Week-11 Challenge: Braking
QUESTION 1 : For a defined driving cycle, calculate the energy required for braking.
ANSWER :
The energy required for acceleration or decelaration is nothing but the kinectic energy required by the vehicle.
Energy required for Acceleration = Energy required for Deceleration = `1/2*m*v^2`
But with respect to calculation of braking energy of a vehicle, we have to consider the initial speed and the final speed of the vehicle to determine the kinetic energy required.
Hence, `BE = 1/2*m*(u^2-v^2)` , where,
BE = Braking Energy (J) = BE/(3600*1000) (kWh)
m = Mass of the vehicle (kg)
u = Initial Velocity (m/s)
v = Final Velocity (m/s)
Now in order to calculate the Braking Energy, we need to create a driving cycle using an Excel sheet as shown below :
The Brake Energy is calculated using an Excel Calculator as shown below :
QUESTION 2 : Why electric motor can’t develop braking torque at high speed similar to starting? How electric and mechanical brakes are coordinated?
ANSWER :
ELECTRIC MOTOR CAN'T DEVELOP BRAKING TORQUE AT HIGH SPEED SIMILAR TO STARTING
The torque developed by an electric motor during deceleration is very low compared to acceleration phase because during acceleration the motor operates at the constant torque region where the starting torque is very high.Whereas during deceleration phase the motor operates in the constant power region where the torque developed is very low.During deceleration ie when the motor is at high speeds ie at speeds greater than base speed, due to the field weakening effect the torque developed will be low.Hence it is not possible to obtain maximum regenerative braking efficiency due to the above explained motor limitation.Another limitation with respect to regenerative braking is due to the battery.The discharge rate of a battery is twice that of its charging rate.Hence during regenerative braking the energy saved cannot be used to charge the battery at the desired rate.The final limitation concerning regenerative braking is associated with the power converter.To supply the energy saved during regeneration from the motor back to the battery, we need a bi-directional power converter, which in turn increases the complexity of the system and also losses during conversion should be considered.
COORDINATION BETWEEN ELECTRIC AND MECHANICAL BRAKES
Coordination between electric and mechanical brakes can be basically obtained using two methods : Series strategy and Parallel strategy.
Under series strategy, both the electric and mechanical brakes are applied one after another.With serial type a blending between friction brake and regenerative brake can be obtained, leading to the vehicle deceleration corresponding to the driver's braking demand. This type offers a high regeneration efficiency and good braking feel, but requests several modifications in the hydraulic brake line to mechanically decouple the connection between the brake pedal and the brake pressure
Under parallel strategy, both the electric and mechanical brakes are applied simultaneously.In parallel systems, the brake pressure is coupled to the brake pedal, and the regenerative braking is added directly to the friction brake operated by brake pedal, which results in a low regeneration efficiency and poor brake comfort.
Electric Driven Intelligent Brake (EDIB) developed by NISSAN is one of the most efficient type of coordinated braking established between electric and mechanical brakes.After stepping on the brake pedal, the system produces natural and adequate braking force that corresponds to the operation. The amount of energy regeneration also needs to be increased as much as possible.EDIB (Electric Driven Intelligent Brake) controls the regenerative brake and friction brake to support both of these requirements. Further, it also controls the reactive force from the pedal in order to unify the feeling when stepping down on the pedal and the sense of deceleration.EDIB is programmed such that during the initial phase of braking ie at high speeds both the regenerative and friction brakes are applied in a pre-determined ratio.Whereas at lower speeds the regenerative brakes are applied pre-dominantly.This is because at lower speeds, large torque is present which increases the efficiency of regenerative braking.
QUESTION 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.
ANSWER :
To plot the efficiency contours of an electric motor, we have to consider the various losses the motor is subjected to.The various losses considered are as follows :
Copper Losses : These losses are due to heat produced by electrical currents in the conductors of transformer windings and other electrical devices.
Copper Loss `prop I^2R`
where R is the armature resistance.Since current is proportinal to torque and by considering a constant for copper loss,we have,
Copper Loss = `kc*T^2`
Iron Losses : These losses are due to the magnetic field effect.Iron losses consist of three components namely eddy current loss, hysteresis loss and anomalous loss.Since these losses vary with the speed of motor and by considering a constant for Iron Losses,we have,
Iron Loss = `ki*omega`
Windage Loss : These are the mechanical and friction losses experienced by the motor.They depend on the size and shape of the motor.They are determined by the folowing equation,
Windage Loss = `kw*omega^3`
Constant Loss : These are determined by various design factors.It is represented by C.
Total Loss : It is the sum of the above four losses
Total Loss = Copper Loss + Iron Loss + Windage Loss + Constant Loss
MATLAB PROGRAM :
% Plotting Efficiency Contours for Electric Motors
clear all
close all
clc
% Determining the Speed Range (rad/s)
x = linspace(1,1000);
% Determing the Torque Range (Nm)
y = linspace(1,250);
% Allocating Motor Loss Constants
% For Copper Losses
kc = 0.2;
% For Iron Losses
ki = 0.008;
% For Windage Losses
kw = 0.00001;
% For Constant Motor Losses
ConL = 20;
% Meshing Speed and Torque ranges to create a 2D Array
[X,Y] = meshgrid(x,y);
% Output Power
Po = (X.*Y);
% Copper Losses
B = kc.*(Y.^2);
% Iron Losses
C = ki.*(X);
% Windage Losses
D = kw.*(X.^3);
% Input Power
Pi = Po + B + C + D + ConL;
% Efficiency
Z = Po./Pi;
% Setting the Efficiencies for which a contour will be plotted
V = [0.70,0.75,0.80,0.85,0.90,0.91,0.92,0.93,0.94,0.95];
% Plotting the Efficiency Contour
box off
grid off
contour(X,Y,Z,V)
xlabel('Speed (rad/s)')
ylabel('Torque (Nm)')
hold on
% Plotting contours of Output Power at 8kW and 10kW
V = [8000,10000];
contour(X,Y,Po,V)CONTOUR PLOT :
