Week-11 Challenge: Braking

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

Answer:

To calculate the energy required for braking, we need a drive cycle with different velocities at different time frames. Here we are considering a drive cycle for 80 secs. The mass of the vehicle is considered to be 1000kg. Below is the excel sheet created for a drive cycle with different velocities. We need to find the braking energy, which happens only when there is a sudden decrease in the speed.

To find the braking energy, the equation is;

Braking Energy = 0.5*m*v²

Where, v is velocity and will have upper limit and lower limit.

actual v = final velocity-initial velocity.

m is the mass of the vehicle, which is m = 1000kg.

Whenever the speed of the vehicle dops to a lower value, there will be an increase in breaking energy. This is shown in the graph attached which shows the velocity and breaking energy plotted against time.

Following is the drive cycle and the braking energy calculated for each point.

The drive cycle excel sheet has been attached. The total braking energy for the above drive cycle is 3125.86 KJ

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

Braking torque in an electrical motor is produced in the opposite direction of the rotation of motor. When the motor speed reduces, there is a breaking energy produced. The braking torque in an electric motor will not be as high as at the time of starting. Regenerative braking is the process where the kinetic energy is transformed to another form of energy or stored for later purpose. When the speed of the motor exceeds synchronous speed, then the regenerative braking mode is activated.

The above figure shows the speed-torque characteristics of a DC motor. It shows that the torque is inversely proportional to the speed of the motor. As the speed increases, torque decreases. When the motor is starting, the speed will be low, but the torque will be at peak value. Since the torque is high, the acceleration will be increased and the vehicle starts to accelerate to higher speeds. However, once the vehicle starts increasing the speed, the torque will be reduced and will not be as high as at the time of starting.

The above figure shows the speed-torque characteristics of an AC induction motor. Here we have constant torque region and constant power region. When the motor is starting until it achieves max speed, the torque will be constant and that area is called constant torque region. Once the max speed is achieved, the torque will be reduced, where we will have constant power. This area is called constant power region. The torque produced in the constant power region will be sufficiently low from the constant torque region, which results in motor not acheiving higher torque at higher speeds.

Electrical braking and Mechanical braking:

In electric braking system, the motor winding is used to produce the braking torque. As we have already mentioned, Braking torque in an electrical motor is produced in the opposite direction of the rotation of motor. So, there will be an opposition faced by the shaft while braking. This is achieved by different electrical connections made according to the speed-torque charateristics the motor will be working. Regenrative braking is the elctric braking system used in EV and HEV.

In mechanical braking, the braking is achieved because of the frictional force between the rotating part and brake drums.Brake shows, btrake lining, brake drum are required for mechanical braking. In this braking, the energy is wasted in the form of heat and we cannot reuse or store the energy.

Electrical and mechanical braking are co-ordinated in 2 configurations:

1. Series 2. Parallel

In series braking, the electric brake will be applied first and then the mechanical brake will be applied. That means when the vehicle is going to be at rest, the regenerative braking will be first applied and then the mechanical brake will be applied to stop the vehicle.

In parallel braking, electrical brake and mechanical brake are applied at the same time parallely. When the brake pedal is pressed, the vehicle comes to halt, and the kinetic energy lost at that time will be stored using electrical braking system.

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.

Answer:

Here we are assuming the values of coefficients of oron loss, windage loss and copper loss. We are defining an array of[1,1500] for speed and [1,350] for torque.

Coefficient of Copper Loss, kc=0.2

Coefficient of Iron loss, ki=0.005

Coefficient of Windage loss, kw=0.000014

Constant motor loss=14

To find copper loss, copper loss= Torque*kc.

To find Iron loss, Iron loss = angular speed*ki.

To find, windage loss, windage loss = (angular speed)^3 *kw.

Input power=(Copper loss)+(Iron loss)+(Windage loss)+(Constant motor loss)+(output power)

Efficiency = Ouput power/input power.

The values of efficiency for which the contour will be plotted is between, 0.6 and 1.

MATLAB code to plot motor speed, torque and efficiency values;

%Speed
w=linspace(1,1500);

%Torque
t=linspace(1,350);

%Constant Coefficient of Losses
%Coefficient of Copper Loss
kc=0.2;
%Coefficient of Iron loss
ki=0.005;
%Coefficient of Windage loss
kw=0.000014;
Constantmotorloss=14;

%Creating mesh
[x,y]=meshgrid(w,t);
Copperloss=(y.^2)*kc;
Ironloss=x.*ki;
Windageloss=(x.^3)*kw;
outputpower=x.*y;
inputpower=(Copperloss)+(Ironloss)+(Windageloss)+(Constantmotorloss)+(outputpower)
efficiency=outputpower./inputpower;
v=[0.6,0.63,0.65,0.68,0.7,0.73,0.75,0.78,0.8,0.83,0.85,0.88,0.9,0.93,0.95,0.98];
box off
grid off
contour(x,y,efficiency,v);
title('Motor-Speed-Torque-characteristics')
xlabel('Speed in rad/s')
ylabel('Torque in Nm')

The resultant plot is given below.

The .m file for the matlab code for plotting the contour is attached.