Week-11 Challenge: Braking

1. For a defined driving cycle, calculate the energy required for braking.
2. Why electric motor can’t develop braking torque at a high speed similar to starting? How electric and mechanical brakes are coordinated?
3. Make a MATLAB program that plots the contour of given motor speed, torque and efficiency values. Attach the code as a .m file attach a screenshot of all the plots

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

Braking: Braking is a condition to slow down or stops a moving vehicle as required. There are several techniques to apply brakes, which can be mechanical or electrical. The most commonly used technique is by applying friction. It consumes energy .The consumed kinetic Energy can be derived as following equation-

Braking Energy = (1/2)*m*v^2

Where, Braking Energy is the kinetic energy in unit of Joule.

m= Total Mass of the Vehicle (Vehicle Mass + Pay Load) in unit of Kg.

v= (Initial Velocity-final Velocity) in unit m/s.

FTP Drive cycle:

data=xlsread('drcycle_week 11'); %initiating the excel file%

speed=data(:,2); %stretching speed values from excel%

time=data(:,1); %stretching time values from excel%

m=800; %Mass of vehicle%

vf=[]; %Assign final velocity array%

vi=[]; %Assign initial velocity array%

for j=1:(length(speed)-1) %for loop starts for every speed values%

if speed(j)>speed(j+1) %if condition for assigning values to vi&vf%

vi=[vi,speed(j)];

vf=[vf,speed(j+1)];

else %if accceleration happen end the if else%

end

energy=(0.5*m*(((vi-vf)*0.277).^2));%calculate Braking energy%

end %end of for loop

x=sprintf('Total breaking energy=%f',sum(energy)); %sum of all energy values%

disp(x); %Display Total breaking energy x value%

Result :

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

electric motor can’t develop braking torque at a high speed similar to starting:

1. At high speeds, the motor is operated above base speed (synchronous speed) and in the field weakening region. Above base speed, voltage is kept at rated value and frequency is increased which increases the speed but decreases the torque value.
2. The peak battery charging rate is not as high as peak battery discharging rate. Therefore, regenerative braking torque cannot be as high as accelerating torque.

• The limitations associated with power converter is another reason. While recharging, the power converter may not have the same efficiency as discharging.
  

This situation occurs only when the vehicle is at high speeds because the motor will be running at the constant power region. In the constant power region, the torque of the motor reduces and so the motor cannot develop the braking torque which is equal to the torque that occurred during low speed at the constant torque range.

The torque vs speed plot of a vector control induction motor is given below.

From the above characteristic curves of the motor. that is seen that the torque of the motor remains high at low speeds and reduces as the speed increases.

Since the electric motor cannot develop braking torque at a high speed similar to starting.

Electrical brake and mechanical brake.

Electrical braking

Electrical braking can be done by many methods such as,

• Unbalancing the input points of current to the motor.
• Using transistors in H-bridge arrangement to reverse the motor direction in order to reduce speed.
• Implementing a regenerative braking system by using bi-directional power converters.
• Regenerative braking converts much of the energy to electrical energy, which may be collected for later use.

Mechanical braking

Mechanical brakes commonly use friction between two surfaces pressed together to convert the kinetic energy of the moving object into heat, though other methods of energy conversion may be employed. The friction between the surfaces is performed using pressurized air or pressurized oil which are commonly known as an air brake and hydraulic brake respectively.

Coordination of Electric and Mechanical Brake :

In a electric Motor the Total Brake distribution is in two aspects which are Electrical Brake which is called Regenerative Brake and Mechanical Brake or Classical Brake with hydraulic cylindrical system. Brake Force distribution must be uniform, If the Regenerative braking power requirement is more than the charging limit of the battery or more than regenerative capacity of the machine we must apply Mechanical brakes as well. In practical case the combination of these two is applied. So usually it is combined response of Mechanical brake and Regenerative brake. The reason for how these brake force are combined, that particular phenomena is called Brake Control Strategies. There are two Brake Control Strategies, which are

Series - In case of Series we apply Mechanical brake and Regenerative electric brake one after another. We can apply only one brake strategy at a time either apply Regenerative or apply Mechanical brake.

The Series Brake Control Strategy has further classification, that is based on whether we want comfort of driver or Brake behaviour compared to energy efficiency.

Parallel - In case of Parallel we apply both brakes simultaneously. This Particular Parallel braking is going to happen both at a time.

Electric Driven Intelligent Brake (EDIB) is a combination of Hydraulic brake or Mechanical brake along with a Regenerative brake. It is a Parallel braking technique. The Control criteria such that at low speed use regenerative braking higher and at high speed or during initial braking period we actually combine the both.

3. Make a MATLAB program that plots the contour of given motor speed, torque and efficiency values. Attach the code as a .m file attach a screenshot of all the plots.

.

clear all

close all

clc

% Speed Array

w = linspace(0, 300, 100);

% Torque Array

T = linspace(0, 2000, 100);

% Meshgrid

[X,Y] = meshgrid(T, w);

% Loss Coefficients

cu = 0.17; % Copper coefficient

fe = 0.006; % Iron coefficient

win = 0.000015; % Windage coefficient

% Losses

b = cu*(Y.^2); % Copper Losses

c = fe*(X.^2); % Iron Losses

d = win*(X.^3); % Windage Losses

f = 18; % Conduction Losses

% Output Power

P_out = (X.*Y);

% Input Power

P_in = P_out + (b) + (c) + (d) + (f);

% Efficiency

eff = ((P_out)./(P_in));

% Efficiency values for contour

v = linspace(0.75, 0.95, 10);

% Contour Plot

box off

grid off

contourf(X,Y,eff,v)

xlabel('Speed, rad/s')

ylabel('Torque, Nm')

title('Speed, Torque and Efficiency characteristics of Motor')