starting torque (at low starting current) and then cut out to obtain optimum running conditions. However, such a procedure cannot be adopted for a squirrel cage motor because its cage is permanently short-circuited. In order to provide high starting torque at low starting current, double-cage construction is used.
Construction As the name suggests, the rotor of this motor has two squirrel-cage windings located one above the other as shown in Fig. (8.35 (i)). (i) The outer winding consists of bars of smaller cross-section short-circuited by end rings. Therefore, the resistance of this winding is high. Since the 223
outer winding has relatively open slots and a poorer flux path around its bars [See Fig. (8.35 (ii))], it has a low inductance. Thus the resistance of the outer squirrel-cage winding is high and its inductance is low. (ii) The inner winding consists of bars of greater cross-section short-circuited by end rings. Therefore, the resistance of this winding is low. Since the bars of the inner winding are thoroughly buried in iron, it has a high inductance [See Fig. (8.35 (ii))]. Thus the resistance of the inner squirrelcage winding is low and its inductance is high.
Fig.(8.35)
Working When a rotating magnetic field sweeps across the two windings, equal e.m.f.s are induced in each. (i) At starting, the rotor frequency is the same as that of the line (i.e., 50 Hz), making the reactance of the lower winding much higher than that of the upper winding. Because of the high reactance of the lower winding, nearly all the rotor current flows in the high-resistance outer cage winding. This provides the good starting characteristics of a high-resistance cage winding. Thus the outer winding gives high starting torque at low starting current. (ii) As the motor accelerates, the rotor frequency decreases, thereby lowering the reactance of the inner winding, allowing it to carry a larger proportion of the total rotor current At the normal operating speed of the motor, the rotor frequency is so low (2 to 3 Hz) that nearly all the rotor current flows in the low-resistance inner cage winding. This results in good operating efficiency and speed regulation. Fig. (8.36) shows the operating characteristics of double squirrel-cage motor. The starting torque of this motor ranges from 200 to 250 percent of full-load torque with a starting current of 4 to 6 times the full-load value. It is classed as a high-torque, low starting current motor.
224
Fig.(8.36)
8.41 Equivalent Circuit of Double Squirrel-Cage Motor Fig. (8.37) shows a section of the double squirrel cage motor. Here Ro and Ri are the per phase resistances of the outer cage winding and inner cage winding whereas Xo and Xi are the corresponding per phase standstill reactances. For the outer cage, the resistance is made intentionally high, giving a high Fig.(8.37) starting torque. For the inner cage winding, the resistance is low and the leakage reactance is high, giving a low starting torque but high efficiency on load. Note that in a double squirrel cage motor, the outer winding produces the high starting and accelerating torque while the inner winding provides the running torque at good efficiency. Fig. (8.38 (i)) shows the equivalent circuit for one phase of double cage motor referred to stator. The two cage impedances are effectively in parallel. The resistances and reactances of the outer and inner rotors are referred to the stator. The exciting circuit is accounted for as in a single cage motor. If the magnetizing current (I0) is neglected, then the circuit is simplified to that shown in Fig. (8.38 (ii)).
Fig.(8.38) 225
From the equivalent circuit, the performance of the motor can be predicted. Total impedance as referred to stator is
Z o1 = R 1 + j X1 +
Z' Z' 1 = R 1 + j X1 + i o 1 Z'i + 1 Z'o Z'i + Z'o
226
Tests for finding parameters of 3Φ induction motor To accurately model the three phase induction motor , the equivalent circuit parameters (R1 , X1 , R2 , X2 , Rc , Xm) should be known. A single-phase equivalent circuit with lumped parameters is the most traditional model of an AC induction motor .The steady-state operating characteristics of a three-phase induction motor are often investigated using a per-phase equivalent circuit . In this circuit R1, and X1 represent stator resistance and leakage reactance, respectively; R2 and X2 denote the rotor resistance and leakage reactance referred to the stator, respectively; Rc resistance stands for core losses; Xm represents magnetizing reactance. The equivalent circuit is used to facilitate the computation of various operating quantities, such as stator current, input power, losses, induced torque, and efficiency. When power aspects of the operation need to be emphasized, the shunt resistance is usually neglected; the core losses can be included in efficiency calculations along with the friction, windage, and stray losses. The parameters of the equivalent circuit can be obtained from the dc, no-load, and blocked-rotor tests .The DC Test is performed to compute the stator winding resistance . A dc voltage is applied to the stator windings of an induction motor. The resulting current flowing through the stator windings is a dc current; thus, no voltage is induced in the rotor circuit, and the motor reactance is zero. The stator resistance is the only circuit parameter limiting current flow. In No-Load Test , a rated, balanced ac voltage at a rated frequency is applied to the stator while it is running at no load, and input power, voltage, and phase currents are measured at the no-load condition. In Blocked-Rotor Test, the rotor of the induction motor is blocked, and a reduced voltage is applied to the stator terminals so that the rated current flows through the stator windings. The input power, voltage, and current are measured.
1
Notes 1– Experience has shown certain Design Types have certain values of X1 and X2 as function of Xbr as follows: • • • • •
Wound rotor : Design A : Design B : Design C : Design D :
X1 = X2 = 0.5Xbr X1 = X2 = 0.5Xbr X1 = 0.4Xbr and X2 = 0.6Xbr X1 = 0.3Xbr and X2 = 0.7Xbr X1 = X2 = 0.5Xbr
2- For some design-class induction motors, a blocked-rotor test is conducted under a test frequency, usually less than the normal operating frequency so as to evaluate the rotor resistance appropriately. If the test frequency is different from the rated frequency, one can compute the total equivalent reactance at the normal operating frequency (Xbr) as follows since the reactance is directly proportional to the frequency:
X br
Where
f rated ' X br f test X´br is blocked-rotor reactance at the test frequency
6
Example : 208 V, 60 Hz, six-pole Y-connected 25-hp design class B induction motor is tested in the laboratory, with the following results: DC test : 13.5 V , 64 A No load test : 208 V , 22.0 A , 1200 W , 60 Hz Blocked-rotor test : 24.6 V , 64.5 A , 2200 W , 15 Hz Find R1 , R2 , X1 , X2 , and Xm of this motor ?
Solution: From DC test : R dc = V dc / I dc = 13.5 V/ 64 A = 0.2109 Ω For Y-connected induction motor R1 = R dc / 2 = 0.2109 / 2 = 0.1054 Ω
From No Load test : I1nl 22 A , V1nl
Q1nl
V1nl I1nl
2
208 120 V , Pnl 1200 W 3
P nl 3
2
7
2
1200 120 22 2609.5 3 2
Q1nl
2
X nl
V1nl Q1nl
X nl
120 2 5.5182 X 1 X m 2609.5
From blocked-rotor test : I1br 64.5 A , V1br Rbr
P1br I 1br
2
24.6 3
14.2 V , P1br
2200 733.333 W 3
733.333 0.1762 64.5 2
R2 Rbr R1 0.1762 0.1054 0.0708 Q1br
V1br I1br 2 P1br 2
Q1br
14.2 64.52 (733.333) 2
548.721
X br
Q1br 548.721 0.1318 2 64.52 I1br
X br
f rated 60 Hz ' X br 0.1318 0.5272 f test 15 Hz
'
at 15 Hz
For design B motor X1 = 0.4 Xbr = 0.4 * 0.5272 = 0.2108 Ω X2 = 0.6 Xbr = 0.6 * 0.5272 = 0.3163 Ω So ,
Xm = Xnl – X1 = 5.5182 – 0.2108 = 5.3074 Ω
8
HOME WORK Q1: Test carried out on 3phase, Y-connected, 5HP ,208V ,4Pole SCIM , with the following results: DC test : 20 V , 25 A No load test : 208 V , 4 A , 250 W , 60 Hz Blocked-rotor test : 35 V , 12 A , 450 W , 15 Hz Find
R1 , R2 , X1 , X2 , and Xm of this motor ?
Q2: Test carried out on 3phase, Δ-connected, 30kW ,415V, 50 Hz SCIM , with the following results: No load test : 415 V , 21 A , 1250 W Blocked-rotor test : 100 V , 45 A , 2730 W Find
X1 , X2 , and Xm of this motor ?
9
Loading , braking & application of a 3phase induction motors Introduction ; • When selecting a 3-phase induction motor for an application several motor types can fill the need , manufactures often specify the motor class best suited to drive the load . • 3-phase induction motors under 500 hp are standardized ,i.e. the frames have standard dimensions. So, A 25 hp, 1725 rpm, 60 Hz motor from one manufacture can be replaced by that of any other manufacturer without having to change the mounting holes, shaft height, or the type of coupling • a motor must satisfy minimum requirements for starting torque, lockedrotor current, overload capacity, and temperature rise.
Motor characteristics under load • High-inertia load stains a motor by prolonging the starting period. – the starting currents in both the rotor and stator are high during starting. – overheating from I²R losses becomes a problems. – prolonged starting of very large motors will overload the utility transmission network. • Induction motors are often started on reduced voltage to:– limits the current drawn by the motor – reduces the heating rate of the rotor • Heat dissipated in the rotor during starting, from zero speed to rated speed, is equal to the final kinetic energy stored in all the revolving components .
LOAD CONSIDERATIONS There are three basic load types, and these types are classified by the relationship of horsepower and speed. 1- CONSTANT TORQUE LOAD. 2- CONSTANT HORSE POWER LOAD. 3- VARIABLE TORQUE LOAD.
1
Constant torque applications are those that have the same torque at all operating speeds, and horsepower varies directly with the speed. About 90% of all applications, other than pumps, are constant torque loads. Examples of this type include conveyors, hoisting loads, surface winding machines, positive displacement pumps and piston and screw compressors.
Constant horsepower applications have higher values of torque at lower speeds, and lower values of torque at higher speeds. Examples include lathes, milling machines, and drill presses .
The drills in the diagram below are an example of a constant horsepower application. When a larger hole is being drilled, the drill is operating at low speed, but it requires a very high torque to turn the large drill in the material. When a small hole is being drilled, the drill is operated at a high
2
speed, but it requires a very low torque to run the small drill in the material.
The last basic load type is variable torque. The torque required varies as the square of its speed, and horsepower requirements increase as the cube of the speed. Examples include centrifugal pumps, turbine pumps, centrifugal blowers, fans and centrifugal compressors.
Motor Classifications according to speed – For a given output power, a high-speed motor compared with a low-speed motor • costs less • is smaller sized • has higher efficiency and power factor –Low speed applications are often best served by using a high speed- motor and a gear-box which is often required to modify operating speed especially for very high-speed applications (> 3600 rpm).
3
Braking of a 3 phase induction motors: 1- Plugging : • In some applications, the motor and its load must come to a quick stop – such braking action can be accomplished by interchanging two stator leads ,the lead switching causes the revolving field to turn in the opposite direction. Kinetic energy is absorbed from the mechanical load causing the speed to fall.The absorbed energy is dissipated as heat in the rotor circuit. • plugging produces I²R losses that exceed the locked rotor losses. • High rotor temperatures will result that may cause the rotor bars to overheat or even melt. The heat dissipated in the rotor during plugging, from initial speed to zero, is three times the original kinetic energy of all revolving parts.
2- DC Braking: • An induction motor with high-inertial load can also come to a quick stop by circulating dc current in the stator winding. – any two stator terminals can be connected to a dc source. – the dc current produces stationary N,S poles in the stator. • the number of stationary poles are the same at the number of rotating poles normally produced with ac currents. – as the rotor bars sweeps past the dc field, an ac rotor voltage is induced • the I²R losses produced in the rotor circuit come at the expense of the kinetic energy stored in the revolving components. – the motor comes to a rest by dissipating as heat all the kinetic energy. • The benefit of dc braking is the far less heat that is produced. – dissipated rotor losses is equal to the kinetic energy of the revolving parts. – energy dissipation is independent of the dc current magnitude – the braking torque is proportional to the square of the dc braking current.
Enclosure considerations The enclosure of the motor must protect the windings, bearings, and other mechanical parts from moisture, chemicals,mechanical damage and abrasion from grit. NEMA standards define more than 20 types of enclosures under the categories of open machines, totally enclosed machines, and machines with encapsulated or sealed windings. The most commonly used motor enclosures are open dripproof, totally enclosed fan cooled and explosionproof.
4
Open Dripproof(ODP): The open dripproof motor has a free exchange of air with the ambient. Drops of liquid or solid particles do not interfere with the operation at any angle from 0 to 15 degrees downward from the vertical. The openings are intake and exhaust ports to accommodate interchange of air. The open dripproof motor is designed for indoor use where the air is fairly clean and where there is little danger of splashing liquid.
ODP MOTOR Totally Enclosed Fan Cooled (TEFC): This type of enclosure prevents the free exchange of air between the inside and outside of the frame, but does not make the frame completely airtight. A fan is attached to the shaft and pushes air over the frame during its operation to help in the cooling process. The ribbed frame is designed to increase the surface area for cooling purposes. There is also a totally enclosed non-ventilated (TENV) design which does not use a fan, but is used in situations where air is being blown over the motor shell for cooling, such as in a propeller fan application. The TEFC style enclosure is the most versatile of all. It is used on pumps, fans, compressors, general industrial belt drive and direct connected equipment. Special protection can be added to the TEFC motor to help it withstand hostile environments such as chemical and pulp and paper applications.
TEFC MOTOR
5
Explosion proof The explosion proof motor is a totally enclosed machine and is designed to withstand an explosion of specified gas or vapor inside the motor casing and prevent the ignition outside the motor by sparks, flashing or explosion. These motors are designed for specific hazardous purposes, such as atmospheres containing gases or hazardous dusts. For safe operation, the maximum motor operating temperature must be below the ignition temperature of surrounding gases or vapors. Explosion proof motors are designed, manufactured and tested under the rigid requirements of the Underwriters Laboratories.
ENVIRONMENTAL CONSIDERATIONS : AC motors that are properly selected and used should give many years of satisfactory service. Motor life is prolonged by keeping the motor cool, dry, clean and lubricated. Choosing the best motor for the application and the environment will insure that long life. Because so many conditions contribute toward short service life, it is impractical to set forth all possibilities, but some discussion will help point out the need for application analysis where trouble is experienced. Overheating: Heat is one of the most destructive stresses causing premature motor failure. Overheating occurs because of motor overloading, low or unbalanced voltage at the motor terminals, excessive ambient temperatures, or poor cooling caused by dirt or lack of ventilation. If the heat is not dissipated, insulation failure and possibly lubrication and bearing failure can damage a motor. Ambient Temperature: For the purposes of standardization, a value of 40°C has been selected as the standard industrial ambient. This value covers a large range of practical ambient temperatures encountered throughout the world. Temperature rise: This aspect is related to the maximum allowable temperature increase above the defined ambient that the winding cannot exceed when the motor is operating at the name plate or rated power output.
6
Rise by Resistance : One method of determining winding temperature rise (Δt)during motor testing is to measure the electrical resistance between the leads of each phase of the winding. Each phase consists of a set of windings (coils) made from a conductor (typically copper) that has a resistance that is measured in ohms. This value is measured and recorded prior to initiating a load test to determine temperature rise, along with the actual ambient temperature, The test is performed by applying a load equal to the designed power capability of the motor by dynamometer or other means . Under test conditions, the winding temperature of a continuous duty motor must achieve steady-state stabilization while operating under the defined power output or loading value. The heating developed under loading results in a resistance change of the winding conductors.
t 235 t Rhot Rcold * 1 Rcold where t1 test room temperatur e Service Factor :NEMA defines maximum allowable temperature rise for power output conditions greater than the nominal rated conditions. The commonly defined service factor is 1.15, but other multipliers can be applied provided the maximum temperature rise is limited to that specified for the service factor condition. A motor with a 1.15 service factor is expected to deliver 15 percent more power than the normal full load condition. Moisture: Moisture should be kept from entering a motor. Water from splashing or condensation seriously degrades an insulation system. The water alone is conducting. The proper type of motor should be chosen for use in a damp environment. Contamination: No conducting contaminants such as factory dust and sand gradually promote over-temperature by restricting cooling air circulation. In addition, these may erode the insulation and the varnish, gradually reducing their effectiveness. Altitude: Standard motor ratings are based on operation at any altitude up to 3300 feet (1000 meters). High altitude derating is required above 3300 feet because of lower air density.Less dense air reduces the ability for heat to be dissipated into the air from the surfaces of equipment With reduced thermal transfer ability, a motor cannot effectively dissipate heat losses, which require adjustments in the temperature rating, and corresponding power capability of a given motor as defined in Table below.
7
Induction motor operating as a generator : When run faster than its synchronous speed , an induction motor runs as a generator called induction generator . It convertsthe mechanical energy it receives into electrical energy and this energy is released by by the stator . As soon as motor speed exceeds its synchronous speed , it starts delivering active power ( P) to the three phase line . However , for creating its own magnetic field ,it absorbs reactive power (Q)from the line to which it is connected.The reactive power can also be supplied by a group of capacitors connected across its terminals.If capacitance is insufficent , the generator voltage will not build up .Induction generator is widely used today in wind turbines ,which convert wind energy into electrical energy.
Complete torque -speed curve of a three phase induction machine
8
Introduction to Synchronous Machines
Definition: A synchronous machine is an ac rotating machine whose speed under steady state condition is proportional to the frequency of the current in its armature. The magnetic field created by the stator currents rotates at the synchronous speed ,and that created by the field current on the rotor is rotating at the synchronous speed also, and a steady torque results. So, these machines are called synchronous machines because they operate at constant speeds and constant frequencies under steadystate conditions. Synchronous machines are commonly used as generators especially for large power systems, such as turbine generators and hydroelectric generators in the grid power supply. Because the rotor speed is equal to the synchronous speed of stator magnetic field, synchronous motors can be used in situations where constant speed drive is required. Since the reactive power generated by a synchronous machine can be adjusted by controlling the magnitude of the rotor field current, unloaded synchronous machines are also often installed in power systems for power factor correction or for control of reactive kVA flow. Such machines, known as synchronous condensers, and may be more economical in the large sizes than static capacitors. The bulk of electric power for everyday use is produced by polyphase synchronous generators( alternators), which are the largest single-unit electric machines in production. For instance, synchronous generators with power ratings of several hundred megavolt-amperes (MVA) are fairly common, and it is expected that machines of several thousand megavolt- amperes will be in use in the near future. Like most rotating machines, synchronous machines are capable of operating both as a motor and as a generator. They are used as motors in constant-speed drives, and where a variable-speed drive is required ,a synchronous motor is used with an appropriate frequency changer such as an inverter. As generators, several synchronous machines often operate in parallel, as in a power station. While operating in parallel, the generators share the load with each other; at a given time one of the generators may not carry any load. In such a case, instead of shutting down the generator, it is allowed to "float" on the line as a synchronous motor on no-load. 1
Construction of synchronous machines : The synchronous machine has 3 phase winding on the stator and a d.c. field winding on the rotor.
1. Stator : It is the stationary part of the machine and is built up of sheet-steel laminations having slots on its inner periphery. A 3-phase winding is placed in these slots. The armature winding is always connected in star and the neutral is connected to ground.
2. Rotor : The rotor carries a field winding which is supplied with direct current through two slip rings by a separate d.c. source. Rotor construction is of two types, namely; (i) Salient (or projecting) pole type . (ii) Non-salient (or cylindrical) pole type . (i) Salient pole type: In this type, salient or projecting poles are mounted on a large circular steel frame which is fixed to the shaft of the alternator as shown in Fig. (1). The individual field pole windings are connected in series in such a way that when the field winding is energized by the d.c. exciter, adjacent poles have opposite polarities. Low-speed alternators (120 - 400 r.p.m.) such as those driven by water turbines have salient pole type rotors due to the following reasons: (a) The salient field poles would cause .an excessive windage loss if driven at high speed and would tend to produce noise. (b) Salient-pole construction cannot be made strong enough to withstand the mechanical stresses to which they may be subjected at higher speeds. Since a frequency of 50 Hz is required, we must use a large number of poles on the rotor of slow-speed alternators. Low-speed rotors always possess a large diameter to provide the necessary space for the poles. Consequently, salient-pole type rotors have large diameters and short axial lengths.
Fig. 1 salient pole rotor
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(ii) Non-salient pole(cylindrical) type : In this type, the rotor is made of smooth solid forged-steel radial cylinder having a number of slots along the outer periphery. The field windings are embedded in these slots and are connected in series to the slip rings through which they are energized by the d.c. exciter. The regions forming the poles are usually left unslotted as shown in Fig. (2). It is clear that the poles formed are non-salient i.e., they do not project out from the rotor surface.
Fig. 2 cylindrical rotor High-speed alternators (1500 or 3000 r.p.m.) are driven by steam turbines and use non-salient type rotors due to the following reasons: (a) This type of construction has mechanical robustness and gives noiseless operation at high speeds. (b) The flux distribution around the periphery is nearly a sine wave and hence a better e.m.f. waveform is obtained than in the case of salient-pole type. Since steam turbines run at high speed and a frequency of 50 Hz is required, we need a small number of poles on the rotor of high-speed alternators (also called turboalternators). We can use not less than 2 poles and this fixes the highest possible speed. For a frequency of 50 Hz, it is 3000 r.p.m. The next lower speed is 1500 r.p.m. for a 4-pole machine. Consequently, turboalternators possess 2 or 4 poles and have small diameters and very long axial lengths.
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Classification of synchronous machines according to the form of excitation: 1. Brushes excitation systems: The field structure is usually the rotating member of a synchronous machine and is supplied with a dc-excited winding to produce the magnetic flux. This dc excitation may be provided by a self-excited dc generator mounted on the same shaft as the rotor of the synchronous machine. This dc generator is known as exciter. The direct current generated inside exciter is fed to the synchronous machine field winding. In slow-speed machines with large ratings, such as hydroelectric generators, the exciter may not be self-excited. Instead, a pilot exciter, which may be self-excited or may have a permanent magnet, activates the main exciter.
Fig 1 Block Diagram of Excitation system by pilot &main
Fig. 1 Brushes Excitation system for a synchronous machine 2.Brushless systems: This type of excitation has a shaft-mounted bridge rectifier, that rotate with the rotor, thus avoiding the need for brushes and slip rings.
Fig.2 Brushless Excitation system for a synchronous machine
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3.static systems: This type of exctation is widely used in small size alternators, in which portion of the AC from each phase of synchronous generator is fed back to the field winding as a DC excitation through a system of transformer ,rectifiers and reactors. External source of a DC is necessary for initial excitation of the field windings.
Advantages of stationary armature ; The field winding of an alternator is placed on the rotor and is connected to d.c. supply through two slip rings. The 3-phase armature winding is placed on the stator. This arrangement has the following advantages: (i) It is easier to insulate stationary winding for high voltages , because they are not subjected to centrifugal forces . (ii) The stationary 3-phase armature can be directly connected to load without going through large, unreliable slip rings and brushes. (iii) Only two slip rings are required for d.c. supply to the field winding on the rotor. Since the exciting current is small, the slip rings and brush gear required are of light construction. (iv) Due to simple and robust construction of the rotor, higher speed of rotating d.c. field is possible. This increases the output obtainable from a machine of given dimensions.
Cooling : Because synchronous machines are often built in extremely large sizes, they are designed to carry very large currents. A typical armature current density may be of the order of 10 A/mm 2 in a well-designed machine. Also, the magnetic loading of the core is such that it reaches saturation in many regions. The severe electric and magnetic loadings in a synchronous machine produce heat that must be appropriately dissipated. Thus the manner in which the active parts of a machine are cooled determines its overall physical structures. In addition to air, some of the coolants used in synchronous machines include water, hydrogen, and helium.
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Damper Bars : So far we have mentioned only two electrical windings of a synchronous machine: the three-phase armature winding and the field winding. We also pointed out that, under steady state, the machine runs at a constant speed, that is, at synchronous speed. However, like other electric machines, a synchronous machine undergoes transients during starting and abnormal conditions. During transients, the rotor may undergo mechanical oscillations and its speed deviates from the synchronous speed, which is an undesirable phenomenon. To overcome this, an additional set of windings, resembling the cage of an induction motor, is mounted on the rotor. When the rotor speed is different from the synchronous speed, currents are induced in the damper windings .the damper winding acts like the cage rotor of an induction motor, producing a torque to restore the synchronous speed. Also, the damper bars provide a means of starting to the synchronous motors, which is otherwise not selfstarting. Fig6 shows the damper bars on a salient rotors
Fig .6 salient pole rotor showing the field winding and damper bars
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Differences between three phase induction machine and synchronous machine
Three phase induction machine Stator phases either star or delta
connected
Synchronous machine Stator phase are star connected only
Rotor windings are not fed by electricity, currents flow through rotor due to induction process.
Rotor windings are fed by dc source.
Run below synchronous speed, as a motor Run above synchronous speed, as a generator
Run at synchronous speed for both motor and generator
Self starting , as a motor
Need damper bars to start , as a motor
Operate with lagging power factor only
Operate with lagging, leading, and unity power factor
Simple in construction , rugged, low maintenance, cheap , especially in squirrel cage type.
Complex in construction , expensive.
Not active in low speed operation, as a motor
Active in both low and high speed operation , as a motor
Not active in generating mode
Used widely in generation of electricity
Difficult in speed regulation, as a motor
Precise speed regulation, as a motor
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THREE PHASE STATOR WINDINGS The stator windings for alternating-current motors and generators are alike. It should be noted that direct-current and alternating-current windings differ essentially by the former being of the closed-circuit type (through commutator) ,while alternating-current windings are of the open-circuit type.
Types of A-C Windings: With reference to the arrangements of coils used in three phase stator, windings may be divided into two general classes as follows: I. Distributed Windings: 1. Spiral or chain. 2. Lap. 3. Wave. II. Concentrated Windings: 1. Lap. 2. Wave.
Distributed Windings: An armature winding which has its conductors of any one phase under a single pole placed in several slots, is said to be distributed. When these conductors are bunched together in one slot per pole per phase, the winding is called concentrated. It is usual in a distributed winding to distribute the series conductors in any phase of the winding among two or more slots under each pole. A distributed winding has two principal advantages, first, a distributed winding generates a voltage wave that is nearly a sine curve, secondly, copper is evenly distributed on the armature surface, therefore, heating is more uniform and this type of winding is more easily cooled.
Concentrated Windings: The concentrated winding gives the largest possible emf from a given number of conductors in the winding. That is for a definite fixed speed and field strength in an alternator, the concentrated winding requires a less number of conductors than a distributed winding, but increases the number of turns per coil. Lap and Wave Windings: Both distributed and concentrated windings make use of lap and wave connections. These arrangements are in principle the same as used in direct-current windings. The diagrams of Figs. 1 and 2 show distributed lap and wave windings . 8
Fig. 1 Single-phase lap winding
Fig 2. Three-phase wave winding , using one slot per pole per phase, star-connected.
Chain Winding : In this winding as shown in Fig. 3 there is only one coil side in a slot. An odd or even number of conductors per slot may be used but several shapes of coils are required since the coils enclose each other. The number of coils required in this winding is also small compared with other windings. This type of winding is mainly used in alternating current generators.
Fig. 3 Three-phase chain winding 9
Single layer and double layer winding : In double layer winding two coil sides are located in each armature slot [Fig. 4 ]. If there is only one coil side located in each armature slot, the winding is called single layer.
Fig. 4 A double layer winding. One side of the coil lies in the top of the slot and the other side in the bottom of the of the slot.
Full pitch and fractional pitch windings: For a three-phase winding the total number of coil sides or the total number of slots should be just divisible by three (the number of phases) and sometimes by the number of poles. This will result in a full pitch winding, that is, a winding in which a coil spans exactly the distance between the centers of adjacent poles. If the coil spans less than this distance, so that its two sides are not exactly under the centers of adjacent poles at the same time, it is said to have a fractional-pitch. When a fractional pitch is used, the total number of slots per phase must be a whole number. The fractional-pitch coils are frequently used in a.c. machines for two main reasons. First, less copper is required per coil and secondly the waveform of the generated voltage is improved .The number of stator slots, divided by the number of poles gives a value of the pole pitch expressed in terms of the slots. Coil pitch is expressed as a fraction of the pole pitch, in slots, or in electrical degrees. In the case of a 6 pole machine having 72 stator slots, and a double-layer winding, the pole pitch would be 12 slots(72 / 6). If the coil pitch were given as 2/3, this would be 120° ( 2/3 * 180°) or 8 slots (2/3 * 12). A full coil pitch for this winding would be 180 degrees, or 12 slots. A full pitch winding is one in which the coil pitch is equal to the pole pitch, and a fractional pitch winding is one in which the coil pitch is not equal to the pole pitch. For the coils used in small machines round insulated wire is most employed. These coils are either wound in the slots by hand or assembled by use of specially formed coils wound in forms and insulated before being placed in the slots. Such formed coils are usually used except in cases where the slots are closed or nearly closed. For large machines where the amperes to be carried in each armature circuit is a large value, copper straps are frequently employed for making up the armature coils. In very large machines a copper bar is used instead of the copper straps. In such a case one bar serves as a conductor of a coil having one turn per slot. The copper bars are connected to the end connections of the coils by brazing, welding or bolting. In all cases, whatever the construction of the coil used, the slots must be properly insulated with mica, polyester film or other suitable insulating material according 10
to the adequate class of insulation. The coil throw refers to the start of coil from one stator slot to the finish of this coil at another stator slot . Figures 5 (a) , (b) , (c), and (d) show four different coils, in each of them the pole pitch is of 12 slots while the coil pitch in (a) is 11 slots , in (b) is 13 slots, in (c) is 9 slots, and in (d) is 15 slots .
Fig. 5. Possible pitch for one coil per slot windings(pole pitch=24 slots/2poles=12) (a) coil throw 1 to 12, coil pitch=12-1=11 (b) coil throw 1 to 14, coil pitch=14-1=13 (c) coil throw 1 to 10, coil pitch=10-1=9 , (d) coil throw 1 to 16, coil pitch=16-1=15 .
Phase belt and phase spread : A group of adjacent slots belonging to one phase under one pole pair is known as phase belt. The angle subtended by a phase belt is known as phase spread. The 3-phase windings are always designed for 60° phase spread . Fig. (6) shows a 2-pole, 3-phase double-layer, full pitch, distributed winding for the stator of an alternator. There are 12 slots and each slot contains two coil sides. The coil sides that are placed in adjacent slots belong to the same phase such as( a1, a3 ) or (a2, a4 )constitute a phase belt. Since the winding has doublelayer arrangement, one side of a coil, such as (a1), is placed at the bottom of a slot and the other side (- a1 ) is placed at the top of another slot spaced one pole pitch apart. Note that each coil has a span of a full pole pitch or 180 electrical degrees. Therefore. the winding is a full-pitch winding. Note that there are 12 total coils and each phase has four coils. The four coils in each phase are connected in series so that their voltages aid. The three phases then may be connected to form Y or -connection. Fig. (7) shows how the coils are 11
connected to form a Y-connection. Any star diagram can be readily changed into a corresponding delta diagram by opening the star points and connecting the inner end of phase A to the outer end of phase B, the inner end of phase B to the outer end of phase C , and the inner end of phase C to the outer end of phase A .
Fig. 6
Fig. 7
Simple rule for checking proper phase relationship in a three-phase winding: A fundamental consideration when checking the instantaneous flow of current in a three phase circuit, is to imagine that when the current flows in the same direction in two legs of the circuit, it flows in the opposite direction in the third leg. This principle can be applied to both motors and generators. In Figure 8, it must be supposed that current flows in all three leads of the star connection toward the point of the star connection. And that in the case of a delta connection the current flows around the three sides of the delta in the same direction. Then in either case for a three-phase winding, the polarity of each of the pole-phase groups will alternate regularly around the winding and can be indicated by arrows as in Fig. 8. By the use of this scheme there is no chance 12
for a reversal of a phase to be passed by not noticed when checking the winding.
. Fig. 8 Simple scheme of alternately reversing arrows of pole-phase groups to check correct phase polarity of a 3-phase winding. It is supposed in this case that current flows in the three leads toward the star points of the winding which are indicated thus ( *) .
Why the alternators have their windings connected in star? Because the star connection has the following advantages over delta connection: (i) To obtain a desired voltage between outside terminals of a generator, the voltage built up in any one phase need be only 58 percent(1 / √3) of the terminal value. Hence only 58 percent of the turns required for a Δ-connected armature are necessary with a consequent lowering of insulation cost. (ii) A star connected winding offers the advantage of a fourth or neutral lead making possible the advantages of a four -wire system, with or without grounded neutral. (iii)The wave shape of a star-connected winding is improved, owing to the elimination of third harmonics and multiples of third harmonics from the terminal voltage. Fig.9 shows a star connected armature( stator), The terminal voltages E12 , E23 and,E 31 are 120 electrical degrees apart. In the third harmonics the e.m.f.'s of three phases (e01 ,e02 ,e03) are being in phase( 3 X 120º = 360º= 0º ),to obtain terminal voltages, the resultant is zero. Hence no third harmonics appears between terminals.The circulating currents from the third harmonics cause unnecessary losses and dangerous heating in Δ-connected alternator, where it is not in the Y-connected alternator case. In addition the use of 13
a(5/6)th pitch winding in three-phase Y-connected generators reduces the fifth and seventh harmonics, if pesent, to almost nil, so the lowest harmonics that can be present is eleventh.
Fig. 9
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Windings factors: The stator winding of synchronous machine is distributed over the entire stator. The distributed winding produces nearly a sine waveform and the heating is more uniform. Likewise, the coils of armature winding are not full-pitched i.e., the two sides of a coil are not at corresponding points under adjacent poles. The fractional pitched armature winding requires less copper per coil and at the same time waveform of output voltage is improved. The distribution and pitching of the coils affect the voltages induced in the coils. We shall discuss two winding factors: (i) Distribution factor (Kd). (ii) Pitch factor (Kp). (i) Distribution factor (Kd) : A winding with only one slot per pole per phase is called a concentrated winding. In this type of winding, the e.m.f. generated/phase is equal to the arithmetic sum of the individual coil e.m.f.s in that phase. However, if the coils/phase are distributed over several slots in space (distributed winding), the e.m.f.s in the coils are not in phase (i.e., phase difference is not zero) but are displaced from each by the slot angle (The angular displacement in electrical agrees between the adjacent slots is called slot angle). The e.m.f./phase will be the phasor sum of coil e.m.f.s. The distribution factor Kd is defined as:
The distribution factor can be determined by constructing a phasor diagram for the coil e.m.f.s. Let m = 3. The three coil e.m.f.s are shown as phasors AB, BC and CD [See Fig. 10 (i)] each of which is a chord of circle with centre at O and subtends an angle at O. The phasor sum of the coil e.m.f.s subtends an angle 15
m(Here m = 3) at O. Draw perpendicular bisectors of each chord such as Ox, Oy etc [See Fig. 10 (ii)].
Note that ( m is the phase spread.
Fig. 10
(ii) Pitch factor (Kp) : A coil whose sides are separated by one pole pitch (i.e., coil span is 180electrical) is called a full-pitch coil. With a full-pitch coil, the e.m.f.s induced in the two coil sides a in phase with each other and the resultant e.m.f. is the arithmetic sum of individual e.m.fs. However the waveform of the resultant e.m.f. can be improved by making the coil pitch less than a pole pitch. Such a coil is called short-pitch coil. This practice is only possible with double-layer type of winding The e.m.f. induced in a short-pitch coil is less than that of a fullpitch coil. The factor by which e.m.f. per coil is reduced is called pitch factor K p. It is defined as:
Consider a coil AB which is short-pitch by an angle Өelectrical degrees as shown in Fig. (11). The e.m.f.s generated in the coil sides A and B differ in phase by an angle Өand can 16
be represented by phasors EA and EB respectively as shown in Fig. (12). The diagonal of the parallelogram represents the resultant e.m.f. ER of the coil.
Fig. 11
Fig. 12
For a full-pitch winding, Kp = 1. However, for a short-pitch winding, Kp < 1. Note that Өis always an integer multiple of the slot angle .
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SYNCHRONOUS
GENERATOR
The machine which produces 3-phase electrical power from mechanical power is called an alternator or synchronous generator. Alternators are the primary source of all the electrical energy we consume. These machines are the largest energy converters found in the world.
Alternator Operation : Like the dc generator, a synchronous generator functions on the basis of Faraday's law, which state if the flux linking the coil changes in time, a voltage is induced in a coil. Stated in another form, a voltage is induced in a conductor if it cuts magnetic flux lines. The rotor winding is energized from the d.c. exciter and alternate N and S poles are developed on the rotor. When the rotor is rotated by a prime mover, the stator or armature conductors are cut by the magnetic flux of rotor poles. Consequently, e.m.f. is induced in the armature conductors due to electromagnetic induction. The induced e.m.f. is alternating since N and S poles of rotor alternately pass the armature conductors. The frequency of induced e.m.f. is given by;
Where
N = speed of rotor in r.p.m. , P = number of rotor poles.
The magnitude of the voltage induced in each phase depends upon the rotor flux, the number and position of the conductors in the phase and the speed of the rotor. [Fig. 1 (i)] shows star-connected armature winding and d.c. field winding. When the rotor is rotated, a 3-phase voltage is induced in the armature winding.. The magnitude of e.m.f. in each phase of the armature winding is the same. However, they differ in phase by 120° electrical as shown in the phasor diagram [ Fig. 1 (ii)].
Fig. 1
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Considering the round rotor synchronous generator in Fig.2,which shows a simple consentrated stator winding ( 1 slot / pole / phase).
Fig .2 (a) A round rotor synchronous generator (b) Air gap flux density distribution produced by the rotor excitation and assuming that the flux density in the air gap is uniform implies that sinu soidally varying voltages will be induced in the three coils aa' ,bb' and cc' if the rotor, rotates at a synchronous speed, Ns. Also if Φр is the flux per pole, ω is the angular frequency, and T is the number of turns in phase a (coil aa'), then the voltage induced in phase a is given as: ea = ω T Φр sin ωt = Em sin ωt Where
E m = ω T Φр , and
ω =2 π f .
Because phases b and c are displaced from phase a by +_ 120º , the corresponding voltages may be written as eb = Em sin (ωt – 120º) ec = Em sin (ωt + 120º) These voltages are sketched in Fig. 3 and correspond to the voltages from a threephase generator.
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Fig. 3 A three-phase voltage produced by a three-phase synchronous generator Generated E.M.F. Equations : The magnetic flux cut by one armature conductor, when the rotor of an alternator is made to revolve through one revolution, is Φр * P, where Φр is the magnetic flux per pole and P is the number of poles. If N is the speed in revolution per minute, then the flux cut per second becomes Φр * P *( N /60) Since 1 volt is generated when 1 weber of flux is cut per second, the average voltage generated in this conductor becomes Eav = Φр * P *( N /60) volts If the total number of conductors on the armature is Z and they are connected into A parallel paths, the average voltage between terminals becomes Eav per conductor * ( Z / A ) In case of alternating current generators the e.m.f. depends not only upon the total flux cut per second but also upon the way in which the flux and the conductors are distributed. A change in distribution of flux changes the relative values of the maximum and effective (r.m.s.) e.m f.'s. In addition, the e.m.f. built up in any one conductor, when considered vectorially, cannot always be added directly to that of another as there may be a phase displacement between them. The instantaneous values, though can be added algebraically, but in adding the effective values it is necessary to consider the phase differences between the different e.m.f.'s to be added. In order to take these factors in account, the flux distribution and winding types must be known. Fig.4(a) shows the space distribution of the airgap flux of an alternator. In this case, the flux density B is assumed to be sinusoidal in space when measured around the inside periphery of the stator. This flux density B can be expressed as B=Bmax sinθ. where θ is measured from the position midway between the poles ; B is the flux density measured in webers per length of the field pole (L ) as shown in Fig.4(b) per length of pole pitch arc ; and Bmax is the maximum flux density produced by a pole.
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The total flux per pole is Φp = ο ∫π LB max sin θ dθ = -| LB max cos θ |ο π =2 LB max Weber ...(i) As this flux wave is moved around the air gap the conductors a and b of the stator coil will have voltages induced in them.
Fig 4 Shapes of flux density waves The voltage induced in conductor a will be e a =BL v ...(ii) where v is the velocity of the flux wave in radians per second, or v = 2 π f , f being the frequency of the flux wave or e.m.f. induced. The voltage in conductor a is ea =B max L 2π f sin θ But θ = ω t where ω is the angular velocity of the rotor in radians per second and t is the time of rotation from the position shown in Fig.4(a). Therefore, e a = B max L 2π f sin ω t ...(iii) Substituting equation (i) in equation (iii) e a = (Φp /2 ) *2π f sin ω t
...(iv)
Likewise, the voltage induced in conductor b e b = (Φp /2 ) *2π f sin ω t
...(v)
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and the voltage at the terminals of the turn made up of the conductors a and b is e turn = Φp 2π f sin ω t ...(vi) If the single turn on the armature (stator) were replaced by a coil of T turns, the voltage per coil would be e eoil = Φp 2π f T sin ω t The effective value of generated e.m.f.
...(vii)
Er.m.s = Emax / √2 = (2π / √2) f Φp T
volts (r.m.s)
Er.m.s =4.44 f Φp T
volts (r.m.s)
...(viii) ...(ix)
As the r.m.s. value is related to the average value so, Form factor (k f ) = (r.m.s. value) / (average value) * For sinusoidal wave of e.m.f., k f = 1.11 E r . m. s . = 4 k f T f Φp
volts
Also, since the induced e.m.f.'s in the conductors of an alternator are not all in phase, the above relation for Er.m.s. must be multiplied by k p and k d , the pitch factor and the distribution (breadth) factor. Therefore, Er.m.s. =4 k f k p k d f T Φp
volts per phase
...(x)
* For full pitched and concentrated windings kd =1 and kP =1. If the alternator is star connected, and neglecting the effect of armature reaction, the alternator line voltage is, EL = √ 3 (Er.m.s. per phase)
, so
EL = √ 3( 4 k f k p k d f T Φp)
volts.
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Synchronous Generator Characteristics : ( i ) Magnetisation curve: A plot of the exciting current versus terminal voltage of alternator is known as the magnetisation curve. This magnetisation curve is obtained by passing different values of currents in exciting windings, thereby giving, correspondingly different values of terminal voltage. The no load magnetisation curve is shown in Fig.5-I] ,which has the same general shape as B-H curve of armature steel. The full load magnetisation characteristics with unity power factor and with 0.8 lagging power factor have been shown at ( II ) and ( III ) in Fig.5.
Fig. 5 Magnetization Curve Fig. 5 Magnetization curves for alternator at different loading conditions
( ii ) Load characteristics : If the speed and exciting current remain constant, the terminal voltage of the alternator changes with the load currents. The plot between the terminal voltage and the load (or armature) current of an alternator is known as load characteristics. An increase in the armature (or load) currents make the terminal voltage drops. This has been shown in Fig. 6. The drop in terminal voltage is attributed to many reasons but primarily because of the following: ( a ) Resistance and reactance of the armature (or stator) winding. ( b ) Armature reaction. The resistance and reactance of the armature winding causes the drop in generated e.m.f. (voltage) ,whereas the armature reaction weakens the magnetic field and thereby decreases the generated e.m.f. (voltage).
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The magnitude of the effect of armature reaction depends upon, the power factor of load i.e. angle of lag or lead of the stator (armature) current. In case of a unity power factor of load, each phase of alternator when connected to the load takes a current which is in phase with its generated voltage. But the magnetic field is strengthened, instead of weakening, if the load (or armature) current, is leading. In the above case, when the power factor is leading, the drop in voltage due to resistance and reactance of stator winding may be less than the increase in voltage due to armature reaction. Thus the terminal voltage on load may be more than that at no-load, if the angle of lead of the load (or armature) current is sufficient.
Fig. 6. Load Characteristic of alternator ( iii ) Effect of variation of power factor on terminal voltage : The load characteristics at different power factors with leading and lagging armature currents are shown in Fig. 7. If the load, current and excitation are kept constant, the terminal voltage falls on changing the power factor from leading to lagging one.This effect is because of the armature flux helping the main flux, in case p.f. is leading, hence generating more e.m,f. and the armature flux, opposing the main flux, in case the p.f. is lagging, hence generating less e.m.f., Therefore, the terminal voltage at lagging power factor decreases from that on leading p.f. because of decrease in generated e.m.f.
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Fig. 7 Armature Leakage & Synchronous Reactances : When the current pass through the stator conductors the flux is set up, and a portion of this flux does not cross the air gap but completes the path inside the stator as shown in Fig. 8. This flux is known as leakage flux,which sets up an e.m.f. in the stator winding .This e.m.f. leads the load current by 90° and proportional to the magnitude of load current. This e.m.f. is due to the leakage inductance of the armature winding. The magnitude of the leakage inductance in practical units,henrys, is given by the general equation : L = (Flux in webers per amperes) * (No. turns) henrys And leakage reactance per phase, Xl = ω L = 2 π f L ohms/ph Xl causes a voltage drop in alternator terminal voltage and this drop is equal to an e.m.f. set up by the leakage flux. Also, there is another source causes voltage drop , that is due to armature reaction which can be represented by a fictitious reactance X a . Xl + X a = Xs Where Xs is the per phase synchronous reactance of armature winding.
Fig 8
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Performance Of A Round-Rotor Synchronous Generator : At the outset we wish to point out that we will study the machine on a per phase basis, implying a balanced operation. Thus let us consider a round-rotor machine operating as a generator on no-load. Variation of the terminal voltage with the exciting current (field current) is shown in Fig. 5-I , and is known as the open-circuit characteristic of a synchronous generator. Let the open-circuit phase voltage be Eo for a certain field current If. Here Eo is the internal voltage of the generator. We assume that If is such that the machine is operating under unsaturated condition. Next we short-circuit the armature at the terminals, keeping the field current unchanged (at If), and measure the armature phase current Ia In this case, the entire internal voltage Eo is dropped across the internal impedance of the machine. In mathematical terms, Eo = IaZs and Zs is known as the synchronous impedance , which is equal to Zs= R a + j XS If the generator operates at a terminal voltage Vt, while supplying a load corresponding to an armature current Ia, then Eo = Vt + Ia (R a + j XS) In an actual synchronous machine, except in very small ones, we almost always have XS >> R a , in which case Zs ≈ j X S. Among the steady-state characteristics of a synchronous generator, its voltage regulation and power-angle characteristics are the most important ones. We define the voltage regulation of a synchronous generator at a given load as percent voltage regulation = (Eo - Vt) / Vt * 100 % where Vt is the terminal voltage on load and Eo is the no-load terminal voltage. The voltage regulation is dependent on the power factor of the load. the voltage regulation for a synchronous generator may even become negative. The angle between Eo and Vt is defined as the power angle, δ. Notice that the power angle, δ, is not the same as the power factor angle, φ. To justify this definition, we consider Fig. 9, from which we obtain I a XS cos φ = Eo sin δ
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…(i)
Fig. 9 Phasor diagram of round rotor generator Now, from the approximate equivalent circuit (assuming that XS >> R a )as shown in Fig 10-a, the power delivered by the generator = power developed, Pd = Vt Ia cos φ , which follows from Fig. 9 also. Hence, in conjunction with the (i) equation, we get Pd = (Eo Vt / XS ) sin δ per phase …(ii) Pd = 3(Eo Vt / XS ) sin δ for three phases Which shows that the internal power of the machine is porportional to sin δ, Equation (ii) is often said to represent the power-angle characteristic of a round rotor synchronous machine. Fig. 10-b shows that for a negative δ, the machine will operate as a motor.
Fig 10 : (a) an approximate equivalent circuit of synchronous machine (b) power-angle characteristics of a round-rotor synchronous machine
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Performance Of A Salient-Pole Synchronous Generator : Because of saliency, the reactance measured at the terminals of a salient-rotor machine will vary as a function of the rotor position. This is not so in a round-rotor machine. Thus a simple definition of the synchronous reactance for a salient-rotor machine is not immediately forthcoming. To overcome this difficulty, we use the two-reaction theory proposed by "Andre Blondel". The theory proposes to resolve the given armature mmf's into two mutually perpendicular components, with one located along the axis of the rotor salient pole, known as the direct (or d )axis and with the other in quadrature and known as the quadrature (or q )axis. Correspondingly, we may define the d-axis and q-axis synchronous reactances, Xd and Xq for a salient-pole synchronous machine. Thus, for generator operation, we draw the phasor diagram of Fig. 11. Notice that Ia has been resolved into its d- and q-axis (fictitious) components, Id and Iq With the help of this phasor diagram, we obtain Id = Ia sin ( δ + φ ) Iq = Ia cos ( δ + φ ) Vt sin δ = Iq Xq = Ia Xq cos ( δ + φ ) From these we get Vt sin δ = Ia Xq cos δ cos φ - Ia Xq sin δ sin φ Or (Vt + Ia Xq sin φ) sin δ = Ia Xq cos δ cos φ Dividing both sides by cos δ and solving for tan δ yields tan δ = (Ia Xq cos φ) / (Vt + Ia Xq sin φ) …(iii) With δ known (in term of φ), the voltage regulation may be computed from Eo = Vt cos δ + Id Xd Percent regulation = (Eo - Vt) / Vt *100%
Fig. 9 Phasor Diagram of a Salient-Pole Generator Fig 11 Phasor diagram of salient-pole generator
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In fact, the phasor diagram depicts the complete performance characteristics of the machine. Let us now use Fig. 11 to drive the power-angle characteristics of a salient-pole generator. If armature resistance is neglected, Pd = Vt Ia cos φ, Now, from Fig. 11, the projection of Ia on Vt is Pd / Vt = Ia cos φ = Iq cos δ + Id sin δ Solving Iq Xq = Vt sin δ and Id Xd = Eo - Vt cos δ For Iq and Id , and substituting in (i), gives Pd = (Eo Vt / Xd ) sin δ + ( Vt2 / 2 ) [ 1/ Xq – 1/ Xd ] sin 2δ Pd = 3 (Eo Vt / Xd ) sin δ + 3 ( Vt2 / 2 ) [ 1/ Xq – 1/ Xd ] sin 2δ
per phase ...(iv) for three phases
The equation can also be established for a salient-pole motor (δ Vt , and the machine becomes overexcited. The voltage-current relationships for the two cases are shown in Fig. 3(a). For Eo > Vt at constant power, δ is greater than the δ for Eo < Vt .Notice that an underexcited motor operates at a lagging power factor (Ia lagging Vt), whereas an overexcited motor operates at a leading power factor. In both cases the terminal voltage and the load on the motor are the same. Thus we observe that the operating power factor of the motor is controlled by varying the field excitation, hence altering Eo. This is a very important property of synchronous motors. The locus of the armature current at a constant load, as given by (i), for varying field current is also shown in Fig 3(a). From this we can obtain the variations of the armature current Ia with the field current, I f (corresponding to Eo ), and this can be done for different loads, as shown in figures 3(b)and(c).
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(c) Fig 3 (a) phasor diagram for motor operation (Eo', Ia', φ', δ')for underexcited operation,( Eo'', Ia'', φ'', δ'')for overexcited operation (b), (c) V-Curves of a synchronous motor These curves are known as the V- Curves of the synchronous motor. One of the applications of a synchronous motor is in power factor correction, as demonstrated by the following examples. In addition to the V curves, we have also shown the curves for constant power factors. These curves are known as compounding curves. In the preceding paragraph, we have discussed the effect of change in the field current on the synchronous machine power factor. However, the load supplied by a synchronous machine cannot be varied by changing the power factor. Rather, the load on the machine is varied by instantaneously changing the speed (in case of a generator, by supplying additional power by the prime mover), and thus changing the power angle corresponding to the new load. In a synchronous motor, a load change results in a change in the power angle. 54
EXAMPLE: A three-phase wye-connected load takes 50 A of current at 0.707 lagging power factor at 220 V between the lines. A three-phase wye-connected, round-rotor synchronous motor, having a synchronous reactance of 1.27 Ω per phase. is connected in parallel with the load. The power developed by the motor is 33 kW at a power angle, δ, of 30°. Neglecting the armature resistance, calculate (a) the reactive kilovolt-amperes (kVAR) of the motor and (b) the overall power factor of the motor and the load. Solution The circuit and the phasor diagram on a per phase basis are shown in figures 4(a)and(b). From power developed equation, we have Pd = 1/3 * 33,000 = 220/√3 * Eo/1.27 * sin 30º Which yield Eo = 220 And from phasor diagram, Ia XS = 127, or Ia = 127/ 1.27 = 100 A, and φa = 30º The reactive kilovolt-amperes of the motor = √3 * Vt Ia sin φa =√3* 220/1000 * 100* sin 30 = 19 VAR Notice that φa is the power-factor angle of the motor, φ L is the power angle of the load, φ is the overall power-factor angle, are shown in fig 4(b). The power angle δ, is also shown in this phasor diagram, from which I = IL + Ia
Fig 4 (a) Circuit diagram (b) Phasor diagram Or algebraically adding the real and reactive components of the currents, we obtain Ireal = Ia cos φa + IL cos φL Ireactive = Ia sin φa – IL sin φL The overall power factor angle, φ, is thue given by tan φ = [Ia sin φa – IL sin φL ] / [Ia cos φa + IL cos φL] = 0.122 Or φ = 7º and cos φ = 0.992 leading. 55
TORQUE – SPEED curve of a synchronous motor: In synchronous motor speed remains constant at Ns for all loads as in figure 5. At any other speed (≠ Ns) ,there is no motor –action , and therefore no torque will produced.
Fig - 5
Starting of Three Phase Synchronous Motor As is clear from the construction of the motor, the stator of the synchronous motor is wound and the winding is connected to a.c. mains while the rotor of the motor is mostly of salient pole field construction except in special high speed two pole motors where it is non-salient pole field construction. The rotor is supplied from a d.c. source. Consider Fig. 5, the stator for convenience of explanation is shown as having salient poles N' and S'. When the rotor is excited from d.c. source, there develop N and S poles on the rotor and this polarity is retained by the rotor throughout but the polarity of stator poles changes because it is connected to a.c. mains and the polarity alternates with the frequency of the a.c. supply. First consider the rotor is stationary and in a position as shown in Fig. 6 (a). At this instant the similar poles of rotor and stator repel each other and the rotor tends to rotate in clockwise direction. But half a cycle later (i.e. 1/2f second) the polarityof the stator poles is reversed [as in Fig 6 (b)] but the polarity of rotor poles remains the same.
(a)
Fig 6
(b)
At this very instant unlike poles of stator and rotor attract each other and, therefore, the rotor tends to rotate back in the anti-clockwise direction. This shows that the torque acting on the rotor of the synchronous motor is not unidirectional but 56
pulsating one. It is because of inertia of the rotor, it will not move in any direction. That is why the synchronous motor is not self-starting one. Now let the rotor (which is yet unexcited) is speeded up to synchronous or near synchronous speed by some external arrangement and then it is excited through the d.c. source. The moment the rotor running at nearly synchronous speed is excited, it is magnetically locked into position with the stator poles which runs synchronously and that both in the same direction. Because of this interlocking between the rotor and stator poles, the synchronous motor has either to run synchronously or not at all.
Methods of Starting Synchronous Motors Since the three phase synchronous motor has no starting torque,so artificial means must be provided for starting it as below: 1. External source. If the d.c. field of a synchronous motor is supplied by a direct-connected exciter, this exciter can be used as a starting motor. This method is now rarely used. 2. Induction-motor start. If a squirrel cage winding is constructed in the pole faces of the synchronous motor (damper bars), it can be used to develop a starting torque (as well as provide damping) similar to that of the ordinary induction motor. As the motor reaches about 95 per cent of synchronous speed (it is operating as an induction motor with 5 percent slip) its field is excited and the motor pulls into step. For such a set up, the problem of limiting the starting current without to low a value of starting torque is met in several ways: . (a) Auto-transformers are used which are similar in design to those employed on induction motors. The stator is connected to the reduced voltage supply of the auto-transformer until synchronous or near synchronous speed is reached, and is then connected to the full voltage. (b) The voltage supplied to the stator can be reduced by using a series reactors in the supply lines. These reactors give a large voltage drop and low P.F. at starting. (c) Various types of rotor windings are used like double squirrel cage, or using special bar cross-sections (T bar, L bar, or simply deep, narrow bars) to give high skin effect to the squirrel cage and thus limit the starting current. (d) Multiple winding. This involves a special arrangement of stator coils so that two or more complete windings are paralleled for normal operation. The paralleled windings have normal values of' reactance. If one winding is left open during starting process the resistance will double and the leakage reactance will be greater than the normal value and will result in a reduced current without the utilization of an autostarter. Switching arrangement is shown in Fig. 7. 57
(a) For starting, the short-circuited switch is open and only winding 1 is utilized (b) The short circuiting switch closes the neutral of the second winding for parallel operation at normal load Fig 7 Arrangement of the double-winding synchronous motor
Synchronous Condenser When a three phase synchronous motor is used for power factor correction with no mechanical output (no load), it is known as a synchronous condenser. The term 'condenser' applied to device that it draws leading current as does a static condenser. There is a considerable increase in leading kilovolt- amperes available when the horsepower load of a synchronous motor is less than its rated load. The synchronous condenser is especially designed so that practically all its rated kVA are available for p.f. correction. Because of the absence of shaft load, the mechanical design is modified from the ordinary motor standards. Because of the p.f. adjustment possible, the synchronous condenser is particularly applicable to transmission line control. The ability to control the power factor of the synchronous motor by simply varying the excitation is of very great practical importance. The importance lies in the fact that the motor can be operated at a leading power factor when desired, and consequently if the remaining installation has a low power factor, the overall. power factor for the installation can be brought close to unity.
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Hunting and Damping of Synchronous Motor When a synchronous motor is operating under steady-load conditions and an additional load is suddenly applied, the developed torque is less than that required by the load and the motor starts to slow down. A slight reduction in speed increases the phase angle of the generated voltage and permits more current to flow through the armature Some of the kinetic energy of the rotating parts is given out to the load during speed reduction . When the motor is slowing down, it cannot cease deceleration at exactly the correct torque angle corresponding to the increased load. It passes beyond this point, develops more than required torque, and increases in speed. This is followed by a reduction in speed and a repetition of the entire cycle. Such a periodic change in speed is called hunting. The mass of the rotating part and the 'spring effect' of the flux lines are the necessary elements which give the rotating field structure a natural oscillating period. If the load varies periodically with this same frequency or some multiple of it, the tendency is for the amplitude of these oscillations to increase cumulatively until the motor is thrown out of step, causing corresponding current and power pulsations. The mechanical stresses are likely to be severe, and the armature current greatly increases when the motor leaves synchronism. A solid pole gives a damping action but has the disadvantage of excessive pole-face losses unless closed slots are used on the armature. The most satisfactory arrangement from the standpoint of simplicity seems to be laminated poles with the squirrel-cage 'amortisseur' or damping winding. The effectiveness of damper depends upon its resistance and to a lesser extent upon the length of the airgap. A low resistance damping winding produces the stronger effect, but if the synchronous motor is to be started by induction-motor action, using this squirrel cage, the winding resistance should be fairly high to produce good starting torque. Because of these opposing tendencies, a compromise usually must be reached in damping winding design.
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SHEET 5 SOLVED EXAMPLES IN SYNCHRONOUS MACHINES Example 1 : Calculate the percent voltage regulation for a three-phase wye-connected 2500 kVA 6600-V turboalternator operating at full-load and 0.8 power factor lagging. The per phase synchronous reactance and the armature resistance are 10.4 Ω and 0.071 Ω, respectively? Solution: Clearly, we have XS >> R a, The phasor diagram for the lagging power factor neglecting the effect of R a is shown in Fig. (a). The numerical values are as follows: Vt = 6600 / √3 = 3810 V Ia = ( 2500 * 1000 ) / ( √3 * 6600) = 218.7 A Eo = 3810 + 218.7( 0.8 – j 0.6) j10.4 = 5485∟19.3º Percentage regulation = ( 5485 – 3810 )/ 3810 *100 = 44% Example 2 : Repeat Ex.1 calculations with 0.8 power factor leading as shown in Fig. (b) ? Solution: Eo = 3810 + 218.7( 0.8 + j 0.6) j10.4 = 3048∟36.6º Percentage regulation = ( 3048 – 3810 )/ 3810 *100 = -20%
Fff H.W. : Repeat Ex.1 calculations with unity power factor , and draw the phasor diagram for this case ?
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Example 3 : A 20-KVA, 220 V, 60 Hz, way-connected three phase salient-pole synchronous generated supplies rated load at 0.707 lagging power factor.The phase constants of the machine are Ra = 0.5 Ω and Xd =2 Xq = 4 Ω. Calculate the voltage regulation at the specified load. ? Solution: Vt = 220 / √3 = 127 V Ia = 20000 / (√3 * 220) = 52.5 A φ = cos-1 0.707 = 45º tan δ = (Ia Xq cos φ) / (Vt + Ia Xq sin φ) = 52.5 * 2 * 0.707 / ( 127 * 52.5 * 2 * 0.707) = 0.37 δ = 20.6º Id = 52.5 sin ( 20.6 + 45 ) = 47.5 A Id Xd = 47.5 * 4 = 190 V Eo = Vt cos δ + Id Xd = 127 cos 20.6 + 190 = 308 V Percent regulation = (Eo – Vt ) / Vt *100% = ( 308 – 127 ) / 127 *100% = 142 % Example 4: The stator core of a 4-pole, 3-phase a.c. machine has 36 slots. It carries a short pitch 3-phase winding with coil span equal to 8 slots. Determine the distribution and coil pitch factors? Solution: Number of slots per pole = 36 / 4 = 9 Angular displacement between slots = β = 180º / 9 = 20º Coil span = 20º * 8 =160º 61
Therefore, pitch factor = k p = cos (θ/2) k p = cos 10º = 0.985
,
θ = 180º – 160º = 20º
Now, number of slots per pole per phase, m = 9 /3 = 3 Distribution factor, kd = ( Length of long chord ) / (Sum of lengths of short chords ) = [sin (mβ/2)] / [m sin (β/2)] = sin 30º / (3 sin 10º) = 0.96 Example 5: 3- phase, 50 Hz generator has 120 turns per phase. The flux per pole is 0.07 weber, assume sinusoidally distributed. Find (a) the e.m f. generated per phase. (b) e.m.f. between the line terminals with star connection. Assume full pitch winding and distribution factor equal to 0.96 ? Solution: (a) E.m.f. generated per phase = 4 k f k p k d f T Φp
volts
= 4 * 1.11 * 0.96 * 1 * 0.07 * 50 * 120 =1792 volts (b) E.m.f. between line terminals= √ 3 E.m.f. generated per phase =√ 3 * 1792 = 3100 volts Example 6: A three phase, 16-pole, star-connected alternator, has 192 stator slots with eight conductors per slot and the conductors of each phase are connected in series. The coil span is 150 electrical degrees. Determine the phase and line voltages if the machine runs at 375 r.p m. and the flux per pole is 64 mWb distributed sinusoidally over the pole? Solution: The flux is sinusoidally distributed over the pole, hence from factor, k f = 1.11 Pitch factor , k p = cos [(180º - 150º) / 2 ] = cos 15º = 0.966 Distribution factor, kd = [sin (mβ/2)] / [m sin (β/2)] m = Number of slots per pole per phase = 192 / (16 * 3) = 4 β = 180º / (No. of slots/pole) = 180º / (192/16) = 15º 62
kd = [ sin (4 * 15º / 2)] / [4 * sin (15º/2)] = sin 30º / (4 * sin 7.5º) = 0.958 Now total number of conductors per phase =Total number of slots per phase * number of conductors per slot. = ( 192 / 3 ) * 8 = 512 Number of turns per phase = T = 512 / 2 = 256 Frequency of generated e.m.f. = f = P N / 120 =[ Number of poles * Synchronous Speed] / 120 =[ 16 * 375] / 120 = 50 cycle per second Flux per pole = Φp=0.064 wb Therefore, e.m.f. per phase = 4 k f k p k d f T Φp volts = 4 * 1.11 * 0.966 * 0.958 * 0.064 * 50 * 256 = 3367 volts Hence for star connected alternator, Line voltage = √3 * 3367 = 5830 volts Example 7: A 2200 volt, star-connected, 50 cycle alternator has 12 poles. The stator has 108 slots each with 5 conductors per slot. Calculate the necessary flux per pole to give 2200 volts on no-load. The winding is concentric and the value of kf can be taken as 1.11 ? Solution: Slots per phase = 108 / 3 = 36 Therefore, conductors per phase = 36 * 5 = 180 and number of turns per phase = 180 / 2 = 90 Slots per pole per phase = 36 / 12 =12 Frequency,f =50 cycle per second kf = 1.11 k p = 1.0 for concentric winding
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kd = [sin (mβ/2)] / [m sin (β/2)] Where β = 180º / (No. Of slots/pole) = 180º / (108/12) = 20º m = Number of voltage vectors= Number of slots per pole per phase = 108 / (12*3) = 3 kd = [sin (3*20º/2)] / [ 3 sin (20º/2)] = [sin 30º] / [3 sin 10º] = 0.96 Also e.m.f. per phase = Eph = 2200 /√3 volts The e.m.f. equation for Alternator is Eph = 4 k f k p k d f T Φp volts Φp = Flux per pole = Eph / [4 k f k p k d f T ] =( 2200 /√3) / [ 4*1.11*1*0.96*50*90] = 0.066 wb Example 8: Find the number of armature conductors in series per phase required for the armature of a 3 phase, 50 Hz, 10 pole alternator with 90 slots. The winding is to be star connected to give a line voltage of 11,000 volts. The flux per pole is 0.16 webers? Solution: Number of slots per pole = 90 / 10 = 9 Therefore,
β = 180º / 9 = 20º
Slots per pole per phase, m = 9 / 3 = 3 Distribution factor,
kd = [sin (mβ/2)] / [m sin (β/2)]
kd = [sin (3*20º/2)] / [ 3 sin (20º/2)] = [sin 30º] / [3 sin 10º] = 0.96 Since there is no mention about the type of winding, the full pitched winding is assumed. Hence , k p = 1.0 Also , Eph = 11000 / √3 = 6352 volts. Now , Eph = 4 k f k p k d f T Φp volts T = [Eph] / [4 k f k p k d f Φp] = [6352] / [4 * 1.11 * 0.96 * 1 * 0.16 * 50] = 1863 Number of Armature conductors in serues per phase = 1863 * 2 = 3726
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Example 9 : A six-pole, 3-phase, 60 cycle alternator has 12 slots per pole and 4 conductors per slot. The winding is 5/6 th pitched. There is 0.025 wb flux entering the armature from each north pole and the flux is sinusoidally distributed along the air gap. The armature coils are all connected in series. The winding is starconnected. Determine the open circuit e.m.f of the alternatorper phase? Solution: Here, number of conductors connected in series per phase =12 * 6 * 4 / 3 = 96 Therefore, number of turns per phase , T = 96 / 2 = 48 Flux per pole , Φp = 0.025 wb Frequency,
f = 5 0 c yc l e p e r s e c o n d .
k f = 1.11 k p = cos [180º(1 – 5/6) / 2] = cos 15º = 0.966. Number of slots per pole per phase, m = 12 / 3 = 4 β = 180º / 12 = 15º Distribution factor , kd = [sin (mβ/2)] / [m sin (β/2)] kd = [sin (4*15º/2)] / [ 4 sin (15º/2)] = [sin 30º] / [4 sin 7.5º] = 0.958 Hence induced e.m.f., Eph = 4 k f k p k d f T Φp volts = 4 * 1.11 * 0.966 * 0.958 * 0.025 * 50 * 48 = 295 volts.
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