Week-11 Challenge: Braking

Week 11 Challenge - Braking

1. For a defined driving cycle, calculate the energy required for braking.
2. Why electric motor can’t develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated?
3. Make a MATLAB program which plots contour of given motor speed, torque and efficiency values. Attach the code as a .m file attach a screenshot of all the plots

ANS 1:

Theory

Braking is a dynamic state of a vehicle where it slows down i.e. decelerates from a certain velocity. The vehicle’s velocity will decrease over a certain period of time and this can occur until the vehicle reaches rest or to another velocity level. In either case, the vehicle is still in motion and as a result, it possesses kinetic energy. When a vehicle accelerates either from rest or while in motion, its kinetic energy will increase and the opposite occurs when a vehicle decelerates.

The kinetic energy carried by an object in motion of mass, m and velocity, v is given by:

`E_k = 1/2 * m*v^2`

If we were to determine the change in kinetic energy, we can find the kinetic energy at the starting velocity and at the final velocity and subtract the former from the latter. This can be simplified by re-writing the kinetic energy equation as:

`deltaE_k = 1/2*m*(v_f - v_i)^2`

Where,

Final velocity - `v_f`

Initial velocity - `v_i`

Program

For this question, I am using the SC03 drive cycle to find the braking power incurred by the vehicle. Firstly, I needed the time and velocity values of the aforementioned drive cycle. The velocity values were in miles per hour (mph) by default, so I had to convert them to metres per second (m/s) in order to keep all values in SI units for the calculations later on. The time and corresponding velocity values in both mph and m/s were entered in to an Excel table.

Next, I started writing the MATLAB program. I needed to extract the values from the table, so I imported the Excel table into the MATLAB workspace as a numeric matrix. This table was saved in the workspace as ‘SC03’. To extract the values from the table, I used indexing to take only the time and velocity data in m/s for the entirety of the drive cycle.

The image below shows the importing of the drive cycle data from the Excel file.

The image below shows the drive cycle values in a table when opened up from the workspace.

The time values were assigned to the variable ‘time’ and the velocity values were assigned to the variable ‘inst_vel’ for the ‘instantaneous velocity’. The vehicle mass was also specified to be 1500 kg and this was assigned to the variable ‘mass’. This will be used for the kinetic energy calculations later on.

m = 1500;                                                                       % Vehicle Mass (kg)
time = SC03(1:601, 1);                                                   % Time array from SC03 drive cyc
inst_vel = SC03(1:601, 3);                                              % Instantaneous velocity array from SC03 

Then, I created a ‘for loop’ which will run for the entirety of the drive cycle minus 1 second. This for loop will run whenever it encounters the deceleration condition. Deceleration is when the final velocity is lesser than the initial velocity. So, I have used the ‘lesser than’ symbol in an ‘if’ statement to create the condition wherein the instantaneous final velocity should be lesser than the instantaneous initial velocity. In this explanation, final and initial are relative terms. They refer to the changing state of the velocity.

Whenever this deceleration condition is met, the ‘for loop’ will perform a calculation to calculate the change in the kinetic energy for the change in the velocity. This calculation will be done using Equation 2 which was previously mentioned. In this way, the ‘for loop’ will go through the entire drive cycle data and look for the instances when deceleration occurred and calculate the change in the kinetic energy that occurred during those instances.

The ‘for loop’ and the ‘if’ statements are both terminated once all the calculations are performed. For reference, I have plotted the drive cycle data to show how the velocity changes with respect to the time. This is done using the ‘plot’ command. Also, the ‘fprintf’ statement to print out the summation of the change in the kinetic energy values.

This is so that the user will know that was the total energy incurred by a vehicle whenever it was in the braking state for the entirety of the SC03 drive cycle. The total energy incurred during braking for the entire drive cycle will be displayed in the command window at the end of the execution of the program and will be in kJ units given to a precision of 2 decimal places.

for i = 1:600
    
    if inst_vel(i+1) < inst_vel(i)                                 % Deceleration condition
        Ek(i) = 0.5*m*(inst_vel(i+1) - inst_vel(i))^2;             % Instantaneous braking energy
        
    end
end

% Plotting

plot(time, inst_vel);
xlabel('Time (s)')
ylabel('Velocity (m/s)')
title('SC03 Drive Cycle')

hold off

fprintf('The total energy required for braking for the SC03 Drive Cycle is %.2f KJ', sum(Ek)/1000) 

The image below, shows the SC03 drive cycle plot as well as the command window showing the total energy incurred in braking the vehicle for the entire drive cycle duration.

ANS 2:

Braking is the change in the motion of a vehicle from a starting speed level to a lower speed level. The act of braking, if sustained for a sufficient time, can either bring a vehicle to a lower speed than it started with or it can bring it to complete rest. So, the act of braking brings about deceleration which is decrease in the speed of the vehicle over a period of time which corresponds to the speed values at those times. Braking is an act which is desired in speed control applications and in the case of vehicles, they are very much necessary to ensure the safety of the driver and the control of the vehicle. There are several methods for performing braking in vehicles whether they are conventional vehicles or HEVs/EVs.

Mechanical Braking

In a conventional IC engine vehicle, there is only one physical braking mechanism that can be performed and that is done with the brake pads/discs. The braking pedal is pressed by the driver, which results in the application of the brake pads/discs onto the wheels of the vehicle. The friction that arises between the rotating wheels and the stationary brake pads/discs will decrease the speed of the vehicle. Another method of braking in conventional vehicles is called "engine braking". In this method, the throttle (accelerator pedal) is released, and the gear is changed to a lower gear (in manual transmission vehicles), which will result in the deceleration of the vehicle. This occurs via the "manifold effect" whereby, the releasing of the accelerator pedal leads to the stopping of the fuel flow to the engine cylinder. This also stops the air flow reslting in a negative traction torque. In heavy duty vehicles e.g. trucks, lorries, engine braking is very useful because it helps to prevent the high heat build up in the brake pads/discs which would be generated if they were only used. So, engine braking is useful as it allows for the desired control while reducing wear and tear on the physical braking mechanism.

Electric Braking

In motors and HEVs/EVs, various methods of braking exist. One method is using a 'brake motor' which consists of a DC rectifier, a magnet, springs and a friction disc. The mechanism of the brake motor involves the rectifier giving a signal to disrupt a magnetic field which will influence the springs to expand and apply a force on the friction discs which will come in to conact with the moving part that is in contact with the rotor. When the vehicle is accelerating or moving with constant speed, the motor still receives its 3 - phase AC supply from the inverter and so the brakes are not applied. This DC rectifier relays no signal to the magnetic field, so that the springs stay in their stationary position and do not force any contact on to the friction discs. However when vehicle is not accelerating, the 3 - phase AC supply from the inverter is cutoff, causing the brakes to be applied. The DC magnetic field is disrupted by an air gap which causes the springs to get excited and expand and apply force on to the friction disc to produce braking action.

EVs and HEVs have the brilliant feature of "regenerative braking" which is where during the act of braking the motor begins to operate as a generator and instead produces electrical energy. Essentially, when a vehicle decelerates, the change in the kinetic energy of that vehicle is used to produce a torque on the motor and thereby producing electrical energy via the generator effect which can be stored by the battery or be diverted for another powering purpose. The electricity generated by the regenerative braking phenomenon is referred to as 'back emf' because it produces a current that travels in the opposite direction to the conventional current. For regenerative braking to occur, the rotor of the motor has to rotate at a speed that is greater than the synchronous speed. This will result in the back emf being greater than the terminal voltage so that the potential difference is strong enough to influence the current to flow in the opposite direction. Using regenerative braking not only decelerates the vehicle, but also increases the efficiency and the energy savings of EVs and HEVs.

Another method of braking in EVs and HEVs is referred to as "dynamic braking". In this method of braking, the supply voltage powering the motor is disconnected and instead a resistor is conected. The absence of the resistor would result in significant frictional and windage losses which would decrease the efficiency of the vehicle and lead to longer braking times. This is not desirable because a good brake needs to bring the vehicle to rest within a certain amount of time and done in a safe manner. With the resistor fitted instead, the braking time can be reduced significantly.

In addition, another braking method in motors and EVs/HEVs is referred to as "terminal voltage reversal". In this method of braking, two separate voltage sources are used with different polarities as well as a switch. When the motor is running normally, the first voltage source is the main supply of power to the motor and creates a current flow in the conventional current. When the braking action is desired, the switch is closed to connect the second voltage supply in to the circuit and disconnect the first voltage source. Because the second voltage source has a reversed polarity, the current flow in the circuit to the motor is reversed thereby slowing the motor down by making its rotor rotate in the reverse direction i.e. negative torque. This method is not always preferred due to the fact that it produces a severe braking and generates high levels of braking torque. So, a braking resistor is also used in the circuit and will come in to use whenver the second voltage source is connected to power the motor.

Control Strategies

From these various braking methods, it is obvious that both mechanical and electrical braking is useful and both have their benefits and drawbacks. In regards to a vehicular application, depending on the type of vehicle either one of both types are utilised. In the case of conventional vehicles, only mechanical braking consisting of friction braking and engine braking is utilised whereas in HEVs/EVs, both are utilised. In HEVs/EVs using both mechanical and electrical braking means that the braking application can get the advantages of both types and minimize the disadvantages as much as possible. HEVs and EVs use controls strategies for controlling the braking method and deciding which brakes to apply and when to apply them. These control strategies are referred to as 'braking modes' and they will find a compromise between the mechanical and electrical braking methods to see which is the most efficient at its job.

For instance, at lower speed levels, it may be more suitable to use mechanical brakes i.e. friction braking because these brakes will be more than capable to brake the vehicle and not generate too much heat and thus will not produce too much wear and tear for the brake pads/discs. However at higher speed levels, it would be more suitable to use electrical braking. This is because electrical braking is more effective, it won't generate nearly as much heat as mechanical brakes would. Electric braking also is more efficient and energy saving because at higher speeds, more kinetic energy will be carried by the vehicle and thus, regenerative braking can occur to convert that kinetic energy in to electrical energy to be stored back in the battery. If mechanical brakes were used instead, high amounts of heat would be generated and would lead to wear and tear of the brake pads/discs over time. The high heat generation also means that the dissipation of the kinetic energy is wasted in to energy that cannot be reused which reiterates its inefficiency compared to the electric brakes. In addition, the electric brakes would also be smoother and more gradual and not sudden like the mechanical brakes would which would make the braking action more comfortable and safer for the passengers.

So these control strategies i.e. braking modes will know from the vehicle data, which type of braking is the best to use depending on its current state of motion. There are various braking modes for different situations the vehicle may face and by utilizing these control strategies, the vehicle can be operating at the highest efficiency at all times and make the best use of the 2 braking mechanisms without damaging or sacrificing one for the other.

Braking Torque

The starting torque of a motor is the torque developed by the motor at no speed and this is a moderately high value. At high speeds, electric motors cannot develop braking torques compared to their starting torques. This is because, the greater the speed of the vehicle, the greater the braking torque that is required to brake the vehicle. This is simply because at higher speeds, more force is needed to oppose the rotation of the rotor of the motor and this means a greater torque i.e. braking torque. So, when a vehicle is travelling at high speeds, a larger braking torque is needed to brake the vehicle. However, when a motor is starting, its starting torque i.e. torque at no speed is the highest torque the motor can achieve. This is because the speed - torque characteristics of a motor are characterized by a linear relationship. If we were to plot a graph of the angular speed of the rotor on the y - axis and the torque on the x - axis, the plot would show a straight line with a negative gradient. The starting torque is represented by the x - intercept of this plot in the first quadrant i.e. the forward motoring region. In this way, the torque generated by the starting of the motor will not be similar to the braking torque that is generated at higher speeds. Also, at higher speeds, the terminal voltage of the voltage source increases, and so the torque values will differ further from the starting torque.

ANS 3:

A contour plot that shows the characteristics of the speed, torque and the efficiency of a motor is referred to as an efficiency map. Contour plots are very useful as they make it very easy to understand which operating states result in the highest efficiencies of the motor. In this way, a manufacturer can design a motor knowing which applications it will be useful for depending on the speed, the torque it achieves at that speed and the efficiency of the motor which operates at those values.

In order to plot this, we need the specifications of a motor. In this question, I am taking generic values for the specifications of the motor as well as for the losses that occur in the motor operation.

In the MATLAB script, I begin with the clear all, close all and clc commands. Then, I create arrays for the speed and the torque of the motor using the ‘linspace’ command. For the speed, I am taking a range of values from 0 to 2000 radians per second whereas for the torque, I am taking a range of values from 0 to 300Nm.

These arrays are then used to create 2D grid coordinates using the ‘meshgrid’ command. These coordinates will help in the creation of the contour plot. In order to find the efficiency of the motor, we need to consider the various losses that occur in the motor’s operation.

clear all
close all
clc

% Speed Array
omega = linspace(0, 300, 100);

% Torque Array
T = linspace(0, 2000, 100);

% Meshgrid
[X,Y] = meshgrid(T, omega);

The main losses that occur are:

• Copper losses
• Iron losses
• Windage losses
• Conduction losses

These various losses are dependent on the speed and the torque of the motor. They each also have a characteristic constant of proportionality which are all assumed for the following calculations and represented as 'K'. In each calculation, the speed and the torque are represented respectively as: `omega and T`

• Copper losses = `K_(Cu) * omega^2`
• Iron losses = `K_(Fe) * T^2`
• Windage losses = `K_w * T^3`
• Conduction losses are assumed to be 18 Joules.

The section of code below shows the loss coefficients of each loss type being assigned to their values.

% Loss Coefficients
cu = 0.17;                              % Copper coefficient
fe = 0.006;                             % Iron coefficient
win = 0.000015;                         % Windage coefficient

This section of code shows the calculations of the losses:

% Losses
cu_loss = cu*(Y.^2);                  % Copper Losses
fe_loss = fe*(X.^2);                  % Iron Losses
win_loss = win*(X.^3);                % Windage Losses
cond_loss = 18;                       % Conduction Losses

We also need to calculate the efficiency of the motor using the data we currently have. Since efficiency is defined as the ratio of the output power to the input power, we can calculate the efficiency of the motor by determining the input and output power of the motor. The output power of the motor can be calculated by multiplying the angular speed with the torque.

`P_(out) = T*omega`

Then, we can calculate the input power supplied to the motor by adding the output power to the energy losses.

`P_(i) = P_(out) + Cu_(loss) + Fe_(loss) + W_(loss) + Cond_(loss)`

Since efficiency is defined as the ratio of the output power to the input power, we can calculate the efficiency of the motor by:

`eta = P_(out) / ((P_i) + Cu_(loss) + Fe_(loss) + W_(loss) + Cond_(loss))`

The section of code below shows the calculations of the output and input power and the efficiency as well. As you can see, the calculations for the output power are performed with the values from the mesh grid and not the individual variables of speed and torque.

% Output Power
P_out = (X.*Y);      

% Input Power
P_in = P_out +  (cu_loss) + (fe_loss) + (win_loss) + (cond_loss);

% Efficiency
eta = ((P_out)./(P_in));

The section of code below shows the plotting of the contour with the 'contour' command. In addition, an array for the efficiency values we want to plot for is created called 'W' and is assigned for values ranging from an efficiency level of 0.75 to 0.95. This array is used in the creation of the contour. Axis labels and a title is also made for better readability.

% Efficiency values for contour
W = linspace(0.75, 0.95, 10);

% Contour Plot

contour(X,Y,eta,W)
xlabel('Angular Speed, rad/s')
ylabel('Torque, Nm')
title('Speed, Torque and Efficiency characteristics of Motor')

The image below shows the contour plot that is achieved when the code is executed:

From this contour plot, we can see that the highest level of efficiency arises in the innermost contour plot and this efficiency decreases as the contours increase in size from the innermost contour. The innermost contour shows that any point on the contour will have an efficiency of 0.9054. The second, third and fourth innermost contour display efficiencies of 0.88, 0.86 and 0.84 respectively anywhere on their respective contour plots.