Week-11 Challenge: Braking

Q1. For a defined driving cycle, calculate the energy required for braking.

Calculating the energy required for braking of UDDS Drive Cycle-

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.

For Example if Initial speed of a vehicle is 20m/s and final speed is 10m/s. The mass of the vehicle is 500kg. Then the required kinetic energy for Braking is (1/2)*500*(20-10)^2 = 25000 J =25KJ

Here for a Drive cycle the required braking energy is calculated for each second of time to slow down or stop the vehicle- Braking Energy Calculation for UDDS Driving Cycle

Q2. Why electric motor can’t develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated? 

DC motor Torque vs Speed Characteristic

Here for both in case of AC or DC there is no torque at maximum speed, gradually the torque decreases at high speed as the speed tends to be maximum speed. This situation does not occur at the beginning of acceleration because most of the torque is used to increase the kinetic energy of the load (and motor). But when the speed is already high and no extra load is applied it is not possible to increase the torque as well as to develop braking torque. The braking operation happens in the motion which is been distributed as per proportion in the Brake Energy and the regenerative braking.

There are two types of braking:

(i)Mechanical Braking: Mechanical brakes all act by generating frictional forces as two surfaces rub against each other. The stopping power or capacity of a brake depends largely on the surface area of frictional surfaces as well as on the actuation force applied. Mechanical brakes are assemblies consisting of mechanical elements for the slowing or stopping of shafts in equipment drives. They use levers or linkages to transmit force from one point to another. Braking slows or stops the movement of the coupled shafts. There are several types of mechanical brakes.

1) Band brakes: These are the simplest brake configuration, have a metal band lined with heat and wear resistant friction material.

2) Drum brakes: These are commonly used on automobile rear wheels, work when shoes press against a spinning surface called a drum.

3) Disc brakes: These are constructed of brake pads, a calliper, and a rotor. During operation, the brake pads are squeezed against the rotor.

(ii) Electrical Braking: Electric brakes are devices that use an electrical current or magnetic actuating force to slow or stop the motion of a rotating component. They are used in industrial and vehicular braking applications that require fast response times and precise tension control. The magnet in the backing plate has 2 conductor wires which tap directly into the wiring. When electricity is on, it magnetizes the brake magnet. The magnet is attracted to the drum face. When it contacts this area, the friction causes it to rotate, which moves the actuating arm, and pushes the shoes out against the drum. Those shoe have a special brake pad material on them that resists the heat caused by that friction. When the shoes press against the inside of the drum, they prevent the hub, and consequently the wheel that's touching the ground.

A very beautiful electric braking scheme called Regenerative Braking is now being employed for electric braking. Here during the braking action, the motor is operated as generator. As we know that during generating action, a counter torque is produced. So this counter torque opposes the motion of motor and motor comes to halt. Moreover, the brake energy is converted to electrical energy which can be fed back to the source using power converters.

Electric and Mechanical brake coordination:

There are mainly 2 ways by which the Electric Vehicles could be braked:

• Series- Regenerative braking and mechanical braking both are applied one after another.

• Parallel- Regenerative braking and mechanical braking both are applied simultaneously.

When brake pedal is pressed the hydraulic fluid reaches the brake pistons and created the braking pressure on the wheels and the same kinetic energy is been used by the Regenerative Braking for generating more power using the Bi-directional motor which operates as a generator while braking.

Q3. 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

MATLAB Code:

clear all
close all
clc
% Speed Array
omega = linspace(0, 300, 100);
% Torque Array
T = linspace(0, 2000, 100);
% Meshgrid
[X,Y] = meshgrid(T, omega);

% Loss Coefficients
cu = 0.17;               % Copper coefficient
fe = 0.006;              % Iron coefficient
win = 0.000015;          % Windage coefficient

% Losses
cu_loss = cu*(Y.^2);     % Copper Losses
fe_loss = fe*(X.^2);     % Iron Losses
win_loss = win*(X.^3);   % Windage Losses
cond_loss = 18;          % Conduction Losses

% Output Power
P_out = (X.*Y);
% Input Power
P_in = P_out + (cu_loss) + (fe_loss) + (win_loss) + (cond_loss);
% Efficiency
n = ((P_out)./(P_in));

% Efficiency values for contour
W = linspace(0.75, 0.95, 10);
% Contour Plot
contour(X,Y,n,W)
xlabel('Angular Speed, rad/s')
ylabel('Torque, Nm')
title('Speed, Torque and Efficiency characteristics of Motor')

Explanation:

In this MATLAB script, we create arrays for the speed and the torque of the motor using the ‘linspace’ command. For the speed, consider a range of values from 0 to 2000 radians per second whereas for the torque, consider a range of values from 0 to 300Nm.These arrays are then used to create 2D grid coordinates using the ‘meshgrid’ command. These coordinates will help in the creation of the contour plot. In order to find the efficiency of the motor, we need to consider the various losses that occur in the motor’s operation. These are:

Copper losses

Iron losses

Windage losses

Conduction losses

These various losses are dependent on the speed and the torque of the motor. They each also have a characteristic constant of proportionality which are all assumed for the following calculations and represented as 'K'. In each calculation, the speed and the torque are represented respectively as: ω and T.

Copper losses = KCu⋅ω2

Iron losses =KFe⋅T2

Windage losses =Kw⋅T3

The values of coefficients are taken as general values. Conduction losses are assumed to be 18 Joules.

Finally we plot the contour with the 'contour' command. In addition, an array for the efficiency values we want to plot for is created called 'W' and is assigned for values ranging from an efficiency level of 0.75 to 0.95. This array is used in the creation of the contour. Axis labels and a title is also made for better readability.

Output:

From this contour plot, we can see that the highest level of efficiency arises in the innermost contour plot and this efficiency decreases as the contours increase in size from the innermost contour.