Challenge

Week-11 Challenge: Braking

The New European Driving Cycle (NEDC) is considered

Lets consider the 800 to 1200km distance in the above cycle,from the drive cycle the maximum velocity during this displacement will be, 120km/h.

Lets take Vi=120km/h and vf=0 and the mass of the vehicle=1000kg

Braking energy required=0.5*1000*(120)^2

=7.2MV is required for braking.

During starting of electric motor, the starting torque required is more to start, so the motor draws a huge amount of current to generate this torque and during the running with respect to the load the torque will be maintained constant irrespective of the speed variation. So the running torque of the motor will be lower than the starting torque of the machine.

electric and mechanical brakes are coordinated due to the following reasons

Peak torque at acceleration will be greater than the peak torque at deceleration, so as a part of compensation both electric and mechanical brakes has to be coordinated.
The battery discharge rate will be 2C whereas the charging rate is only 1C so the remaining has to be compensated using mechanical brake.
The usage of bidirectional controller control system will be complex it need to be simplified.

So the strategies which can be applied for correlating will be

i). Series- Regenerative braking and mechanical braking will be applied one after another.

ii). Parallel- Both Regenerative and mechanical braking applied simultaneously.

3.x=linspace(1,1000);

y=linspace(1,250);

Kc=0.2;

Ki=0.008;

Kw=0.00001;

ConL=20;

[X,Y]=meshgrid(x,y);

outputpower=(X.*Y);

B=(Y.^2)*Kc;

C=(X.*Ki);

D=((X.^3)*Kw);

inputpower=outputpower+B+C+D+ConL;

Z=outputpower./inputpower;

V=[0.70,0.80,0.85,0.90,0.91,0.92,0.925,0.93];

box off

grid off

contourf(X,Y,Z,V);

colorbar

title('Efficiency contour');

xlabel('Speed in rad/s');

ylabel('Torque Nm');

hold on;

V=[8000,10000];

contour(X,Y,outputpower,V);

Ship work like this
in your own course.