Learn everything about the Induction Motor

Week 9 Challenge - Induction Motor 2

1. Discuss about the equivalent circuit network of induction motor MATLAB model.
2. 3 phase, 50 Hz induction motor, represented by equivalent circuit constants X1 = X2 = 0.1 ohm and R1 = R2 = 0.2 ohm is operated at half of rated voltage and frequency. Calculate the ratio of starting torque at half voltage & frequency to rated values.
3. Make a MATLAB script file which will plot speed torque characteristics for frequency control method.
4. Explain in brief about operation of BLDC motor.

ANS 1:

In MATLAB, the block for an induction motor powered by an AC supply is shown below:

The eqivalent circuit network of induction motor block in MATLAB is shown below:

In the above circuit, the various terms are as follows:

• Stator resistance - R1
• Rotor resistance - R2
• Stator inductance - L1
• Rotor inductance - L2
• Magnetizing inductance - Lm
• Rotor slip - s
• Sinsuoidal supply voltage - `barV`
• Sinusoidal current phasors - `bar I`
• Rotor Current - `I_2`

Working Principle:

An induction motor consists of a stationary component called the stator and a rotary component called the rotor. The stator consists of 3 coil windings supplied with 3 phase AC power which are wound around stacks of thin, highly permeable laminations in a cast iron or steel frame. Changing the frequency of the 3 phase supply power supply (i.e. the sinusoidal supply voltage and current phasors) will generate a rotating magnetic field. This induces an emf and therefore a current in the rotor when the magnetic field interacts with the rotor via eletromagnetic induction. But now that the rotor is a current carrying conductor in the rotating magnetic field, it will experience an electromagnetic force producing a torque via the motor effect.

Since an emf is induced in the rotor, there will be a current flowing through it and this is denoted as: `I_2`

The induced emf in the rotor as a result of electromagnetic induction is also referred to as the 'back emf' and it is shown in the equivalent circuit diagram as well. This emf is referred to as 'back emf' because it acts in the opposite direction to the direction of the 3 phase voltage across the coil windings in the stator.

The magnetizing inductance, Lm is required at the air gap because across it is the boundary between the stator and the rotor. The stator will produce the rotary magnetic field and it needs to cross the air gap to interact with the rotor to excite it to produce a torque. So the magnetizing inductance, Lm is needed to cross this air gap.

Xm, X1 and X2 all represent reactance values. Reactance is a form of opposition that electronic components exhibit to the passage of current through them due to their capacitance or inductance properties.

Operation:

The speed of the rotating magnetic field is also known as the synchronous speed and it is dependent on the frequency of the 3 phase power supply and the number of poles of the induction motor. So, the frequency of the 3 phase power suppply is proportional to the synchronous speed and also the motor speed.

The synchronous speed is determined by the formula below:

`omega_(ms) = (120 * f) / p`

where,

Frequency of the 3 phase AC power supply - f

Number of Poles - p

This synchronous speed needs to be greater than the motor speed because if the synchronous speed is greater than the rotor speed, there will be an interaction between the magnetic field and the rotor bringing about excitation. It is this excitation that induces the emf in the rotor, causing it to experience the electromagnetic force thus producing the torque. The slip is basically the phase difference between the synchronous speed and the rotor speed.

The slip varies between 0 and 1 and it is a maximum when the rotor speed is equal to the synchronous speed. In the former case, there will be no torque because there will be no interaction between the rotating magnetic field and the rotor. In the latter case, this is referred to as the starting torque and this is the highest torque that is produced by the induction motor.

For this model, the slip is mathematically expressed in terms of the motor speed, the number of poles and the frequency of the 3 phase power supply as so:

`s = 1 - (omega_m*p)/f`

where,

motor speed - `omega_m`

number of poles - p

frequency of 3 phase power supply - f

This can also be expressed as:

`s = (omega_(ms) - omega_m) / omega_(ms)`

where,

synchronous speed - `omega_(ms)`

motor speed - `omega_m`

The torque that is produced by the rotor of the induction motor can be expressed mathematically as so:

`T = (n*p*R_2)/(s*f) * V_(rms)^2/[(R_1 + R_2 + (1-s)/s*R_2)^2 + (X_1 + X_2)^2]`

where,

number of phases - n

Stator Inductance - X1

Rotor Inductance - X2

Root mean square Voltage - `V_(rms)`

The block can be modified in terms of the parameters mentioned below the equivalent circuit model or in terms of the motor ratings. From the block we can see that there are 2 mechanical ports, C and R. When the block is used in a simulation, a positive torque is produced acting from those ports.

The block also has 2 thermal ports, one for each winding and they are mainly used to simulate the copper resistance losses that occur which are in the form of heat energy. To see these ports, we need to right click on the block, and from the options select 'Simscape', then 'Block Choices' and then 'Show Thermal Port'.

While the simplified induction motor block can model the torque speed characteristics of an induction motor there are still some limitations that it has. For instance, it does not model the starting mechanism for a single phase induction motor. Another limitation is that if the block is modelled in terms of the motor rating parameters, then the equivalent circuit model parameters are derived by assuming that the magnetizing inductance term, Lm is negligible. Hence, the effect of the magnetizing inductance term will not be included in the simulation.

ANS 2:

Given Data:

Number of Phases = 3

Rated Frequency = 50 Hz

Stator Inductance, X1 = 0.1 Ohms

Rotor Inductance, X2 = 0.1 Ohms

Stator Resistance, R1 = 0.2 Ohms

Rotor Resistance, R2 = 0.2 Ohms

To find:

We need to find the ratio of the starting torque at the half rated voltage and frequency to the same at the rated voltage and frequency.

The equation relating all the given data is: `T = 3/omega_(ms) * [(V^2 * R_2/s) / ((R_1 + R_2/s)^2 + (X_1 + X_2)^2)]`

Since we are only required to find the starting torque we can eliminate the slip term, 's'. This is because at the starting torque, the motor speed is 0 and so the synchronous speed terms will cancel each other out meaning the slip will take a maximum value of 1.

The equation for slip is: `slip, s = (omega_(ms) - omega_(m)) / omega_(ms)`

So, we can rewrite the torque equation omitting the slip term, 's' since it is equal to 1.

The starting torque is given by the equation: `T_s = 3/omega_(ms) * [(V^2 * R_2) / ((R_1 + R_2)^2 + (X_1 + X_2)^2)]`

Now, we can use this equation to find the starting torque at the half rated and full rated conditions.

Synchronous Speed

The formula to find the synchronous speed is: `omega_(ms) = (120 * f)/p`

where,

f - Frequency

p - Number of Poles

Let us consider the number of poles = n

Let the half rated synchronous speed = `omega_(ms1)`

Let the rated synchronous speed = `omega_(ms2)`

Half rated Frequency = Rated Frequency / 2 = 50/2 = 25Hz

`omega_(ms1) = (120 * 25) / n`

`omega_(ms1) = 3000/n`

Rated Frequency = 50Hz

`omega_(ms2) = (120 * f)/p`

`omega_(ms2) = 6000/n`

Since, we know that the rated synchronous speed is double the half rated synchronous speed, we can express the latter in terms of the former.

`omega_(ms1) = omega_(ms2)/2`

Calculation of Starting Torque at Half Rated Condition:

Now that we have expressed the half rated synchronous speed in terms of the rated synchronous speed, we will subsitute this expression in the place of the synchronous speed term when calculating the half rated starting torque.

Since we are also varying the voltage between the 2 conditions, let us assume that the rated voltage is V and that the half rated voltage is V/2.

Let the half rated starting torque be, `T_(s1)`

`T_(s1) = 3/omega_(ms) * [(V^2 * R_2) / ((R_1 + R_2)^2 + (X_1 + X_2)^2)]`

Substituting the expression for synchronous speed in terms of the rated synchronous speed:

`T_(s1) = 3/(omega_(ms2)/2) * [((V/2)^2 * 0.2) / ((0.2 + 0.2)^2 + (0.1 + 0.1)^2)]`

`T_(s1) = 6/omega_(ms2) * [(V^2/4 * 0.2) / (0.04 + 0.08 + 0.04 + 0.04)]`

`T_(s1) = 6/omega_(ms2) * [(0.05*V^2) / 0.2]`

`T_(s1) = (1.5*V^2) / omega_(ms2)`

Calculation of Starting Torque at Full Rated Condition:

In this condition, we will use the full rated voltage, V, the full rated frequency of 50Hz and the full rated synchronous speed to find the starting torque. Here we will substitute the synchronous speed at the full rated condition: `w_(ms2)`

Let the staring torque under full rated condition be: `T_(s2)`

`T_(s2) = 3/omega_(ms2) * [(V^2 * R_2) / ((R_1 + R_2)^2 + (X_1 + X_2)^2)]`

`T_(s2) = 3/omega_(ms2) * [(V^2*0.2 / 1)/0.2]`

`T_(s2) = (3*V^2)/omega_(ms2)`

Comparison of Starting Torques at both conditions:

We are required to find the ratio of the starting torque at the half rated condition to the full rated condition.

The starting torque at the half rated condition is: `T_(s1) = (1.5*V^2) / omega_(ms2)`

The starting torque at the full rated condition is: `T_(s2) = (3*V^2)/omega_(ms2)`

So, the ratio of: `T_(s1) / T_(s2) = (1.5*V^2)/omega_(ms2) / (3*V^2)/omega_(ms2)`

`T_(s1) / T_(s2) = (1.5*V^2)/omega_(ms2) * (omega_(ms2))/ (3*V^2)`

`T_(s1) / T_(s2) = 1/2`

Therefore, the ratio of the starting torque at the half rated condition to the full rated condition is 1/2 or 0.5.

ANS 3:

The frequency control method is one of the few ways in which the speed of an inducton motor can be controlled in its operation. Varying the frequency will vary the speed of the rotating magnetic field i.e. the synchronous speed and this will in turn vary the speed of the rotor i.e. the motor speed. Different frequencies of operation will have different torque-speed characteristics because each frequency will have a different synchronous speed. Each synchronous speed will correspond to a different motor speed and hence a different slip between itself and the motor speed. This in turn will result in different torques.

In the MATLAB script, I have used the same given data as was provided in the previous question. These are as follows:

• Rated Freqency - 50Hz
• Number of Phases - 3
• Stator Inductance, X1 = 0.1 Ohms
• Rotor Inductance, X2 = 0.1 Ohms
• Stator Resistance, R1 = 0.2 Ohms
• Stator Resistance, R2 = 0.2 Ohms

clear all
close all
clc

% Frequency Array
freq = linspace(10, 50, 5);

% No. of Phases
phase = 3;

% Rated Voltage 
v = 230/ sqrt(phase);

% No. of Poles
pol_num = 4;

x1 = 0.1;       % Stator Inductance
x2 = 0.1;       % Rotor Inductance
r1 = 0.2;       % Stator Resistance
r2 = 0.2;       % Rotor Resistance

In the above section of code, I have entered the given data in to their respective variables. Since we are varying the frequency, I have created an array containing 5 frequency values equally spaced from 10 Hertz to 50 Hertz. Each frequency will correspond to its own speed-torque characteristic. Next, I have set the numer of phases to 3 since most electric powered devices operate on a 3 phase AC supply. Then, I have calculated the rated voltage by dividing 230V (the voltage at which electricity is supplied to domestic areas) by the sqare root of the number of phases. The number of poles of the induction motor is also specified as 4 because there are 2 pole pairs. Finally, the stator and rotor inductance and resistance value are entered.

% Slip Array
s = linspace(1, 0, 100);

for i = 1 : length(freq)
    
    % Synchronous Speed Calculation
    w_ms(i) = (120*freq(i))/(pol_num);
    
    % Starting Torque
    
    T_start(i) = (3 / w_ms(i)) * [(v^2 * r2) / (r1 + r2)^2 + (x1 + x2)^2]
    
    
    for j = 1:length(s)
        
        % Torque Calculation
      
        T(j) = (3 ./ w_ms(i)) * [(v^2 * r2/s(j)) / (r1 + r2/s(j))^2 + (x1 + x2)^2];
        
        % Motor Speed Calculation
        w_m(j) = w_ms(i)*(1-s(j));
        
    end
    
    % Plotting of Motor Speed vs. Torque
    plot(w_m, T, 'linewidth', 2);
    hold on
    
end

In this section of code, I have programmed a for loop which will perform the calculations for the torque and the motor speed. This is done by first creating a slip array. The slip is the discrepancy between the synchronous and motor speeds. The slip is measured by taking the ratio of the difference between the synchronous and motor speeds to the synchronous speed. The slip array will help in the calculations for the torque. The highest slip that can occur in an induction motor is when the motor speed is 0 and this condition is referred to as the starting torque. The lowest slip that can occur is 0 and this is when the synchronous speed and motor speed are the same which will result in no torque produced. In this way, the slip array ranges from 1 to 0 with 100 equally spaced increments.

Next, the synchronous speed is calculated using the formula: `omega_(ms) = (120 * f)/p`

where,

f - Frequency (Hz)

p - Number of Poles

A for loop is used for this calculation because we have an array of 5 frequency values and each frequency will correspond to a specific synchronous speed. So the for loop will range from 1 to the length of the frequency array which is 5 which will output 5 synchronous speeds. But this for loop is not given an end command just yet because we will need it in the torque calculations.

Next, the starting torque is calculated with the for loop using the formula:`T_s = 3/omega_(ms) * [(V^2 * R_2) / ((R_1 + R_2)^2 + (X_1 + X_2)^2)]`

This formula is adapted from the original torque formula: `T = 3/omega_(ms) * [(V^2 * R_2/s) / ((R_1 + R_2/s)^2 + (X_1 + X_2)^2)]`

The only difference is that the slip term is omitted form the starting torque formula because as mentioned earlier, the starting torque is produced when the motor speed is 0 which results in a slip of 1. The 5 synchronous speeds calculated previously are used to output 5 different starting torques.

Then, the main torque values are calculated using the original torque formula and in it, the slip term is included. This calculation will use another for loop within the first one and so will use another counter, 'j'. The for loop will range from 1 to the length of the slip array which means we should get an ouput of 100 torque values with each value corresponding to a slip value. The slip terms in the torque formula will be replaced with the values from the slip array and the synchronous speeds calculated in the first for loop will also be included. In this for loop, we can also find the motor speed with the slip array and the synchronous speed.

The slip is given by the equation: `s = (w_(ms) - omega_m) / omega_(ms)`

This equation can be rearranged to make the motor speed the subject to give us: `omega_m = omega_(ms)(1-s)`

Here, we will use the synchronous speed values calculated from the first for loop and the slip values from the slip array. Since this calculation is in the second for loop it also ranges from 1 to the length of the slip array meaning this will output 100 values of the motor speed. The second for loop is terminated with the end command.

Now that we have the torque and motor speed values, I have used the plot command to plot the torque on the y-axis and the motor speed on the x-axis. A linewidth of 2 is provided for easier readability. The 'hold on' command is used so that each of the 5 torque - speed characteristics are plotted in the same figure. Here, the first for loop is terminated.

title('Torque - Speed Characteristics of Induction Motor')
ylabel('Torque, Nm')
xlabel('Motor Speed, rad/s')
legend('10Hz', '20Hz', '30Hz', '40Hz', '50Hz')
grid on

At last, modifications are made to the plot such as a title, axis labels and a legend key to show that each torque - speed characteristic corresponds to a particular freqency. The 'grid on' command is used to activate the grid of the plot space. The plot of the torque - speed characteristics for each frequency from 10Hz to 50Hz is shown below:

ANS 4:

A BLDC Motor stands for Brushless DC Motor. It is a synchronous motor that is powered by a DC power supply via an inverter which produces an AC power supply. Conventional DC motors were constructed with brush contacts found on the outside of the commutator. Over time, the brush contacts of conventional DC motors experience wear and tear and can even cause sparking due to high friction between the brush and the commutator. So, the brush type DC motor was considered to be unsuitable for applications requiring high durability and long life. It was eventually replaced with the improved design of the BLDC motor.

Construction:

The rotor of the BLDC motor is a permanent magnet with a north and south pole. The stator consists of 3 sets of 2 coil windings, which for example can be called A, B and C. BLDC motors can be of outrunner design of inrunner design. In the outrunner type, the rotor surrounds the stator, whereas in the inrunner type, the stator surrounds the rotor. A DC supply is given to one set of the coil windings causing them to get energized and this produces a magnetic field. One end of the set of windings will be the North pole and the other end becomes the South pole. The interaction between this electromagnetic field and the rotor magnetic field, exerts a force on the rotor to rotate producing a torque.

Working:

The image above shows the outrunner type of BLDC. As we can see, initially the rotor is kept in line with coil set C and coil set A is made to be energized. The rotor's north pole and south pole will be attracted to the south pole and north pole of coil set A respectively. This will cause the rotor to move toward coil set A. When the rotor arrives at coil set A, coil set B is then energized by passing the DC supply through it and the rotor will move toward coil set B. When the rotor arrives at coil set B, coil set C is made to energize and the rotor moves toward coil set C.

At this point, the rotor has made a rotation of 180 degrees from its initial position, and now the magnetic polarity of the rotor is reversed. So, to ensure the continuous rotation of the rotor in the same direction, the DC supply provided to the coils is reversed throughout the next 180 degrees of rotation. This process is repeated, with the energizing of the coils occurring in sequence to ensure the continuous rotation of the rotor to produce torque in the same direction.

This kind of operation is similar to that of an induction motor whereby there is a slip between the stator flux and rotor speed. The rotor speed will always be lesser than the stator flux and this is what causes the rotor to experience a force. The stator flux has to be quicker in order to attract the rotor behind it. If the stator flux and rotor speed were the same, there would be no interaction between the rotor magnetic field and stator magnetic field. This would result in no force and hence no torque produced by the rotor.

The graph below shows how the DC supply is varied in coil sets A, B and C during the operation of the BDLC motor.

The drawback of this operation is that at any given instant, only one set of coils is energized while the remaining 2 sets of coils are inactive. This greatly reduces the potential power output of the motor and this can be overcome by energizing 2 sets of coils at any instant.

Consider the rotor is in line with coil set C and coil set A is energized in such a way that the rotor will move toward it. When coil set A is energized, the coil set behind the rotor's current position (which in this case is coil set B) should also be energized at the same instant. But, the DC supply given to coil set B should be in the reverse direction in order to create a repulsion with the rotor magnetic field and push the rotor forward. This will give the rotor more thrust to move toward coil set A and in effect will increase the speed of rotation and the torque as well to produce a higher power output. This operation can be seen in the image below:

The rotor's north pole is currently at coil set C. Coil set A is energized so that the south pole of the electromagnetic field attracts the north pole of the rotor (shown by the green arrow). But when coil set A is energized, coil set B is also energized but with the reverse DC supply so that its north pole will repel the north pole of the rotor (shown by the red arrow).

This method of energizing 2 separate sets of coils together can be simplified by connecting one free end of the coils together. For instance, when the rotor's north pole is at coil set C, the south pole of coil A and north pole of coil B need to be energized at the same time to create the attractive and repulsive forces respectively. So the free end of coil set A and be linked to coil set B. Similarly, the other pairs of coil sets can be connected by the same coil at their free ends. The image below shows how these free ends are connected together in comparison with the coils being separately energized.

A hall effect sensor or a rotary encoder is used to measure the position of the rotor at any given instant. The sensor sends this information to an electronic controller which gives commands to energize and de-energize a set of coils. So, the electronic contoller with the help of the hall effect sensor will know when to energize and de-energize a set of coils based on the rotor's instantaneous position and will also know which specific coils to energize and de-energize.

Performance:

The two main performance parameters of a BLDC are the back - emf constant i.e. speed constant and the motor constant i.e. torque constant. The amount of power that can be supplied to a BLDC is limited by the amount of heat dissipation. This is because the high heat can damage the winding insulation of the coils and can also weaken the permanent magnets of the rotor.

The BLDC motor generates maximum torque under when stationary and this decreases linearly as the rotor velocity increases. The efficiency of the BLDC motor is greatest in the no load and low load conditions. However, the conversion from electrical energy to mechanical energy is more efficient in a BLDC motor for several reasons.

One reason is that the frequency of switching electricity is solely dependent on the position of the rotor so that power is supplied at the right time at the right place. So, the electronic control of a BLDC makes them more efficient and power saving than the brush type DC motor.

Also, the absence of the brush contacts makes the BLDC motor have lesser mechanical losses due to the friction (and less heat and noise) that would otherwise exist between the brushes and commutator. This also makes them lighter than the brush type DC motor and provide the same power output giving the BLDC a higher power to weight ratio.

In addition, as there are no windings on the rotor, the windings are supported by the housing and they can be cooled by conduction when heat builds up. This can prevent the BLDC motor from getting too hot and damaging the winding insulation and the weakening the permanent magnets. Also, the internal construction of the BDLC motor can be protected from the entry of dirt or other foreign matter.

Under high mechanical loads, BLDC motors are more comparable in efficiency and as a result they are more reliable and less noisy than the brush type DC motors. This means they have a longer working life and can operate at higher speeds with maintenance-free operation.

Applications:

BLDC motors are used in a variety of applications such as:

• computer hard drives
• CD/DVD players
• in transport e.g. HEVs, and EVs
• cordless electronics e.g. drills, blowers
• in heating and ventilation for HVAC equipment
• in manufacturing applications e.g. motion control systems, positioning and actuation

Despite BLDC motors being better than the brush type DC motors in every single aspect, they are still more expensive than the latter. This is understandable given the numerous construction and working improvements that the BLDC motor offers over the brush type. Due to this, DC Brush type motors are still in use in certain applications where it is not financially feasible to implement the BLDC motor.