Week-11 Challenge: Braking

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

A driving cycle is a series of data points representing the speed of a vehicle versus time.Driving cycles are produced by different countries and organizations to assess the performance of vehicles in various ways, as for example fuel consumption and polluting emissions.

Considering UDDS drive cycle

Calculating total energy for braking using MATLAB code

clear all;
close all;
clc;
E=0;
P=0;
a=zeros(1,475);
b=zeros(1,475);
c=zeros(1,475);
n=1;
m=2100;
data=xlsread('UDDS.xlsx');
time=data(:,1);
vel=0.44704.*data(:,2);
figure(1)
plot(time,vel);
xlabel('Time (sec)');
ylabel('Velocity (m/s)');
for i=2:1369
    if vel(i)>vel(i+1)
        a(1,n)=vel(i);
        b(1,n)=vel(i+1);
        c(1,n)=time(i+1)-time(i);
        n=n+1;
    end
end
for j=1:475
    %Total Energy(KJ)
    E=E+0.5*m*(a(j)^2-b(j)^2)/1000;
    %Total Power(KW)
    P=P+(0.5*m*(a(j)^2-b(j)^2))/(1000*c(j));
end

Total braking enegy required in entire drive cycle is 4406.7 Kw

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

Torque-speed characteritics of electric motor

The electric motor starts with low speed which is in constant tourque region. So, high torque can be developed by motor at low speed. At high speed, motor runs in constant power region where high speed is achieved by compromising torque.

In constant power region, `Power=T*omega`

If `omega`is increased `T` has to decreased to maintain contant `Power` .

At the time of forward braking, speed is high and torque is low. 

Cordination of Regenerative and Mechanical Braking

There are limitation of regenerative braking. Some of the limitation are 

1. Battery can accept charging current of particular range.
2. Power conversion is a great concern due to it's complexity. Bidirectional powerconverter is required.  

The limitation of mechanical brake is it's low braking torque at high speed.

Combination of regenerative and mechanical braking is used for better comfort and enegry  recovery.

The types of brake control strategies are

a) Series with optimal feel:- The control objective is to minimize the stopping distance and optimize the driver’s feel. When the braking force demanded is less than the maximum braking force that the electric motor can produce, only electrically regenerative braking will apply. When the commanded braking force is greater than the available regenerative braking force, the electric motor will operate to produce its maximum braking torque, and the remaining braking force is met by the mechanical brake system.

b) Series with optimal energy recovery:- The principle of the series braking system with optimal energy recovery is to recover the braking energy as much as possible in the condition of meeting the total braking force demanded for the given deceleration.

b)Parallel:- The parallel brake system includes both an electrical (regenerative) brake and a
mechanical brake, which produce braking forces is parallel and simultaneously. The operating principle.

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

MATLAB Code

close all;
clear all;
clc;
Velocity = linspace(1,1000);               % speed in rad/s
Torque = linspace(1,250);                  % Torque in Nm 0 to 40 Nm
t=linspace(-1,-250);
[X,y]=meshgrid(Velocity,t);
kc =0.2;                                   % kc - copper losses coefficient
ki = 0.008;                                % ki – iron losses coefficient
kw = 0.00001;                              % kw – windage losses coefficient
conL = 20;                                 % Constant motor losses

[X,Y] = meshgrid(Velocity,Torque); 
Output_power_1 =(X.*Y);
Output_power_2=(X.*y);

Copper_losses = (Y.^2)*kc;                     
Iron_losses = X.*ki;                         
Windage_losses = (X.^3)*kw;                      

Input_power_1 = Output_power_1+Copper_losses+Iron_losses+Windage_losses+conL;
Input_power_2 = Output_power_2-Copper_losses-Iron_losses-Windage_losses-conL;
Efficiency_1  =  Output_power_1./Input_power_1;
Efficiency_2  =  Output_power_2./Input_power_2;
V = [0.7,0.75,0.8,0.85,0.875,0.9,0.91,0.92,0.93]; % Efficiency curves to be plotted

box off
grid off
contourf(X,Y,Efficiency_1,V);
hold on
contourf(X,y,Efficiency_2,V)
colorbar
xlabel('Rotor Speed (rpm)')                      
ylabel('Torque induced N.m')                     
zlabel('Efficiency')                             
title('Torque speed curve of induction motor with Efficiency values')   

Contour Plot