Week-11 Challenge: Braking

1.

driving cycle considered is IDC (Indian Driving Cycle)

braking energy consumed in braking, while braking the Kinetic enrgy of the vehicle is absorbed.

K.E = 1/2*m*v^2

assume m=500kg

so constant k=1/2*m = 225kg

in IDC,

Brake energy @ t = 33sec = k*(6^2-4^2)=4500

Brake energy @ t = 51sec = k*(8^2-6^2)=6300

Brake energy @ t = 78sec = k*(11^2-9^2)=9000

Brake energy @ t = 108sec=k*(12^2-0^2)=32400

total brake energy in cycle is = 52200 J = 52.2 kJ

2.

it should be noted that maximum regenerative braking force produced by an electric motor is closely related to the electric motors speed.

at low speed, the maximum torque is constant, however at high speed the maximum torque decreases hyperbolically with its speed. there fore the mechanical brake torque at a given vehicle deceleartion varies with vehicle speed.

Regenerative braking in EV and HEV:

regenerative braking adds complexity to braking system design

1) One is how to distribute the total braking forces required between the regenerative braking and mechanical friction brake so as to recover the kinetic energy of the vehicle as much as possible.

2) the second is how to distribute the braking force between front and rear wheels, to achieve a steady state braking.

usually regenerative braking is effective only for the driven axle. The traction motor must be controlled to produce the proper amount of the braking force for recovering the kinetic energy as much as possible.

the machine must be controlled to meet the braking force command from the driver.

in order to meet these requirements, we have three control strategies.

a) series braking with optimum braking

b) series braking with optimal energy recovery

c) parallel braking

3.

the DC motor has various losses assosiated with it

copper losses

Armature copper losses = I_a²R_a where I_a is armature current, and Ra is the armature resistance. can be expressed as a function of torque K*T^2

iron losses

since at various speeds the back emf will be different, the iron losses will be a function of the rotational speed.

Ki * w

windage losses

K * w^3

constant losses

it comprises of other losses other than described above

% A program for plotting efficiency contours
% for electric motors
% X-axis - W (speed)
% Y-axis - T

x = linspace (1, 1000); % speed -> rad/s
y = linspace (1, 250);% 0-40 Nm

Kc = 0.2; % Cu loss
Ki = 0.008; % Iron loss
Kw = 0.00001; % Windage loss
L = 20; % constant motor loss
% Now make mesh,
[X, Y] = meshgrid (x, y);
Outputpower = (X.*Y); %T*W=P
B = (Y.^2)*Kc; % Copper loss
C = X*Ki; % Iron loss
D = (X.^3)*Kw; % Windage loss

Inputpower = Outputpower + B + C + D + L
Z = Outputpower./Inputpower;

% we know efficiencies for which a contour will be plotted

V = [0.7, 0.8, 0.85, 0.9, 0.91, 0.92, 0.925, 0.93];

box off

grid off
contour(X,Y,Z,V);

xlabel ('speed/rad.s');
ylabel ('Torque/N.m');

hold on

% now plot a contour of power output
% the array of output power has already been calculated

% We draw contours at 3 & 5hz

v = [8000,10000];
contour (X,Y, Outputpower, V);