3-phase Induction Machines

Lecture 24 - EE743

3-Phase Induction Machines Torque Speed characteristics

Professor: Ali Keyhani

Importance of T-speed characteristics ■

Torque speed characteristic of a motor is important from the point of view of its applications to specific situations. To calculate the torque produced by the machine, first, we compute the motor power. The motor power, or the mechanical power supplied to the load is Pem = Pin − Plosses The electrical power inputted to the machine can be calculated from the eq. circuit given below

2

Importance of T-speed characteristics

3

Machine Torque R2' = a 2 R2

 2 ~ ~* P1 = 3 Re V1 I1 = 3 I1  R1 +  s   ■ Power dissipated is

{

}

2

(

R2'

)

2

X 2' = a 2 X 2

(

PΩ = 3 I1 R1 + a 2 R2 = 3 I1 R1 + R2' ■

Motor power Pem is 2 

■

)

R2 = Rrotor + Rext

R2'  1− s ' 2 '  Pem = 3 I1  R1 +  − R1 + R2  = 3 R2 I1 s  s   Motor torque can be calculated from 3 1 − s  ' 2 Pem = Temω m Tem =   R2 I1 ωm  s 

(

)

4

Machine Torque ■

Since ω m = ω sync ( 1 − s ) Tem =

■

3

R2'

ω sync s

I1

Tm can be written as V1

2

2

I1 =

2

2

   R1 +  + X 1 + X 2' s   R2'

(

)

2

Substituting for I1 in equation (3), we will have Tem =

3

R2'

V1

2

' 2 ω sync s  R2  ' 2  R1 +  + ( X 1 + X 2 ) s  

5

Power calculations ■

Figure 2 shows the plot of equation (3) for the values of slip ' 2 from zero to unity. In Fig.1 R2 = a ( R2 + Rext ) . This corresponds to the normal range of the speed of an induction motor from starting (ωm=0, s=1) to the synchronous speed (ωm= ωsync, s=0)

6

smax and Tmax calculation ■

To find smax first set derivative of (4) with respect to s equal to zero. dTm =0 ds

■

Which will results in R2' smax = ± 2 ' 2 R1 + X 1 + X 2 substituting smax in (4) will result in Tmax 2 V 3 Tmax = ± 2ω sync  R + R 2 + X + X ' 2  1 1 2   1 

(

■

)

(

)

7

Torque-speed characteristic ■

Equation (5) shows that the slip at which the maximum torque occurs is proportional to the rotor resistance. Equation (6) shows that the maximum torque is independent of the rotor resistance

8

Torque-speed characteristic ■

The machine is operating as a motor for the range of s for which the torque-speed curves are shown in fig.1 and fig.2. In this range, torque is positive and ωsync is greater than rotor speed ωm. Note that the torque is zero at slip equal to zero. As was state before, slip is given by ω sync − ω m s= ω sync

■

Let us consider three different cases:

9

Torque-speed characteristic ■

Case 1: ωsync and ωm are rotating in the same direction and ω sync is rotating faster than ωm.

■

Case 2: ωsync and ωm are rotating in the same direction and ω sync is rotating slower than ωm.

■

Case 3: ωsync and ωm are rotating in different directions.

10

Torque-speed characteristic. Case 1. ■

Case 1: ωsync and ωm are rotating in the same direction and ω sync is rotating faster than ωm. –

This case is the normal operation of the induction machine. Machine operates as a motor. Note also in equation (7) ωsync and ωm are in the same direction and the slip is positive for motor operation

11

Torque-speed characteristic. Case 2. ■

Case 2: ωsync and ωm are rotating in the same direction and ω sync is rotating slower than ωm. –

If the speed of the machine is increased beyond its synchronous speed by an external prime motor, but still rotated in the same direction as the stator field, the slip will be negative (s<0). This region (s<0) corresponds to the generator operation. For this region, torque is negative. This means that the mechanical power is required to drive the machine, which in turn delivers electric power at the stator terminals

12

Torque-speed characteristic. Case 3. ■

Case 3: ωsync and ωm are rotating in different directions. –

Suppose an induction motor is operating under normal conditions at the same value of positive slip in stable region (0<s<smax). Now we interchange any two terminals of the stator. This reverses the direction of the stator rotating field. The rotor speed ωm may now be considered as a negative with respect to that of the stator 1 −field. s ' For this case, s>1 and power loss in the variable resistance s R2 is negative, indicating that mechanical energy is being converted to electric energy. Both the power fed from stator and power fed from rotor are lost as heat in the rotor resistance. This region is called the braking region

13

Example 1 ■

On no-load a 3-phase delta-connected induction motor takes 6.8A and 390W at 220V line to line. R1=0.1Ω/phase, friction and windage losses are 120W. Determine Xm,and Rm of the motor equivalent circuit. I line 6.8 I phase = = = 3.926 A 3 3

V phase = 220V = VL− L

I 2phase R phase = 1.54W 2 V phase

390 − 120 Core losses = − 1.54 = 88.46W = 3 Rm 1 −3 1 220 2 G = = 1 . 828 ⋅ 10 Rm = = 547.14Ω m Rm Ω 88.46 14

Example 1 Ym =

I m 3.926 1 = = 17.84 ⋅ 10 −3 V 220 Ω

1 2 2 −3 1 Bm = = Ym − Gm = 17.746 ⋅ 10 Xm Ω

X m = 56.35Ω

15

Example 2 ■

The motor of the previous example take 30A and 480W at 36V line-to-line, when the rotor is blocked. Determine the complete equivalent circuit of the motor. Assume that X1=X’2.

( R1 + ) = R2'

R1 = 0.1Ω

( X 1 + X 2' ) =

480 / 3

( 30 / 3) so

2

= 0.533Ω = Req R2' = 0.433Ω 2

36  2 2 2 Z eq − Req =   − 0.533 = 2.0Ω = X eq  30 / 3  X eq X1 = X 2 ≅ = 1Ω 2 16

Example 3 ■

An induction motor has an output of 30kW at η=0.86. For this operating condition Pcoil,1=Pcoil,2=Pcore=Prot. Determine the slip M P0 Eff = Pin ,3φ P0 30,000 Pin ,3φ = = = 34,884 W Eff 0.86

Ploss ,3φ = 34,884 − 30,000 = 4,884 W 17

Example 3

PAG = P0 + PR' + Pcore + Prot = 33,663W

s=

PR' ,3φ 2

2

PR' =

4,884 = 1,221W 4

PAG ,3φ 2 1,221 s= = 0.036 or 3.6% 33,663

18

Example 4 ■

A wound rotor six-pole 60Hz ind. motor has R’2=0.8Ω and runs at 1152 rpm (s=0.04) at a given load. The load torque remains constant at all speeds. How much resistance must be inserted in the rotor circuit to change the speed to 960 rpm (s=0.2). Neglect the motor leakage reactance, X1 and X2. The air gap power function is: ' ' 2 2 R R V 1 Pg = 3I 2' 2 ≅ 3 2 ' 2 s s  R2   R1 +  + X 1 + X 2' s  

(

)

2

19

Example 4 ■

If

R2' = cons tan t (see eq. circuit ), the voltage, current, s

air-gap power, and torque conditions remain the same, i.e R2' = Const. s

then

T = Const.

' R2' + Rinsert R2' = 0.04 0.2 ' Rinsert

0.2 ' = R2 − R2' = 5(0.8) − 0.8 = 3.2Ω 0.04

20

Example 5 ■

A 400V, 3-phase WYE connected motor has ' Z1 = ( 0.6 + j1.2 ) Ω / phase , and Z 2 = ( 0.5 + j1.3) Ω / phase using the approx. eq. circuit. Determine the max. electromagnetic power, Pd V1 V1 = = Z1 + Z 2 ' 2  R2   R1 +  + X 1 + X 2' s   231 I 2' = 2 0 . 5  0.6 +  + ( 2.5) 2   s   I 2'

(

)

2

V1 =

400 = 231V 3

21

Example 5 Pd = Pg −

'2 ' I 2 R2

(

)

1− s ' ' 2 = ( 1 − s ) Pg = R2 I 2 = function ( s ) s

A s − s2 Pd = 2 s + Bs + C

(

)

dPd A( 1 − 2 s ) A s − s2 =0= 2 − ds s + Bs + C s 2 + Bs + C Pd ,max, phase =

( A( s − s 2 )

( s 2 + Bs + C )

)

( 2s + B ) 2

s at max P = 0.155 d

= 6,962 W

2 s =0.155

Pd ,max,3 phase = 3 × 6,962 = 20,886W 22