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POLYPHASE INDUCTION MACHINES Both DC and Synchronous machines seen were of the type “double fed”, that is one has to provide the excitation as well as the armature. The induction machines receive their excitation by INDUCTION, hence we need only one source of power. They are called “singly-fed” machines. There are no commutators, slip rings or brushes. Hence this is a most rugged and maintenance free machine. Because there is no losses in brush contacts or mechanical friction, it is of a high efficiency. Principle of operation: The stator is similar to the stator of a synchronous machine. It is fed with a 3 phase alternating current and provides a rotating flux. This flux rotates at “synchronous speed” Ns =

120 f (in rpm) where (f) if the source frequency and (p) the number of poles on the p

machine. or: ω s =

4πf p

rad / s

The rotor has either a 3 phase short circuited winding, or any conducting elements. In a first study we will coonsider a 3 phase winding on the rotor (wound rotor machine). The revolving field induces a flux in the rotor windings. Since the windings are short circuited, a current is flowing and that current produces a flux of its own. According to Lenz’s law, this flux will oppose to the flux which created it, and hence a torque is developed on the shaft. If this torque is higher than the load torque, the rotor will start to rotate. Under no load, the rotor current will have a very low frequency. This also called an “ASYNCHRONOUS MACHINE”. At operating speed, the rotor currents will have the frequency of the RELATIVE MOTION between rotor and stator rotating speed. Hence the rotor frequency is: N r = N s − N shaft The relative speed ids called the SLIP

s=

ω r N s − N shaft ω s − ω shaft = = ωs Ns ωs

When the rotor is stationary (start), the slip is 1 and we have the equivallent of a short circuited transformer!... (primary is the stator, secondary is the rotor). Equivalent Circuit: The induction machine can be looked upon as transformer with a rotating secondary winding. The only difference would be that the secondary induced currents have a frequency dependant upon the slip. From the equivalent circuit of a 3 phase balnced transformer (per phase) with a short circuited


Induction Machine pg 2

secondary winding, one can derive:

jX 1

R1 V1

Rr

Ir

I1

Rc

jX m E1

Ic

Im

jX r

Er

R1 , X 1 are the per phase stator resistance and leakage reactance Rc , X m are the per phase equivalent core loss resistance and magnetizing reactance

Rr , X r are the per phase rotor winding resistance and leakage reactance E1 is the per phase induced emf in the stator winding. we have already established for a tranformer: E1 = 4.44 fN1k w1 Φ m The induced emf in the secondary (locked rotor condition) is the same as a transformer: Eb = 4.44 fN 2 k w2 Φ m

(b for ‘blocked’)

If the shaft rotates at slip (s) Er = sEb

Hence one can derive from the circuit:

per phase induced emf in the rotor winding Ir =

Er sEb Eb = = Rr + jX r Rr + jsX b (Rr / s + jX b )

this shows a rotor “effective resistance”

Rr and we can simplify the per phase s

equivalent circuit as:

jX 1

R1 V1

Ir

I1

R r/s

Rc

jX m E 1

Ic

Im

Eb

jX b


Induction Machine pg 3

Now we can refer the secondary to the primary as we did with a transformer R2 = a 2 Rr and X 2 = a 2 X b leading to the equivalent circuit refered to the stator side:

jX 1

R1 V1

I2

R2/ s

I1

Rc

jX m

Ic

Im

jX 2

POWER RELATIONS The defintion of power carrid by a wave with voltage and current is:

p (t ) = v (t ).i (t )

If we have 2 sinusoidal voltage and current at same frequency, the instantaneous power carried by the wave is:

p (t ) = V m sin (ω t ) × I m sin (ω t − θ )

where θ is the angle between the current and the voltage. The power in that circuit will be the average power or: p (t ) =

V m I m cos θ V m I m − cos (2ω t − θ ) 2 2

If we integrate over one cycle of the waveform: P = the second term is null, and the 1st term is but if we consider for a sine wave

1 T

T

0

p (t )dt

Vm I m cos θ 2

Vm = 2Vrms

and

I m = 2 I rms

the power in a circuit is: P = VI cosθ

cosθ is called the “power factor” In a balanced 3 phse circuit the total power is Pt = 3 Pphase or

Pt = 3Vl − n I cosθ = 3Vl −l I cosθ


Induction Machine pg 4

R2 (1 − s ) s s This way we have split it into a regular secondary resistance (R2) in series with a resistance

We can manipulate the apparent resistance R2

by replacing it with R2 +

R2 (1 − s ) s the equivalent circuit is now:

which depends upon the slip

jX 1

R1 V1

I2

R2

jX 2

I1

Rc

jX m

Ic

Im

R2 (1-s) s

and this looks much like a transformer equivalent circuit with winding resistance R2 and a load resistance in the secondary !!!! Hence, the power DISSIPATED in that load resistance should be the MECHANICAL POWER transformed by the circuit. The stator copper losses in the machine are:

Pstator copper = 3I12 R1

the core loss is:

Pcore = 3 I c2 Rc

and therefore the net power crossing the air gap is: Pag = Pin − PR1 − Pcore 3I 22 R2 this air gap power must be delivered to the hyothetical resistance R2/s, that is Pag = s But the electrical loss in the secondary circuit is:

P2− Joule = 3I 22 R2 = sPag

hence the power developed by the motor should be: simplifying gives:

Pd = Pag − Protorlosse s

3I 22 (1 − s )R2 = (1 − s )Pag s The electromagnetic torque developed by the motor is now: Pd =

Td =

P Pd R = ag = 3I 22 2 sω s ωm ωs


Induction Machine pg 5

Speed - Torque Characteristics One of the most important characteristics of the motor is it’s speed/torque characteristic. From the equivalent circuit we can derive E1

I2 =

but because R2 << X 2 the starting torque developed by the motor

R22 + X 22

Tds = 3I 22

becomes

R2 ωs

when the slip falls below a cetain value, R2 / s >> X 2 and the rotor current approximates to I2 ≈

sE1 R2

this is the “breakdown slip”

The typical characteristic looks like below;

T Torque

Breakdown torque

No-load slip

starting torque

slip 0

0.5

1

Typical Torque Speed Characteristic of a 3 phase Induction Motor

Power Developed If one neglects the effect of the magnetizing reactance and core loss resistance, the current in the equivalent circuit is: I2 =

V1 (R1 + R2 ) + j ( X 1 + X 2 ) + R2 (1 − s ) / s

one usually lumps resistances and reactances as: hence:

I2 =

Re = R1 + R2

and

X e = X1 + X 2

V1 Re + jX e + R2 (1 − s ) / s

The power developed by the motor being Pd =

Pd =

3I 22 (1 − s )R2 s

3V12 R2 (1 − s ) / s 2 Re2 + X e2 + [R2 (1 − s ) / s ] + 2 Re R2 (1 − s ) / s

we can substitute and obtain


Induction Machine pg 6

for a given source voltage, the power developed is a function of the slip If one derive the equation w.r.t. (s), and set the derivative to null, we have the point where the slip gives the maximum power: 2

R R  Z e = 2 (1 − sm ) R + X =  2 (1 − sm ) or sm  sm  This states that the power is maximum when the equivalent load resistance (dynamic of course), is equal to the magnitude of the standstill impedance of the motor. This is really the “maximum power theorem”. 2 e

At maximum power:

2 e

sm =

R2 R2 + Z e

and the maximum power developed becomes: Pdm =

3  V12    2  Re + Z e 

Pd Pd From the general formula of power, we can extract the torque developed: Td = ω = (1 − s )ω m s

3V12 R2 / s Td = 2 2 Re + X e2 + [R2 (1− s) / s] + 2Re R2 (1− s) / s ωs

{

}

Similarly we can derive the maximum torque at the breakdown slip by differentiating and setting to null:

sb =

R2

R + (X1 X 2 ) 2 1

2

 1   R1 + R12 + ( X 1 + X 2 )2  (note: the maximum torque is independent of the rotor resistance!) The rotor resistance affects only the breakdown torque. 3V12 T = and the maximum torque at that point is: dm 2ω s

   

SOME CONCLUSIONS When the motor operates near its rated slip (typically a few %) the hypothetical resistance is greater than the leakage reactance. R2 / s >> X 2 and therefore a first approximation of the torque is:

3V12 s Td ≈ ω s R2 V1s similarly the current can be approximated as: I 2 ≈ R 2


Induction Machine pg 7

EQUIVALENT CIRCUIT PARAMETERS Stator Resistance Test: Measured with a DC source, voltage is applied to each phase R1 = 0.5 R measured for Y connection

R1 = 1.5 R measured for ∆ connection These values have to be further correctd by factors of 1.05 to 1.25 to account for the skin effect.

Blocked rotor test (or LOCKED) Similar to the short circuit test in transformers. Since s=1, the rotor circuit impedance is relatively small. The applied voltage is much lower than the rated voltage. Thus the excitation current is small and can be neglected. Z e = R1 + R2 + j ( X 1 + X 2 ) = Re + jX e

Re =

therefore:

and

Ze =

Vl .r I l .r

leads to

Plocked rotor 2 I locked rotor

hence R2 = Re − R1

X e = Z e2 − Re2

for all practical purposes X 1 = X 2 = 0.5 X e NO LOAD TEST In this case rated voltage is applied to the stator. The motor is not loaded, hence this is equivalent to the open circuit test for a transformer. However, here we must know the friction/ windage losses. Poc = Wmeasured − Pfriction − windage

Woc with cos θ oc = V I oc oc

hence Rc =

leading to

Xm =

Voc2 Poc

Voc I oc sin θ oc

Torque Speed In order to plot the torque speed characteristic, a dynamometer is used and the motor is loaded


Induction Machine pg 8

STARTING of INDUCTION MOTORS At start the rotor speed is zero, so the starting current is I 2s =

V1 Re + jX e

and Tds =

3V12 R2 ω s Re2 + X e2

[

]

Since Re is very small compared to the R2/s value, the starting current may be as much 400 to 800% of full rated current. The starting torque however is only in the 200 to 350% of full load torque. 1) To limit this current one could start the motor under reduced voltage. However this causes the starting torque to be very small. Hence this can be used only for loads which need very little torque at starting (i.e. fans) 2) We could also increase the value of the rotor resistance, however: - this reduces the torque developed at full laod - produces high copper losses - causes a reduction in motor efficiency at full load. One application however is the “wound rotor machine” where one can introduce a series resistance at start, and short the resistance when the machine is running. This has the advantage of a high starting torque, and good efficiency at full load. However one needs slip rings on the rotor. In case we use a squirrel cage motor, one can use electromagnetic field behaviour to realize variations in the effective resistance of the rotor. This is obtained by using the skin effect. Here we see a rotor bar which has an elongated shape. When the rotor frequency is high (start), the skin effect is pronounced, and therefore the effective resistance is very high. When the slip decreases, the frequency nears zero hence the skin effect is practically null, and therefore the effective resistance is small. leakage flux

One can obtain the same effect using “double cage” construction. Again, at start the current stays on the outer cage therefore a high resistance, but as the speed increases, the current moves also in the deep bar, hence the resistance decreases considerably.

outer cage

inner cage

3) Y-Delta If one starts with a Y connected stator, the starting current sees 3R resistance. As the motor increases its speed, one connects in delta, and the effective resistance drops to R. Hence a limit in starting current


Induction Machine pg 9

SPEED CONTROL Induction motors are cheap and rugged, however their speed is difficult to control. 1) VOLTAGE CONTROL Unfortunately this is not an effective control. As the voltage decreses, the torque decreases (see the square of the voltage on the numerator of the torque equation). Practically this is confined to 80100% control. 2) FREQUENCY CONTROL This is by far the most efficient way to control the speed. However, one has to make sure that the machine does not saturate. since the flux is proportional to V/f, this control has to assure that the magnitude of the voltage is proportional to the speed. Power electronic circuits are best suited for this kind of control. 3) CHANGING STATOR POLES For a stator wich has several independent windings, one can connect them is series for starting, essentially building N*poles. The speed of the machine will be reduced by the same factor. As the machine speed increases, one can switch the stator connection to a parallel connection, hence reducing the amount of poles and hence accelerating the machine. This method is simple, but can really accomodate only 2 speeds. 4) ROTOR RESISTANCE As seen for the starting, one can insert a variable resistance in the rotor (slip rings) and hence cause the developed torque to vary, hence control the speed. 5) DOUBLY FED MOTOR A special application can be to inject a current in the rotor. Hence the air gap flux will depend upon the difference of frequency between stator and rotor currents, and therefore the speed can be varied by varying the rotor frequency. 6) KRAMER CIRCUIT (recovery) With the method of variable resistor in the rotor circuit, a lot of power is disspated in this additional resistor. With the Kramer method, one takes the the rotor windings, and feed a 3 phase rectifier. This DC voltage is then fed through an inverter back to the source. Here only the component losses are accounted for, the excess power not transformed in mechanical torque will be fed back to the source.


Induction Machine pg 10

TYPES of MOTORS (NEMA) (National Electrical Manufacturers Association) Class A: Standard motor suitable for constant speed. Motor can be started by applying rated voltage. It develops a starting torque of 125% to 175% of full load torque. The starting current is limited to 5 tp 7 times the rated value. Full load slip is of the order of 5% >> low inertia, high acceleration (fans, blowers, centrifugal pumps and machine tools) Class B: General purpose motor that can be started with full line voltage. The starting torque is about the same as class A, but starting currents are lower (4 times rated). Their speed regulation is slightly better. Same applications as class A Class C: Usually a double cage motor. Full voltage starting, and current limited to 3-5 times rated. Starting torque 200 to 275% of rated. Regulation 4-5% applications: >>> high starting torqus, compressors crushers, boring mills, conveyor equipment, textile machinery, wood working equipment. Class D High resistance capable of 250 - 300% starting torques. Efficiency lower and speed regulation up to 10%. >> bulldozers, shearing machines, punch presses, stamping machines, laudry equipment, hoists. Class E: Low starting torques and operates at low slip rate. Starting current is low and can be started at full voltage. However, above 7hp might require low voltage starting. Class F Double cage usually, it is a low torque motor and requires low starting current. Starting torque is usually 1.5 rated and starting current 2-4 times rated value. Speed regulation around 5%. They are designed to replace class B motors above 25hp.


POLYPHASE INDUCTION MACHINES