Week-11 Challenge: Braking

Q1.DRIVING CYCLE FOR BRAKING

STEP 1:Now we are going to calculate braking power required for the vechicle with the given diven cycle

STEP 2:Taken an driving cycle from the net and wee are going to callculate the braking energy with 1/2*M*V^2

STEP 3: The velocity is measured as V=Vf-Vi(final-initial)

STEP 4:the masss value is taken as 1000 kg and velocity is taken which ienergy is calculated using the above formula

STEP 5:Then it converted from watt to kilowatt and the todal energy used for the barking will be calculted and given in the XLs sheet below.

Q2. ELECTRIC MOTOR CANT DEVELOP AN BRAKING TORQUE

The electric motor can not develop braking torque at high speed similar to the starting torque because if the electric supply is cut off to the motor then it act as an generator hence due to the flux weakening the generator produce an low torque.

STARTING TORQUE: This occurs during the starting stage. the speed values are very low and starts rising so the torque value will be much greater

BRAKING TORQUE: During the braking time already the motor will run in an higher speed hence due to the high speed the torque value will be very low.

REGENERATIVE BRAKING:Regenerative braking is a mechanism found on most hybrid and electric vehicle. It captures the kinetic energy from braking and converts it into the electrical power that charges the vehicle’s high voltage battery. Regenerative braking also slows the car down, which assists the use of traditional brakes.

Factors affect the braking power during the regenerative braking

*The controller should check weather the battery s full or not at the case of fully charged condition regenerative braking will produce an extra energy which will leads to rise the battery to unsafe level

*braking power is not only limited by the motors power rating but by the charging power limit of the battery pack. this limit depends on the battey type.

COORDINATION OF ELECTRICAL AND MECHANICAL BRAKING

From the above explanation we clearly got that electric motors doesn’t have an high torque capacity to handle all the vehicle braking.hence we are using the mechanical braking in larger proportion than the regenerative braking

We have three sifferent barake controls

*series breaking with optimal braking

*series braking with optimal energy recovery

*parallel braking

Q3.CONTOUR OF GIVEN PARAMETERS OF MOTOR

STEP 1: Open the MATLAB and create the new script for the contour program

STEP 2: create the script with initial parameters as given velocity and torque in vector quality also enter the copper lose,iron ;ose,windage loses values as given

STEP 3:darw an meshgrid with velocity and torque also find the output power using the formula from which the efficiency is calculated

STEP 4:Now the torque speed efficiency plots are taken and the program for the above motor perfomance is given. and the plot are given as screensheet

Velocity = linspace(1,500); % speed in rad/s

Torque = linspace(1,150); % Torque in Nm 0 to 40 Nm

kc =0.2; % kc - copper losses coefficient

ki = 0.008; % ki – iron losses coefficient

kw = 0.00001; % kw – windage losses coefficient

conL = 20; % Constant motor losses

[X,Y] = meshgrid(Velocity,Torque);

Output_power =(X.*Y);

Copper_losses = (Y.^2)*kc;

Iron_losses = X.*ki;

Windage_losses = (X.^3)*kw;

Input_power = Output_power+Copper_losses+Iron_losses+Windage_losses+conL;

Efficiency = Output_power./Input_power;

V = [0.7,0.75,0.8,0.85,0.875,0.9,0.91,0.92,0.93]; % Efficiency curves to be plotted

box off

grid off

contourf(X,Y,Efficiency,V);

colorbar

xlabel('Rotor Speed (rpm)')

ylabel('Torque induced N.m')

zlabel('Efficiency')

title('Torque speed curve of induction motor with Efficiency values')

figure

surf(X,Y,Efficiency)

colorbar

xlabel('Rotor Speed (rpm)')

ylabel('Torque induced N.m')

zlabel('Efficiency')

title('Efficiency values for various operating points of motor')