Electrical Machines: IV. Three-Phase Induction Machines

IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl

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IV. Three-Phase Induction Machines Induction Machines


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Example 1: A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor has the following parameters:

R1=0.461 Ω, R2=0.258 Ω, X1=0.507 Ω, X2=0.309 Ω, Xm

i. Synchronous speed in rpm

=30.74 Ω Rotational losses are 2450W. The motor drives a mechanical load at a speed of 1170 rpm. Calculate the following information:

ii. slip iii. Line Current iv. Input Power v. Airgap Power

vi. Torque Developed vii. Output Power in Hp

viii. Efficiency

This machine has no iron loss resistance, so the equivalent circuit is as follows:

i. Synchronous speed is given by:

ii. Slip is given by

Using the rpm equation,

iii. Now, phase current is given by

s = (1200-1170)/1200 = 0.025


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where phase impedance is given by

Using the above equation, Zin = 9.57 + j3.84 Ω

And noting that the machine is delta connected, V1 = VLL = 480V

I1 = 43.1 - j17.4 A. |I1| =46.6 A, θ = -21.9° Therefore IL

iv. Input power is given by: = √3 × 46.6 = 80.6 A

Pin

v. Airgap power is the input power minus stator losses. In this case the core losses are grouped

= 62.2 kW

with rotational loss. Therefore

Pgap = 62.2 kW - 3× 46.62 × 0.461 Pgap

vi. Torque developed can be found from

= 59.2 kW

where synchronous speed in radians per second is given by

giving

Vii.Output power in horsepower is the output power in Watts divided by 746. (there are 746 W in one Hp).

τ = 471 Nm

and


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Therefore output power in Watts is: Pout

Viii. Efficiency is given by

= 55.3kW

η = 55.3/62.2 = 88.9%


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Using the induction machine power and torque equations it is possible to produce the torque speed curve shown below.

Operating Regions The torque-speed curve brakes down into three operating regions:

1. Braking, nm

it can be seen that if the power converted is negative (from

< 0, s > 1 Torque is positive whilst speed is negative. Considering the power conversion equation

P = τ ω) then the airgap power is positive. i.e. the power is flowing from the stator to the rotor and also into the rotor from the mechanical system. This operation is also called plugging

2. Motoring,

. This mode of operation can be used to quickly stop a machine. If a motor is travelling forwards it can be stopped by interchanging the connections to two of the three phases. Switching two phases has the result of changing the direction of motion of the stator magnetic field, effectively putting the machine into braking mode in the opposite direction.

0 < nm < ns, 1 > s > 0 Torque and motion are in the same direction. This is the most common mode of operation.

http://www.ece.ualberta.ca/~knight/ee332/induction/basics/pandt.html�


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3. Generating, nm > ns

indicates that if the power converted is negative, so is the air gap power. In this case, power flows from the mechanical system, to the rotor circuit, then across the air gap to the stator circuit and external electrical system.

, s < 0 In this mode, again torque is positive whilst speed is negative. However, unlike plugging,

Motoring Torque Characteristic The motoring region of the induction machine torque-speed curve is the region of greatest interest.

• Starting torque – Torque at zero speed – Typically 1.5 times the full-load torque • Pull-up torque – The minimum torque developed by the motor while accelerating from zero speed – Greater than full-load torque; less than starting torque • Breakdown torque – The maximum torque that the motor can develop – Typically 2.5 times the full-load torque • Normal operation – At full-load, the motor runs at n rpm – Rotor speed decreases slightly from synchronous speed with increasing load torque – Motor will stall when the load torque exceeds the breakdown torque


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The torque equation Using the equation

Multiple solutions of the above equation for torque at different slips can be made simpler by simplifying the equivalent circuit model. Consider the diagram below:

The stator part of the equivalent circuit (together with the magnetising branch) can be replaced by a Thevenin equivalent circuit. In the Thevenin circuit, the stator phase voltage has been replaced by its Thevenin equivalent,


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and the impedances have been replaced by Thevenin equivalent impedances.

Incorporating the Thevenin model into the circuit model results in the Thevenin equivalent circuit model of an induction machine.

In the above circuit, the calculation of rotor current is greatly simplified

The above expression for rotor current can be squared and substituted into the torque equation

Using the above equation, the variation of torque with slip can be plotted directly . Note that if power or efficiency calculations are needed, the full equivalent circuit model should be used (not the Thevenin version). since synchonous speed is constant, maximum torque occurs at the same slip as maximum airgap power. Considering the Thevenin circuit, and applying maximum power transfer theory, maximum airgap power and maximum torque will occur when

𝑅𝑅2

𝑠𝑠= �𝑅𝑅𝑇𝑇𝑇𝑇2 + (𝑋𝑋𝑇𝑇𝑇𝑇 + 𝑋𝑋2)2


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Re-arranging it is possible to obtain the slip for maxiumum torque, or pullout torque

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 =3𝑉𝑉𝑇𝑇𝑇𝑇2

2𝜔𝜔𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �𝑅𝑅𝑇𝑇𝑇𝑇 + �𝑅𝑅𝑇𝑇𝑇𝑇2 + (𝑋𝑋𝑇𝑇𝑇𝑇 + 𝑋𝑋2)2�

.

𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 =𝑅𝑅2

�𝑅𝑅𝑇𝑇𝑇𝑇2 + (𝑋𝑋𝑇𝑇𝑇𝑇 + 𝑋𝑋2)2

Substituting the pullout slip into the Thevenin torque equation:

• The starting torque is proportional to 𝑉𝑉𝑇𝑇𝑇𝑇2 𝑇𝑇𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠 = 3𝑉𝑉𝑇𝑇𝑇𝑇2 𝑅𝑅2

𝜔𝜔𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 [(𝑅𝑅𝑇𝑇𝑇𝑇 +𝑅𝑅2)2+(𝑋𝑋𝑇𝑇𝑇𝑇 +𝑋𝑋2)2]

The Starting torque

• The starting torque will be reduced if the stator and rotor leakage inductances are increased

• The starting torque will be reduced if the stator frequency is increased

• When the rotor resistance is increased, the starting torque will first increase and then decrease. 𝑇𝑇𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑇𝑇 𝑚𝑚𝑚𝑚𝑚𝑚 when 𝑅𝑅2=𝑋𝑋1 + 𝑋𝑋2


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Equivalent Circuit Model Analysis Example A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor has the following parameters:

R1=0.461 Ω, R2=0.258 Ω, X1=0.507 Ω, X2=0.309 Ω, Xm

i. Thevenin circuit parameters and Thevenin voltage

=30.74 Ω Rotational losses are 2450W. The motor drives a mechanical load at a speed of 1170 rpm. Find:

ii. Pullout slip iii. Pullout Torque iv. Start Torque

Solution:

Using Matlab or Excel (or another computer program) plot the torque speed curve for slip in the range 0 to 1

i. Thevenin circuit parameters and Thevenin voltage:

The Thevenin voltage is the voltage applied to the rotor assuming that the rotor current is zero. Thevenin impedance is the impedance of the stator part of the circuit, seen from the rotor, assuming that the stator supply is short circuited.

i.

Substituting the equivalent circuit parameters in to the above equations gives: VTH = 475.2 V, RTH = 0.452Ω, XTH

i. Pullout slip

= 0.313Ω

The slip at which maximum torque occurs can be found from maximum power transfer theory. Maximum torque and maximum airgap power occur at the same slip, therefore maximum torque occurs when


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ii. Pullout Torque

Pullout torque can be found by substituting the above pullout slip into the Thevenin torque equation

or from the maximum torque equation directly

Substituting into the above equation:

i. Start Torque

Start torque can be found by setting s=1 in the above equation for torque.


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Effect of Rotor Resistance


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Investigating the torque-speed curve it is apparent that the rotor circuit resistance has significant impact on speed at which maximum torque occurs. The plots below illustrate two cases, with low rotor resistance on the left and high rotor resistance on the right.

Example 1: A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-rotor induction motor has the following impedances in ohms per phase referred to the stator circuit:

𝑹𝑹𝟏𝟏 = 𝟎𝟎.𝟔𝟔𝟔𝟔𝟏𝟏𝛀𝛀 𝑹𝑹𝟐𝟐 = 𝟎𝟎.𝟑𝟑𝟑𝟑𝟐𝟐𝛀𝛀

𝑿𝑿𝟏𝟏 = 𝟏𝟏.𝟏𝟏𝟎𝟎𝟔𝟔𝛀𝛀 𝑿𝑿𝟐𝟐 = 𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔𝛀𝛀 𝑿𝑿𝑴𝑴 = 𝟐𝟐𝟔𝟔.𝟑𝟑𝛀𝛀

a) What is the maximum torque of this motor? At what speed and slip does it occur? b) What is the starting torque of this motor? c) When the rotor resistance is doubled, what is the speed at which the maximum torque now occurs?

What is the new starting torque of the motor?

Solution: The Thevenin voltage of this motor is 𝑉𝑉𝑇𝑇𝑇𝑇 = 𝑉𝑉𝜙𝜙𝑋𝑋𝑀𝑀

�𝑅𝑅12+(𝑋𝑋1+𝑋𝑋𝑀𝑀 )2

= (266 𝑉𝑉)(26.3 𝛀𝛀 )�(𝟎𝟎.𝟔𝟔𝟔𝟔𝟏𝟏𝛀𝛀 )2+(𝟏𝟏.𝟏𝟏𝟎𝟎𝟔𝟔𝛀𝛀 +𝟐𝟐𝟔𝟔.𝟑𝟑𝛀𝛀 )2 = 255.2𝑉𝑉

The Thevenin resistance is 𝑅𝑅𝑇𝑇𝑇𝑇 ≈ 𝑅𝑅1 �𝑋𝑋𝑀𝑀

𝑋𝑋1+𝑋𝑋𝑀𝑀�

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≈ (𝟎𝟎.𝟔𝟔𝟔𝟔𝟏𝟏𝛀𝛀) � 𝟐𝟐𝟔𝟔.𝟑𝟑𝛀𝛀 𝟏𝟏.𝟏𝟏𝟎𝟎𝟔𝟔𝛀𝛀+𝟐𝟐𝟔𝟔.𝟑𝟑𝛀𝛀

�2

= 0.590𝛀𝛀

The Thevenin reactance is 𝑋𝑋𝑇𝑇ℎ ≈ 𝑋𝑋1 ≈ 1.106𝛀𝛀

(a) The slip at which maximum torque occurs is given by Equation (2):

𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑅𝑅2

�𝑅𝑅𝑇𝑇𝑇𝑇2 +(𝑋𝑋𝑇𝑇𝑇𝑇 +𝑋𝑋2)2

= 𝟎𝟎.𝟑𝟑𝟑𝟑𝟐𝟐𝛀𝛀 �(0.590𝛀𝛀 )2+(1.106𝛀𝛀+𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔𝛀𝛀 )2 = 0.198

This corresponds to a mechanical speed of

𝑠𝑠𝑚𝑚 = (1 − 𝑠𝑠)𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ℎ = (1 − 0.198)(1800 𝑠𝑠/𝑚𝑚𝑚𝑚𝑠𝑠) = 1444 𝑠𝑠 𝑚𝑚𝑚𝑚𝑠𝑠⁄

http://www.ece.ualberta.ca/~knight/ee332/induction/basics/trq_speed.html�

http://www.ece.ualberta.ca/~knight/ee332/induction/basics/peak_trq.html�


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The torque at this speed is

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 = 3𝑉𝑉𝑇𝑇𝑇𝑇2

2𝜔𝜔𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �𝑅𝑅𝑇𝑇𝑇𝑇 +�𝑅𝑅𝑇𝑇𝑇𝑇2 +(𝑋𝑋𝑇𝑇𝑇𝑇 +𝑋𝑋2)2�

= 3(255.2 𝑉𝑉)2

2(188.5 𝑠𝑠𝑚𝑚𝑟𝑟 /𝑠𝑠)[(0.590𝛀𝛀)2+(1.106𝛀𝛀+𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔𝛀𝛀)2] = 229 𝑁𝑁.𝑚𝑚

(b) The starting torque of this motor is found by setting s=1 in Equation (1) :

𝑇𝑇𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠 =3𝑉𝑉𝑇𝑇𝑇𝑇2 𝑅𝑅2

𝜔𝜔𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 [(𝑅𝑅𝑇𝑇𝑇𝑇 + 𝑅𝑅2)2 + (𝑋𝑋𝑇𝑇𝑇𝑇 + 𝑋𝑋2)2]

=3(255.2 𝑉𝑉)2(𝟎𝟎.𝟑𝟑𝟑𝟑𝟐𝟐𝛀𝛀)

(188.5 𝑠𝑠𝑚𝑚𝑟𝑟 𝑠𝑠⁄ )[(0.590𝛀𝛀+ 𝟎𝟎.𝟑𝟑𝟑𝟑𝟐𝟐𝛀𝛀)2 + (1.106𝛀𝛀+ 𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔𝛀𝛀)2] = 104 𝑁𝑁.𝑚𝑚

(c ) If the rotor resistance is doubled, then the slip at maximum torque doubles, too. Therefore,

𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 = 0.396

and the speed at maximum torque is

𝑠𝑠𝑚𝑚 = (1 − 𝑠𝑠)𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = (1 − 0.396)(1800 𝑠𝑠 𝑚𝑚𝑚𝑚𝑠𝑠⁄ ) = 1087 𝑠𝑠 𝑚𝑚𝑚𝑚𝑠𝑠⁄

The maximum torque is still

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 = 229 𝑁𝑁.𝑚𝑚

The starting torque is now

𝑇𝑇𝑠𝑠𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠 =3(255.2𝑉𝑉)2(0.664𝛀𝛀)

(188.5 𝑠𝑠𝑚𝑚𝑟𝑟 𝑠𝑠⁄ )[(0.590𝛀𝛀+ 𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔𝛀𝛀)2 + (1.106𝛀𝛀+ 𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔𝛀𝛀)2] = 170𝑁𝑁.𝑚𝑚


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When the switch S is in the START position, the stator windings are connected in STAR

When the motor picks up speed, say 80% of its rated value, the switch S is thrown quickly to the RUN position which connects the stator windings in DELTA.

In induction motors that are designed to operate with delta stator connection it is possible, during starting, to reduce the phase voltage by switching to Y- connection During Y- connection, the phase voltage Vs

becomes

so the phase current, for same slip, IsY

, is reduced 3 times


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Now the line current in Δ connection IlΔ

is

So the line current is three times smaller for Y- connection. The torque is proportional to phase voltage squared

Therefore, the Y-delta starting is equivalent to a reduction of phase voltage and a 3 to 1 reduction in torque.


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Fig.1 Slip ring IM

Fig.2 Torque-slip curves


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Fig.3 Torque-speed curves

Documents

Electrical Machines: IV. Three-Phase Induction Machines