Week-11 Challenge: Braking

Question - 1

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

Answer -

The driving cycle used in the given example is the UDDS Driving Cycle

Link - https://drive.google.com/file/d/16-Nckl3aUGQ81ov_yVemkWNHHMIZ_7oe/view?usp=sharing

As we know that the energy can never be created or be destroyed with the today's development we have been able to utilize the enegry which is been wasted at the time of braking. The kinetic energy which assists 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 & Degree of Brake Pedal been pressed
• Road Condition
• Temperature
• Brake Pad and Brake Disc Condition
• Mass of the vehicle

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

BE = 1/2 * Mass of the vehicle * (Velocity ^ 2)

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

Braking Energy Required = 160 kJ for the speed of 57 kmph

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Question - 2

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

Answer -

Why electric motor can’t develop braking torque at high speed similar to starting

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.

Electrical Braking and Mechanical Braking

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 many 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 pads come closer to the disc connected with the driving disc. The brake pads usually help in efficient braing,

There are 2 generic methods by which the vehicle could be braked in the Electric Brakes

• Series - The Regenerative Braking or the Mechanical braking both are applied one after another
• Parellel - The Regenerative Braking or the Mechanical braking both are applied parellely.

When the brake pedal is pressed the hydraulic fluid reaches the brake pistons 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 Intelligent Braking)

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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 contour is a graphical technique of representing 3-dimensional surface. The same is mentioned below for the given details,

clear all
close all
clc

%speed array
w = linspace(0,1000)
%Torque Array
T = linspace(0,300)

%Loss coefficients
kc = 0.18;%copper losses
ki = 0.006;%Iron Losses
kw = 0.000015;%Windage Losses

%Meshgrid
[X,Y] = meshgrid(w,T)
copper_loss = (Y.^2)*kc
iron_loss =X*ki
windage_loss=(X.^3)*kw
cons=20

o_p=X.*Y

input_power=(copper_loss)+(iron_loss)+(windage_loss)+(o_p)+cons
Eff=(o_p)./(input_power)
n=linspace(0.7,0.95,10)
box off
grid off
contour(X,Y,Eff,n)
xlabel ('speed rad/s')
ylabel ('Torque N/m')

The contour for the code,

.m file - https://drive.google.com/file/d/1yOF_iJWY6DE5V-9KiYuXxSn0qeDfDmed/view?usp=sharing