Week-11 Challenge: Braking

- Aim:

- To calculate the energy required for braking.

- To understand the electric and mechanical braking system.

- Code for efficiency and speed and torque behaviour.

- Objectives:

- Calculation of braking energy considering the cycle.

- Understanding the working of the electric and mechanical braking system.

- Calculation of efficiency for the motor, and understanding the behaviour of torque, speed and efficiency.

- Method:

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

- In order to calculate the braking energy, the following formula has been used:

Braking energy = 1/2*m*V^2

where,

m = mass considering the inertia effect as well.

V = velocity difference = Vfinal - Vinitial

- In order to calculate the total braking energy it is necessary to consider all the deceleration processes and take the difference of Vfinal & Vinitial for all cycles.

- In order to simplify, as the number of deceleration cycles was much in number, fewer cycles has been considered for the study. It can be seen from the image given below, there are several cycles but for study initial 10 has been taken.

- The image is showing the cycle of a test, from which the initial 10 cycles has been taken in order to calculate the braking energy. The deceleration cycles are as per the following for the test:

Braking energy = 1/2*m*(Vfinal - Vinitial)^2

= 1/2*m*(SUM(Vfinal - Vinitial)^2)

- SUM is mentioned so to add all the velocity differences for the cycles taken into consideration. Mass m is taken common along with 1/2.

= 1/2 * 1575 * 7755.9

= 6107.00Kw

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

- In case of a HEV or EV, there are basically two types of the braking system. One is Regenerative braking and other is mechanical braking which is also called as friction braking.

- DC motors are relatively simple machines: when the load on the motor is constant, speed is proportional to supply voltage. And when the supply voltage is constant, speed is inversely proportional to the load on the motor. This second relationship between speed and load (or torque) is typically shown on the motor’s torque-speed curve.

- The inverse relationship between speed and torque means that an increase in the load (torque) on the motor will cause a decrease in speed. This can be demonstrated by the DC motor torque equation:

- Brake Control Strategies:

- These strategies are used in order to apply the brakes. As the acceleration torque is higher than deceleration torque, only electric braking is not sufficient to stop the vehicle. Also while reducing the speed, the speed is not directly reduced, as a process, the regeneration takes place, where the curve enters into 2nd quadrant and again enter to 3rd to a reduced speed.

- In such cases to obtain the reduction in speed, series and parallel strategies are used.

- In series, mechanical and electric brakes are applied one after the another, whereas, in parallel brakes, these are applied at a time.

- In order to control the system, Electic Driven Intelligent Brake system is used in order to control the vehicle. For low speed generally, only one braking system is used, as a regenerative brake system, whereas for high-speed operations combination is used.

- From the plot, it can be seen that the characteristics curve for the speed and torque is inversely proportional. 

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.

- In order to plot the speed, torque and efficiency plot, the following code has been done.

- Separate variables have been taken into consideration, like for speed X, for torque Y.

- Efficiency is the ratio of output to input while calculating the input, the losses also need to take into consideration.

- The losses are iron losses, copper losses, windage losses.

- Copper losses = I^2.R = Kc.T^2, where Kc is the resistance of brushes coil.

- Iron losses = Ki.w, this varies with speed. As at various speed, we have different values of back emf and that will affect these losses.

- Windage losses = Kw.W^3, depends on the size and shape of the motor.

- Constant losses c.

- Total losses = Kc.T^2 + Ki.w + Kw.W^3 + c

- Variable V is efficiency points we are giving.

% A program for plotting efficiency contours for
% electric motors.
% The x-axis corresponds to motor speed (w),
% and the y axis to torque T.
% First, set up arrays for range.
x=linspace(1,180);% speed, omega, N.B. rad/s NOT rpm
y=linspace(1,40); % 0 to 40 N.m
% Allocate motor loss constants.
kc=1.5; % For copper losses
ki=0.1; % For iron losses
kw=0.00001; % For windage losses
ConL=20; % For constant motor losses
% Now make mesh
[X,Y]=meshgrid(x,y);
Outputpower=(X.*Y); % Torque x speed = power
B=(Y.^2)*kc; % Copper losses
C=X*ki; % Iron losses
D=(X.^3)*kw; % Windage losses
Inputpower = Outputpower + B + C + D + ConL;
Z = Outputpower./Inputpower;
% We now set the efficiencies for which a contour
% will be plotted.
V=[0.5,0.6,0.7,0.75,0.8,0.85,0.88];
box off
grid off
contour(X,Y,Z,V);
xlabel('Speed/rad.s^-^1'), ylabel('Torque/N.m');
hold on
% Now plot a contour of the power output
% The array Output Power has
% already been calculated. We draw contours at
% 3 and 5 kW.
V=[3000,5000];
contour(X,Y,Outputpower,V);

- From the figure, it can be observed that, the behaviour of torque, speed and efficiency.

- Learning outcome:

- Performing calculations for braking energy.

- Understanding of electric and mechanical braking.

- Code for the plotting of torque, speed and efficiency behaviour.