Week-11 Challenge: Braking

Question 1 For a defined driving cycle, calculate the energy required for braking.

As we know that the energy can never be created or niether destroyed with the today's development

we have been able to utilize the energy which is been wasted at the time of braking. The kinetic energy which assist the vehicle to move also gets dissipated while the braking is done. Though it depends on multiple parameters on how efficient the braking would be it can be depended upon.

Number of Brake Pistons

Duration and Degree of Brake Pedal been Pressed

Road Condition

Temperature

Brake Pad and Disc Condition

Mass of the Vehicle

Still with a basic operation where the vehicle in motion and the brakes are applied the same efficiency could be calculated by the given equations.

BE= ½ mv^2

BE= braking energy

m-mass of the vehicle

V = velocity

In this manner the same calculation is been carried out for calculating the brake efficiency that would be required at the peak point. It bieng a drive cycle with speed variations the braking power required would vary but still the highest braking power required is mentioned below

Question 2 :-Why electric motor can’t develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated?

When the motor is at rest and it has to operate from standstill position to the reach the moderate speed the higher torque is required to bring the motor in operation. Whereas when the vehicle is already at high speed the capping or the limit is reached by the motor operation and thus any further higher torque could not be obtained while braking. The braking operation happens in the motion which is been distributed as per proportion in the BRAKE ENERGY and the REGENERATIVE BRAKING.

• How electric and mechanical brakes are coordinated?

In Mechanical braking which we usually see and experience in all vehicles the brake pedal in the vehicle cabin helps us to engage or disengage the brakes Currently the braking technology have done wonders by inventing life saving braking technologies such as ABS, EBD, BA and mny others. All these technologies play vital role in stopping the vehicle and coming back to rest from a high speed. In all these vehicles when the driver applies the brakes the brake peds come closer to the disc connected with the driving disc. The brake pads usually help in efficient braking

There are two generic methods by which the vehicle could be brakedin the ELECTRIC BRAKES.

Series - The Regenerative Braking or the Mechanical Braking noth are applied one after another

Parallel - The Regenerative Braking or the Mechnical Braking both are applied Parallely.

When the brake pedal is pressed the hydraulic fluid reaches the brake piston and created the braking pressure on the wheels and the same kinetic energy is been used by the Regenerative Braking for generating more power using the Bi-directional motor which operates as a generator while braking.

An example of the proposition of Regenerative Braking and the Mechanical Braking

NISSAN LEAD - EDIB (ELECTRONIC DRIVEN INTELLEGENT BRAKING)

EDIB (Electric Driven Intelligent Brake) controls the regenerative brake and friction brake to support both of these requirements. Further, it also controls the reactive force from the pedal in order to unify the feeling when stepping down on the pedd and the sense of deceleration.

Question 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.

As we know that contour is a graphical technique of representing 3- dimensional surface . the same is mentioned below for the given detail

clear all

close all

clc

x=linspace(1,1000,250); %Speed in rad/sec

y=linspace(1,250,250); %Torque in Nm

%Motor loss constants

kc=0.2; %Copper loss constant

ki=0.008; %Iron loss constant

kw=0.00001; %Windage loss constant

Const_Loss=20; %Constant motor loss

[X,Y]=meshgrid(x,y); %Converting to 2D

Output_power=(X.*Y); %Power = Torque x Speed

B= (Y.^2)*kc; %Copper loss

C= X.*ki; %Iron loss

D= (X.^3)*kw; %Windage loss

Input_power = Output_power + B + C + D + Const_Loss;

Efficiency= Output_power./Input_power; %Eff of the motor

V = [0.70,0.80,0.90,0.91,0.92,0.925,0.93]; %Efficiencies to be plotted

box off

grid off

contourf(X,Y,Efficiency,V);

contourcbar

title('Speed-Torque-Efficiency Characteristics of Motor');

xlabel('Speed (rad/sec)');

ylabel('Torque (Nm)');