Week-11 Challenge: Braking

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

Temprature

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=12mv2BE=12mv2

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-

Braking Energy Require 

To reduce speed from 80 to 40 kmph in 20 sec =292 KJ

To reduce speed from 80 to 0 kmph in 0 sec =560 KJ

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 a standstill position to reach the moderate speed the higher torque is required to bring the motor into 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, braking technology has done wonders by inventing life-saving braking technologies such as ABS, EBD, BA and many others. All these technologies play a vital role in stopping the vehicle and coming back to rest at a high speed. In all these vehicles when the driver applies the brakes the brake pads

come closer to the disc connected with the driving disc. The brake pads usually help inefficient braking.

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

1.Series - The Regenerative Braking of the Mechanical Braking are applied one after another.

2.Parallel - The Regenerative Braking or the Mechanical 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 LEAF EDIB (ELECTRONIC-DRIVEN INTELLIGENT 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 pedal and the sense of deceleration

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.

a MATLAB program that plots the contour of given motor speed, torque and efficiency values

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)');

Explanation:

In this program, we are plotting speed Vs torque characteristics with the efficiency values. For this, first, we are defining the speed and torque values of the motor.

Then as we need to draw a contour, we convert this ID array to a 2D array using the "MESH GRID" command.

Then we are defining the Cu constants for losses, like Cu loss constants, iron loss constant and windage loss constant.

Pi=Pout+KcT^2+Ki*w+Kw*^3+Const

Using this formula, we calculate the efficiency and then plot the contour.