UNIT - IV

Design of Three Phase Induction Motor

CONSTRUCTIONAL DETAILS OF THREE PHASE INDUCTION MOTOR· A poly phase induction motor consists essentially of two major parts,

the stator and rotor.· The construction of each one is basically a laminated core provided

with slots which house windings.· The stator core internal diameter and the length are main dimensions

of induction motor.· The two different types of rotor of three phase induction motors are

squirrel cage rotor and slip ring rotor.· In squirrel cage construction the rotor consists of core, copper or

aluminium bars and end-rings. The end-rings are provided to short circuit the rotor bars at both the ends.

· The rotor core is made of laminated sheet steel with thickness 0.5 mm.· The aluminium bars and end-rings are casted directly over the rotor core.

· When copper is employed, the rotor bars are inserted on the slots fromend of the rotor and the end rings are joined to them by bracing.

· The wound rotor consists of a core, winding slip rings and brushes. Therotor core is made of laminations and it carries three phase winding.

· One end of each phase are connected to from a star point. The other end of each phase are connected to three slip rings.

· The slip-rings are mounted on the rotor shaft and they are insulated from the rotor and from each other.

· Carbon brushes are mounted over the slip-rings which offers the facility of connecting the external resistance to rotor winding.

OUTPUT EQUATION OF THREE PHASE INDUCTION MOTOR· Consider an 'm' phase machine having one circuit per phase.

· kVA rating of machine Q = number of phases x output voltage per phase x cUrrent per phase x 10- 3 = mEphlph x 10-3

Terminal voltage of each phase may be taken equal to the induced emf per phase.

= 1.11Kw (PO) (2m1phT ph) n, x 1 0 -s

· Now current in each conductor I= 'ph

· Total number of armature conductor z = number of phases x 2 x turns phase.

= 2mT ph Tph

· Total electric loading = izz = 2rniph Q =

1.111c (P) (Izz) 5 x 10- 3

= 1.11K (total magnetic loading) xW

(total electric loading) (synchronous speed) x 3

= TrDLBav

I,Z = rOcic

Q = 1.11K. (r:DLB„,) (rOac) n, x 10- 3

= (11BavacKw x 10- 3) D2Ln,

Design of Rotor Bars and Slots

1. Rotor bar current Current in each barFor a three phase machine tn2 = 3.

61sT,lb = Kw, cos (1)

'Dr

--- 0.85 xSr

The above relation may be interpreted as that the rotor mmf is about 85 per cent of stator mmf.

2. Area of rotor barsThe performance of an induction motor is greatly influenced by the

resistance of rotor. A motor designed with high rotor resistance has the advantage that it has a high starting torque. However, a rotor with a highresistance has the disadvantage that its 12R is greater and therefore its efficiency is lower under running conditions.

The value of rotor resistance depends upon the current density used for rotor conductors, the higher the current density, the lower is the conductor area and greater the resistance. Therefore, a rotor designed with a high value of current density results in high starting torque and a lower efficiency for the machine.

The rotor resistance is the sum of the resistance of the bars and the end rings. The cross-section of the bars and the end rings must be so selected that a proper value of rotor resistance is obtained i.e. a value of rotor resistance which meets both the requirements of starting torque as will as the efficiency.

It is desirable to have a compromise between a high resistance rotor which gives a good starting torque and a low resistance rotor which gives a high value of efficiency under running conditions.

Current density in the rotor bars may be taken between 4 to 7 Aimm2.

61,T,

Area of each bar ab

Shape and size of rotors slotsThe rotor slots for squirrel cage rotor may either be closed or semi-enclosed

types.Closed slots are preferred for small size machines because the reluctance of the air gap is not large owing to absence of slot opening. This gives a reduced value of magnetising current.

· As the surface of the rotor is smooth, the operation of the machine isquieter. The biggest advantage is that the leakage reactance with closed slots is large and therefore the current at starting can be limited.

· This is very useful in the case of machines which are started with

direct on-line starters. But the disadvantages is that the increased value of reactance results in reduction of overload capacity. A semi-enclosed slot gives a better overload capacity.

· The rectangular shaped bars and slots are generally preferred to circular bars and slots as the higher leakage reactance of the lower part of the rectangular bars, during starting, forces most of the current through the top of the bar.

· This increases the rotor resistance at starting and improves the starting torque. Deep slots, however, give an increased leakage reactance and a high flux density at the root of the teeth.

Rotor slot insulationNo insulation is used between bars and rotor core. A clearance of 0.15 to

0.4 mm can be left between rotor bars and the core depending upon whether slots are skewed or not. Higher clearances have to be left for the skewed slots.

Design of End Rings

1. End ring currentThe distribution of current in the bars and end rings of a squirrel cage

motor is complicated.The stator winding is a 3 phase distributed winding and thus produces a

revolving field. This field may be considered as sinusoidally distributed in space as the harmonics in most cases are small and produce only secondary effects.

This revolving field produces emfs of fundamental in the bars.This is dear from the phasor diagram for the nthe harmonic. The nth

harmonic emf is reduced to zero since its phase spread is 2n.

2. Area of end ringsThe value of current density chosen for the end rings should be such that

the desired value of rotor resistance is obtained.The ventilation is generally better for end rings and therefore a slightly

higher value of current density than that obtaining in rotor bars can be taken.

4.10.4 Full Load Slip

The value of slip at full load is determined by the rotor resistance. A reasonable value of rotor resistance to be incorporated in the rotor can be obtained by the knowledge of reasonable values of full load slip. The value of slip, a, is derived from the following relationship

rotor copper loss rotor output 1 - s

where s is the per unit slip.

Table gives the usual values of per cent slip at full load for various ratings.

Total copper loss = 457 + 264 = 721 W .

Synchronous speed - 120 x 50 - 1.00 r.p.m.6

Rotor speed = (1 - 0.58) x 1000 954.2 r.p.m

This is nearly equal to the specified speed.

DESIGN OF WOUND ROTOR

Number of rotor slots

· The rotor winding for wound rotor motors are 3 phase windings, and the number of rotor slots should be such that a balanced winding is obtained. Generally windings with an integral number of slots per pole

per phase are used for the rotor.

· Fractional slot windings may also be used. It is preferable to use a number of slots which are a multiple of phases and pair of poles in the case of fractional slot windings.

Number of rotor turns

· The rotor represents the secondary of a transformer and the voltage between slip rings is maximum when the rotor is at rest.

· Therefore, to keep the rotor voltage to an acceptable level the ratio of effective stator to rotor turns must be properly adjusted.

· The choice of this turns ratio is arbitrary and is controllable by the designer.

· The rotor voltage on open circuit between slip rings should not exceed 100 V for small machines where hand operated starters and switch gear are employed.

· The voltage is limited to a small value in order to protect persons working the motor if the brush gear is not perfectly protected. More-over with small rotor voltages, it is easier to insulate the rotor windings.

· In case of high voltage machines and also in the case of large machines, the rotor voltages should be high because in such cases if the rotor voltage is kept small, the rotor current becomes large involving use of large conductor sections.

· Large rotor currents complicate the design of slip rings, brush gear and starter contacts. For large size machines, voltages upto 1000 - 2000 V can be used and there seems to be no objection to rather higher voltages for very large motors where it is worth while to completely protect the brush gear.

· Let Ts, Tt = number of turns per phase for stator and rotor respectively,

K u,s, Ki„= winding factor for stator and rotor respectively E s =

stator voltage per phase

E r = rotor voltage per phase at standstill

E r I c r T , E s Ku i s T ,

In case of small machines:E r should not exceed 500 V and 500/= 290 V for delta star connected

machines respectively.

Area of rotor conductors

The full load rotor mmf is taken as 85 per cent of stator mmf.

The area of the rotor conductors is found out by assuming a suitable value for current density. In order to avoid excessive rotor copper loss, the current density in the rotor is chosen almost equal to that in the stator.

Round conductors are used for small motors. But for large motors, it becomes necessary to use bar conductors.4.11.4 Rotor windings

· For small induction motors of slip ring type, it is a normal practice to use much windings for rotor housed in semi-enclosed slots.

· The coils are roughly formed outside the machine and dropped into the slots opening one by one.

· It is usual to use several wires in parallel per turn to keep the conductor small to go through the narrow slot opening. The rotor is invariably star connected and three leads are brought through the shaft to the slip rings.

· For larger motors, a double layer bar type wave winding is used. This winding has generally two bars per slot. The bars are pushed through partially closed slots and are bent to shape at the other end. In motors of outputs of about 750 KW and over, we have to use 5 bars sometimes. The use of 4 bars slot is made to reduce the current handled by each slip ring. The winding with more than 2 bars per slit is called a barrel winding and is usually wave wound.

Rotor Teeth

The width of rotor slot should be such that the flux density in the rotor

Now

teeth does not exceed about 1.7 Wb/m2. The maximum flux density for rotor teeth occurs at their root as their section is minimum there.

4)/7/ Minimum width of rotor teeth Wir,min) =

1.7 x (Srl p) x Lt

It should be checked that the value of minimum tooth width actually provided in the machine is higher than the value given by equation. Minimum width of tooth actually provided.

It (Dr — 2d„)Wtr= rotor slot pitch at the root — rotor slot width =___________ War Sr

whered„= depth of rotor slot and W„ = width of rotor slot.

Rotor Core

The flux density in the rotor is generally equal to stator core density.

Depth of rotor core dc,. = 2 xB xL crt

whereBer = flux density in the rotor core.

Inside diameter of rotor lamination Di= Dr — 2 (d„ + dcr)

• The flux density in rotor teeth and core can be taken slightly higher than those in the stator and core. This is because the iron losses in the rotor are very small owing to small value of frequency of rotor currents.

DISPERSION COEFFICIENT

There are two factors which influence the power factor of an inductionmotor.

(ii) Ideal short circuit current:· Ideal short circuit current is defined as the current drawn by the motor

at standstill if its resistance is neglected. If the ideal short circuit current of a machine is large it indicates that its leakage reactance is small.

· A small value of leakage reactance means that the power factor of the machine is good. Thus a large value of short circuit current (a small value of leakage reactance) indicates a good power factor.

· Here was introduce a co-efficient called 'Dispersion co-efficient'. It is defined as the ratio of magnetising current to ideal short circuit current. Thus dispersion co-efficient.

· If Im and Xs are small, the value of power factor will be good. Thus a small value of dispersion co-efficient indicates a good power factor while a large value of dispersion co-efficient means a poor power factor.

· Let us evaluate the dispersion co-efficient.·Consider a machine with a given core length and diameter. We find that

as we increase the number of poles, the dispersion co-efficient increases and an increase in the value of dispersion co-efficient means a poor power factor.

·Thus induction motor with large number of poles (slow speed machines) have inherently a poor power factor.

·Let us examine the same equation in another form. As the number of poles increases, the Dol e pitch 't decreases and the number of slots per pole per phase qs alt.o decreases and hence there is a large increase in

the value of dispen on co-efficient with increase in number of poles. Therefore, with increase in number of poles it becomes difficult to get a good power factor.

· It is clear that the dispersion co-efficient can be decreased and hence power factor increased by using small gap length. But small air gaps give trouble owing to increased zigzag leakage. Therefore, length of air gap should not be unduly decreased.

REDUCTION OF HARMONIC TORQUES

Following are some of the methods used for reduction/elimination of harmonic torques.

(i) Chording: The simplest way to eliminate the harmonic induction torques is to weaken the stator winding mmf harmonics. In order to achieve this, chorded windings with integral number of slots per pole phase are used.

(ii) Integral slot windings: Windings with fractional number of slots per pole per phase create asymmetrical mmf distribution around the air gap and favour the creation of noise in the motor. Therefore, fractional slot windings are not used for induction motor stators and only integral slot windings are used.

(iii) Skewing: The motor noise and vibrations, clogging and synchronous cusps can be reduced or even entirely eliminated by skewing either the stator or the rotor.

• The practice generally followed in India is to skew the rotor. If either stator or rotor slots are skewed, the variations in flux density, magnetic pull, and torque due to the slot openings will be displaced in time phase along the core length, resulting in more uniform torque, loss noise, and better voltage wave form.

In order to eliminate the effect of any harmonic, the rotor bars should be skewed thorough an angle so that the bars lie under alternate harmonic poles of the same polarity or in other words, bars must be skewed through two pitches,

SUMMARY

1. Output coefficient, Co 11K„,sBrivac x 10- 3

2. Input KVA, Q = C0D2Lns (or) Q - HP / 0.746

3. For best power factor, t = J0.18/.

4. Maximum flux, Om B„on.DL

5. Stator phase voltage, Es= 4.44 Ics f0mTs

Q = C0D2Lns

C o = 11B o va cKw x1 0 - 3

equation (5) is known as the output equation of an a.c machine:Co is called the output co-efficient.

MAIN DIMENSIONS

· The product D2L is obtained from input kVA. The main dimensions of induction motor are the diameter of stator bore, D and length of stator core.

· The separation of D and L from the product D2L depends on the ratioTrD

— I where T = is the pole pitch.

· The ratio of core length to pole pitch (ratio L / t) for various design feature is.

Minimum cost — = 1.5 to 2

Good power factor, — = 1.0 to 1.25

Good efficiency - 1.5T

Good overall design— = 1

X cos 0

...

· Generally —L

lies between 0.6 to 2. It can be shown that, for best power

factor for the pole pitch is given by equationT = 1/0.18L

· For small motors, high values of —L

result in small diameter which may

not be able to accommodate even a small number of slots.

· In such cases the above values are not practicable and so lower values down to 0.6 may be taken.

Estimate the stator core dimensions, number of stator slots and number of stator conductors per slot for a 100 KW, 3300 V, 50 Hz

12 pole, star connected slip ring induction motor. Bay = 0.4 Whim2,

cu. = 25000 amp.condlm, n = 0.9, pf= 0.9. Choose main dimensions to give best power factor. The slot loading should not exceed 500 amp. conductors.

Given Data

100 KW

50 Hz= 0.9

star connected

Solution:

3300 V

12 polep1= 0.93 phase,

Bay = 0.4 WI/m2

ac = 25000 amp.cond/m.slot loading 500 amp.cond.

Let, K = 0.96

Output coefficient, Co = 11BavacKws x 10- 3

= 11 x 0.4 x 25000 x 0.96 x 10- 3

= 105.6

it2

put D2 = 2.6262L in the equation for D2L

...D2L = 2.6262L x L = 0.1403

2.6262 = 0.2311 m = 0.23 m

D2, 2.6262L, ...D = .V2.6262 x 0.23 = 0.772 = 0.78 m. L = 0.23 and

D = 0.78 in

1905.256 s 4.44 x 50 x 0.0187x 0.96 - 478 The stator

slot pitch should lie between 15 to 25 mm 7-cD

Stator slots, Ss =78

n W he n y s s = 15 m m, S s - x 0 . - 16 3

15 x 10-3

it x 0.78 When yss = 25 mm, Ss - 3 - 9825 x 10-

Ss should lie between 98 to 163.

The stator slots be multiple of q, where q is slot/pole/phase.When, q = 2, Ss= 3 x 12 x 2 = 72

When, q = 3, Ss= 3 x 12 x 3 = 108

When, q = 4, Ss = 3 x 12 x 4 = 144

When, q = 5, Ss = 3 x 12 x 5 = 180

The Ss values of 108 and 144 lie in the range of 98 to 163.

Ss can be either 108 or 144.

Y S S

Determine the approximate diameter length of stator core, the number of stator slots and the number of conductors for a 11 KW, 400 V, 30 , 4-pole, 1425 rpm, delta connected induct ion motor.B ay = 0.45 Wb1m2, ac = 23000 amp.condlni, full load efficiency =0.85,

pf= 0.88, LIT =1. The stator employs a double layer winding. Given

Data

11 KW 1425 rpm

Bay = 0.45 Whim2

4 - pole ac = 23000 amp.cond/m.400 V = 0.85delta connected pf= 0.88double layer winding Lit = 1

Solution

Let, Kws= 0.955

Cy = 11K0sBayac x 10- 3

= 11 x 0.955 x 0.45 x 23000.x 10-3 = 108.7267

KVA input, Q = C0D2Ln,

2 Q 14.7 ...D L = Cons

= 108.7267 x 25 = 0.0054 m3

A 15 kW, 440 V, 4 pole, 50 Hz, 3 phase induction motor is built with a stator bore 0.25 m and a core length of 0.16. The specific electric loading is 23000 ampere conductor per metre. Using Mt data of this machine, determine the core dimensions, number of stator slots and number of stator conductors for a 11 kW, 460 V, 6 pole, 50 Hz motor. Assume a full load efficiency of 84 per cent and power factor of 0.82 for each machine. The winding factor is 0.955.

Solution:15 kW Motor:

15 kVA input = 0.84 x 0.82 - 21.8. Synchronous speed ns = 2 x 50/4 = 25 r.p.s

21.8 Now, Co - 2

Q - - 87.2.D Lns (0.25)2 x 0.16 x 25

Co = 11 Ku,Bovac x 10- 3 = 11 x 0.955 x Bay x 23000 x 1(1 3 =

241.6B

Average flux density in the air gap Bay = 87.2/241.6 = 0.36 Wb/m2.

Pole pitch = x 0.25/4 = 0.196 m

Ratio Lit = 01.6/0.196 -7 0.815

11 kW Motor:

We have to use the same data for the 11 kW machine as is calculated above for 15 kW machine.

Bay = 0.36 Wb/m2 ; ac = 23000 Aim;

L/T = 0.815 and C0= 87.2

Using 34 conductors per slot, total conductors = 1836.Stator turns per phase Ts = 1836/6 = 306.

A 3 phase, 440 V, 750 r.p.m. 50 Hz star connected induction motor has a stator with an internal diameter of 0.25 m and an axial length of 0.15 m. It has 48 slots with 24 conductors per slot. Calculate the air gap flux per pole. The area of each stator conductor is to be3 mm2. Calculate the width and the depth of the slot to accommodate

the stator conductors. The maximum flux density in the teeth is to be 1.7 VVb/m2. Conductor insulation is 0.08 mm thick and slot insulation is 0.8 mm thick. Make other suitable assumptions.: A 11 Kw, 3 phase, 6 pole, 50 Hz, 220 V, star connected induction motor has 54 stator slots, each containing 9 conductors. Calculate the values of bar and end ring currents. The number of rotor bars is 64. The machine has an efficiency of 0.86 and a power factor of 0.86. The rotor mmf may be assumed as 85 per cent of stator mmf Also f ind the bar and the end ring sect ions if the current densi ty is 5 A/mm2

Stator current per phase

11 x 1000 Is NIT x 220 x 0.86 x 0.5 40 A.

Number of stator conductors = 54 x 9 = 486.

Stator turns/phase Ts = 486/6.= 81

Stator mmf = sT = 3 x 40 x 81 = 9720 A

Rotor mmf = 0.85 x 9720 = 8250 A.

But rotor mmf = S,I5I2 = 32/b

32Is= 8250 or current in rotor bars Ib = 258 A

E n d r i n g c u r r e n t I s - Srlb - 64 x 258 x 6 - 8 8 3 A icp

S r l b 64 x 258 End ring current Is = - x 6 - 883 A

rcp

Area of each bar ab = 258/5 = 51.6 mm2.and of each end ring as= 883i 2

5 = 176.6 mm