Challenge

Week-11 Challenge: Braking

1)

Aim: To Find the breaking force required for a defined driving cycle:

Driving cycle and its calcualtion Is in this excel sheet:

https://docs.google.com/spreadsheets/d/13gVpftNeHabmYDSkCvPE8It7R7BEZDlDYUX5Ny7FWBQ/edit?usp=sharing

A Brief about "Braking Energy" :

The energy required to move the vehicle is as equal to the energy that is required for braking.

In mechanical brakes, the energy which is spent during braking is lost as heat due to friction. This energy can be used upto a certain limit by using electrical brakes(Ex: dynamic braking or regenerative braking).

By, Newton's 3rd law," Every action will always have an equal and opposite reaction ". Similarly by using regenerative braking the amount of energy which can be restored is limited, based on the charging capacity of the battery. Due to which, the total kinetic energy of the vehicle during braking is not drained, so naturally the mechanical brakes come into the play.

2)

Why motor can’t develop braking torque at high speed similar to starting ?

Figure a.

From figure (a), we can understand that a motor can be operated in 2 regions namely, Constant torque region followed by constant power region. The motor will attain its maximum torque in “Constant Torque region” that is reason to have high starting torque. So, as the speed of the motor increases, it moves into constant power region, where its torque starts to reduce. At this instant when the brakes are applied by the driver, the Torque value will be lesser than the maximum torque of the motor, which is required to stop the motor (bringing speed to 0 RPM).

So, to compensate this deficient braking torque, the mechanical brakes are used along with Regenerative braking(Electric Braking).

Coordinating of Mechanical and Electric Brakes :

This is done In two ways , i.e. in parallel or series method.

In parallel method, both the brakes Is applied together, by distributing the braking torque between them. So, whenever higher braking torque is required at high speed, mechanical brakes is used more than electrical brakes. Similarly, at low speed, high braking torque is sufficiently provided by Electric brakes itself.

During braking, as times lapses, the speed reduces and even if the torque requirements increase, the motor will be able to manage it, since it will start operating in its constant torque region, but in a different quadrant of its operation.

Example graph:

Hydraulic Brake = Mechanical Brake.

Regenerative brake = Electric Brake

In Series Method, the application of brakes is in a sequence, for example: Electric brakes followed by mechanical brakes or vice versa. In this method, the driver can choose which one to be applied 1st, based on the required braking torque, State of charge of the battery, comfortness,etc.

References :

a) http://sites.uci.edu/haghpanah/files/2016/12/Effect-of-Different-Regenerative-Braking-Strategies-on-Braking-Performance-and-Fuel-Economy-in-a-Hybrid-Electric-Bus-Employing-CRUISE-Vehicle-Simulation.pdf

3)

The contour plot gives the idea of the machine's performance at various available efficiency. It is clear that, at all instances the machine can't operate at maximum efficiency. So, based on requirements, the operating conditions of the machine can be verified with the help of contour graphs, in order to utilise the suitable machine by satisfying the requirements.

Matlab code for plotting a contour of a motor, considering its speed, torque and efficiency values.

% efficiency contour plot 

x = linspace(1,1000); %speed in radians per seconds
y = linspace(1,250); % Torque in N/m

% Values of constants for losses
KC=0.2; % for copper loss
KI=0.008; % for Iron loss
kW=0.00001; % for windage loss
cont = 20 % constant motor  loss

[X,Y]= meshgrid(x,y); % to find the intermediate points
op = (X.*Y); % output is caluculated at all points of(X,Y)

%calculations of losses in the motor
a=(KC*(Y.^2)); % Copper Loss
b= (KI*X) ; % Iron Loss
c=(kW*(X.^3)); % Windage Loss

% Total input power
ip = (op + a + b + c +cont);

eff= op./ip ; % efficiency of the motor

v=[.1,.2,.3,.4,.5,.6,.7,.8,.9,.92,.93,.95,1]; % variable used for plotting efficiecny curve

contour(Y,X,eff,v)
hold on

The torque and speed values are assumed as arrays. The constant for the losses is assumed and the losses occuring during functioning is found out. All the 2D points are calculated using "Meshgrid" command. After which, the output power, Input power, Efficiency is calculated and plotted respectively.

Output Graph:

This graph shows the performance contour of a motor at various operating conditions, which has varying efficiency values.

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