Week-11 Challenge: Braking

Q1)

• In urban area driving, a significant amount of energy is consumed in braking. The braking performance is one of the most importance in vehicle safety.
• Electrification of vehicles makes it feasible to recover some of the energy lost in braking.
• A hybrid braking system was developed to ensure the efficient braking of the vehicle and recovering the braking energy.
• The design and control objectives of such a hybrid braking system are to ensure sufficient braking force to quickly reduce the vehicle speed, properly distributing the brake force to the front and rear axle, and also recover as much braking energy as possible.

By calcualting we get the total braking energy= 794.47kwh

clc
 %Program to find the Braking energy for the drive cycle
>> drive_cycle=xlsread('Book1'); %Loading the drive cycle
>> speed=drive_cycle(:,2); %Speed values extracted
>> time=drive_cycle(:,1); %Time values extracted
>> vf=[]; %Final speed array
>> vi=[]; %Initial speed array
>> m=600; %Mass of vehicle 
>> %For loop for picking the initial and final speed of decelaration
>> for i=1:(length(speed)-1);
if speed(i)>speed(i+1)&&speed(i+2)<speed(i+1)&&speed(i-1)<speed(i)
vi=[vi,speed(i)];
elseif speed(i)<speed(i+1)&&speed(i+2)>speed(i+1)&&speed(i-1)>speed(i)
vf=[vf,speed(i)];
end
end
Index exceeds the number of array elements. Index
must not exceed 31.
 
>> vf=[vf,0]; %Adding zero to the last value of Final speed
>> B=(0.5.*m.*(vi-vf).^2)/1000; %Calculating the Braking energy for each step down
>> x=sprintf('Total Braking Energy = %0.2fkwh',sum(B)); %Formating the data into string.
>> disp(x) %Displaying the total braking energy
Total Braking Energy = 794.47kwh
>> %Plotting drive cycle
>> plot(time,speed,'linewidth',2)
>> title('DRIVE CYCLE')
>> xlabel('TIME')
>> ylabel('SPEED')
>> grid on

Q2)

• The motor operation where in the first quadrant both torque and speed are +ve and in the second quadrant the speed is +ve and the torque is –ve.
• At high speed, the torque will be less & operates at a constant power hence it is difficult to bring the vehicle to rest or decelerate immediately.
• The peak acceleration torque is greater than the peak decelaration torque. Hence braking torque of the EV motor is not as high as the acceleration Torque.
• Mechanical Braking is mostly used in scooters, motor vehicles, and motorcycles where small power is required.
• It is essential in manufacturing power transmission applications, material handling, etc. It delivers forces to the axle or a wheel to stop motion.
• It helps to reduce the speed of the system slowly by the mechanical process when compared to electrical braking.
• Types of mechanical braking:- (a) Drum Braking (b) Disk Braking (c) Band Braking (d) Pawl and Ratchet braking.
• Electrical Braking is used to reduce the speed of the machine depending on flux and torque.
• This type of braking is mainly used for functional braking to control the speed of the machine. It is easy to handle and comfortable. But it cannot be used for emergency braking and parking braking.
• Types of Electrical Braking:- (a) Plugging type braking (b) DC injection type braking (c) Eddy current braking (d) Dynamic resistor type braking (e) Regenerative braking (f) Sharing DC bus type braking.
• Electrical and mechanical braking systems can are coordinated by using a brake strategy subsystem (1) Serial Braking Strategy (2) Parallel Braking Strategy.

Q3)

• In this MATLAB script, we create arrays for the speed and the torque of the motor using the ‘linspace’ command.
• For the speed, consider a range of values from 0 to 2000 radians per second whereas for the torque, consider 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.
• To find the efficiency of the motor, we need to consider the various losses that occur in the motor’s operation.
• These are:- (a) Copper losses (b) Iron losses (c) Windage losses (d) Conduction losses.
• Whereas (a) Copper losses = KCu⋅ω2 (b) Iron losses =KFe⋅T (c)Windage losses =Kw⋅
• The values of coefficients are taken as general values. Condition losses are assumed to be 18 Joules.

 %Speed Array
>> omega= linespace(0, 300, 100);
Unrecognized function or variable 'linespace'.
 
Did you mean:
>> omega= linspace(0, 300, 100);
>> %Torque Array
>> T= linspace(0, 2000, 100);
>> %Meshgrid
>> [X,Y] = meshgrid(T, omega);
>> %Loss Coefficients
>> cu = 0.17; %Copper coefficient
>> fe = 0.006; %Iron coefficient
>> win = 0.000015; %Windage coefficient
>> %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
>> %Output Power
>> P_out = (X.*Y);
>> %Input Power
>> P_in = P_out + (cu_loss) + (fe_loss) + (cond_loss);
>> %Efficiency
>> n = ((P_out)./(P_in));
>> %Efficiency values for contour
>> W = linespace(0.75, 0.95, 10);
Unrecognized function or variable 'linespace'.
 
Did you mean:
>> W = linspace(0.75, 0.95, 10);
>> %Contour Plot
>> contour(X,Y,n,W)
>> xlabel('Angular Speed, rad/s')
>> ylabel('Torque, Nm')
>> title('Speed, Torque and Efficiency characteristics of motor')
>> grid on