Equivalent Circuit of Induction Motor

Equivalent Circuit of Induction Motor  AU : May-06, 09, 12, 13, 16,18, Oct.-97, Dec -05, 07, 08, 11, 12, 16

We have already seen that the induction motor can be treated as generalized transformer. Transformer works on the principle of electromagnetic induction. The induction motor also works on the same principle. The energy transfer from stator to rotor of the induction motor takes place entirely with the help of a flux mutually linking the two. Thus stator acts as a primary while the rotor acts as a rotating secondary when induction motor is treated as a transformer.

If       E₁ = Induced voltage in stator per phase

E₂ = Rotor induced e.m.f. per phase on standstill 

K = Rotor turns / Stator turns

then   K = E₂ / E₁

Thus if V₁ is the supply voltage per phase to stator, it produces the flux which links with both stator and rotor. Due to self induction, E₁ is the induced e.m.f. in stator per phase while E₂ is the induced e.m.f. in rotor due to mutual induction, at standstill. In running condition the induced e.m.f. in rotor becomes E_2r which is sE₂.

Now  E_2r = Rotor induced e.m.f. in running condition per phase

R₂ = Rotor resistance per phase

X_2r = Rotor reactance per phase in running condition

R₁ = Stator resistance per phase

X₁ = Stator reactance per phase

So induction motor can be represented as a transformer as shown in the Fig. 5.18.1.

When induction motor is on no load, it draws a current from the supply to produce the flux in air gap and to supply iron losses.

This current Io has two components.

1. I_c = Active component which supplies no load losses

2. I_m = Magnetising component which sets up flux in core and air gap

These two currents give us the elements of an exciting branch as,

R_o = Representing no load losses = V₁ / I_c 

And X_o = Representing flux set up = V₁ / I_m

Thus, 

The equivalent circuit of induction motor thus can be represented as shown in the Fig. 5.18.2

The stator and rotor sides are shown separated by an air gap.

I_2r = Rotor current in running condition 

It is important to note that as load on the motor changes, the motor speed changes. Thus slip changes. As slip changes the reactance X_2r changes. Hence X_2r = sX₂ is shown variable

Representation of rotor impedance :

It is known that, 

So it can be assumed that equivalent rotor circuit in the running condition has fixed reactance X₂, fixed voltage E₂ but a variable resistance R₂/s, as indicated in the above equation.

Now 

So the variable rotor resistance R2/s has two parts,

1. Rotor resistance R2 itself which represents copper loss.

2. R₂ (1 - s)/s which represents load resistance R_L. So it is electrical equivalent of mechanical load on the motor.

Key Point Thus mechanical load on the motor is represented by the pure resistance of value R₂ (1 - s) /s.

So rotor equivalent circuit can be shown as,

Now let us obtain equivalent circuit referred to stator side.

Equivalent circuit referred to stator :

Transfer all the rotor parameters to stator,

K = E₂ / E₁ = Transformation ratio

E₂ = E₂ / K

The rotor current I2r has its reflected component on the stator side which is I_2r.

Thus R_L is reflected mechanical load on stator.

So equivalent circuit referred to stator can be shown as in the Fig. 5.18.4.

The resistance R'₂ (l - s)/s = R'_L is fictitious resistance representing the mechanical load on the motor.

 

1. Approximate Equivalent Circuit

Similar to the transformer the equivalent circuit can be modified by shifting the exciting circuit (R_o and X_o) purely across the supply, to the left of R₁ and X₁. Due to this, we are neglecting the drop across R₁ and X₁ due to I_o , which is very small. Hence the circuit is called approximate equivalent circuit. The circuit is shown in the Fig. 5.18.5.

Now the resistances R₁ and R₂ while reactance X₁ and X₂ can be combined. So we

R_le = Equivalent resistance referred to stator = R₁ + R₂

X_le = Equivalent reactance referred to stator = X₁ + X₂

Thus the equivalent circuit can be shown in the Fig. 5.18.6.

 

2. Power Equations from Equivalent Circuit

With reference to approximate equivalent circuit shown in the Fig. 5.18.6, we can write various power equations as,

P_in = Input power = 3 V₁ I₁ cos ϕ

Where V₁ = Stator voltage per phase

I₁ = Current drawn by stator per phase

cos ϕ = Power factor of stator

Stator core loss = I_m² R_o

Stator copper loss = 3 I₁² R₁

where R₁ = Stator resistance per phase

P₂ = Rotor input = 3(I_2r)² R₂ / s

P_c = Rotor copper loss = 3(I_2r)² R₂

Then P_c = s P₂

P_m = Gross mechanical power developed

Key Point Remember that in all the above formulae all the values are per phase values.

 

3. Maximum Power Output

Consider the approximate equivalent circuit as shown in the Fie. 5.18.7.

In this circuit, the exciting current Io is neglected hence the exciting no load branch is not shown.

I₁ = I_2r

The total impedance is given by,

The power supplied to the load i.e. P_out per phase is,

Per phase P_out = I¹₂ R'_L watts per phase

Total P_out = 3 I²₁ R'_L

To obtain maximum output power, differentiate the equation of total P_out with respect to variable R_L and equate to zero.

Thus the mechanical load on the induction motor should be such that the equivalent load resistance referred to stator is equal to the total leakage impedance of motor referred to stator.

Expression for maximum P_out : Using the condition obtained in expression of total P_out , we can get maximum P_out

 

4. Maximum Torque

In case of induction motor, the speed of the motor decreases with increase in load. Thus the maximum power output is not obtained at a slip which corresponds to maximum torque. In the previous section we have seen the condition for maximum power output. In this section we will find the condition which gives maximum torque.

The expression for torque is given by,

 The condition for maximum torque can be. obtained from maximum power transfer theorem. When (I’_2r)² R’₂ / is maximum consider the approximate equivalent circuit of induction motor as shown in the Fig. 5.18.8.

The value of R₀ is assumed to be negligible. Hence the circuit will be reduced as shown in the Fig. 5.18.9.

The The venin's equivalent circuit for the V1 above network is shown in the Fig. 5.18.10 across the terminals x and y.

The mechanical torque developed by rotor is maximum if there is maximum power transfer to the resistor R’₂ / s. This takes place when R’₂ / s equals to impedance looking back into the supply source.

From the above expression, it can be seen that the maximum torque is independent of rotor resistance R₂.

 

5. Synchronous Watt

The torque produced in the induction motor is given by,

Thus torque is directly proportional to the rotor input. By defining new unit of torque which is synchronous watt we can write,

T = P₂ synchronous-watts

If torque is given in synchronous-watts then it can be obtained in N-m as,

Key Point Unit synchronous watt can be defined as the torque developed by the motor such that the power input to the rotor across the air gap is 1 W while running at synchronous speed.

 

Example 5.18.1 The following data refers to a 12 pole, 420 V, 50 Hz, three phase mesh connected induction motor :

r₁ = 2.95 Ω, x₁ = 6.82 Ω, r₂ = 2.08 Ω, x₂ = 4.11 Ω, per phase.

On no load, the line value of magnetizing current is 6.7 A and the total core loss is 269 W. Determine the power factor, input current, equivalent rotor current and torque developed by the motor at a slip of 3 % using exact equivalent circuit. Determine the maximum torque developed and the corresponding speed. AU : Dec.- 08, Marks 16

Solution : On no load, the equivalent circuit is as shown in the Fig. 5.18.11.

I_m(Line) = 6.7 A

As motor is delta connected,

The exact equivalent circuit is shown is the Fig. 5.18.12.

The equivalent rotor current I_2r can be obtained by using current distribution rule in parallel circuit as,

The torque developed is,

To find the maximum torque T_m, first calculate Thevenin's equivalent R_TH and X _TH

 

Example 5.18.2 A 3 phase, 200 V, induction motor has stator impedance of (0.07 +j 0.4) Ω /ph while equivalent rotor impedance referred to stator is (0.08+j 0.2) Ω /ph. Neglecting no load current, calculate the maximum mechanical power output and the slip corresponding to maximum output condition. Assume star connected stator.

Solution : The given values are

 

Example 5.18.3 A 3 phase, star connected 400 V, 50 Hz, 4 pole induction motor has the following per phase parameters in ohms, referred to the stator.

R₁ = 0.15, X₁ = 0.45, R'₂ = 0.12, X’₂ = 0.45, X_m = 28.5

Compute the stator current and power factor when the motor is operated at rated voltage and frequency with s = 0.04.

Solution : The equivalent circuit referred to stator is shown in the Fig. 5.18.13.

 

Example 5.18.4 A 400 V, 50 Hz, 3 phase, 6 pole star connected induction motor has the following values for the various parameters of its equivalent circuit.

Stator impedance = (0.3 + j 0.4) Ω

Equivalent rotor impedance = (0.2 + j 0.4) Ω

Magnetizing reactance = 20 Ω

Resistance to account fore core loss = 100 Ω

Using the approximate equivalent circuit calculate for a slip of 4 percent a) Motor speed b) Stator current c) Power factor d) Motor output and efficiency. Assume stator losses, mechanical losses to be 2 kW each. 

Solution The given values are

R₁ = 0.3 Ω ,          X₁ = 0.4 Ω   Mechanical losses = 2 kW

R₂ = 0.2 Ω ,          X’₂ = 0.4 Ω

X_o = 20 Ω,  s = 0.04

R_o = 100 Ω,          Stator loss = 2 kW

The approximate equivalent circuit is shown in Fig. 5.18.14.

 

Example 5.18.5 A 4 pole, 50 Hz, 3 phase induction motor, when loaded has total rotor input of 3000 W. Determine the torque developed in i) N-m and ii) Synchronous watts.

Solution : The given values are,

 

Example 5.18.6 A 50 HP, 420 V, 3 phase, 10 pole, delta connected induction motor has r1 - 0.19 ohm, r₂ - 0.29 ohm, X₁ = X₂ = 1.12 ohm rm - 143 ohm, X_m - 16.8 ohm. Full load slip - 2.6 %. Use exact equivalent circuit and calculate i) Rotor Input ii) Rotor Copper loss per phase iii) Stators current and power factor.

Solution :

 

Example 5.18.7 A 3 phase, 25 kW, 400 V, 50 Hz, 8-pole induction motor has rotor resistance of 0.08 ohm and standstill resistance of 0.4 ohm. The effective stator/rotor turn ratio is 2.5. The motor is to drive a constant-torque load of 250 N-m . Neglect stator impedance : 1) Calculate the minimum resistance to be added in rotor circuit for the motor to start up on load. 2) At what speed would the motor run, if the added rotor resistance is (A) left in the circuit, and (B) subsequently short circuited.

Solution :

Neglecting stator impedance, the equivalent circuit on rotor side is shown in the Fig. 5.18.6

 

Examples for Practice

Example 5.18.8 A three phase, star connected 400 volt, 50 Hz, 4 pole induction motor has following parameters referred to stator ;

r₁ = 0.15 ohm, x₁= 0.44 ohm,

r'₂ = 0.12 ohm, x'₂ = 0.44 ohm,

x_m - 30 ohm neglect core loss resistance (r_c), find stator current, power factor at rated voltage and slip of 4 %.

[Ans.: 73.045 Z- 21.42° A, 0.9309 lagging]

Example 5.18.9 A 440 V, 50 Hz, 4-pole, 3-phase delta connected induction motor has a leakage impedance of (0.3 + 7 5.5 + 0.25 /s) ohm / phase (delta phase) referred to the stator. The stator to rotor voltage ratio is 2.5. Determine the external resistance to be inserted in each star-phase of the rotor winding such that the motor develops a gross torque of 150 N-m at a speed of 1250 r.p.m. UPTU : 2006-07

[Ans.: R_x= 0.189 Ω/ph]

Example 5.18.10 Consider thr equivalent circuit diagram of 3 Y- connected, 440 V, shown Fig.5.18.17:

Where r₁ = 0.209Ω x₁ = 0.503 Ω x_m = 13.25 Ω r₂ = 0.209 Ω x₂ = 0.144Ω

i) Determine the stator current and power factor (motor runs at N_r = 1460 r.p.m.)

ii) Determine the air gap power and rotor copper losses (motor  runs at Nr = 1450 r.p.m.)

iii) What is the value of motor speed at which it takes 30 ampere current at 0.8 p.f. lagging from supply mains ?

iv) What is value of slip when motor runs at N_r = 1480 rpm. UPTU : 2011-12

[Ans.: i) 35.233∠-33.573°A,0.8331 lagging, ii) 867.247 W, 26 kW, iii) 1467.75 r.p.m., iv) 1.333 %]

Example 5.18.11 A 4 pole, 50 Hz, 3 phase induction motor, when loaded has total rotor input of 3000 W. Determine the torque developed in i) N-m and ii) Synchronous watts.

[Ans.: i) 19.098 N-m, ii) 3000 syn-watt]

Example 5.18.12 For a 3ϕ IMs, show that :

Example 5.18.13 A 400 V, 1450 r.p.m., 50 Hz, wound rotor induction motor has the following circuit modal parameters.

R₁ = 0.3 Ω, R₂ = 0.25 Ω, X₁ = X₂= 0.6 Ω, X_m = 35 Ω,

Rotational loss = 1500 W.

i) Calculate the starting torque and current when the motor is started direct on full voltage.

ii) Find the slip for maximum torque and the value of maximum torque. VTU : Feb -06

[Ans.: 176.6407∠- 65.528 ° A, 143.9 Nm, 0.2041 i.e. 20.41 %, 324.922 Nm]

Example 5.18.14 The following data refers to a 12 pole, 420 V, 50 Hz, three phase mesh connected induction motor :

r₁ = 2.95 Ω,          x₁ = 6.82Ω ,  r'₂ = 2.08 Ω, x'₂ = 4.11Ω, per phase.

On no load, the line value of magnetizing current is 6.7 A and the total core loss is 269 W. Determine the power factor, input current, equivalent rotor current and torque developed by the motor at a slip of 3 % using exact equivalent circuit. Determine the maximum torque developed and the corresponding speed.

[Ans.: 0.7655 lagging, 6.732 ∠-40M° A, 5.413 ∠- 6.811“ A, 116.405 Nm, 331.35 Nm, 404.4 r.p.m.]

Review Questions

1. Develop an equivalent circuit of a 3 phase induction motor. What do the various paramenters represent ? Represent the approximate equivalent circuit and state its significance.  AU : May-06, 16,18, Dec.-05, 07, 11, 12, Marks 8

2. Starting from the equivalent circuit, derive the various power equations of an induction motor. AV : May-12, Marks 8

3. Show how a 3 phase induction motor can be represented by an approximate equivalent circuit. AU : May-13, Dec.-16, Marks 8