induction machines fundamentals

Induction Machines 1

Induction Motor Stator-2 pole

concentrated winding


Induction motor stator- Distributed

4 pole lap winding

Ref: Electrical Machine Design by A.K. Sawhney

& A. Chakrabarti

Induction Machines 3 1 Cycle

Amp

time t0

t1 t2 t3 t4

t01 t12 Currents in different phases of AC Machine


Axis of phase a

a’ a’

-90 -40 10 60 110 160 210 260 -1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Fa

Space angle (Ө) in degrees

t0

t01

t12

t2

a

MMF Due to ‘a’ phase current


a Fc

-93 10 113 216 -1.5

-1

-0.5

0

0.5

1

1.5

a’

c’ b’

b c

a

a’

c’ b’

b c

a

a’

c’ b’

b c

a

a’

c’ b’

b c

Fb

Fa F

Fb Fc

F

Fa

F

Fb

Fc Fc Fb

F

Space angle () in degrees

F Fa Fc

Fb

t = t0= t4

t = t1 t = t2 t = t3

t = t0= t4

RMF(Rotating Magnetic Field)


Mathematical expression for RMF

;sec/2

);(#

2

3

;)2

cos(),(

radinvoltageappliedoffrequencyf

bookthewithconsistentbetospoleofP

INF

tP

FtF

or

msm

m

3

2

2cos

3

2cos

3

2

2cos

3

2cos)

2cos()cos(),(

PtIN

PtIN

PtINtF

ms

msms

Speed of RMF (synchronous speed)

The RMF will become constant if

02

tP

That can happen if the rate of change of

.sec/.2

1 radPdt

d

is

This is called the synchronous speed.

rpmP

f

P

f

Pdt

dn

120

2*

60*2*2

2*

60*2

2

601

In rpm,

7 Induction Machines


Mathematical expression and speed of

RMF with space and time harmonics

nP

h

hn

thnP

FtF

nh

mnh

2by given is speed RMF thegeneralIn

6. slidein described procedure thefollowingsign out the findcan Onenegative are they etc. 5,11,17For

positive arethey etc.1,7,13,For.harmonicsondependentis RMF theofrotationofspeedThe

...,5,7,6,11.harmonics1timeoforder....1,5,7,6,11 harmonicsspaceoforder

;)2

cos(),(


RMF- How it looks for different poles


# of poles –versus synchronous speed in

rpm for a machine excited with 60 Hz

n1


Induction Motor

•Most popular motor today in the low and medium horsepower range

•Very robust in construction

•Speed easily controllable using V/f or Field Oriented Controllers

•Have replaced DC Motors in areas where traditional DC Motors

cannot be used such as mining or explosive environments

•Of two types depending on motor construction: Squirrel Cage

or Slip Ring

•Only Disadvantage: Most of them run with a lagging power factor


Squirrel Cage Rotor


Slip Ring Rotor

•The rotor contains windings similar to stator.

•The connections from rotor are brought out using slip rings that

are rotating with the rotor and carbon brushes that are static.


Torque Production in an Induction Motor

•In a conventional DC machine field is stationary and the

current carrying conductors rotate.

•We can obtain similar results if we make field structure

rotating and current carrying conductor stationary.

•In an induction motor the conventional 3-phase winding

sets up the rotating magnetic field(RMF) and the rotor

carries the current carrying conductors.

•An EMF and hence current is induced in the rotor due to

the speed difference between the RMF and the rotor,

similar to that in a DC motor.

•This current produces a torque such that the speed

difference between the RMF and rotor is reduced.


Slip in Induction Motor

•However, this speed difference cannot become zero because that

would stop generation of the torque producing current itself.

•The parameter slip ‘s’ is a measure of this relative speed difference

1

1

1

1

mm

n

nns

where n1,1 are the speeds of the RMF in RPM and rad/sec

respectively

nm,m are the speeds of the motor in RPM and rad/sec respectively

•The angular slip frequency and the slip frequency at which voltage

is induced in the rotor is given by

sff,s slipslip


Induction Motor Example 1

A 100 hp, 8 pole, 60 Hz, 3 phase induction motor runs at 891 rpm

under full load. Determine the i) synchronous speed in rpm, ii) slip, iii)

slip frequency at full load. Also, iv) estimate speed if load torque becomes

half of full load torque, given the fact that torque is proportional to

slip in the region between breakdown torque or zero torque.

Solution:

i) Synchronous speed= 𝑛1 = 120 𝑓

𝑃=

120 . 60

8 = 900

rpm.

ii) Slip = 𝑠 =𝑛1−𝑛𝑚

𝑛1=

900−891

900=

9

900= 0.01

iii) Slip frequency = 𝑠. 𝑓 = 0.01 .60 = 0.6 Hz.

Example I: Solution


T

S, nm

Tstart

Tmax

}Toruqe is proportional to

slip between maximum

(break down) torque and zero torque

Tfull load

iv) Following the figure above

𝑇𝐹𝐿 = 𝐾𝑠𝐹𝐿 and 𝑇𝐹𝐿/2 = 𝐾𝑠𝐹𝐿/2 where ‘FL’ implies full load and ‘K’ is the constant of proportionality.

∴ 𝑠𝐹𝐿/2 =𝑇𝐹𝐿/2

𝑇𝐹𝐿𝑠𝐹𝐿 =

𝑠𝐹𝐿2

= 0.005

∴ speed at half load torque= 1 − 𝑠𝐹𝐿2

𝑛𝑠 = 1 −𝑠𝐹𝐿

2 𝑛𝑠 = 1 − 0.005 900 = 895.5 rpm.


Rotor Equivalent Circuit

R2

sE2

jsX2

I2

E2

jX2

R2/sI2

The equation for the circuit on the left is

sE2=I2(R2+jsX2)

Dividing both sides by s we can obtain

E2=I2(R2/s+jX2)


Induction Motor Equivalent Circuit

Note: Prot includes the iron losses along with the rotational losses. The sum of the

iron and rotational losses and hence Prot is constant over the normal operating range at rated

frequency of operation. This is because as the speed increases the rotational losses increase, but

the rotor iron losses decrease as the slip frequency is lower. The converse is true when speed

decreases.

V1

I1

R1

jX1 jX2'

R2’

((1-s)/s)R2’jXm

I2’

Im

Pin=3|V1||I1|cosf Pag

P1=3|I1|2R1

Stator copper

loss

Pdev =

3|I2'|2{(1-s)/s}R2

'Pout

P2=3|I2'|2R2

'

Rotor copper

loss

Prot

Rotational loss

Power Relationships

ssPPP

ssPP

PPP

devag

dev

devag

1::1::

1::

2

2

2


Induction Motor Example 2 A, 3 phase, 15 hp, 460V, 4 pole, 60 Hz,1728 rpm IM delivers full

output power to a load connected to its shaft. Windage and friction

losses amount to 750W. Determine i) gross output(mechanical)

power, ii) Air-gap power and iii) Rotor copper loss.

Solution:

(i) Full load shaft power, 𝑃𝑜𝑢𝑡 = 15 hp = 15 * 746 =11190 W.

Gross ouput power = 𝑃𝑑𝑒𝑣 = Shaft power + windage and friction loss = 11190+ 750 = 11940 W.

(ii) Synchronous speed= 𝑛1 = 120 𝑓

𝑃=

120∗ 60

4 = 1800 rpm.

Slip = 𝑠 =𝑛1−𝑛𝑚

𝑛1=

1800−1728

1800=

72

1800= 0.04

𝑃𝑎𝑔 =𝑃𝑑𝑒𝑣1 − 𝑠

=11940

1 − 0.04= 12437.5 W

(iii) Rotor copper loss = 𝑃2 = 𝑠𝑃𝑎𝑔 = 0.04 ∗ 12437.5 = 497.5 W.

OR

𝑃2 = 𝑃𝑎𝑔 − 𝑃𝑑𝑒𝑣 = 12437.5 −11940 W =497.5 W.


Induction Motor Example 3


A 30 hp, 4 pole, 440V,60 Hz delta connected induction motor

has a rotational loss of 900W at1764 rpm. Find the pf, line

current, output power, air-gap power, copper loss, output torque,

efficiency at 1764 rpm. R1 =1.2 ohms,

R2'=0.6 ohms X1=2 ohms, X2'=0.8 ohms , Xm=50 ohms.

Solutions:

V1 =440<00V

I1

R1=1.2 Ω

jX1=j2 Ω jX2’=j0.8 Ω

R2’=0.6 Ω

((1-s)/s)R2’ =29.4ΩjXm= j50 Ω

I2’

Im

Induction Motor Example 3 (contd..)


Synchronous speed= 𝑛𝑠 = 120 𝑓

𝑃=

120 . 60

4 = 1800 rpm.

Slip = 𝑠 =𝑛𝑠−𝑛𝑚

𝑛𝑠=

1800−1764

1800=

36

1800= 0.02.

∴1−𝑠

𝑠𝑅2′ =

1−0.02

0.02 . 0.6 = 29.4 Ω.

From the above figure

𝑍1 = 𝑅1 + 𝑗𝑋1 + 𝑗𝑋𝑚 . 1 − 𝑠

𝑠𝑅2′ + 𝑅2

′ + 𝑗𝑋2′ / 𝑗𝑋𝑚 +

1 − 𝑠

𝑠𝑅2′ + 𝑅2

′ + 𝑗𝑋2′

=22.75+j 15.51 Ω =27.53∠34.290 Ω

∴ 𝐼1 = 𝑉1

𝑍1= 440∠00

27.53∠34.290 = 15.98∠− 34.290A

∴ 𝑝𝑓 = 𝑐𝑜𝑠𝜙 = cos 34.29 = 0.83 lag.

Line current = 3 𝐼1 = 27.68 A.

Induction Motor Example 3 (contd..)


Input power = 𝑃𝑖𝑛 = 3 𝑉1 𝐼1 𝑐𝑜𝑠𝜙 = 3. 440. 15.98 . 0.83 = 17.43 kW.

𝐼2′ =

𝑉1−𝐼1(𝑅1+𝑗𝑋1)

𝑅2′

𝑠+𝑗𝑋2

′=

440∠00−15.98∠−34.290 .(1.2+𝑗2)

(30+𝑗0.8)= 13.55∠ − 3.730 A

Output power =𝑃𝑜𝑢𝑡 = 𝑃𝑑𝑒𝑣 − 𝑃2 = 3 𝐼2′ 2 1−𝑠

𝑠𝑅2′ − 900 = 16193.74 − 900 = 15293.74 W

Copper loss = 𝑃1 + 𝑃2 = 3 𝐼12𝑅1 + 3 𝐼2

′ 2𝑅2′ = 3 ∗ 15.982 ∗ 1.2 + 3 ∗ 13.552 ∗ 0.6 = 919.30 +

330.48 = 1249.78 W

Output Torque = 𝑇𝑜𝑢𝑡 =𝑃𝑜𝑢𝑡

𝜔𝑚=

15293 .74

2𝜋∗ 1764

60

= 82.79 𝑁 −𝑚.

% Efficiency =𝜂 =𝑃𝑜𝑢𝑡

𝑃𝑖𝑛. 100 =

15293 .74

17430. 100 = 87.74 %

𝑃𝑎𝑔 = 𝑃𝑖𝑛 − 𝑃2=17430-3 ∗ 15.982 ∗ 1.2 =17430-919.30=16510.7 W.

Three Phase Table


Relationships of balanced three phase circuits

Star Delta

Voltage (V) 𝑉𝐿−𝐿 = 3𝑉𝑝ℎ 𝑉𝐿−𝐿 = 𝑉𝑝ℎ

Current (A) 𝐼𝐿 = 𝐼𝑝ℎ 𝐼𝐿 = 3𝐼𝑝ℎ

Active Power (P) (W) 3𝑉𝐿−𝐿𝐼𝐿 cos𝜑

=3𝑉𝑝ℎ 𝐼𝑝ℎ cos𝜑 3𝑉𝐿−𝐿𝐼𝐿 cos𝜑

=3𝑉𝑝ℎ 𝐼𝑝ℎ cos𝜑

Reactive Power (Q)

(VAR) 3𝑉𝐿−𝐿𝐼𝐿 sin𝜑

=3𝑉𝑝ℎ 𝐼𝑝ℎ sin𝜑

3𝑉𝐿−𝐿𝐼𝐿 sin𝜑

=3𝑉𝑝ℎ 𝐼𝑝ℎ sin𝜑

Apparent Power(VA) 3𝑉𝐿−𝐿𝐼𝐿=3𝑉𝑝ℎ𝐼𝑝ℎ 3𝑉𝐿−𝐿𝐼𝐿=3𝑉𝑝ℎ𝐼𝑝ℎ

Note: 𝑉𝐿−𝐿= Line-Line Voltage, 𝐼𝐿= Line Current, 𝑉𝑝ℎ= Phase

voltage, 𝐼𝑝ℎ= Phase Current, cos𝜑 = Power Factor, 𝜑 = Power Factor

Angle.